Introduction
In 1994, the seminal result of Diaconis and Shahshahani [16 ] partially solved the question of averages of powers of traces of unitary matrices using an elegant method from representation theory. Averages of traces appear in many active research directions, as is seen in the diversity of work citing Diaconis and Shahshahani. There are multiple links to number theory; for example, these averages can relate to counting almost-primes in short intervals [1 , 25 ] and provide insight into the analysis of the Riemann zeta function, as well as related L L L [32 , 17 , 29 ] ), where quantum quenches use them to characterise how complex many-body systems thermalise, and in quantum information, where the long-term spectrum of states produced by quantum channels is governed by a trace average over unitary matrices.
However, the elegant result of Diaconis and Shahshahani holds only in a specific range when the matrix size is large enough.
The compact space of unitary matrices of dimension N N N U ( N ) U(N) U ( N ) U U U e i θ 1 , … , e i θ N e^{i\theta_1},\ldots,e^{i\theta_N} e i θ 1 , … , e i θ N
Tr U = e i θ 1 + ⋯ + e i θ N and Tr ( U m ) = e i m θ 1 + ⋯ + e i m θ N , \operatorname{Tr} U=e^{i\theta_1}+\cdots+e^{i\theta_N}\quad\text{and}\quad\operatorname{Tr}(U^m)=e^{im\theta_1}+\cdots+e^{im\theta_N}, Tr U = e i θ 1 + ⋯ + e i θ N and Tr ( U m ) = e im θ 1 + ⋯ + e im θ N , where m m m
The theorem of Diaconis and Shahshahani (see [16 ] for the original paper, but also [15 ] which corrects an error in the statement of the theorem) states that for vectors a = ( a 1 , … , a l ) a=(a_1,\ldots,a_l) a = ( a 1 , … , a l ) b = ( b 1 , … , b l ) b=(b_1,\ldots,b_l) b = ( b 1 , … , b l ) a j a_j a j b j b_j b j N N N N N N δ a b \delta_{ab} δ ab a a a b b b
∫ U ( N ) ∏ j = 1 l ( Tr M j ) a j ( Tr M j ‾ ) b j d U = δ a b ∏ j = 1 l j a j a j ! (1) \int_{U(N)} \prod_{j=1}^{l} \left(\operatorname{Tr} M^j\right)^{a_j} \left(\overline{\operatorname{Tr} M^j}\right)^{b_j}\,\mathrm{d}U = \delta_{ab} \prod_{j=1}^{l} j^{a_j}a_j! \tag*{(1)} ∫ U ( N ) j = 1 ∏ l ( Tr M j ) a j ( Tr M j ) b j d U = δ ab j = 1 ∏ l j a j a j ! (1) but only when N ≥ max ( ∑ j = 1 l j a j , ∑ j = 1 l j b j ) N \ge\max\left(\sum_{j=1}^{l} ja_j,\sum_{j=1}^{l} jb_j\right) N ≥ max ( ∑ j = 1 l j a j , ∑ j = 1 l j b j ) N N N [31 ] giving some insight into the problem if a j = b j = 0 a_j=b_j=0 a j = b j = 0 j < N j<N j < N a j = b j = 1 a_j=b_j=1 a j = b j = 1 j j j a a a b b b N N N [18 ] , and is further explained at equation (3.5) of [22 ] . Pastur and Vasilchuk [30 ] calculated a specific average in the case that only one of the a j a_j a j b j b_j b j N N N [25 ] , in Section 7.3, consider averages of four traces in a range where 2 N + 1 ≥ max ( ∑ j = 1 l j a j , ∑ j = 1 l j b j ) 2N+1 \ge\max\left(\sum_{j=1}^{l} ja_j,\sum_{j=1}^{l} jb_j\right) 2 N + 1 ≥ max ( ∑ j = 1 l j a j , ∑ j = 1 l j b j ) [34 ] , that the averages of traces are zero unless the sum of the exponents on the traces of U U U Tr M j ‾ \overline{\operatorname{Tr} M^j} Tr M j Tr ( ( M † ) j ) \operatorname{Tr}\left((M^\dagger)^j\right) Tr ( ( M † ) j )
∑ j j a j = ∑ j j b j . (2) \sum_j ja_j = \sum_j jb_j. \tag*{(2)} j ∑ j a j = j ∑ j b j . (2) Beyond this, much literature has concentrated on the large N N N [20 , 21 , 13 , 19 ] ). Recently, Irfan has provided concrete formulae for averages of traces for any matrix size [19 ] . Weingarten Calculus (see [5 , 6 ] , but particularly [4 ] , for example in Proposition 3.2) also provides a way of writing down formulae for averages of traces. Soshnikov and Wu look instead at cumulants of traces in [33 ] . These formulae fully solve the problem via the moment-cumulant formula, but highlight the intense complexity of the combinatorics of the trace averages.
In this paper, we use the fact that averages, with respect to Haar measure, of logarithmic derivatives of characteristic polynomials of matrices from U ( N ) U(N) U ( N )
Our main theorems are: Theorem 4.1, for the case of one trace of a power of U † U^\dagger U † U U U N N N U † U^\dagger U † U U U N N N U U U 2 N + 2 2N+2 2 N + 2
Ratios Theorem and the logarithmic derivative of the characteristic polynomial
We define the characteristic polynomial of a matrix U ∈ U ( N ) U \in U(N) U ∈ U ( N ) e i θ 1 , … , e i θ N e^{i\theta_1},\ldots,e^{i\theta_N} e i θ 1 , … , e i θ N
Λ U ( s ) = ∏ j = 1 N ( 1 − s e − i θ j ) . (3) \Lambda_U(s)=\prod_{j=1}^{N}(1-se^{-i\theta_j}). \tag*{(3)} Λ U ( s ) = j = 1 ∏ N ( 1 − s e − i θ j ) . (3) The Ratios Theorem, first proposed by Conrey, Farmer and Zirnbauer in Theorem 4.1 of [7 ] , revisited by Conrey, Forrester and Snaith in Proposition 2.1 of [8 ] , and written in a slightly more intuitive set notation in Theorem 3.2 of [10 ] , gives a formula for ratios of products of characteristic polynomials, each evaluated at different points in the complex plane, of matrices drawn from U ( N ) U(N) U ( N ) [10 ] . We restate this Theorem here.
Theorem 2.1 (Averages of logarithmic derivatives: Theorem 3.3 [10 ] ). If R a j > 0 \mathfrak{R}a_j>0 R a j > 0 R b j > 0 \mathfrak{R}b_j>0 R b j > 0 a j ∈ A a_j\in\mathcal{A} a j ∈ A b j ∈ B b_j\in\mathcal{B} b j ∈ B U † U^\dagger U † U U U J ( A ; B ) = J ⋆ ( A ; B ) J(\mathcal{A};\mathcal{B})=J^\star(\mathcal{A};\mathcal{B}) J ( A ; B ) = J ⋆ ( A ; B )
J ( A ; B ) : = ∫ U ( N ) ∏ a ∈ A ( − e − a ) Λ U ′ Λ U ( e − a ) ∏ b ∈ B ( − e − b ) Λ U † ′ Λ U † ( e − b ) d U , (4) J(\mathcal{A};\mathcal{B}) := \int_{U(N)}\prod_{a\in\mathcal{A}}(-e^{-a})\frac{\Lambda_U'}{\Lambda_U}(e^{-a})\prod_{b\in\mathcal{B}}(-e^{-b})\frac{\Lambda_{U^\dagger}'}{\Lambda_{U^\dagger}}(e^{-b})\,dU, \tag*{(4)} J ( A ; B ) := ∫ U ( N ) a ∈ A ∏ ( − e − a ) Λ U Λ U ′ ( e − a ) b ∈ B ∏ ( − e − b ) Λ U † Λ U † ′ ( e − b ) d U , (4) J ⋆ ( A ; B ) : = ∑ S ⊂ A , T ⊂ B ∣ S ∣ = ∣ T ∣ e − N ( ∑ a ^ ∈ S a ^ + ∑ b ^ ∈ T b ^ ) Z ( S , T ) Z ( S − , T − ) Z † ( S , S − ) Z † ( T , T − ) ∑ A − S + B − T = U 1 + ⋯ + U R ∣ U r ∣ ≤ 2 ∏ r = 1 R H S , T ( U r ) , (5) J^\star(\mathcal{A};\mathcal{B}):=
\sum_{\substack{S\subset\mathcal{A},T\subset\mathcal{B}\\|S|=|T|}}
e^{-N\left(\sum_{\hat a\in S}\hat a+\sum_{\hat b\in T}\hat b\right)}
\frac{Z(S,T)Z(S^{-},T^{-})}{Z^\dagger(S,S^{-})Z^\dagger(T,T^{-})}
\sum_{\substack{\mathcal{A}-S+\mathcal{B}-T\\=U_1+\cdots+U_R\\|U_r|\leq2}}
\prod_{r=1}^{R}H_{S,T}(U_r), \tag*{(5)} J ⋆ ( A ; B ) := S ⊂ A , T ⊂ B ∣ S ∣ = ∣ T ∣ ∑ e − N ( ∑ a ^ ∈ S a ^ + ∑ b ^ ∈ T b ^ ) Z † ( S , S − ) Z † ( T , T − ) Z ( S , T ) Z ( S − , T − ) A − S + B − T = U 1 + ⋯ + U R ∣ U r ∣ ≤ 2 ∑ r = 1 ∏ R H S , T ( U r ) , (5) and
H S , T ( W ) = { ∑ a ^ ∈ S z ′ z ( a − a ^ ) − ∑ b ^ ∈ T z ′ z ( a + b ^ ) if W = { a } ⊂ A − S , ∑ b ^ ∈ T z ′ z ( b − b ^ ) − ∑ a ^ ∈ S z ′ z ( b + a ^ ) if W = { b } ⊂ B − T , ( z ′ z ) ( a + b ) if W = { a , b } with a ∈ A − S , b ∈ B − T , 0 otherwise. (6) H_{S,T}(W)=
\begin{cases}
\sum_{\hat a\in S}\frac{z'}{z}(a-\hat a)-\sum_{\hat b\in T}\frac{z'}{z}(a+\hat b) & \text{if }W=\{a\}\subset\mathcal{A}-S,\\
\sum_{\hat b\in T}\frac{z'}{z}(b-\hat b)-\sum_{\hat a\in S}\frac{z'}{z}(b+\hat a) & \text{if }W=\{b\}\subset\mathcal{B}-T,\\
\left(\frac{z'}{z}\right)(a+b) & \text{if }W=\{a,b\}\text{ with }a\in\mathcal{A}-S,\ b\in\mathcal{B}-T,\\
0 & \text{otherwise.}
\end{cases} \tag*{(6)} H S , T ( W ) = ⎩ ⎨ ⎧ ∑ a ^ ∈ S z z ′ ( a − a ^ ) − ∑ b ^ ∈ T z z ′ ( a + b ^ ) ∑ b ^ ∈ T z z ′ ( b − b ^ ) − ∑ a ^ ∈ S z z ′ ( b + a ^ ) ( z z ′ ) ( a + b ) 0 if W = { a } ⊂ A − S , if W = { b } ⊂ B − T , if W = { a , b } with a ∈ A − S , b ∈ B − T , otherwise. (6) The definition of z ( x ) z(x) z ( x )
z ( x ) = 1 1 − e − x . (7) z(x)=\frac{1}{1-e^{-x}}. \tag*{(7)} z ( x ) = 1 − e − x 1 . (7) Also, Z ( A , B ) = ∏ a ∈ A ∏ b ∈ B z ( a + b ) Z(\mathcal{A},\mathcal{B})=\prod_{a\in\mathcal{A}}\prod_{b\in\mathcal{B}}z(a+b) Z ( A , B ) = ∏ a ∈ A ∏ b ∈ B z ( a + b ) Z † ( S , S − ) Z^\dagger(S,S^{-}) Z † ( S , S − ) z ( x ) z(x) z ( x ) A − S \mathcal{A}-S A − S A \mathcal{A} A S S S ∣ T ∣ |T| ∣ T ∣ T T T
There is a convention to call the terms in (5) “zero-swap” terms where ∣ S ∣ = ∣ T ∣ = 0 |S|=|T|=0 ∣ S ∣ = ∣ T ∣ = 0 ∣ S ∣ = ∣ T ∣ = 1 |S|=|T|=1 ∣ S ∣ = ∣ T ∣ = 1 n n n a a a n n n b b b ∣ S ∣ = ∣ T ∣ = n |S|=|T|=n ∣ S ∣ = ∣ T ∣ = n n n n
The trace of U j U^j U j j j j Tr ( U j ) = e i j θ 1 + e i j θ 2 + ⋯ + e i j θ N \operatorname{Tr}(U^j)=e^{ij\theta_1}+e^{ij\theta_2}+\cdots+e^{ij\theta_N} Tr ( U j ) = e ij θ 1 + e ij θ 2 + ⋯ + e ij θ N U j U^j U j Tr ( ( U † ) j ) = e − i j θ 1 + e − i j θ 2 + ⋯ + e − i j θ N \operatorname{Tr}((U^\dagger)^j)=e^{-ij\theta_1}+e^{-ij\theta_2}+\cdots+e^{-ij\theta_N} Tr (( U † ) j ) = e − ij θ 1 + e − ij θ 2 + ⋯ + e − ij θ N ∣ y ∣ < 1 \lvert y\rvert< 1 ∣ y ∣ < 1
y Λ U ′ ( y ) Λ U ( y ) = y ∑ n = 1 N − e − i θ n 1 − y e − i θ n = − y ∑ n = 1 N e − i θ n ∑ j = 0 ∞ ( y e − i θ n ) j = − ∑ j = 1 ∞ y j ∑ n = 1 N e − i j θ n = − ∑ j = 1 ∞ y j Tr ( ( U † ) j ) , (8) \begin{aligned}
y\frac{\Lambda'_U(y)}{\Lambda_U(y)} = y\sum_{n=1}^{N}\frac{-e^{-i\theta_n}}{1-ye^{-i\theta_n}} \\
&= -y\sum_{n=1}^{N}e^{-i\theta_n}\sum_{j=0}^{\infty}(ye^{-i\theta_n})^j \\
&= -\sum_{j=1}^{\infty}y^j\sum_{n=1}^{N}e^{-ij\theta_n} \\
&= -\sum_{j=1}^{\infty}y^j\operatorname{Tr}((U^\dagger)^j), \tag*{(8)}
\end{aligned} y Λ U ( y ) Λ U ′ ( y ) = y n = 1 ∑ N 1 − y e − i θ n − e − i θ n = − y n = 1 ∑ N e − i θ n j = 0 ∑ ∞ ( y e − i θ n ) j = − j = 1 ∑ ∞ y j n = 1 ∑ N e − ij θ n = − j = 1 ∑ ∞ y j Tr (( U † ) j ) , (8) meaning that we can connect averages of products of traces of U † U^\dagger U † U U U
∫ U ( N ) ∏ i = 1 n ( − A i Λ U ′ Λ U ( A i ) ) ∏ j = 1 m ( − B j Λ U † ′ Λ U † ( B j ) ) d U = ∑ i 1 , … , i n = 1 ∞ ∑ j 1 , … , j m = 1 ∞ A 1 i 1 ⋯ A n i n B 1 j 1 ⋯ B m j m ∫ U ( N ) Tr ( ( U † ) i 1 ) ⋯ Tr ( ( U † ) i n ) Tr ( U j 1 ) ⋯ Tr ( U j m ) d U . (9) \begin{aligned}
\int_{U(N)}\prod_{i=1}^{n}\left(-A_i\frac{\Lambda'_U}{\Lambda_U}(A_i)\right)\prod_{j=1}^{m}\left(-B_j\frac{\Lambda'_{U^\dagger}}{\Lambda_{U^\dagger}}(B_j)\right)dU \\
&= \sum_{i_1,\ldots,i_n=1}^{\infty}\sum_{j_1,\ldots,j_m=1}^{\infty}A_1^{i_1}\cdots A_n^{i_n}B_1^{j_1}\cdots B_m^{j_m}\int_{U(N)}\operatorname{Tr}((U^\dagger)^{i_1})\cdots\operatorname{Tr}((U^\dagger)^{i_n})\operatorname{Tr}(U^{j_1})\cdots\operatorname{Tr}(U^{j_m})dU. \tag*{(9)}
\end{aligned} ∫ U ( N ) i = 1 ∏ n ( − A i Λ U Λ U ′ ( A i ) ) j = 1 ∏ m ( − B j Λ U † Λ U † ′ ( B j ) ) d U = i 1 , … , i n = 1 ∑ ∞ j 1 , … , j m = 1 ∑ ∞ A 1 i 1 ⋯ A n i n B 1 j 1 ⋯ B m j m ∫ U ( N ) Tr (( U † ) i 1 ) ⋯ Tr (( U † ) i n ) Tr ( U j 1 ) ⋯ Tr ( U j m ) d U . (9) We will see a simple case of this in the next section.
Second moment of the logarithmic derivative and two-trace averages
Averages of two traces are completely and simply understood [18 , 16 , 31 ] , but we obtain the known results here using the Ratios Theorem as a demonstration of the simplest case of this method.
Let a , b > 0 a,b>0 a , b > 0 Theorem 2.1 ,
∫ U ( N ) e − a Λ U ′ Λ U ( e − a ) e − b Λ U † ′ Λ U † ( e − b ) d U = ( z ′ z ) ′ ( a + b ) + e − N ( a + b ) z ( a + b ) z ( − a − b ) , (10) \begin{aligned}
\int_{U(N)}e^{-a}\frac{\Lambda'_U}{\Lambda_U}(e^{-a})e^{-b}\frac{\Lambda'_{U^\dagger}}{\Lambda_{U^\dagger}}(e^{-b})dU \\
&=\left(\frac{z'}{z}\right)'(a+b)+e^{-N(a+b)}z(a+b)z(-a-b), \tag*{(10)}
\end{aligned} ∫ U ( N ) e − a Λ U Λ U ′ ( e − a ) e − b Λ U † Λ U † ′ ( e − b ) d U = ( z z ′ ) ′ ( a + b ) + e − N ( a + b ) z ( a + b ) z ( − a − b ) , (10) where
z ( x ) = 1 1 − e − x . (11) z(x)=\frac{1}{1-e^{-x}}. \tag*{(11)} z ( x ) = 1 − e − x 1 . (11) For use in this and later calculations, if e − x = X e^{-x}=X e − x = X ∣ X ∣ < 1 \lvert X\rvert<1 ∣ X ∣ < 1
z ( x ) z ( − x ) = − e − x ( 1 − e − x ) 2 = − X ( 1 − X ) 2 = − ∑ n = 0 ∞ n X n z(x)z(-x)=\frac{-e^{-x}}{(1-e^{-x})^2}=-\frac{X}{(1-X)^2}=-\sum_{n=0}^{\infty}nX^n z ( x ) z ( − x ) = ( 1 − e − x ) 2 − e − x = − ( 1 − X ) 2 X = − n = 0 ∑ ∞ n X n z ′ z ( x ) = − e − x 1 − e − x = − X 1 − X = − ∑ n = 0 ∞ X n + 1 and \frac{z'}{z}(x)=\frac{-e^{-x}}{1-e^{-x}}=-\frac{X}{1-X}=-\sum_{n=0}^{\infty}X^{n+1}\quad\text{and} z z ′ ( x ) = 1 − e − x − e − x = − 1 − X X = − n = 0 ∑ ∞ X n + 1 and ( z ′ z ) ′ ( x ) = e − x ( 1 − e − x ) 2 = X ( 1 − X ) 2 = ∑ n = 0 ∞ n X n . (12) \left(\frac{z'}{z}\right)'(x)=\frac{e^{-x}}{(1-e^{-x})^2}=\frac{X}{(1-X)^2}=\sum_{n=0}^{\infty}nX^n. \tag*{(12)} ( z z ′ ) ′ ( x ) = ( 1 − e − x ) 2 e − x = ( 1 − X ) 2 X = n = 0 ∑ ∞ n X n . (12) On the other hand, if e − x = X e^{-x}=X e − x = X ∣ X ∣ > 1 |X|>1 ∣ X ∣ > 1
z ′ z ( x ) = − e − x 1 − e − x = − X 1 − X = 1 1 − 1 / X = ∑ n = 0 ∞ X − n . (13) \frac{z'}{z}(x)=\frac{-e^{-x}}{1-e^{-x}}=\frac{-X}{1-X}=\frac{1}{1-1/X}=\sum_{n=0}^{\infty}X^{-n}. \tag*{(13)} z z ′ ( x ) = 1 − e − x − e − x = 1 − X − X = 1 − 1/ X 1 = n = 0 ∑ ∞ X − n . (13) Returning to (10), for convenience we will write e − a = A e^{-a}=A e − a = A e − b = B e^{-b}=B e − b = B
∫ U ( N ) A Λ U ′ Λ U ( A ) B Λ U † ′ Λ U † ( B ) d U = A B ( 1 − A B ) 2 − A N + 1 B N + 1 ( 1 − A B ) 2 , (14) \int_{U(N)}A\frac{\Lambda'_U}{\Lambda_U}(A)B\frac{\Lambda'_{U^\dagger}}{\Lambda_{U^\dagger}}(B)dU
=\frac{AB}{(1-AB)^2}-\frac{A^{N+1}B^{N+1}}{(1-AB)^2}, \tag*{(14)} ∫ U ( N ) A Λ U Λ U ′ ( A ) B Λ U † Λ U † ′ ( B ) d U = ( 1 − A B ) 2 A B − ( 1 − A B ) 2 A N + 1 B N + 1 , (14) where the first term is a zero-swap term and the second is a one-swap term.
Since ∣ A B ∣ < 1 |AB|<1 ∣ A B ∣ < 1
∫ U ( N ) A Λ U ′ Λ U ( A ) B Λ U † ′ Λ U † ( B ) d U = ∑ n = 0 ∞ n ( A B ) n − ∑ j = 0 ∞ j ( A B ) N + j = ∑ n = 1 N n ( A B ) n + ∑ n = N + 1 ∞ n ( A B ) n − ∑ j = N + 1 ∞ ( j − N ) ( A B ) j = ∑ n = 1 N n ( A B ) n + N ∑ j = N + 1 ∞ ( A B ) j . (15) \begin{aligned}
\int_{U(N)}A\frac{\Lambda'_U}{\Lambda_U}(A)B\frac{\Lambda'_{U^\dagger}}{\Lambda_{U^\dagger}}(B)dU
&=\sum_{n=0}^{\infty}n(AB)^n-\sum_{j=0}^{\infty}j(AB)^{N+j}\\
&=\sum_{n=1}^{N}n(AB)^n+\sum_{n=N+1}^{\infty}n(AB)^n-\sum_{j=N+1}^{\infty}(j-N)(AB)^j\\
&=\sum_{n=1}^{N}n(AB)^n+N\sum_{j=N+1}^{\infty}(AB)^j. \tag*{(15)}
\end{aligned} ∫ U ( N ) A Λ U Λ U ′ ( A ) B Λ U † Λ U † ′ ( B ) d U = n = 0 ∑ ∞ n ( A B ) n − j = 0 ∑ ∞ j ( A B ) N + j = n = 1 ∑ N n ( A B ) n + n = N + 1 ∑ ∞ n ( A B ) n − j = N + 1 ∑ ∞ ( j − N ) ( A B ) j = n = 1 ∑ N n ( A B ) n + N j = N + 1 ∑ ∞ ( A B ) j . (15) Now, we relate this to traces. Using (9),
∫ U ( N ) A Λ ′ Λ ( A ) B Λ ′ Λ ( B ) d U = ∑ m , k = 1 ∞ A m B k ∫ U ( N ) Tr ( ( U † ) m ) Tr ( U k ) d U . (16) \begin{aligned}
\int_{U(N)}A\frac{\Lambda'}{\Lambda}(A)B\frac{\Lambda'}{\Lambda}(B)dU
&=\sum_{m,k=1}^{\infty}A^mB^k\int_{U(N)}\operatorname{Tr}((U^\dagger)^m)\operatorname{Tr}(U^k)dU. \tag*{(16)}
\end{aligned} ∫ U ( N ) A Λ Λ ′ ( A ) B Λ Λ ′ ( B ) d U = m , k = 1 ∑ ∞ A m B k ∫ U ( N ) Tr (( U † ) m ) Tr ( U k ) d U . (16) Comparing (16) with (15), we can read off that
∫ U ( N ) Tr ( U j ) Tr ( ( U † ) m ) d U H a a r = 0 if j ≠ m (17) \int_{U(N)}\operatorname{Tr}(U^j)\operatorname{Tr}((U^\dagger)^m)dU_{\mathrm{Haar}}=0\quad\text{if }j\ne m \tag*{(17)} ∫ U ( N ) Tr ( U j ) Tr (( U † ) m ) d U Haar = 0 if j = m (17) and
∫ U ( N ) Tr ( U j ) Tr ( ( U † ) j ) d U H a a r = j if j < N (18) \int_{U(N)}\operatorname{Tr}(U^j)\operatorname{Tr}((U^\dagger)^j)dU_{\mathrm{Haar}}=j\quad\text{if }j<N \tag*{(18)} ∫ U ( N ) Tr ( U j ) Tr (( U † ) j ) d U Haar = j if j < N (18) (which is a case proven by Diaconis and Shahshahani [16 ] ) and
∫ U ( N ) Tr ( U j ) Tr ( ( U † ) j ) d U H a a r = N if j ≥ N , (19) \int_{U(N)}\operatorname{Tr}(U^j)\operatorname{Tr}((U^\dagger)^j)dU_{\mathrm{Haar}}=N\quad\text{if }j\ge N, \tag*{(19)} ∫ U ( N ) Tr ( U j ) Tr (( U † ) j ) d U Haar = N if j ≥ N , (19) which is the result due to Rains (see just before Theorem 1.2 in [31 ] ). All of the above were known to Dyson [18 ] . Notice that only the zero-swap terms contribute to the Diaconis and Shahshahani range of trace averages j < N j<N j < N N N N A A A B B B j = N j=N j = N
Averages of traces with one U † U^\dagger U † U U U
We now analyse the structure of Theorem 2.1 in the case of one U † U^\dagger U † U U U (5) . We will now prove the following theorem:
Theorem 4.1. [One dagger trace moment formula] Let n ≥ 2 n \ge2 n ≥ 2 β 1 , … , β n ≥ 1 \beta_1,\ldots,\beta_n \ge1 β 1 , … , β n ≥ 1
∫ U ( N ) Tr ( ( U † ) α ) ∏ j = 1 n Tr ( U β j ) d U = { ∑ ∅ ≠ T ⊆ { 1 , … , n } ( − 1 ) ∣ T ∣ ( ∑ j ∈ T β j − N ) + , α = ∑ j = 1 n β j , 0 , otherwise . (20) \int_{U(N)} \operatorname{Tr}\left((U^\dagger)^\alpha\right)\prod_{j=1}^{n}\operatorname{Tr}\left(U^{\beta_j}\right)\,\mathrm{d}U =
\begin{cases}
\sum_{\varnothing\ne T\subseteq\{1,\ldots,n\}}(-1)^{\lvert T\rvert}\left(\sum_{j\in T}\beta_j-N\right)_{+}, & \alpha=\sum_{j=1}^{n}\beta_j,\\
0, & \text{otherwise}.
\end{cases} \tag*{(20)} ∫ U ( N ) Tr ( ( U † ) α ) j = 1 ∏ n Tr ( U β j ) d U = ⎩ ⎨ ⎧ ∑ ∅ = T ⊆ { 1 , … , n } ( − 1 ) ∣ T ∣ ( ∑ j ∈ T β j − N ) + , 0 , α = ∑ j = 1 n β j , otherwise . (20) where the notation ( x ) + (x)_{+} ( x ) + x x x x x x
Proof. The proof proceeds by expanding J ⋆ ( A ; B ) J^\star(\mathcal{A};\mathcal{B}) J ⋆ ( A ; B ) Theorem 2.1 , in the case ∣ A ∣ = 1 \lvert\mathcal{A}\rvert=1 ∣ A ∣ = 1 ∣ B ∣ = n \lvert\mathcal{B}\rvert=n ∣ B ∣ = n A = { a } \mathcal{A}=\{a\} A = { a } B = { b 1 , … , b n } \mathcal{B}=\{b_1,\ldots,b_n\} B = { b 1 , … , b n } J ⋆ ( A 1 ; B n ) J^\star(\mathcal{A}_1;\mathcal{B}_n) J ⋆ ( A 1 ; B n ) A = e − a A=e^{-a} A = e − a B j = e − b j B_j=e^{-b_j} B j = e − b j A α ∏ j = 1 n B j β j A^\alpha\prod_{j=1}^{n}B_j^{\beta_j} A α ∏ j = 1 n B j β j ∣ A ∣ , ∣ B 1 ∣ , ∣ B 2 ∣ , … , ∣ B n ∣ < 1 \lvert A\rvert,\lvert B_1\rvert,\lvert B_2\rvert,\ldots,\lvert B_n\rvert<1 ∣ A ∣ , ∣ B 1 ∣ , ∣ B 2 ∣ , … , ∣ B n ∣ < 1 ∣ B n ∣ < ∣ B n − 1 ∣ < ⋯ < ∣ B 1 ∣ \lvert B_n\rvert<\lvert B_{n-1}\rvert<\cdots<\lvert B_1\rvert ∣ B n ∣ < ∣ B n − 1 ∣ < ⋯ < ∣ B 1 ∣
The outer sum in (5) for this case now only contains the zero-swap and one-swap terms. Firstly when ∣ S ∣ = ∣ T ∣ = 0 \lvert S\rvert=\lvert T\rvert=0 ∣ S ∣ = ∣ T ∣ = 0 S = T = ∅ S=T=\varnothing S = T = ∅ Z Z Z
∑ U 1 + ⋯ + U R = A + B ∣ U r ∣ ≤ 2 ∏ r = 1 R H ∅ , ∅ ( U r ) . (21) \sum_{\substack{\mathcal{U}_1+\cdots+\mathcal{U}_R=\mathcal{A}+\mathcal{B}\\ \lvert\mathcal{U}_r\rvert\le2}}\prod_{r=1}^{R}H_{\varnothing,\varnothing}(\mathcal{U}_r). \tag*{(21)} U 1 + ⋯ + U R = A + B ∣ U r ∣ ≤ 2 ∑ r = 1 ∏ R H ∅ , ∅ ( U r ) . (21) In this case, if n ≥ 2 n\ge2 n ≥ 2 A + B \mathcal{A}+\mathcal{B} A + B A \mathcal{A} A B \mathcal{B} B (6) that for any singleton, H ∅ , ∅ = 0 H_{\varnothing,\varnothing}=0 H ∅ , ∅ = 0 ∅ \varnothing ∅ n ≥ 2 n\ge2 n ≥ 2
The one-swap term is when ∣ S ∣ = ∣ T ∣ = 1 \lvert S\rvert=\lvert T\rvert=1 ∣ S ∣ = ∣ T ∣ = 1 S = { a } S=\{a\} S = { a } (5) runs over the cases T = { b 1 } , T = { b 2 } , … , T = { b n } T=\{b_1\},T=\{b_2\},\ldots,T=\{b_n\} T = { b 1 } , T = { b 2 } , … , T = { b n }
J ⋆ ( A 1 ; B n ) = ∑ i = 1 n e − N ( a + b i ) Z ( a , b i ) Z ( − a , − b i ) Z † ( a , − a ) Z † ( b i , − b i ) ∏ j ∈ { 1 , … , n } ∖ { i } H { a } , { b i } ( { b j } ) = ∑ i = 1 n e − N ( a + b i ) z ( a + b i ) z ( − a − b i ) ∏ j ∈ { 1 , … , n } ∖ { i } ( z ′ z ( b j − b i ) − z ′ z ( a + b j ) ) . (22) \begin{aligned}
J^\star(\mathcal{A}_1;\mathcal{B}_n)
&=\sum_{i=1}^{n}e^{-N(a+b_i)}
\frac{Z(a,b_i)Z(-a,-b_i)}{Z^\dagger(a,-a)Z^\dagger(b_i,-b_i)}
\prod_{j\in\{1,\ldots,n\}\setminus\{i\}}H_{\{a\},\{b_i\}}(\{b_j\})\\
&=\sum_{i=1}^{n}e^{-N(a+b_i)}z(a+b_i)z(-a-b_i)
\prod_{j\in\{1,\ldots,n\}\setminus\{i\}}
\left(\frac{z'}{z}(b_j-b_i)-\frac{z'}{z}(a+b_j)\right). \tag*{(22)}
\end{aligned} J ⋆ ( A 1 ; B n ) = i = 1 ∑ n e − N ( a + b i ) Z † ( a , − a ) Z † ( b i , − b i ) Z ( a , b i ) Z ( − a , − b i ) j ∈ { 1 , … , n } ∖ { i } ∏ H { a } , { b i } ({ b j }) = i = 1 ∑ n e − N ( a + b i ) z ( a + b i ) z ( − a − b i ) j ∈ { 1 , … , n } ∖ { i } ∏ ( z z ′ ( b j − b i ) − z z ′ ( a + b j ) ) . (22) where the product is over the numbers 1 to n n n i i i
We can use (12) and (13) and change into the variables A = e − a A=e^{-a} A = e − a B j = e − b j B_j=e^{-b_j} B j = e − b j
J ⋆ ( A 1 ; B n ) = ∑ i = 1 n A N B i N − A B i ( 1 − A B i ) 2 ∏ j ∈ { 1 , … , n } ∖ { i } ( − B j / B i 1 − B j / B i − − A B j 1 − A B j ) = ∑ i = 1 n A N B i N − A B i ( 1 − A B i ) 2 ∏ 1 ≤ p < i ( 1 1 − B i / B p + A B p 1 − A B p ) ∏ i < q ≤ n ( − B q / B i 1 − B q / B i + A B q 1 − A B q ) . (23) \begin{aligned}
J^\star(\mathcal{A}_1;\mathcal{B}_n)
&=\sum_{i=1}^{n}A^NB_i^N\frac{-AB_i}{(1-AB_i)^2}
\prod_{j\in\{1,\ldots,n\}\setminus\{i\}}
\left(\frac{-B_j/B_i}{1-B_j/B_i}-\frac{-AB_j}{1-AB_j}\right)\\
&=\sum_{i=1}^{n}A^NB_i^N\frac{-AB_i}{(1-AB_i)^2}
\prod_{1\le p<i}\left(\frac{1}{1-B_i/B_p}+\frac{AB_p}{1-AB_p}\right)
\prod_{i<q\le n}\left(\frac{-B_q/B_i}{1-B_q/B_i}+\frac{AB_q}{1-AB_q}\right). \tag*{(23)}
\end{aligned} J ⋆ ( A 1 ; B n ) = i = 1 ∑ n A N B i N ( 1 − A B i ) 2 − A B i j ∈ { 1 , … , n } ∖ { i } ∏ ( 1 − B j / B i − B j / B i − 1 − A B j − A B j ) = i = 1 ∑ n A N B i N ( 1 − A B i ) 2 − A B i 1 ≤ p < i ∏ ( 1 − B i / B p 1 + 1 − A B p A B p ) i < q ≤ n ∏ ( 1 − B q / B i − B q / B i + 1 − A B q A B q ) . (23) Now, since we have A B i AB_i A B i B i / B j B_i/B_j B i / B j i > j i>j i > j (12) and (13) to obtain the following,
J ∗ ( A 1 ; B n ) = ∑ i = 1 n ∑ k = 0 ∞ ( − k ) A N + k B i N + k ∏ 1 ≤ p < i ( ∑ l p = 0 ∞ ( B i l p B p − l p + A l p + 1 B p l p + 1 ) ) × ∏ i < q ≤ n ( ∑ m q = 0 ∞ ( − B q m q + 1 B i − m q − 1 + A m q + 1 B q m q + 1 ) ) . (24) J^*(A_1;B_n)=\sum_{i=1}^{n}\sum_{k=0}^{\infty}(-k)A^{N+k}B_i^{N+k}\prod_{1\leq p<i}\left(\sum_{l_p=0}^{\infty}\left(B_i^{l_p}B_p^{-l_p}+A^{l_p+1}B_p^{l_p+1}\right)\right)
\newline{}\times\prod_{i<q\leq n}\left(\sum_{m_q=0}^{\infty}\left(-B_q^{m_q+1}B_i^{-m_q-1}+A^{m_q+1}B_q^{m_q+1}\right)\right). \tag*{(24)} J ∗ ( A 1 ; B n ) = i = 1 ∑ n k = 0 ∑ ∞ ( − k ) A N + k B i N + k 1 ≤ p < i ∏ l p = 0 ∑ ∞ ( B i l p B p − l p + A l p + 1 B p l p + 1 ) × i < q ≤ n ∏ m q = 0 ∑ ∞ ( − B q m q + 1 B i − m q − 1 + A m q + 1 B q m q + 1 ) . (24) Each factor inside the products contains two terms. When we expand the products into one sum, each term corresponds to one choice from each factor. We encode these choices using binary vectors σ = ( σ p ) p < i ∈ { 0 , 1 } i − 1 \sigma=(\sigma_p)_{p<i}\in\{0,1\}^{i-1} σ = ( σ p ) p < i ∈ { 0 , 1 } i − 1 τ = ( τ q ) q > i ∈ { 0 , 1 } n − i \tau=(\tau_q)_{q>i}\in\{0,1\}^{n-i} τ = ( τ q ) q > i ∈ { 0 , 1 } n − i
σ p = { 0 if we choose B i l p B p − l p , 1 if we choose A l p + 1 B p l p + 1 , \sigma_p=
\begin{cases}
0 & \text{if we choose }B_i^{l_p}B_p^{-l_p},\\
1 & \text{if we choose }A^{l_p+1}B_p^{l_p+1},
\end{cases} σ p = { 0 1 if we choose B i l p B p − l p , if we choose A l p + 1 B p l p + 1 , τ q = { 0 if we choose − B q m q + 1 B i − m q − 1 , 1 if we choose A m q + 1 B q m q + 1 . (25) \tau_q=
\begin{cases}
0 & \text{if we choose }-B_q^{m_q+1}B_i^{-m_q-1},\\
1 & \text{if we choose }A^{m_q+1}B_q^{m_q+1}.
\end{cases} \tag*{(25)} τ q = { 0 1 if we choose − B q m q + 1 B i − m q − 1 , if we choose A m q + 1 B q m q + 1 . (25) Each complete selection of terms corresponds uniquely to a pair ( σ , τ ) (\sigma,\tau) ( σ , τ ) 1 ≤ p < i 1\leq p<i 1 ≤ p < i i < q ≤ n i<q\leq n i < q ≤ n ( σ , τ ) (\sigma,\tau) ( σ , τ )
J ∗ ( A 1 ; B n ) = ∑ i = 1 n ∑ k = 0 ∞ ∑ l 1 , … , l i − 1 = 0 ∞ ∑ m i + 1 , … , m n = 0 ∞ ∑ σ ∈ { 0 , 1 } i − 1 ∑ τ ∈ { 0 , 1 } n − i ( − k ) ( − 1 ) ( n − i ) − ∑ q > i τ q A E A ∏ r = 1 n B r E B r . (26) J^*(A_1;B_n)=\sum_{i=1}^{n}\sum_{k=0}^{\infty}\sum_{l_1,\ldots,l_{i-1}=0}^{\infty}\sum_{m_{i+1},\ldots,m_n=0}^{\infty}\sum_{\sigma\in\{0,1\}^{i-1}}\sum_{\tau\in\{0,1\}^{n-i}}(-k)(-1)^{(n-i)-\sum_{q>i}\tau_q}A^{E_A}\prod_{r=1}^{n}B_r^{E_{B_r}}. \tag*{(26)} J ∗ ( A 1 ; B n ) = i = 1 ∑ n k = 0 ∑ ∞ l 1 , … , l i − 1 = 0 ∑ ∞ m i + 1 , … , m n = 0 ∑ ∞ σ ∈ { 0 , 1 } i − 1 ∑ τ ∈ { 0 , 1 } n − i ∑ ( − k ) ( − 1 ) ( n − i ) − ∑ q > i τ q A E A r = 1 ∏ n B r E B r . (26) Here for the A A A
E A = N + k + ∑ p = 1 i − 1 σ p ( l p + 1 ) + ∑ q = i + 1 n τ q ( m q + 1 ) , (27) E_A=N+k+\sum_{p=1}^{i-1}\sigma_p(l_p+1)+\sum_{q=i+1}^{n}\tau_q(m_q+1), \tag*{(27)} E A = N + k + p = 1 ∑ i − 1 σ p ( l p + 1 ) + q = i + 1 ∑ n τ q ( m q + 1 ) , (27) and for the B B B
for r = i r=i r = i
E B i = N + k + ∑ p = 1 i − 1 ( 1 − σ p ) l p − ∑ q = i + 1 n ( 1 − τ q ) ( m q + 1 ) , (28) E_{B_i}=N+k+\sum_{p=1}^{i-1}(1-\sigma_p)l_p-\sum_{q=i+1}^{n}(1-\tau_q)(m_q+1), \tag*{(28)} E B i = N + k + p = 1 ∑ i − 1 ( 1 − σ p ) l p − q = i + 1 ∑ n ( 1 − τ q ) ( m q + 1 ) , (28) for r = p < i r=p<i r = p < i
E B p = { − l p if σ p = 0 l p + 1 σ p = 1 , (29) E_{B_p}=
\begin{cases}
-l_p & \text{if }\sigma_p=0\\
l_p+1 & \sigma_p=1,
\end{cases} \tag*{(29)} E B p = { − l p l p + 1 if σ p = 0 σ p = 1 , (29) for r = q > i r=q>i r = q > i
E B q = m q + 1 , (independent of τ q ) . (30) E_{B_q}=m_q+1,\qquad\text{(independent of }\tau_q\text{)}. \tag*{(30)} E B q = m q + 1 , (independent of τ q ) . (30) We aim to extract the coefficient of a monomial
A α ∏ r = 1 n B r β r with α ≥ 1 , β r ≥ 1. (31) A^\alpha\prod_{r=1}^{n}B_r^{\beta_r}\qquad\text{with }\alpha\geq1,\ \beta_r\geq1. \tag*{(31)} A α r = 1 ∏ n B r β r with α ≥ 1 , β r ≥ 1. (31) From (9) , we see that there must be an overall coefficient of zero on any term in the expansion in which some B r B_r B r ≤ 0 \leq0 ≤ 0 p < i p<i p < i (29) , we see that choosing σ p = 0 \sigma_p=0 σ p = 0 E B p = − l p ≤ 0 E_{B_p}=-l_p\leq0 E B p = − l p ≤ 0 B p β p B_p^{\beta_p} B p β p β p ≥ 1 \beta_p\geq1 β p ≥ 1 p < i p<i p < i σ p = 1 \sigma_p=1 σ p = 1 B B B
E B p = l p + 1 ( p < i ) , E B q = m q + 1 ( q > i ) . (32) E_{B_p}=l_p+1 \quad(p<i), \qquad E_{B_q}=m_q+1 \quad(q>i). \tag*{(32)} E B p = l p + 1 ( p < i ) , E B q = m q + 1 ( q > i ) . (32) Therefore, matching E B r = β r E_{B_r}=\beta_r E B r = β r
l p = β p − 1 ( p < i ) , m q = β q − 1 ( q > i ) , (33) l_p=\beta_p-1 \quad(p<i), \qquad m_q=\beta_q-1 \quad(q>i), \tag*{(33)} l p = β p − 1 ( p < i ) , m q = β q − 1 ( q > i ) , (33) which are automatically non-negative because β r ≥ 1 \beta_r\geq1 β r ≥ 1 i i i β 1 , … , β n \beta_1,\ldots,\beta_n β 1 , … , β n l p l_p l p m q m_q m q k k k τ = ( τ q ) q > i ∈ { 0 , 1 } n − i \tau=(\tau_q)_{q>i}\in\{0,1\}^{n-i} τ = ( τ q ) q > i ∈ { 0 , 1 } n − i l p l_p l p m q m_q m q E A E_A E A E B i E_{B_i} E B i
E A = α = N + k + ∑ p = 1 i − 1 β p + ∑ q = i + 1 n τ q β q (34) E_A=\alpha=N+k+\sum_{p=1}^{i-1}\beta_p+\sum_{q=i+1}^{n}\tau_q\beta_q \tag*{(34)} E A = α = N + k + p = 1 ∑ i − 1 β p + q = i + 1 ∑ n τ q β q (34) and
E B i = β i = N + k − ∑ q = i + 1 n ( 1 − τ q ) β q . (35) E_{B_i}=\beta_i=N+k-\sum_{q=i+1}^{n}(1-\tau_q)\beta_q. \tag*{(35)} E B i = β i = N + k − q = i + 1 ∑ n ( 1 − τ q ) β q . (35) We now also can see from (34) and (35) that α = ∑ r = 1 n β r \alpha=\sum_{r=1}^{n}\beta_r α = ∑ r = 1 n β r k k k β \beta β β i \beta_i β i
k = β i − N + ∑ q > i ( 1 − τ q ) β q . (36) k=\beta_i-N+\sum_{q>i}(1-\tau_q)\beta_q. \tag*{(36)} k = β i − N + q > i ∑ ( 1 − τ q ) β q . (36) So for a fixed i i i β \beta β τ ∈ { 0 , 1 } n − i \tau\in\{0,1\}^{n-i} τ ∈ { 0 , 1 } n − i k k k k ≥ 1 k\geq1 k ≥ 1
Therefore the contribution from the i i i A α ∏ r = 1 n B r β r A^\alpha\prod_{r=1}^{n}B_r^{\beta_r} A α ∏ r = 1 n B r β r k k k A α ∏ r = 1 n B r β r A^\alpha\prod_{r=1}^{n}B_r^{\beta_r} A α ∏ r = 1 n B r β r
∑ i = 1 n ∑ τ ∈ { 0 , 1 } n − i ( − 1 ) 1 + ( n − i ) − ∑ q > i τ q ( β i − N + ∑ q > i ( 1 − τ q ) β q ) + . (37) \sum_{i=1}^{n}\sum_{\tau\in\{0,1\}^{n-i}}(-1)^{1+(n-i)-\sum_{q>i}\tau_q}\left(\beta_i-N+\sum_{q>i}(1-\tau_q)\beta_q\right)_+. \tag*{(37)} i = 1 ∑ n τ ∈ { 0 , 1 } n − i ∑ ( − 1 ) 1 + ( n − i ) − ∑ q > i τ q ( β i − N + q > i ∑ ( 1 − τ q ) β q ) + . (37) Now instead of summing over τ \tau τ { i + 1 , … , n } \{i+1,\ldots,n\} { i + 1 , … , n } j j j i + 1 ≤ j ≤ n i+1\leq j\leq n i + 1 ≤ j ≤ n τ j \tau_j τ j i i i { 1 , … , n } \{1,\ldots,n\} { 1 , … , n } i i i 1 + ( n − i ) − ∑ q > i τ q 1+(n-i)-\sum_{q>i}\tau_q 1 + ( n − i ) − ∑ q > i τ q Γ \Gamma Γ
∫ U ( N ) Tr ( ( U † ) α ) ∏ j = 1 n Tr ( U β j ) d U = { ∑ ∅ ≠ Γ ⊆ { 1 , … , n } ( − 1 ) ∣ Γ ∣ ( ∑ r ∈ Γ β r − N ) + , if α = ∑ r = 1 n β r , 0 , otherwise. (38) \int_{U(N)}\operatorname{Tr}\left((U^\dagger)^\alpha\right)\prod_{j=1}^{n}\operatorname{Tr}\left(U^{\beta_j}\right)\,dU
=
\begin{cases}
\sum_{\varnothing\neq\Gamma\subseteq\{1,\ldots,n\}}(-1)^{|\Gamma|}\left(\sum_{r\in\Gamma}\beta_r-N\right)_+, & \text{if }\alpha=\sum_{r=1}^{n}\beta_r,\\
0, & \text{otherwise.}
\end{cases} \tag*{(38)} ∫ U ( N ) Tr ( ( U † ) α ) j = 1 ∏ n Tr ( U β j ) d U = { ∑ ∅ = Γ ⊆ { 1 , … , n } ( − 1 ) ∣Γ∣ ( ∑ r ∈ Γ β r − N ) + , 0 , if α = ∑ r = 1 n β r , otherwise. (38) Negative values of averages of traces with one U † U^\dagger U † U U U
While all averages involving just 2 or 3 traces are non-negative, numerically it can be seen that averages with one power of the trace of U † U^\dagger U † U U U Theorem 4.1 to describe the region in which these negative values arise.
Unitary matrices weighted by Haar measure can be generated numerically by following [28 ] . We show an example in Figure 1 , where values of
Figure 1. Numerical values of trace averages (39) for matrices from U ( 5 ) U(5) U ( 5 )
∫ U ( N ) Tr ( ( U † ) β 1 + β 2 + β 3 ) ∏ j = 1 3 Tr ( U β j ) d U (39) \int_{U(N)} \operatorname{Tr}\left((U^\dagger)^{\beta_1+\beta_2+\beta_3}\right)\prod_{j=1}^{3}\operatorname{Tr}\left(U^{\beta_j}\right)\,\mathrm{d}U \tag*{(39)} ∫ U ( N ) Tr ( ( U † ) β 1 + β 2 + β 3 ) j = 1 ∏ 3 Tr ( U β j ) d U (39) are plotted. Here the exponent on the trace of U † U^\dagger U † α \alpha α β 1 + β 2 + β 3 \beta_1+\beta_2+\beta_3 β 1 + β 2 + β 3 (2) if we want the average to be something other than zero.
A clear region of negativity (depicted in blue) is visible when all three parameters β 1 , β 2 , β 3 \beta_1,\beta_2,\beta_3 β 1 , β 2 , β 3 N N N ( β 1 , β 2 ) (\beta_1,\beta_2) ( β 1 , β 2 ) β 3 \beta_3 β 3 β 3 \beta_3 β 3 N N N N N N β 1 \beta_1 β 1 β 2 \beta_2 β 2 ( β 1 , β 2 ) (\beta_1,\beta_2) ( β 1 , β 2 ) β 1 + β 2 = k \beta_1+\beta_2=k β 1 + β 2 = k k k k β 3 = 2 \beta_3=2 β 3 = 2 β 3 = 3 \beta_3=3 β 3 = 3
The subset-sum formula, see Theorem 4.1 , expresses the 1–3 trace average as a linear combination of positive-part terms:
∫ U ( N ) Tr ( ( U † ) α ) ∏ j = 1 3 Tr ( U β j ) d U = { C ( β 1 , β 2 , β 3 ; N ) , α = β 1 + β 2 + β 3 , 0 , α ≠ β 1 + β 2 + β 3 . (40) \int_{U(N)}\operatorname{Tr}\left((U^\dagger)^\alpha\right)\prod_{j=1}^{3}\operatorname{Tr}\left(U^{\beta_j}\right)\,\mathrm{d}U =
\begin{cases}
C(\beta_1,\beta_2,\beta_3;N), & \alpha=\beta_1+\beta_2+\beta_3,\\
0, & \alpha\ne\beta_1+\beta_2+\beta_3.
\end{cases} \tag*{(40)} ∫ U ( N ) Tr ( ( U † ) α ) j = 1 ∏ 3 Tr ( U β j ) d U = { C ( β 1 , β 2 , β 3 ; N ) , 0 , α = β 1 + β 2 + β 3 , α = β 1 + β 2 + β 3 . (40) where
C ( β 1 , β 2 , β 3 ; N ) = − ( β 1 + β 2 + β 3 − N ) + + ( β 1 + β 2 − N ) + + ( β 1 + β 3 − N ) + + ( β 2 + β 3 − N ) + − ( β 1 − N ) + − ( β 2 − N ) + − ( β 3 − N ) + . (41) C(\beta_1,\beta_2,\beta_3;N)=-(\beta_1+\beta_2+\beta_3-N)_+ +(\beta_1+\beta_2-N)_+ +(\beta_1+\beta_3-N)_+ +(\beta_2+\beta_3-N)_+\\
-(\beta_1-N)_+-(\beta_2-N)_+-(\beta_3-N)_+. \tag*{(41)} C ( β 1 , β 2 , β 3 ; N ) = − ( β 1 + β 2 + β 3 − N ) + + ( β 1 + β 2 − N ) + + ( β 1 + β 3 − N ) + + ( β 2 + β 3 − N ) + − ( β 1 − N ) + − ( β 2 − N ) + − ( β 3 − N ) + . (41) Each term of the form ( ⋅ ) + (\cdot)_+ ( ⋅ ) + ( β 1 , β 2 , β 3 ) (\beta_1,\beta_2,\beta_3) ( β 1 , β 2 , β 3 ) ( β 1 , β 2 , β 3 ) (\beta_1,\beta_2,\beta_3) ( β 1 , β 2 , β 3 ) α = β 1 + β 2 + β 3 \alpha=\beta_1+\beta_2+\beta_3 α = β 1 + β 2 + β 3 α \alpha α
( x , y , z ) : = ( β 1 , β 2 , β 3 ) , (42) (x,y,z):=(\beta_1,\beta_2,\beta_3), \tag*{(42)} ( x , y , z ) := ( β 1 , β 2 , β 3 ) , (42) and define
F ( x , y , z ) : = ∑ ∅ ≠ S ⊂ { 1 , 2 , 3 } ( − 1 ) ∣ S ∣ ( ∑ j ∈ S β j − N ) + . (43) F(x,y,z):=\sum_{\varnothing\ne S\subset\{1,2,3\}}(-1)^{\lvert S\rvert}\left(\sum_{j\in S}\beta_j-N\right)_+. \tag*{(43)} F ( x , y , z ) := ∅ = S ⊂ { 1 , 2 , 3 } ∑ ( − 1 ) ∣ S ∣ j ∈ S ∑ β j − N + . (43) We now determine precisely where F ( x , y , z ) F(x,y,z) F ( x , y , z ) x , y , z > 0 x,y,z>0 x , y , z > 0
Step 1: localisation to the cube via singleton terms. If x ≥ N x \ge N x ≥ N x x x F ( x , y , z ) F(x,y,z) F ( x , y , z )
F ( x , y , z ) = ( y + z − N ) + − ( y − N ) + − ( z − N ) + ≥ 0. (44) F(x,y,z) = (y+z-N)_{+}-(y-N)_{+}-(z-N)_{+} \ge0. \tag*{(44)} F ( x , y , z ) = ( y + z − N ) + − ( y − N ) + − ( z − N ) + ≥ 0. (44) The same holds if y ≥ N y \ge N y ≥ N z ≥ N z \ge N z ≥ N
1 ≤ x , y , z ≤ N − 1. (45) 1 \le x,y,z \le N-1. \tag*{(45)} 1 ≤ x , y , z ≤ N − 1. (45) In this region, the singleton terms vanish, and so F F F
F ( x , y , z ) = − ( x + y + z − N ) + + ( x + y − N ) + + ( x + z − N ) + + ( y + z − N ) + . (46) F(x,y,z) = -(x+y+z-N)_{+}+(x+y-N)_{+}+(x+z-N)_{+}+(y+z-N)_{+}. \tag*{(46)} F ( x , y , z ) = − ( x + y + z − N ) + + ( x + y − N ) + + ( x + z − N ) + + ( y + z − N ) + . (46) Step 2: the lower plane x + y + z = N x+y+z=N x + y + z = N (45) , if x + y + z ≤ N x+y+z \le N x + y + z ≤ N ( x + y + z − N ) + = 0 (x+y+z-N)_{+}=0 ( x + y + z − N ) + = 0
x + y + z ≤ N ⟹ F ( x , y , z ) = 0. (47) x+y+z \le N \quad\Longrightarrow\quad F(x,y,z)=0. \tag*{(47)} x + y + z ≤ N ⟹ F ( x , y , z ) = 0. (47) Therefore, negativity (or any nonzero value) requires crossing the lower diagonal plane
x + y + z ≥ N + 1. (48) x + y + z \ge N + 1. \tag*{(48)} x + y + z ≥ N + 1. (48) In this region, we have ( x + y + z ) + = x + y + z − N (x + y + z)_{+} = x + y + z - N ( x + y + z ) + = x + y + z − N
F ( x , y , z ) = − ( x + y + z − N ) + ( x + y − N ) + + ( x + z − N ) + + ( y + z − N ) + . (49) F(x,y,z) = -(x + y + z - N) + (x + y - N)_{+} + (x + z - N)_{+} + (y + z - N)_{+}. \tag*{(49)} F ( x , y , z ) = − ( x + y + z − N ) + ( x + y − N ) + + ( x + z − N ) + + ( y + z − N ) + . (49) Step 3: the upper plane x + y + z = 2 N x + y + z = 2N x + y + z = 2 N x + y + z ≥ 2 N x + y + z \ge2N x + y + z ≥ 2 N
x + y = ( x + y + z ) − z ≥ 2 N − ( N − 1 ) = N + 1 , (50) x + y = (x + y + z) - z \ge2N - (N - 1) = N + 1, \tag*{(50)} x + y = ( x + y + z ) − z ≥ 2 N − ( N − 1 ) = N + 1 , (50) and similarly for x + z x + z x + z y + z y + z y + z
F ( x , y , z ) = − ( x + y + z ) + ( x + y − N ) + ( x + z − N ) + ( y + z − N ) (51) F(x,y,z) = -(x + y + z) + (x + y - N) + (x + z - N) + (y + z - N) \tag*{(51)} F ( x , y , z ) = − ( x + y + z ) + ( x + y − N ) + ( x + z − N ) + ( y + z − N ) (51) = x + y + z − 2 N ≥ 0. (52) = x + y + z - 2N \ge0. \tag*{(52)} = x + y + z − 2 N ≥ 0. (52) Therefore, negativity cannot occur above the upper diagonal plane
x + y + z ≥ 2 N ⟹ F ( x , y , z ) ≥ 0. (53) x + y + z \ge2N \quad\Longrightarrow\quad F(x,y,z) \ge0. \tag*{(53)} x + y + z ≥ 2 N ⟹ F ( x , y , z ) ≥ 0. (53) Step 4: negativity in the slab between the planes. Assume now that we are inside the cube (45) and in the slab
N + 1 ≤ x + y + z ≤ 2 N − 1. (54) N + 1 \le x + y + z \le2N - 1. \tag*{(54)} N + 1 ≤ x + y + z ≤ 2 N − 1. (54) We claim that F ( x , y , z ) < 0 F(x,y,z) < 0 F ( x , y , z ) < 0 x + y x + y x + y x + z x + z x + z y + z y + z y + z N N N
(i) All three pair sums exceed N N N
F ( x , y , z ) = − ( x + y + z − N ) + ( x + y − N ) + ( x + z − N ) + ( y + z − N ) = x + y + z − 2 N . (55) F(x,y,z) = -(x + y + z - N) + (x + y - N) + (x + z - N) + (y + z - N) = x + y + z - 2N. \tag*{(55)} F ( x , y , z ) = − ( x + y + z − N ) + ( x + y − N ) + ( x + z − N ) + ( y + z − N ) = x + y + z − 2 N . (55) and so F < 0 F < 0 F < 0 x + y + z = 2 N x + y + z = 2N x + y + z = 2 N
(ii) Exactly two pair sums exceed N N N x x x ( x + y − N ) + = x + y − N (x + y - N)_{+} = x + y - N ( x + y − N ) + = x + y − N ( x + z − N ) + = x + z − N (x + z - N)_{+} = x + z - N ( x + z − N ) + = x + z − N ( y + z − N ) + = 0 (y + z - N)_{+} = 0 ( y + z − N ) + = 0
F = − ( x + y + z − N ) + ( x + y − N ) + ( x + z − N ) = x − N ≤ − 1 , (56) F = -(x + y + z - N) + (x + y - N) + (x + z - N) = x - N \le-1, \tag*{(56)} F = − ( x + y + z − N ) + ( x + y − N ) + ( x + z − N ) = x − N ≤ − 1 , (56) since x ≤ N − 1 x \le N - 1 x ≤ N − 1
(iii) Exactly one pair sum exceeds N N N ( x + y − N ) + = x + y − N (x + y - N)_{+} = x + y - N ( x + y − N ) + = x + y − N
F = − ( x + y + z − N ) + ( x + y − N ) = − z ≤ − 1. (57) F = -(x + y + z - N) + (x + y - N) = -z \le-1. \tag*{(57)} F = − ( x + y + z − N ) + ( x + y − N ) = − z ≤ − 1. (57) In each case F ( x , y , z ) < 0 F(x,y,z) < 0 F ( x , y , z ) < 0
∫ U ( N ) Tr ( ( U † ) α ) ∏ j = 1 3 Tr ( U β j ) d U < 0 ⟺ 1 ≤ β 1 , β 2 , β 3 ≤ N − 1 and N + 1 ≤ β 1 + β 2 + β 3 ≤ 2 N − 1. (58) \int_{U(N)} \operatorname{Tr}\left((U^{\dagger})^{\alpha}\right) \prod_{j=1}^{3} \operatorname{Tr}\left(U^{\beta_j}\right)\,\mathrm{d}U < 0 \Longleftrightarrow1 \le\beta_1,\beta_2,\beta_3 \le N - 1 \text{ and } N + 1 \le\beta_1 + \beta_2 + \beta_3 \le2N - 1. \tag*{(58)} ∫ U ( N ) Tr ( ( U † ) α ) j = 1 ∏ 3 Tr ( U β j ) d U < 0 ⟺ 1 ≤ β 1 , β 2 , β 3 ≤ N − 1 and N + 1 ≤ β 1 + β 2 + β 3 ≤ 2 N − 1. (58) This matches the blue region in Figure 1 .
Averages of traces with two U U U U † U^\dagger U †
In Theorem 2.1 , we write e − a 1 = A 1 e^{-a_1}=A_1 e − a 1 = A 1 e − a 2 = A 2 e^{-a_2}=A_2 e − a 2 = A 2 e − b 1 = B 1 e^{-b_1}=B_1 e − b 1 = B 1 e − b 2 = B 2 e^{-b_2}=B_2 e − b 2 = B 2 ∣ A 1 ∣ , ∣ A 2 ∣ , ∣ B 1 ∣ , ∣ B 2 ∣ < 1 |A_1|,|A_2|,|B_1|,|B_2|<1 ∣ A 1 ∣ , ∣ A 2 ∣ , ∣ B 1 ∣ , ∣ B 2 ∣ < 1 ∣ A 2 ∣ < ∣ A 1 ∣ |A_2|<|A_1| ∣ A 2 ∣ < ∣ A 1 ∣ ∣ B 2 ∣ < ∣ B 1 ∣ |B_2|<|B_1| ∣ B 2 ∣ < ∣ B 1 ∣
∫ U ( N ) A 1 Λ U N Λ U ( A 1 ) A 2 Λ U N Λ U ( A 2 ) B 1 Λ U † N Λ U † ( B 1 ) B 2 Λ U † N Λ U † ( B 2 ) d U = A 1 B 1 ( 1 − A 1 B 1 ) 2 A 2 B 2 ( 1 − A 2 B 2 ) 2 + A 1 B 2 ( 1 − A 1 B 2 ) 2 A 2 B 1 ( 1 − A 2 B 1 ) 2 + A 1 N B 1 N − A 1 B 1 ( 1 − A 1 B 1 ) 2 [ ( − B 2 / B 1 1 − B 2 / B 1 + A 1 B 2 1 − A 1 B 2 ) ( − A 2 / A 1 1 − A 2 / A 1 + A 2 B 1 1 − A 2 B 1 ) + A 2 B 2 ( 1 − A 2 B 2 ) 2 ] + A 1 N B 2 N − A 1 B 2 ( 1 − A 1 B 2 ) 2 [ ( 1 1 − B 2 / B 1 + A 1 B 1 1 − A 1 B 1 ) ( − A 2 / A 1 1 − A 2 / A 1 + A 2 B 2 1 − A 2 B 2 ) + A 2 B 1 ( 1 − A 2 B 1 ) 2 ] + A 2 N B 1 N − A 2 B 1 ( 1 − A 2 B 1 ) 2 [ ( − B 2 / B 1 1 − B 2 / B 1 + A 2 B 2 1 − A 2 B 2 ) ( 1 1 − A 2 / A 1 + A 1 B 1 1 − A 1 B 1 ) + A 1 B 2 ( 1 − A 1 B 2 ) 2 ] + A 2 N B 2 N − A 2 B 2 ( 1 − A 2 B 2 ) 2 [ ( 1 1 − B 2 / B 1 + A 2 B 1 1 − A 2 B 1 ) ( 1 1 − A 2 / A 1 + A 1 B 2 1 − A 1 B 2 ) + A 1 B 1 ( 1 − A 1 B 1 ) 2 ] + A 1 N A 2 N B 1 N B 2 N − A 1 B 1 ( 1 − A 1 B 1 ) 2 − A 1 B 2 ( 1 − A 1 B 2 ) 2 − A 2 B 1 ( 1 − A 2 B 1 ) 2 − A 2 B 2 ( 1 − A 2 B 2 ) 2 ( 1 − A 2 / A 1 ) 2 ( 1 − B 2 / B 1 ) 2 ( − A 2 / A 1 ) ( − B 2 / B 1 ) . (59) \begin{aligned}
\int_{U(N)} A_1\frac{\Lambda_{U}^{N}}{\Lambda_U}(A_1)A_2\frac{\Lambda_{U}^{N}}{\Lambda_U}(A_2)B_1\frac{\Lambda_{U^\dagger}^{N}}{\Lambda_{U^\dagger}}(B_1)B_2\frac{\Lambda_{U^\dagger}^{N}}{\Lambda_{U^\dagger}}(B_2)\,dU
={}&\frac{A_1B_1}{(1-A_1B_1)^2}\frac{A_2B_2}{(1-A_2B_2)^2}
+\frac{A_1B_2}{(1-A_1B_2)^2}\frac{A_2B_1}{(1-A_2B_1)^2}\\
&+A_1^N B_1^N\frac{-A_1B_1}{(1-A_1B_1)^2}
\left[\left(\frac{-B_2/B_1}{1-B_2/B_1}+\frac{A_1B_2}{1-A_1B_2}\right)
\left(\frac{-A_2/A_1}{1-A_2/A_1}+\frac{A_2B_1}{1-A_2B_1}\right)
+\frac{A_2B_2}{(1-A_2B_2)^2}\right]\\
&+A_1^N B_2^N\frac{-A_1B_2}{(1-A_1B_2)^2}
\left[\left(\frac{1}{1-B_2/B_1}+\frac{A_1B_1}{1-A_1B_1}\right)
\left(\frac{-A_2/A_1}{1-A_2/A_1}+\frac{A_2B_2}{1-A_2B_2}\right)
+\frac{A_2B_1}{(1-A_2B_1)^2}\right]\\
&+A_2^N B_1^N\frac{-A_2B_1}{(1-A_2B_1)^2}
\left[\left(\frac{-B_2/B_1}{1-B_2/B_1}+\frac{A_2B_2}{1-A_2B_2}\right)
\left(\frac{1}{1-A_2/A_1}+\frac{A_1B_1}{1-A_1B_1}\right)
+\frac{A_1B_2}{(1-A_1B_2)^2}\right]\\
&+A_2^N B_2^N\frac{-A_2B_2}{(1-A_2B_2)^2}
\left[\left(\frac{1}{1-B_2/B_1}+\frac{A_2B_1}{1-A_2B_1}\right)
\left(\frac{1}{1-A_2/A_1}+\frac{A_1B_2}{1-A_1B_2}\right)
+\frac{A_1B_1}{(1-A_1B_1)^2}\right]\\
&+A_1^N A_2^N B_1^N B_2^N
\frac{-A_1B_1}{(1-A_1B_1)^2}
\frac{-A_1B_2}{(1-A_1B_2)^2}
\frac{-A_2B_1}{(1-A_2B_1)^2}
\frac{-A_2B_2}{(1-A_2B_2)^2}
\frac{(1-A_2/A_1)^2(1-B_2/B_1)^2}{(-A_2/A_1)(-B_2/B_1)}. \tag*{(59)}
\end{aligned} ∫ U ( N ) A 1 Λ U Λ U N ( A 1 ) A 2 Λ U Λ U N ( A 2 ) B 1 Λ U † Λ U † N ( B 1 ) B 2 Λ U † Λ U † N ( B 2 ) d U = ( 1 − A 1 B 1 ) 2 A 1 B 1 ( 1 − A 2 B 2 ) 2 A 2 B 2 + ( 1 − A 1 B 2 ) 2 A 1 B 2 ( 1 − A 2 B 1 ) 2 A 2 B 1 + A 1 N B 1 N ( 1 − A 1 B 1 ) 2 − A 1 B 1 [ ( 1 − B 2 / B 1 − B 2 / B 1 + 1 − A 1 B 2 A 1 B 2 ) ( 1 − A 2 / A 1 − A 2 / A 1 + 1 − A 2 B 1 A 2 B 1 ) + ( 1 − A 2 B 2 ) 2 A 2 B 2 ] + A 1 N B 2 N ( 1 − A 1 B 2 ) 2 − A 1 B 2 [ ( 1 − B 2 / B 1 1 + 1 − A 1 B 1 A 1 B 1 ) ( 1 − A 2 / A 1 − A 2 / A 1 + 1 − A 2 B 2 A 2 B 2 ) + ( 1 − A 2 B 1 ) 2 A 2 B 1 ] + A 2 N B 1 N ( 1 − A 2 B 1 ) 2 − A 2 B 1 [ ( 1 − B 2 / B 1 − B 2 / B 1 + 1 − A 2 B 2 A 2 B 2 ) ( 1 − A 2 / A 1 1 + 1 − A 1 B 1 A 1 B 1 ) + ( 1 − A 1 B 2 ) 2 A 1 B 2 ] + A 2 N B 2 N ( 1 − A 2 B 2 ) 2 − A 2 B 2 [ ( 1 − B 2 / B 1 1 + 1 − A 2 B 1 A 2 B 1 ) ( 1 − A 2 / A 1 1 + 1 − A 1 B 2 A 1 B 2 ) + ( 1 − A 1 B 1 ) 2 A 1 B 1 ] + A 1 N A 2 N B 1 N B 2 N ( 1 − A 1 B 1 ) 2 − A 1 B 1 ( 1 − A 1 B 2 ) 2 − A 1 B 2 ( 1 − A 2 B 1 ) 2 − A 2 B 1 ( 1 − A 2 B 2 ) 2 − A 2 B 2 ( − A 2 / A 1 ) ( − B 2 / B 1 ) ( 1 − A 2 / A 1 ) 2 ( 1 − B 2 / B 1 ) 2 . (59) The first line after the equal sign is the zero-swap term, the four similar lines in the middle are one-swap terms, and the final line is a two-swap term. We will see that the two-swap term becomes combinatorially much more complicated and this limits using Theorem 2.1 for progressively more complicated trace averages.
The zero- and one-swap terms in this case were explored in [25 ] , but here we adopt the notation that unifies them with all the other results in this paper. Identifying the two-swap terms is new and they are crucial for producing Theorem 6.1 that holds for unrestricted N N N
The contributions from the different swap terms are evaluated in the sections below, but here we will state the theorem:
Theorem 6.1. Given integers α 1 , α 2 , β 1 , β 2 \alpha_1,\alpha_2,\beta_1,\beta_2 α 1 , α 2 , β 1 , β 2 let α 1 + α 2 = β 1 + β 2 \alpha_1+\alpha_2=\beta_1+\beta_2 α 1 + α 2 = β 1 + β 2 Then the following formula holds for any N N N
∫ U ( N ) Tr ( ( U † ) α 1 ) Tr ( ( U † ) α 2 ) Tr ( U β 1 ) Tr ( U β 2 ) d U = β 1 β 2 ( δ ( α 1 = β 1 , α 2 = β 2 ) + δ ( α 1 = β 2 , α 2 = β 1 ) ) − ( β 1 + β 2 − N ) + + ( β 2 − N ) + + ( β 1 − N ) + + ( β 1 + β 2 − α 2 − N ) + − ( β 2 − α 2 − N ) + − ( β 1 − α 2 − N ) + + ( β 1 + β 2 − α 1 − N ) + − ( β 2 − α 1 − N ) + − ( β 1 − α 1 − N ) + − β 2 ( β 1 − N ) + ( δ ( α 1 = β 1 , α 2 = β 2 ) + δ ( α 1 = β 2 , α 2 = β 1 ) ) − β 1 ( β 2 − N ) + ( δ ( α 1 = β 1 , α 2 = β 2 ) + δ ( α 1 = β 2 , α 2 = β 1 ) ) + 1 β 2 > N + 1 , α 2 > N + 1 ( min { α 1 , β 1 } − N ) + + ∑ r = − 1 1 ( − 2 ) 1 − ∣ r ∣ ( ∣ α 1 − β 1 ∣ − r ) + ( β 2 − ( α 1 − β 1 ) + − N − 1 + r ) + + { 0 α 1 < β 2 or α 2 ≤ N or β 2 ≤ N ( max { α 2 , β 2 } − N − 2 ) + α 1 = β 2 and one or both of α 2 or β 2 is N + 1 ( α 2 − N ) ( β 2 − N ) α 1 = β 2 and α 2 > N + 1 and β 2 > N + 1 α 1 − β 2 − 1 α 1 > β 2 and α 2 = β 2 = N + 1 − 2 α 1 > β 2 and min { α 2 , β 2 } = N + 1 and max { α 2 , β 2 } > N + 1 2 ( α 1 − β 2 ) α 1 > β 2 and α 2 > N + 1 and β 2 > N + 1. (60) \begin{aligned}
\int_{U(N)} \operatorname{Tr}\left((U^\dagger)^{\alpha_1}\right)\operatorname{Tr}\left((U^\dagger)^{\alpha_2}\right)\operatorname{Tr}\left(U^{\beta_1}\right)\operatorname{Tr}\left(U^{\beta_2}\right)dU
={}&\beta_1\beta_2\left(\delta(\alpha_1=\beta_1,\alpha_2=\beta_2)+\delta(\alpha_1=\beta_2,\alpha_2=\beta_1)\right)\\
&-(\beta_1+\beta_2-N)_++(\beta_2-N)_++(\beta_1-N)_+\\
&+(\beta_1+\beta_2-\alpha_2-N)_+-(\beta_2-\alpha_2-N)_+-(\beta_1-\alpha_2-N)_+\\
&+(\beta_1+\beta_2-\alpha_1-N)_+-(\beta_2-\alpha_1-N)_+-(\beta_1-\alpha_1-N)_+\\
&-\beta_2(\beta_1-N)_+\left(\delta(\alpha_1=\beta_1,\alpha_2=\beta_2)+\delta(\alpha_1=\beta_2,\alpha_2=\beta_1)\right)\\
&-\beta_1(\beta_2-N)_+\left(\delta(\alpha_1=\beta_1,\alpha_2=\beta_2)+\delta(\alpha_1=\beta_2,\alpha_2=\beta_1)\right)\\
&+\mathbf{1}_{\beta_2>N+1,\alpha_2>N+1}\left(\min\{\alpha_1,\beta_1\}-N\right)_+\\
&+\sum_{r=-1}^{1}(-2)^{1-|r|}\left(|\alpha_1-\beta_1|-r\right)_+\left(\beta_2-(\alpha_1-\beta_1)_+-N-1+r\right)_+\\
&+\begin{cases}
0 & \alpha_1<\beta_2\ \text{or}\ \alpha_2\leq N\ \text{or}\ \beta_2\leq N\\
\left(\max\{\alpha_2,\beta_2\}-N-2\right)_+ & \alpha_1=\beta_2\ \text{and one or both of }\alpha_2\text{ or }\beta_2\text{ is }N+1\\
(\alpha_2-N)(\beta_2-N) & \alpha_1=\beta_2\ \text{and }\alpha_2>N+1\ \text{and }\beta_2>N+1\\
\alpha_1-\beta_2-1 & \alpha_1>\beta_2\ \text{and }\alpha_2=\beta_2=N+1\\
-2 & \alpha_1>\beta_2\ \text{and }\min\{\alpha_2,\beta_2\}=N+1\\
&\quad\text{and }\max\{\alpha_2,\beta_2\}>N+1\\
2(\alpha_1-\beta_2) & \alpha_1>\beta_2\ \text{and }\alpha_2>N+1\ \text{and }\beta_2>N+1.
\end{cases} \tag*{(60)}
\end{aligned} ∫ U ( N ) Tr ( ( U † ) α 1 ) Tr ( ( U † ) α 2 ) Tr ( U β 1 ) Tr ( U β 2 ) d U = β 1 β 2 ( δ ( α 1 = β 1 , α 2 = β 2 ) + δ ( α 1 = β 2 , α 2 = β 1 ) ) − ( β 1 + β 2 − N ) + + ( β 2 − N ) + + ( β 1 − N ) + + ( β 1 + β 2 − α 2 − N ) + − ( β 2 − α 2 − N ) + − ( β 1 − α 2 − N ) + + ( β 1 + β 2 − α 1 − N ) + − ( β 2 − α 1 − N ) + − ( β 1 − α 1 − N ) + − β 2 ( β 1 − N ) + ( δ ( α 1 = β 1 , α 2 = β 2 ) + δ ( α 1 = β 2 , α 2 = β 1 ) ) − β 1 ( β 2 − N ) + ( δ ( α 1 = β 1 , α 2 = β 2 ) + δ ( α 1 = β 2 , α 2 = β 1 ) ) + 1 β 2 > N + 1 , α 2 > N + 1 ( min { α 1 , β 1 } − N ) + + r = − 1 ∑ 1 ( − 2 ) 1 − ∣ r ∣ ( ∣ α 1 − β 1 ∣ − r ) + ( β 2 − ( α 1 − β 1 ) + − N − 1 + r ) + + ⎩ ⎨ ⎧ 0 ( max { α 2 , β 2 } − N − 2 ) + ( α 2 − N ) ( β 2 − N ) α 1 − β 2 − 1 − 2 2 ( α 1 − β 2 ) α 1 < β 2 or α 2 ≤ N or β 2 ≤ N α 1 = β 2 and one or both of α 2 or β 2 is N + 1 α 1 = β 2 and α 2 > N + 1 and β 2 > N + 1 α 1 > β 2 and α 2 = β 2 = N + 1 α 1 > β 2 and min { α 2 , β 2 } = N + 1 and max { α 2 , β 2 } > N + 1 α 1 > β 2 and α 2 > N + 1 and β 2 > N + 1. (60) Here the delta functions and the indicator function are 1 if their conditions are true and zero otherwise. The positivity notation ( x ) + (x)_+ ( x ) + x ≤ 0 x\leq0 x ≤ 0 x x x
The following table compares (60) to a numerical evaluation, following [28 ] , of the various trace averages with two powers of Tr U \operatorname{Tr}U Tr U Tr ( U † ) \operatorname{Tr}(U^\dagger) Tr ( U † )
Table 1. Averages of traces calculated by numerically generating matrices from U ( 8 ) U(8) U ( 8 ) α 1 \alpha_1 α 1 α 2 \alpha_2 α 2 β 1 \beta_1 β 1 β 2 \beta_2 β 2
Zero-swap terms
With the usual method of expanding as geometric series, the zero-swap term becomes
A 1 B 1 ( 1 − A 1 B 1 ) 2 A 2 B 2 ( 1 − A 2 B 2 ) 2 + A 1 B 2 ( 1 − A 1 B 2 ) 2 A 2 B 1 ( 1 − A 2 B 1 ) 2 = ∑ n = 0 ∞ n ( A 1 B 1 ) n ∑ m = 0 ∞ m ( A 2 B 2 ) m + ∑ n = 0 ∞ n ( A 1 B 2 ) n ∑ m = 0 ∞ m ( A 2 B 1 ) m = ∑ n , m = 0 ∞ n m A 1 n A 2 m B 1 n B 2 m + ∑ n , m = 0 ∞ n m A 1 n A 2 m B 1 m B 2 n . (61) \begin{aligned}
\frac{A_{1}B_{1}}{(1-A_{1}B_{1})^{2}}\frac{A_{2}B_{2}}{(1-A_{2}B_{2})^{2}}+\frac{A_{1}B_{2}}{(1-A_{1}B_{2})^{2}}\frac{A_{2}B_{1}}{(1-A_{2}B_{1})^{2}} \\
&=\sum_{n=0}^{\infty}n(A_{1}B_{1})^{n}\sum_{m=0}^{\infty}m(A_{2}B_{2})^{m}+\sum_{n=0}^{\infty}n(A_{1}B_{2})^{n}\sum_{m=0}^{\infty}m(A_{2}B_{1})^{m} \\
&=\sum_{n,m=0}^{\infty}nmA_{1}^{n}A_{2}^{m}B_{1}^{n}B_{2}^{m}+\sum_{n,m=0}^{\infty}nmA_{1}^{n}A_{2}^{m}B_{1}^{m}B_{2}^{n}. \tag*{(61)}
\end{aligned} ( 1 − A 1 B 1 ) 2 A 1 B 1 ( 1 − A 2 B 2 ) 2 A 2 B 2 + ( 1 − A 1 B 2 ) 2 A 1 B 2 ( 1 − A 2 B 1 ) 2 A 2 B 1 = n = 0 ∑ ∞ n ( A 1 B 1 ) n m = 0 ∑ ∞ m ( A 2 B 2 ) m + n = 0 ∑ ∞ n ( A 1 B 2 ) n m = 0 ∑ ∞ m ( A 2 B 1 ) m = n , m = 0 ∑ ∞ nm A 1 n A 2 m B 1 n B 2 m + n , m = 0 ∑ ∞ nm A 1 n A 2 m B 1 m B 2 n . (61) In line with (9), we look for the coefficient of A 1 α 1 A 2 α 2 B 1 β 1 B 2 β 2 A_{1}^{\alpha_{1}}A_{2}^{\alpha_{2}}B_{1}^{\beta_{1}}B_{2}^{\beta_{2}} A 1 α 1 A 2 α 2 B 1 β 1 B 2 β 2 ∫ U ( N ) Tr ( ( U † ) α 1 ) Tr ( ( U † ) α 2 ) Tr ( U β 1 ) Tr ( U β 2 ) d U \int_{U(N)}\operatorname{Tr}((U^{\dagger})^{\alpha_{1}})\operatorname{Tr}((U^{\dagger})^{\alpha_{2}})\operatorname{Tr}(U^{\beta_{1}})\operatorname{Tr}(U^{\beta_{2}})\,dU ∫ U ( N ) Tr (( U † ) α 1 ) Tr (( U † ) α 2 ) Tr ( U β 1 ) Tr ( U β 2 ) d U
β 1 β 2 [ δ ( α 1 = β 1 , α 2 = β 2 ) + δ ( α 1 = β 2 , α 2 = β 1 ) ] , (62) \beta_{1}\beta_{2}\left[\delta(\alpha_{1}=\beta_{1},\alpha_{2}=\beta_{2})+\delta(\alpha_{1}=\beta_{2},\alpha_{2}=\beta_{1})\right], \tag*{(62)} β 1 β 2 [ δ ( α 1 = β 1 , α 2 = β 2 ) + δ ( α 1 = β 2 , α 2 = β 1 ) ] , (62) where the delta functions are 1 when all the conditions are satisfied and zero otherwise.
One-swap terms
The first one-swap term is
A 1 N B 1 N − A 1 B 1 ( 1 − A 1 B 1 ) 2 [ ( − B 2 / B 1 1 − B 2 / B 1 + A 1 B 2 1 − A 1 B 2 ) ( − A 2 / A 1 1 − A 2 / A 1 + A 2 B 1 1 − A 2 B 1 ) + A 2 B 2 ( 1 − A 2 B 2 ) 2 ] = A 1 N B 1 N ( − ∑ n = 1 ∞ n ( A 1 B 1 ) n ) [ ∑ m = 1 ∞ ( ( A 1 B 2 ) m − ( B 2 / B 1 ) m ) ∑ j = 1 ∞ ( ( A 2 B 1 ) j − ( A 2 / A 1 ) j ) + ∑ m = 1 ∞ m ( A 2 B 2 ) m ] = − ∑ n , m , j = 1 ∞ n ( A 1 N + n + m A 2 j B 1 N + n + j B 2 m − A 1 N + n + m − j A 2 j B 1 N + n + j B 2 m − A 1 N + n A 2 j B 1 N + n + j − m B 2 m + A 1 N + n − j A 2 j B 1 N + n − m B 2 m ) \begin{aligned}
A_{1}^{N}B_{1}^{N}\frac{-A_{1}B_{1}}{(1-A_{1}B_{1})^{2}}\left[\left(\frac{-B_{2}/B_{1}}{1-B_{2}/B_{1}}+\frac{A_{1}B_{2}}{1-A_{1}B_{2}}\right)\left(\frac{-A_{2}/A_{1}}{1-A_{2}/A_{1}}+\frac{A_{2}B_{1}}{1-A_{2}B_{1}}\right)+\frac{A_{2}B_{2}}{(1-A_{2}B_{2})^{2}}\right] \\
&=A_{1}^{N}B_{1}^{N}\left(-\sum_{n=1}^{\infty}n(A_{1}B_{1})^{n}\right)\left[\sum_{m=1}^{\infty}\left((A_{1}B_{2})^{m}-(B_{2}/B_{1})^{m}\right)\sum_{j=1}^{\infty}\left((A_{2}B_{1})^{j}-(A_{2}/A_{1})^{j}\right)+\sum_{m=1}^{\infty}m(A_{2}B_{2})^{m}\right] \\
&=-\sum_{n,m,j=1}^{\infty}n\left(A_{1}^{N+n+m}A_{2}^{j}B_{1}^{N+n+j}B_{2}^{m}-A_{1}^{N+n+m-j}A_{2}^{j}B_{1}^{N+n+j}B_{2}^{m}-A_{1}^{N+n}A_{2}^{j}B_{1}^{N+n+j-m}B_{2}^{m}+A_{1}^{N+n-j}A_{2}^{j}B_{1}^{N+n-m}B_{2}^{m}\right)
\end{aligned} A 1 N B 1 N ( 1 − A 1 B 1 ) 2 − A 1 B 1 [ ( 1 − B 2 / B 1 − B 2 / B 1 + 1 − A 1 B 2 A 1 B 2 ) ( 1 − A 2 / A 1 − A 2 / A 1 + 1 − A 2 B 1 A 2 B 1 ) + ( 1 − A 2 B 2 ) 2 A 2 B 2 ] = A 1 N B 1 N ( − n = 1 ∑ ∞ n ( A 1 B 1 ) n ) [ m = 1 ∑ ∞ ( ( A 1 B 2 ) m − ( B 2 / B 1 ) m ) j = 1 ∑ ∞ ( ( A 2 B 1 ) j − ( A 2 / A 1 ) j ) + m = 1 ∑ ∞ m ( A 2 B 2 ) m ] = − n , m , j = 1 ∑ ∞ n ( A 1 N + n + m A 2 j B 1 N + n + j B 2 m − A 1 N + n + m − j A 2 j B 1 N + n + j B 2 m − A 1 N + n A 2 j B 1 N + n + j − m B 2 m + A 1 N + n − j A 2 j B 1 N + n − m B 2 m ) − ∑ n , m = 1 ∞ n m A 1 N + n A 2 m B 1 N + n B 2 m . (63) -\sum_{n,m=1}^{\infty}nmA_{1}^{N+n}A_{2}^{m}B_{1}^{N+n}B_{2}^{m}. \tag*{(63)} − n , m = 1 ∑ ∞ nm A 1 N + n A 2 m B 1 N + n B 2 m . (63) Again, we look for the coefficient of A 1 α 1 A 2 α 2 B 1 β 1 B 2 β 2 A_{1}^{\alpha_{1}}A_{2}^{\alpha_{2}}B_{1}^{\beta_{1}}B_{2}^{\beta_{2}} A 1 α 1 A 2 α 2 B 1 β 1 B 2 β 2 α 1 , α 2 , β 1 , β 2 \alpha_{1},\alpha_{2},\beta_{1},\beta_{2} α 1 , α 2 , β 1 , β 2
α 1 = N + n + m , α 2 = j , β 1 = N + n + j , β 2 = m , (64) \alpha_{1}=N+n+m,\qquad\alpha_{2}=j,\qquad\beta_{1}=N+n+j,\qquad\beta_{2}=m, \tag*{(64)} α 1 = N + n + m , α 2 = j , β 1 = N + n + j , β 2 = m , (64) which has exactly one integer solution:
n = β 1 − α 2 − N , j = α 2 , m = β 2 . (65) n=\beta_{1}-\alpha_{2}-N,\qquad j=\alpha_{2},\qquad m=\beta_{2}. \tag*{(65)} n = β 1 − α 2 − N , j = α 2 , m = β 2 . (65) However, this is only an allowable solution if all of n n n j j j m m m
− ( β 1 − α 2 − N ) + . (66) -(\beta_{1}-\alpha_{2}-N)_{+}. \tag*{(66)} − ( β 1 − α 2 − N ) + . (66) Note that due to the condition for a non-zero average, α 1 + α 2 = β 1 + β 2 \alpha_1+\alpha_2=\beta_1+\beta_2 α 1 + α 2 = β 1 + β 2 (66) , and we could equally well have written − ( α 1 − β 2 − N ) + -(\alpha_1-\beta_2-N)_+ − ( α 1 − β 2 − N ) +
The other sums in (63) are dealt with in exactly the same manner. The total contribution of the first one-swap term in (59) is
− ( β 1 − α 2 − N ) + + ( β 1 − N ) + + ( β 1 + β 2 − α 2 − N ) + − ( β 1 + β 2 − N ) + − δ ( α 1 = β 1 , α 2 = β 2 ) β 2 ( β 1 − N ) + . (67) -(\beta_1-\alpha_2-N)_+ +(\beta_1-N)_+ +(\beta_1+\beta_2-\alpha_2-N)_+-(\beta_1+\beta_2-N)_+-\delta(\alpha_1=\beta_1,\alpha_2=\beta_2)\beta_2(\beta_1-N)_+. \tag*{(67)} − ( β 1 − α 2 − N ) + + ( β 1 − N ) + + ( β 1 + β 2 − α 2 − N ) + − ( β 1 + β 2 − N ) + − δ ( α 1 = β 1 , α 2 = β 2 ) β 2 ( β 1 − N ) + . (67) The second one-swap term is
A 1 N B 2 N − A 1 B 2 ( 1 − A 1 B 2 ) 2 [ ( 1 1 − B 2 / B 1 + A 1 B 1 1 − A 1 B 1 ) ( − A 2 / A 1 1 − A 2 / A 1 + A 2 B 2 1 − A 2 B 2 ) + A 2 B 1 ( 1 − A 2 B 1 ) 2 ] = − ∑ n , m , j = 1 ∞ n ( A 1 N + n A 2 j B 1 m + 1 B 2 N + n + m + j − 1 − A 1 N + n − j A 2 j B 1 m + 1 B 2 N + n + m − 1 \begin{aligned}
A_1^N B_2^N\frac{-A_1B_2}{(1-A_1B_2)^2}\left[\left(\frac{1}{1-B_2/B_1}+\frac{A_1B_1}{1-A_1B_1}\right)\left(\frac{-A_2/A_1}{1-A_2/A_1}+\frac{A_2B_2}{1-A_2B_2}\right)+\frac{A_2B_1}{(1-A_2B_1)^2}\right] \\
&=-\sum_{n,m,j=1}^{\infty}n\left(A_1^{N+n}A_2^jB_1^{m+1}B_2^{N+n+m+j-1}-A_1^{N+n-j}A_2^jB_1^{m+1}B_2^{N+n+m-1}\right.
\end{aligned} A 1 N B 2 N ( 1 − A 1 B 2 ) 2 − A 1 B 2 [ ( 1 − B 2 / B 1 1 + 1 − A 1 B 1 A 1 B 1 ) ( 1 − A 2 / A 1 − A 2 / A 1 + 1 − A 2 B 2 A 2 B 2 ) + ( 1 − A 2 B 1 ) 2 A 2 B 1 ] = − n , m , j = 1 ∑ ∞ n ( A 1 N + n A 2 j B 1 m + 1 B 2 N + n + m + j − 1 − A 1 N + n − j A 2 j B 1 m + 1 B 2 N + n + m − 1 + A 1 N + n + m A 2 j B 1 m B 2 N + n + j − A 1 N + n + m − j A 2 j B 1 m B 2 N + n ) − ∑ n , m = 1 ∞ n m A 1 N + n A 2 m B 1 m B 2 N + n . (68) \left.+A_1^{N+n+m}A_2^jB_1^mB_2^{N+n+j}-A_1^{N+n+m-j}A_2^jB_1^mB_2^{N+n}\right)-\sum_{n,m=1}^{\infty}nmA_1^{N+n}A_2^mB_1^mB_2^{N+n}. \tag*{(68)} + A 1 N + n + m A 2 j B 1 m B 2 N + n + j − A 1 N + n + m − j A 2 j B 1 m B 2 N + n ) − n , m = 1 ∑ ∞ nm A 1 N + n A 2 m B 1 m B 2 N + n . (68) We see that the first two sums in (68) cannot contribute to the coefficient of A 1 α 1 A 2 α 2 B 1 β 1 B 2 β 2 A_1^{\alpha_1}A_2^{\alpha_2}B_1^{\beta_1}B_2^{\beta_2} A 1 α 1 A 2 α 2 B 1 β 1 B 2 β 2 β 1 \beta_1 β 1 − m + 1 -m+1 − m + 1 B 1 B_1 B 1 (68) contribute
− ( β 2 − α 2 − N ) + + ( β 2 − N ) + − δ ( α 1 = β 2 , α 2 = β 1 ) β 1 ( β 2 − N ) + . (69) -(\beta_2-\alpha_2-N)_+ +(\beta_2-N)_+-\delta(\alpha_1=\beta_2,\alpha_2=\beta_1)\beta_1(\beta_2-N)_+. \tag*{(69)} − ( β 2 − α 2 − N ) + + ( β 2 − N ) + − δ ( α 1 = β 2 , α 2 = β 1 ) β 1 ( β 2 − N ) + . (69) to the coefficient of the A 1 α 1 A 2 α 2 B 1 β 1 B 2 β 2 A_1^{\alpha_1}A_2^{\alpha_2}B_1^{\beta_1}B_2^{\beta_2} A 1 α 1 A 2 α 2 B 1 β 1 B 2 β 2
The third one-swap term in (59) is
A 2 N B 1 N − A 2 B 1 ( 1 − A 2 B 1 ) 2 [ ( − B 2 / B 1 1 − B 2 / B 1 + A 2 B 2 1 − A 2 B 2 ) ( 1 1 − A 2 / A 1 + A 1 B 1 1 − A 1 B 1 ) + A 1 B 2 ( 1 − A 1 B 2 ) 2 ] = − ∑ n , m , j = 1 ∞ n ( A 1 j + 1 A 2 N + n + m + j − 1 B 1 N + n B 2 m + A 1 j A 2 N + n + m B 1 N + n + j B 2 m \begin{aligned}
A_2^N B_1^N\frac{-A_2B_1}{(1-A_2B_1)^2}\left[\left(\frac{-B_2/B_1}{1-B_2/B_1}+\frac{A_2B_2}{1-A_2B_2}\right)\left(\frac{1}{1-A_2/A_1}+\frac{A_1B_1}{1-A_1B_1}\right)+\frac{A_1B_2}{(1-A_1B_2)^2}\right] \\
&=-\sum_{n,m,j=1}^{\infty}n\left(A_1^{j+1}A_2^{N+n+m+j-1}B_1^{N+n}B_2^m+A_1^jA_2^{N+n+m}B_1^{N+n+j}B_2^m\right.
\end{aligned} A 2 N B 1 N ( 1 − A 2 B 1 ) 2 − A 2 B 1 [ ( 1 − B 2 / B 1 − B 2 / B 1 + 1 − A 2 B 2 A 2 B 2 ) ( 1 − A 2 / A 1 1 + 1 − A 1 B 1 A 1 B 1 ) + ( 1 − A 1 B 2 ) 2 A 1 B 2 ] = − n , m , j = 1 ∑ ∞ n ( A 1 j + 1 A 2 N + n + m + j − 1 B 1 N + n B 2 m + A 1 j A 2 N + n + m B 1 N + n + j B 2 m − A 1 j + 1 A 2 N + n + j − 1 B 1 N + n − m B 2 m − A 1 j A 2 N + n B 1 N + n − m + j B 2 m ) − ∑ n , m = 1 ∞ n m A 1 m A 2 N + n B 1 N + n B 2 m . (70) \left.-A_1^{j+1}A_2^{N+n+j-1}B_1^{N+n-m}B_2^m-A_1^jA_2^{N+n}B_1^{N+n-m+j}B_2^m\right)-\sum_{n,m=1}^{\infty}nmA_1^mA_2^{N+n}B_1^{N+n}B_2^m. \tag*{(70)} − A 1 j + 1 A 2 N + n + j − 1 B 1 N + n − m B 2 m − A 1 j A 2 N + n B 1 N + n − m + j B 2 m ) − n , m = 1 ∑ ∞ nm A 1 m A 2 N + n B 1 N + n B 2 m . (70) Two of the sums have a negative exponent on A 1 A_1 A 1
− ( β 1 − α 1 − N ) + + ( β 1 + β 2 − α 1 − N ) + − δ ( α 1 = β 2 , α 2 = β 1 ) β 2 ( β 1 − N ) + . (71) -(\beta_1-\alpha_1-N)_+ +(\beta_1+\beta_2-\alpha_1-N)_+-\delta(\alpha_1=\beta_2,\alpha_2=\beta_1)\beta_2(\beta_1-N)_+. \tag*{(71)} − ( β 1 − α 1 − N ) + + ( β 1 + β 2 − α 1 − N ) + − δ ( α 1 = β 2 , α 2 = β 1 ) β 2 ( β 1 − N ) + . (71) The fourth one-swap term is
A 2 N B 2 N − A 2 B 2 ( 1 − A 2 B 2 ) 2 [ ( 1 1 − B 2 / B 1 + A 2 B 1 1 − A 2 B 1 ) ( 1 1 − A 2 / A 1 + A 1 B 2 1 − A 1 B 2 ) + A 1 B 1 ( 1 − A 1 B 1 ) 2 ] = − ∑ n , m , j = 1 ∞ n ( A 1 − j + 1 A 2 N + n + j − 1 B 1 − m + 1 B 2 N + n + m − 1 + A 1 j A 2 N + n B 1 − m + 1 B 2 N + n + m + j − 1 \begin{aligned}
A_{2}^{N}B_{2}^{N}\frac{-A_{2}B_{2}}{(1-A_{2}B_{2})^{2}}\left[\left(\frac{1}{1-B_{2}/B_{1}}+\frac{A_{2}B_{1}}{1-A_{2}B_{1}}\right)\left(\frac{1}{1-A_{2}/A_{1}}+\frac{A_{1}B_{2}}{1-A_{1}B_{2}}\right)+\frac{A_{1}B_{1}}{(1-A_{1}B_{1})^{2}}\right] \\
&= -\sum_{n,m,j=1}^{\infty}n\left(A_{1}^{-j+1}A_{2}^{N+n+j-1}B_{1}^{-m+1}B_{2}^{N+n+m-1}+A_{1}^{j}A_{2}^{N+n}B_{1}^{-m+1}B_{2}^{N+n+m+j-1}
\right.
\end{aligned} A 2 N B 2 N ( 1 − A 2 B 2 ) 2 − A 2 B 2 [ ( 1 − B 2 / B 1 1 + 1 − A 2 B 1 A 2 B 1 ) ( 1 − A 2 / A 1 1 + 1 − A 1 B 2 A 1 B 2 ) + ( 1 − A 1 B 1 ) 2 A 1 B 1 ] = − n , m , j = 1 ∑ ∞ n ( A 1 − j + 1 A 2 N + n + j − 1 B 1 − m + 1 B 2 N + n + m − 1 + A 1 j A 2 N + n B 1 − m + 1 B 2 N + n + m + j − 1 + A 1 − j + 1 A 2 N + n + m + j − 1 B 1 m B 2 N + n + A 1 j A 2 N + n + m B 1 m B 2 N + n + j ) (72) \left.
+A_{1}^{-j+1}A_{2}^{N+n+m+j-1}B_{1}^{m}B_{2}^{N+n}+A_{1}^{j}A_{2}^{N+n+m}B_{1}^{m}B_{2}^{N+n+j}\right) \tag*{(72)} + A 1 − j + 1 A 2 N + n + m + j − 1 B 1 m B 2 N + n + A 1 j A 2 N + n + m B 1 m B 2 N + n + j ) (72) − ∑ n , m = 1 ∞ n m A 1 m A 2 N + n B 1 m B 2 N + n . -\sum_{n,m=1}^{\infty}nmA_{1}^{m}A_{2}^{N+n}B_{1}^{m}B_{2}^{N+n}. − n , m = 1 ∑ ∞ nm A 1 m A 2 N + n B 1 m B 2 N + n . Here only the final two sums have no negative exponents, and these contribute
− ( β 2 − α 1 − N ) + − δ ( α 1 = β 1 , α 2 = β 2 ) β 1 ( β 2 − N ) + . (73) -(\beta_{2}-\alpha_{1}-N)_{+}-\delta(\alpha_{1}=\beta_{1},\alpha_{2}=\beta_{2})\beta_{1}(\beta_{2}-N)_{+}. \tag*{(73)} − ( β 2 − α 1 − N ) + − δ ( α 1 = β 1 , α 2 = β 2 ) β 1 ( β 2 − N ) + . (73) So we find that the contribution to the trace average ∫ U ( N ) Tr ( ( U † ) α 1 ) Tr ( ( U † ) α 2 ) Tr ( U β 1 ) Tr ( U β 2 ) d U \int_{U(N)}\operatorname{Tr}((U^{\dagger})^{\alpha_{1}})\operatorname{Tr}((U^{\dagger})^{\alpha_{2}})\operatorname{Tr}(U^{\beta_{1}})\operatorname{Tr}(U^{\beta_{2}})\,dU ∫ U ( N ) Tr (( U † ) α 1 ) Tr (( U † ) α 2 ) Tr ( U β 1 ) Tr ( U β 2 ) d U
− ( β 1 + β 2 − N ) + + ( β 2 − N ) + + ( β 1 − N ) + + ( β 1 + β 2 − α 2 − N ) + − ( β 2 − α 2 − N ) + − ( β 1 − α 2 − N ) + + ( β 1 + β 2 − α 1 − N ) + − ( β 2 − α 1 − N ) + − ( β 1 − α 1 − N ) + \begin{aligned}
-(\beta_{1}+\beta_{2}-N)_{+}+(\beta_{2}-N)_{+}+(\beta_{1}-N)_{+} \\
&+(\beta_{1}+\beta_{2}-\alpha_{2}-N)_{+}-(\beta_{2}-\alpha_{2}-N)_{+}-(\beta_{1}-\alpha_{2}-N)_{+} \\
&+(\beta_{1}+\beta_{2}-\alpha_{1}-N)_{+}-(\beta_{2}-\alpha_{1}-N)_{+}-(\beta_{1}-\alpha_{1}-N)_{+}
\end{aligned} − ( β 1 + β 2 − N ) + + ( β 2 − N ) + + ( β 1 − N ) + + ( β 1 + β 2 − α 2 − N ) + − ( β 2 − α 2 − N ) + − ( β 1 − α 2 − N ) + + ( β 1 + β 2 − α 1 − N ) + − ( β 2 − α 1 − N ) + − ( β 1 − α 1 − N ) + − [ β 2 ( β 1 − N ) + + β 1 ( β 2 − N ) + ] [ δ ( α 1 = β 1 , α 2 = β 2 ) + δ ( α 1 = β 2 , α 2 = β 1 ) ] . (74) -\left[\beta_{2}(\beta_{1}-N)_{+}+\beta_{1}(\beta_{2}-N)_{+}\right]\left[\delta(\alpha_{1}=\beta_{1},\alpha_{2}=\beta_{2})+\delta(\alpha_{1}=\beta_{2},\alpha_{2}=\beta_{1})\right]. \tag*{(74)} − [ β 2 ( β 1 − N ) + + β 1 ( β 2 − N ) + ] [ δ ( α 1 = β 1 , α 2 = β 2 ) + δ ( α 1 = β 2 , α 2 = β 1 ) ] . (74) or, to write it in a way that might suggest a generalisation,
− [ β 2 ( β 1 − N ) + + β 1 ( β 2 − N ) + ] [ δ ( α 1 = β 1 , α 2 = β 2 ) + δ ( α 1 = β 2 , α 2 = β 1 ) ] + ∑ S ⊂ { 1 , 2 } , T ⊆ { 1 , 2 } ∣ S ∣ ≤ 2 0 < ∣ T ∣ ( − 1 ) ∣ T ∣ + ∣ S ∣ + 1 ( ∑ t ∈ T β t − ∑ s ∈ S α s − N ) + . (75) \begin{aligned}
-\left[\beta_{2}(\beta_{1}-N)_{+}+\beta_{1}(\beta_{2}-N)_{+}\right]\left[\delta(\alpha_{1}=\beta_{1},\alpha_{2}=\beta_{2})+\delta(\alpha_{1}=\beta_{2},\alpha_{2}=\beta_{1})\right] \\
&+\sum_{\substack{S\subset\{1,2\},\,T\subseteq\{1,2\}\\ |S|\le2\\ 0<|T|}}(-1)^{|T|+|S|+1}\left(\sum_{t\in T}\beta_{t}-\sum_{s\in S}\alpha_{s}-N\right)_{+}. \tag*{(75)}
\end{aligned} − [ β 2 ( β 1 − N ) + + β 1 ( β 2 − N ) + ] [ δ ( α 1 = β 1 , α 2 = β 2 ) + δ ( α 1 = β 2 , α 2 = β 1 ) ] + S ⊂ { 1 , 2 } , T ⊆ { 1 , 2 } ∣ S ∣ ≤ 2 0 < ∣ T ∣ ∑ ( − 1 ) ∣ T ∣ + ∣ S ∣ + 1 ( t ∈ T ∑ β t − s ∈ S ∑ α s − N ) + . (75) where ∣ T ∣ |T| ∣ T ∣ T T T
Two-swap terms
The two-swap term in (59) is
S = A 1 N A 2 N B 1 N B 2 N − A 1 B 1 ( 1 − A 1 B 1 ) 2 − A 1 B 2 ( 1 − A 1 B 2 ) 2 − A 2 B 1 ( 1 − A 2 B 1 ) 2 − A 2 B 2 ( 1 − A 2 B 2 ) 2 ( 1 − A 2 / A 1 ) 2 − A 2 / A 1 ( 1 − B 2 / B 1 ) 2 − B 2 / B 1 . (76) S=A_{1}^{N}A_{2}^{N}B_{1}^{N}B_{2}^{N}
\frac{-A_{1}B_{1}}{(1-A_{1}B_{1})^{2}}
\frac{-A_{1}B_{2}}{(1-A_{1}B_{2})^{2}}
\frac{-A_{2}B_{1}}{(1-A_{2}B_{1})^{2}}
\frac{-A_{2}B_{2}}{(1-A_{2}B_{2})^{2}}
\frac{(1-A_{2}/A_{1})^{2}}{-A_{2}/A_{1}}
\frac{(1-B_{2}/B_{1})^{2}}{-B_{2}/B_{1}}. \tag*{(76)} S = A 1 N A 2 N B 1 N B 2 N ( 1 − A 1 B 1 ) 2 − A 1 B 1 ( 1 − A 1 B 2 ) 2 − A 1 B 2 ( 1 − A 2 B 1 ) 2 − A 2 B 1 ( 1 − A 2 B 2 ) 2 − A 2 B 2 − A 2 / A 1 ( 1 − A 2 / A 1 ) 2 − B 2 / B 1 ( 1 − B 2 / B 1 ) 2 . (76) Expanding as geometric series in the usual way, gives us the rearrangement
S = ∑ n , m , j , k = 1 ∞ n m j k A 1 N + n + m A 2 N + j + k B 1 N + n + j B 2 N + m + k M , (77) S=\sum_{n,m,j,k=1}^{\infty}nmjkA_{1}^{N+n+m}A_{2}^{N+j+k}B_{1}^{N+n+j}B_{2}^{N+m+k}M, \tag*{(77)} S = n , m , j , k = 1 ∑ ∞ nmj k A 1 N + n + m A 2 N + j + k B 1 N + n + j B 2 N + m + k M , (77) where M M M I = { − 1 , 0 , 1 } I=\{-1,0,1\} I = { − 1 , 0 , 1 }
M = ( 2 − A 1 − 1 A 2 − A 1 A 2 − 1 ) ( 2 − B 1 − 1 B 2 − B 1 B 2 − 1 ) = ( ∑ p ∈ I 2 1 − ∣ p ∣ ( − 1 ) p A 1 p A 2 − p ) ( ∑ q ∈ I 2 1 − ∣ q ∣ ( − 1 ) q B 1 q B 2 − q ) = ∑ p , q ∈ I g p , q A 1 p A 2 − p B 1 q B 2 − q . (78) \begin{aligned}
M=(2-A_{1}^{-1}A_{2}-A_{1}A_{2}^{-1})(2-B_{1}^{-1}B_{2}-B_{1}B_{2}^{-1}) \\
&=\left(\sum_{p\in I}2^{1-|p|}(-1)^{p}A_{1}^{p}A_{2}^{-p}\right)\left(\sum_{q\in I}2^{1-|q|}(-1)^{q}B_{1}^{q}B_{2}^{-q}\right) \\
&=\sum_{p,q\in I}g_{p,q}A_{1}^{p}A_{2}^{-p}B_{1}^{q}B_{2}^{-q}. \tag*{(78)}
\end{aligned} M = ( 2 − A 1 − 1 A 2 − A 1 A 2 − 1 ) ( 2 − B 1 − 1 B 2 − B 1 B 2 − 1 ) = p ∈ I ∑ 2 1 − ∣ p ∣ ( − 1 ) p A 1 p A 2 − p q ∈ I ∑ 2 1 − ∣ q ∣ ( − 1 ) q B 1 q B 2 − q = p , q ∈ I ∑ g p , q A 1 p A 2 − p B 1 q B 2 − q . (78) where g p , q = ( − 2 ) 2 − ∣ p ∣ − ∣ q ∣ = 2 2 − ∣ p ∣ − ∣ q ∣ ( − 1 ) p + q g_{p,q}=(-2)^{2-|p|-|q|}=2^{2-|p|-|q|}(-1)^{p+q} g p , q = ( − 2 ) 2 − ∣ p ∣ − ∣ q ∣ = 2 2 − ∣ p ∣ − ∣ q ∣ ( − 1 ) p + q
Now we can plug this in to S S S
S = ∑ p , q ∈ I n , m , j , k = 1 ∞ g p , q n m j k A 1 N + n + m + p A 2 N + j + k − p B 1 N + n + j + q B 2 N + m + k − q . (79) S=\sum_{\substack{p,q\in I\\ n,m,j,k=1}}^{\infty}g_{p,q}nmjkA_1^{N+n+m+p}A_2^{N+j+k-p}B_1^{N+n+j+q}B_2^{N+m+k-q}. \tag*{(79)} S = p , q ∈ I n , m , j , k = 1 ∑ ∞ g p , q nmj k A 1 N + n + m + p A 2 N + j + k − p B 1 N + n + j + q B 2 N + m + k − q . (79) This labelling gives us a systematic way to look for coefficients that match the term A 1 α 1 A 2 α 2 B 1 β 1 B 2 β 2 A_1^{\alpha_1}A_2^{\alpha_2}B_1^{\beta_1}B_2^{\beta_2} A 1 α 1 A 2 α 2 B 1 β 1 B 2 β 2 α 1 , α 2 , β 1 \alpha_1,\alpha_2,\beta_1 α 1 , α 2 , β 1 β 2 \beta_2 β 2 N + 1 N+1 N + 1
Matching exponents gives the equations
N + n + m + p = α 1 (80) N+n+m+p=\alpha_1 \tag*{(80)} N + n + m + p = α 1 (80) N + j + k − p = α 2 (81) N+j+k-p=\alpha_2 \tag*{(81)} N + j + k − p = α 2 (81) N + n + j + q = β 1 (82) N+n+j+q=\beta_1 \tag*{(82)} N + n + j + q = β 1 (82) N + m + k − q = β 2 . (83) N+m+k-q=\beta_2. \tag*{(83)} N + m + k − q = β 2 . (83) Subtracting (82) from (80) and subtracting (83) from (81) and subtracting (83) from (80) and rearranging gives (respectively)
m = α 1 − N − n − p . (84) m=\alpha_1-N-n-p. \tag*{(84)} m = α 1 − N − n − p . (84) j = β 1 − N − n − q . (85) j=\beta_1-N-n-q. \tag*{(85)} j = β 1 − N − n − q . (85) k = n − α 1 + β 2 + p + q . (86) k=n-\alpha_1+\beta_2+p+q. \tag*{(86)} k = n − α 1 + β 2 + p + q . (86) Summing over n n n ( x ) + = max ( x , 0 ) (x)_+=\max(x,0) ( x ) + = max ( x , 0 ) (77) containing A 1 α 1 A 2 α 2 B 1 β 1 B 2 β 2 A_1^{\alpha_1}A_2^{\alpha_2}B_1^{\beta_1}B_2^{\beta_2} A 1 α 1 A 2 α 2 B 1 β 1 B 2 β 2
∑ p , q ∈ I n = 1 ∞ g p , q n ( α 1 − N − n − p ) + ( β 1 − N − n − q ) + × ( n − α 1 + β 2 + p + q ) + A 1 α 1 A 2 α 2 B 1 β 1 B 2 β 2 . (87) \sum_{\substack{p,q\in I\\ n=1}}^{\infty}g_{p,q}n(\alpha_1-N-n-p)_+(\beta_1-N-n-q)_+
\times(n-\alpha_1+\beta_2+p+q)_+A_1^{\alpha_1}A_2^{\alpha_2}B_1^{\beta_1}B_2^{\beta_2}. \tag*{(87)} p , q ∈ I n = 1 ∑ ∞ g p , q n ( α 1 − N − n − p ) + ( β 1 − N − n − q ) + × ( n − α 1 + β 2 + p + q ) + A 1 α 1 A 2 α 2 B 1 β 1 B 2 β 2 . (87) Hence, from (9) , the contribution to
∫ U ( N ) Tr ( ( U † ) α 1 ) Tr ( ( U † ) α 2 ) Tr ( U β 1 ) Tr ( U β 2 ) d U (88) \int_{U(N)}\operatorname{Tr}((U^\dagger)^{\alpha_1})\operatorname{Tr}((U^\dagger)^{\alpha_2})\operatorname{Tr}(U^{\beta_1})\operatorname{Tr}(U^{\beta_2})\,dU \tag*{(88)} ∫ U ( N ) Tr (( U † ) α 1 ) Tr (( U † ) α 2 ) Tr ( U β 1 ) Tr ( U β 2 ) d U (88) from two-swap terms is
∑ n = 1 ∞ ∑ p , q ∈ I g p , q n ( α 1 − N − n − p ) + ( β 1 − N − n − q ) + ( n − α 1 + β 2 + p + q ) + . (89) \sum_{n=1}^{\infty}\sum_{p,q\in I}g_{p,q}n(\alpha_1-N-n-p)_+(\beta_1-N-n-q)_+(n-\alpha_1+\beta_2+p+q)_+. \tag*{(89)} n = 1 ∑ ∞ p , q ∈ I ∑ g p , q n ( α 1 − N − n − p ) + ( β 1 − N − n − q ) + ( n − α 1 + β 2 + p + q ) + . (89) We find that for many values of n n n
For a given n n n p , q ∈ I p,q\in I p , q ∈ I
If all three expressions inside the positive part functions are non-negative for all p , q ∈ I p,q\in I p , q ∈ I n n n
∑ p , q ∈ I g p , q ( A − p ) ( B − q ) ( C + p + q ) (90) \sum_{p,q\in I}g_{p,q}(A-p)(B-q)(C+p+q) \tag*{(90)} p , q ∈ I ∑ g p , q ( A − p ) ( B − q ) ( C + p + q ) (90) for some A A A B B B C C C p p p q q q
We rearrange (90) as a quadratic in p p p
∑ p , q ∈ I g p , q ( B − q ) ( A ( C + q ) + A p − p ( C + q ) − p 2 ) . (91) \sum_{p,q\in I} g_{p,q}(B-q)\left(A(C+q)+Ap-p(C+q)-p^{2}\right). \tag*{(91)} p , q ∈ I ∑ g p , q ( B − q ) ( A ( C + q ) + A p − p ( C + q ) − p 2 ) . (91) Now let p i = ∑ p ∈ I g p , q p i p_i=\sum_{p\in I}g_{p,q}p^i p i = ∑ p ∈ I g p , q p i i = 0 , 1 , 2 i=0,1,2 i = 0 , 1 , 2
∑ q ∈ I ( B − q ) ( A ( C + q ) p 0 + ( A − C − q ) p 1 − p 2 ) . (92) \sum_{q\in I}(B-q)\left(A(C+q)p_0+(A-C-q)p_1-p_2\right). \tag*{(92)} q ∈ I ∑ ( B − q ) ( A ( C + q ) p 0 + ( A − C − q ) p 1 − p 2 ) . (92) But we can see that p 0 p_0 p 0 p 1 p_1 p 1
p 0 = ( − 2 ) 1 − ∣ q ∣ + ( − 2 ) 2 − ∣ q ∣ + ( − 2 ) 1 − ∣ q ∣ = ( − 2 ) 1 − ∣ q ∣ ( 1 − 2 + 1 ) = 0 (93) p_0=(-2)^{1-\lvert q\rvert}+(-2)^{2-\lvert q\rvert}+(-2)^{1-\lvert q\rvert}=(-2)^{1-\lvert q\rvert}(1-2+1)=0 \tag*{(93)} p 0 = ( − 2 ) 1 − ∣ q ∣ + ( − 2 ) 2 − ∣ q ∣ + ( − 2 ) 1 − ∣ q ∣ = ( − 2 ) 1 − ∣ q ∣ ( 1 − 2 + 1 ) = 0 (93) p 1 = ( − 2 ) 1 − ∣ q ∣ ( − 1 ) + 0 + ( − 2 ) 1 − ∣ q ∣ ( 1 ) = 0 , (94) p_1=(-2)^{1-\lvert q\rvert}(-1)+0+(-2)^{1-\lvert q\rvert}(1)=0, \tag*{(94)} p 1 = ( − 2 ) 1 − ∣ q ∣ ( − 1 ) + 0 + ( − 2 ) 1 − ∣ q ∣ ( 1 ) = 0 , (94) and also that p 2 = ( − 1 ) 2 g − 1 , q + 0 + g 1 , q = 2 g 1 , q p_2=(-1)^2g_{-1,q}+0+g_{1,q}=2g_{1,q} p 2 = ( − 1 ) 2 g − 1 , q + 0 + g 1 , q = 2 g 1 , q g − p , q = g p , q g_{-p,q}=g_{p,q} g − p , q = g p , q (92) reduces to
∑ q ∈ I 2 g 1 , q ( q − B ) . (95) \sum_{q\in I}2g_{1,q}(q-B). \tag*{(95)} q ∈ I ∑ 2 g 1 , q ( q − B ) . (95) By the symmetry of p p p q q q (93) and (94) .
The previous conclusions relied on the fact that, for a given n n n (89) as p p p q q q
For a bracket like ( A − p ) + (A-p)_+ ( A − p ) + p ∈ { − 1 , 0 , 1 } p\in\{-1,0,1\} p ∈ { − 1 , 0 , 1 } A − 1 A-1 A − 1 A A A A + 1 A+1 A + 1 A = − 1 A=-1 A = − 1 ( A − p ) + (A-p)_+ ( A − p ) + p p p (89) would go to zero if α 1 − N − n ≤ − 1 \alpha_1-N-n\leq-1 α 1 − N − n ≤ − 1 A = 1 A=1 A = 1 ( A − p ) + = ( A − p ) (A-p)_+=(A-p) ( A − p ) + = ( A − p ) α 1 − N − n ≥ 1 \alpha_1-N-n\geq1 α 1 − N − n ≥ 1 (89) would be zero too, by the argument starting at (90) , as long as neither of the other expressions in the positivity functions in (89) change sign as p p p q q q A = 0 A=0 A = 0 A − p A-p A − p p p p { − 1 , 0 , 1 } \{-1,0,1\} { − 1 , 0 , 1 } α 1 − N − n = 0 \alpha_1-N-n=0 α 1 − N − n = 0
Similarly for the factor ( β 1 − N − n − q ) + (\beta_1-N-n-q)_+ ( β 1 − N − n − q ) + β 1 − N − n ≤ 1 \beta_1-N-n\leq1 β 1 − N − n ≤ 1 (89) is zero. If β 1 − N − n ≥ 1 \beta_1-N-n\geq1 β 1 − N − n ≥ 1 p p p q q q β 1 − N − n = 0 \beta_1-N-n=0 β 1 − N − n = 0
For the ( C + p + q ) + (C+p+q)_+ ( C + p + q ) + C − 2 C-2 C − 2 C − 1 C-1 C − 1 C C C C + 1 C+1 C + 1 C + 2 C+2 C + 2 C = − 2 C=-2 C = − 2 n − α 1 + β 2 ≤ − 2 n-\alpha_1+\beta_2\leq-2 n − α 1 + β 2 ≤ − 2 (89) goes to zero. If n − α 1 + β 2 ≥ 2 n-\alpha_1+\beta_2\geq2 n − α 1 + β 2 ≥ 2 p p p q q q n − α 1 + β 2 ∈ { − 1 , 0 , 1 } n-\alpha_1+\beta_2\in\{-1,0,1\} n − α 1 + β 2 ∈ { − 1 , 0 , 1 }
We will start with α 1 − N − n = 0 \alpha_1-N-n=0 α 1 − N − n = 0 n = α 1 − N n=\alpha_1-N n = α 1 − N n n n n n n n = β 1 − N n=\beta_1-N n = β 1 − N n − α 1 + β 2 ∈ { − 1 , 0 , 1 } n-\alpha_1+\beta_2\in\{-1,0,1\} n − α 1 + β 2 ∈ { − 1 , 0 , 1 } n ∈ { α 1 − β 2 − 1 , α 1 − β 2 , α 1 − β 2 + 1 } n\in\{\alpha_1-\beta_2-1,\alpha_1-\beta_2,\alpha_1-\beta_2+1\} n ∈ { α 1 − β 2 − 1 , α 1 − β 2 , α 1 − β 2 + 1 } (89) are
∑ n ∈ { ( α 1 − N ) + , ( β 1 − N ) + , ( α 1 − β 2 − 1 ) + , ( α 1 − β 2 ) + , ( α 1 − β 2 + 1 ) + } ∑ p , q ∈ { − 1 , 0 , 1 } g p , q n ( α 1 − N − n − p ) + ( β 1 − N − n − q ) + ( n − α 1 + β 2 + p + q ) + . (96) \sum_{\substack{n\in\{(\alpha_1-N)_+,\\(\beta_1-N)_+,\\(\alpha_1-\beta_2-1)_+,\\(\alpha_1-\beta_2)_+,\\(\alpha_1-\beta_2+1)_+\}}}
\sum_{p,q\in\{-1,0,1\}}g_{p,q}n(\alpha_1-N-n-p)_+(\beta_1-N-n-q)_+(n-\alpha_1+\beta_2+p+q)_+. \tag*{(96)} n ∈ {( α 1 − N ) + , ( β 1 − N ) + , ( α 1 − β 2 − 1 ) + , ( α 1 − β 2 ) + , ( α 1 − β 2 + 1 ) + } ∑ p , q ∈ { − 1 , 0 , 1 } ∑ g p , q n ( α 1 − N − n − p ) + ( β 1 − N − n − q ) + ( n − α 1 + β 2 + p + q ) + . (96) We have a maximum of 45 terms, but we will see that only a few can be non-zero. We also note that if any of the n n n α 1 − N = α 1 − β 2 + 1 \alpha_1-N=\alpha_1-\beta_2+1 α 1 − N = α 1 − β 2 + 1
Finally, the use of the positive part notation for the n n n α 1 − N \alpha_1-N α 1 − N n = ( α 1 − N ) + = 0 n=(\alpha_1-N)_+=0 n = ( α 1 − N ) + = 0
For the n = ( α 1 − N ) + n=(\alpha_1-N)_+ n = ( α 1 − N ) + p = − 1 p=-1 p = − 1
( α 1 − N ) + ∑ p , q ∈ I g p , q ( − p ) + ( β 1 − α 1 − q ) + ( β 2 − N + p + q ) + = ( α 1 − N ) + ∑ q ∈ I g 1 , q ( β 1 − α 1 − q ) + ( β 2 − N − 1 + q ) + (97) \begin{aligned}
(\alpha_1-N)_+\sum_{p,q\in I}g_{p,q}(-p)_+(\beta_1-\alpha_1-q)_+(\beta_2-N+p+q)_+ \\
&=(\alpha_1-N)_+\sum_{q\in I}g_{1,q}(\beta_1-\alpha_1-q)_+(\beta_2-N-1+q)_+ \tag*{(97)}
\end{aligned} ( α 1 − N ) + p , q ∈ I ∑ g p , q ( − p ) + ( β 1 − α 1 − q ) + ( β 2 − N + p + q ) + = ( α 1 − N ) + q ∈ I ∑ g 1 , q ( β 1 − α 1 − q ) + ( β 2 − N − 1 + q ) + (97) For the n = ( β 1 − N ) + n=(\beta_1-N)_+ n = ( β 1 − N ) +
( β 1 − N ) + ∑ p ∈ I g p , 1 ( α 1 − β 1 − p ) + ( α 2 − N − 1 + p ) + (98) (\beta_1-N)_+\sum_{p\in I}g_{p,1}(\alpha_1-\beta_1-p)_+(\alpha_2-N-1+p)_+ \tag*{(98)} ( β 1 − N ) + p ∈ I ∑ g p , 1 ( α 1 − β 1 − p ) + ( α 2 − N − 1 + p ) + (98) In order for (97) to contribute, it needs for ( β 1 − α 1 − q ) + (\beta_1-\alpha_1-q)_+ ( β 1 − α 1 − q ) + q q q β 1 − α 1 + 1 > 0 \beta_1-\alpha_1+1>0 β 1 − α 1 + 1 > 0 β 1 ≥ α 1 \beta_1\geq\alpha_1 β 1 ≥ α 1 (98) to contribute, we must have α 1 ≥ β 1 \alpha_1\geq\beta_1 α 1 ≥ β 1
If α 1 ≠ β 1 \alpha_1\neq\beta_1 α 1 = β 1 (97) or (98) will contribute, and if α 1 = β 1 \alpha_1=\beta_1 α 1 = β 1 n = N − α 1 = N − β 1 n=N-\alpha_1=N-\beta_1 n = N − α 1 = N − β 1 n n n (96) .
Since only one of these terms will show up at a time, we can combine them by replacing ( α 1 − N ) + (\alpha_1-N)_+ ( α 1 − N ) + ( β 1 − N ) + (\beta_1-N)_+ ( β 1 − N ) + ( min { α 1 , β 1 } − N ) + (\min\{\alpha_1,\beta_1\}-N)_+ ( min { α 1 , β 1 } − N ) + (97) and (98) with ( ∣ α 1 − β 1 ∣ − r ) + (\lvert\alpha_1-\beta_1\rvert-r)_+ (∣ α 1 − β 1 ∣ − r ) + β 1 ≥ α 1 \beta_1\geq\alpha_1 β 1 ≥ α 1 ( β 2 − N − 1 + q ) + (\beta_2-N-1+q)_+ ( β 2 − N − 1 + q ) + ( α 2 − N − 1 + p ) + (\alpha_2-N-1+p)_+ ( α 2 − N − 1 + p ) + ( β 2 − ( α 1 − β 1 ) + − N − 1 + r ) + (\beta_2-(\alpha_1-\beta_1)_+-N-1+r)_+ ( β 2 − ( α 1 − β 1 ) + − N − 1 + r ) + n = ( α 1 − N ) + n=(\alpha_1-N)_+ n = ( α 1 − N ) + n = ( β 1 − N ) + n=(\beta_1-N)_+ n = ( β 1 − N ) +
( min { α 1 , β 1 } − N ) + ∑ r ∈ I g r , 1 ( ∣ α 1 − β 1 ∣ − r ) + ( β 2 − ( α 1 − β 1 ) + − N − 1 + r ) + (99) (\min\{\alpha_1,\beta_1\}-N)_+\sum_{r\in I}g_{r,1}(\lvert\alpha_1-\beta_1\rvert-r)_+(\beta_2-(\alpha_1-\beta_1)_+-N-1+r)_+ \tag*{(99)} ( min { α 1 , β 1 } − N ) + r ∈ I ∑ g r , 1 (∣ α 1 − β 1 ∣ − r ) + ( β 2 − ( α 1 − β 1 ) + − N − 1 + r ) + (99) This term only exists when all α 1 , α 2 , β 1 , β 2 > N \alpha_1,\alpha_2,\beta_1,\beta_2>N α 1 , α 2 , β 1 , β 2 > N α \alpha α β \beta β N N N (76) . We can see this condition in some of the positivity notation in (99) .
Now for the n = ( α 1 − β 2 + r ) + n=(\alpha_1-\beta_2+r)_+ n = ( α 1 − β 2 + r ) + r ∈ I r\in I r ∈ I E E E
E = ∑ r ∈ I ( α 1 − β 2 + r ) + ∑ p , q ∈ I g p , q ( β 2 − N − r − p ) + ( α 2 − N − r − q ) + ( p + q + r ) + (100) E=\sum_{r\in I}(\alpha_1-\beta_2+r)_+\sum_{p,q\in I}g_{p,q}(\beta_2-N-r-p)_+(\alpha_2-N-r-q)_+(p+q+r)_+ \tag*{(100)} E = r ∈ I ∑ ( α 1 − β 2 + r ) + p , q ∈ I ∑ g p , q ( β 2 − N − r − p ) + ( α 2 − N − r − q ) + ( p + q + r ) + (100) We can write this expression in terms of A = α 1 − β 2 A=\alpha_1-\beta_2 A = α 1 − β 2 B = β 2 − N B=\beta_2-N B = β 2 − N C = α 2 − N C=\alpha_2-N C = α 2 − N
E = ∑ p , q , r ∈ I g p , q ( A + r ) + ( B − r − p ) + ( C − r − q ) + ( p + q + r ) + (101) E=\sum_{p,q,r\in I}g_{p,q}(A+r)_+(B-r-p)_+(C-r-q)_+(p+q+r)_+ \tag*{(101)} E = p , q , r ∈ I ∑ g p , q ( A + r ) + ( B − r − p ) + ( C − r − q ) + ( p + q + r ) + (101) If A < 0 A<0 A < 0 B < 1 B<1 B < 1 C < 1 C<1 C < 1 A < 0 A<0 A < 0 ( A + 1 ) + (A+1)_+ ( A + 1 ) + ( A ) + (A)_+ ( A ) + ( A − 1 ) + (A-1)_+ ( A − 1 ) +
We look first at the case A = 0 A=0 A = 0 r = 1 r=1 r = 1
E A = 0 = ∑ p , q ∈ I g p , q ( B − 1 − p ) + ( C − 1 − q ) + ( p + q + 1 ) + (102) E_{A=0}=\sum_{p,q\in I}g_{p,q}(B-1-p)_+(C-1-q)_+(p+q+1)_+ \tag*{(102)} E A = 0 = p , q ∈ I ∑ g p , q ( B − 1 − p ) + ( C − 1 − q ) + ( p + q + 1 ) + (102) If, in addition, B = 1 B=1 B = 1 ( B − 1 − p ) + (B-1-p)_+ ( B − 1 − p ) + p = − 1 p=-1 p = − 1 p + q > − 1 p+q>-1 p + q > − 1 q = 1 q=1 q = 1 E A = 0 ; B = 1 = g − 1 , 1 ( C − 2 ) + = ( C − 2 ) + E_{A=0;B=1}=g_{-1,1}(C-2)_+=(C-2)_+ E A = 0 ; B = 1 = g − 1 , 1 ( C − 2 ) + = ( C − 2 ) + A = 0 A=0 A = 0 C = 1 C=1 C = 1 E A = 0 ; C = 1 = ( B − 2 ) + E_{A=0;C=1}=(B-2)_+ E A = 0 ; C = 1 = ( B − 2 ) + B B B C C C A = 0 A=0 A = 0 B B B C C C 1 1 1 ( max { B , C } − 2 ) + (\max\{B,C\}-2)_+ ( max { B , C } − 2 ) +
If A = 0 A=0 A = 0 B , C > 1 B,C>1 B , C > 1
E A = 0 ; B , C > 1 ∑ p , q ∈ I g p , q ( B − 1 − p ) ( C − 1 − q ) ( p + q + 1 ) + = B C . (103) E_{A=0;B,C>1}\sum_{p,q\in I}g_{p,q}(B-1-p)(C-1-q)(p+q+1)_+=BC. \tag*{(103)} E A = 0 ; B , C > 1 p , q ∈ I ∑ g p , q ( B − 1 − p ) ( C − 1 − q ) ( p + q + 1 ) + = B C . (103) Now we consider the case that A > 0 A>0 A > 0 A − 1 ≥ 0 A-1\geq0 A − 1 ≥ 0 r r r ( A + r ) + = ( A + r ) (A+r)_+=(A+r) ( A + r ) + = ( A + r )
E A > 0 = ∑ p , q , r ∈ I g p , q ( A + r ) ( B − r − p ) + ( C − r − q ) + ( p + q + r ) + . (104) E_{A>0}=\sum_{p,q,r\in I}g_{p,q}(A+r)(B-r-p)_+(C-r-q)_+(p+q+r)_+. \tag*{(104)} E A > 0 = p , q , r ∈ I ∑ g p , q ( A + r ) ( B − r − p ) + ( C − r − q ) + ( p + q + r ) + . (104) If A > 0 A>0 A > 0 B = C = 1 B=C=1 B = C = 1 r = − 1 r=-1 r = − 1 r = 0 r=0 r = 0 1 − r − p > 0 1-r-p>0 1 − r − p > 0 p < 1 p<1 p < 1 ( C − r − q ) + (C-r-q)_+ ( C − r − q ) + q < 1 q<1 q < 1 ( p + q + r ) + (p+q+r)_+ ( p + q + r ) + p < 1 p<1 p < 1 q < 1 q<1 q < 1 r = 0 r=0 r = 0 r = 1 r=1 r = 1 ( B − r − p ) + (B-r-p)_+ ( B − r − p ) + p < 0 p<0 p < 0 p = − 1 p=-1 p = − 1 ( C − r − q ) + (C-r-q)_+ ( C − r − q ) + q < 0 q<0 q < 0 r = − 1 r=-1 r = − 1 ( p + q + r ) + (p+q+r)_+ ( p + q + r ) + p = q = 1 p=q=1 p = q = 1 E A > 0 ; B = C = 1 = g 1 , 1 ( A − 1 ) ( 1 + 1 − 1 ) ( 1 + 1 − 1 ) ( 1 + 1 − 1 ) = ( A − 1 ) E_{A>0;B=C=1}=g_{1,1}(A-1)(1+1-1)(1+1-1)(1+1-1)=(A-1) E A > 0 ; B = C = 1 = g 1 , 1 ( A − 1 ) ( 1 + 1 − 1 ) ( 1 + 1 − 1 ) ( 1 + 1 − 1 ) = ( A − 1 )
If A > 0 A>0 A > 0 B = 1 B=1 B = 1 C > 1 C>1 C > 1
E A > 0 ; B = 1 ; C > 1 = ∑ p , q , r ∈ I g p , q ( A + r ) ( 1 − r − p ) + ( C − r − q ) + ( p + q + r ) + . (105) E_{A>0;B=1;C>1}=\sum_{p,q,r\in I}g_{p,q}(A+r)(1-r-p)_+(C-r-q)_+(p+q+r)_+. \tag*{(105)} E A > 0 ; B = 1 ; C > 1 = p , q , r ∈ I ∑ g p , q ( A + r ) ( 1 − r − p ) + ( C − r − q ) + ( p + q + r ) + . (105) If r = 1 r=1 r = 1 ( 1 − r − p ) + (1-r-p)_+ ( 1 − r − p ) + p = − 1 p=-1 p = − 1 ( p + q + r ) + (p+q+r)_+ ( p + q + r ) + q = 1 q=1 q = 1 g 1 , 1 ( A + 1 ) ( 1 ) ( C − 2 ) ( 1 ) = ( A + 1 ) ( C − 2 ) g_{1,1}(A+1)(1)(C-2)(1)=(A+1)(C-2) g 1 , 1 ( A + 1 ) ( 1 ) ( C − 2 ) ( 1 ) = ( A + 1 ) ( C − 2 ) r = 0 r=0 r = 0 1 − r − p 1-r-p 1 − r − p p = 0 , − 1 p=0,-1 p = 0 , − 1 p + q > 0 p+q>0 p + q > 0 p = 0 p=0 p = 0 q = 1 q=1 q = 1 g 0 , 1 A ( 1 ) ( C − 1 ) ( 1 ) = A ( − 2 C + 2 ) g_{0,1}A(1)(C-1)(1)=A(-2C+2) g 0 , 1 A ( 1 ) ( C − 1 ) ( 1 ) = A ( − 2 C + 2 ) r = − 1 r=-1 r = − 1 ( p + q + r ) + (p+q+r)_+ ( p + q + r ) + p = q = 1 p=q=1 p = q = 1 g 1 , 1 ( A − 1 ) ( 1 ) ( C + 1 − 1 ) ( 1 ) = ( A − 1 ) C g_{1,1}(A-1)(1)(C+1-1)(1)=(A-1)C g 1 , 1 ( A − 1 ) ( 1 ) ( C + 1 − 1 ) ( 1 ) = ( A − 1 ) C E A > 0 ; B = 1 ; C > 1 = ( A + 1 ) ( C − 2 ) + A ( − 2 C + 2 ) + ( A − 1 ) C = − 2 E_{A>0;B=1;C>1}=(A+1)(C-2)+A(-2C+2)+(A-1)C=-2 E A > 0 ; B = 1 ; C > 1 = ( A + 1 ) ( C − 2 ) + A ( − 2 C + 2 ) + ( A − 1 ) C = − 2 E A > 0 ; B > 1 ; C = 1 = − 2 E_{A>0;B>1;C=1}=-2 E A > 0 ; B > 1 ; C = 1 = − 2
If A > 0 A>0 A > 0 B , C > 1 B,C>1 B , C > 1
E A > 0 ; B , C > 1 = ∑ p , q , r ∈ I g p , q ( A + r ) ( B − r − p ) ( C − r − q ) ( p + q + r ) + (106) E_{A>0;B,C>1}=\sum_{p,q,r\in I}g_{p,q}(A+r)(B-r-p)(C-r-q)(p+q+r)_+ \tag*{(106)} E A > 0 ; B , C > 1 = p , q , r ∈ I ∑ g p , q ( A + r ) ( B − r − p ) ( C − r − q ) ( p + q + r ) + (106) = ( A + 1 ) [ ( B − 2 ) C + B ( C − 2 ) + 4 ( B − 1 ) ( C − 1 ) − 4 ( B − 2 ) ( C − 1 ) − 4 ( B − 1 ) ( C − 2 ) + 3 ( B − 2 ) ( C − 2 ) ] + A [ − 2 B ( C − 1 ) − 2 ( B − 1 ) C + 2 ( B − 1 ) ( C − 1 ) ] + ( A − 1 ) B C (107) \begin{aligned}
={}&(A+1)[(B-2)C+B(C-2)+4(B-1)(C-1)\\
&\quad-4(B-2)(C-1)-4(B-1)(C-2)+3(B-2)(C-2)]\\
&+A[-2B(C-1)-2(B-1)C+2(B-1)(C-1)]\\
&+(A-1)BC \tag*{(107)}
\end{aligned} = ( A + 1 ) [( B − 2 ) C + B ( C − 2 ) + 4 ( B − 1 ) ( C − 1 ) − 4 ( B − 2 ) ( C − 1 ) − 4 ( B − 1 ) ( C − 2 ) + 3 ( B − 2 ) ( C − 2 )] + A [ − 2 B ( C − 1 ) − 2 ( B − 1 ) C + 2 ( B − 1 ) ( C − 1 )] + ( A − 1 ) B C (107) = 2 A . (108) =2A. \tag*{(108)} = 2 A . (108) So in summary,
E = { 0 A < 0 or B < 1 or C < 1 ( max { B , C } − 2 ) + A = 0 and one or both of B or C is 1 B C A = 0 and B > 1 and C > 1 A − 1 A > 0 and B = C = 1 − 2 A > 0 and min { B , C } = 1 , max { B , C } > 1 2 A A > 0 and B > 1 and C > 1. (109) E=
\begin{cases}
0 & A<0\text{ or }B<1\text{ or }C<1\\
(\max\{B,C\}-2)_+ & A=0\text{ and one or both of }B\text{ or }C\text{ is }1\\
BC & A=0\text{ and }B>1\text{ and }C>1\\
A-1 & A>0\text{ and }B=C=1\\
-2 & A>0\text{ and }\min\{B,C\}=1,\max\{B,C\}>1\\
2A & A>0\text{ and }B>1\text{ and }C>1.
\end{cases} \tag*{(109)} E = ⎩ ⎨ ⎧ 0 ( max { B , C } − 2 ) + B C A − 1 − 2 2 A A < 0 or B < 1 or C < 1 A = 0 and one or both of B or C is 1 A = 0 and B > 1 and C > 1 A > 0 and B = C = 1 A > 0 and min { B , C } = 1 , max { B , C } > 1 A > 0 and B > 1 and C > 1. (109) Recalling definitions of A A A B B B C C C n = α 1 − β 2 − 1 , α 1 − β 2 n=\alpha_1-\beta_2-1,\alpha_1-\beta_2 n = α 1 − β 2 − 1 , α 1 − β 2 α 1 − β 2 + 1 \alpha_1-\beta_2+1 α 1 − β 2 + 1
E ( α 1 , α 2 , β 1 , β 2 , N ) = E = { 0 α 1 < β 2 or α 2 ≤ N or β 2 ≤ N , ( max { α 2 , β 2 } − N − 2 ) + α 1 = β 2 and one or both of α 2 or β 2 is N + 1 , ( α 2 − N ) ( β 2 − N ) α 1 = β 2 and α 2 > N + 1 and β 2 > N + 1 , α 1 − β 2 − 1 α 1 > β 2 and α 2 = β 2 = N + 1 , − 2 α 1 > β 2 and min { α 2 , β 2 } = N + 1 and max { α 2 , β 2 } > N + 1 , 2 ( α 1 − β 2 ) α 1 > β 2 and α 2 > N + 1 and β 2 > N + 1. (110) E(\alpha_1,\alpha_2,\beta_1,\beta_2,N)=E=
\begin{cases}
0 & \alpha_1<\beta_2\text{ or }\alpha_2\le N\text{ or }\beta_2\le N,\\
(\max\{\alpha_2,\beta_2\}-N-2)_+ & \alpha_1=\beta_2\text{ and one or both of }\alpha_2\text{ or }\beta_2\text{ is }N+1,\\
(\alpha_2-N)(\beta_2-N) & \alpha_1=\beta_2\text{ and }\alpha_2>N+1\text{ and }\beta_2>N+1,\\
\alpha_1-\beta_2-1 & \alpha_1>\beta_2\text{ and }\alpha_2=\beta_2=N+1,\\
-2 & \alpha_1>\beta_2\text{ and }\min\{\alpha_2,\beta_2\}=N+1\\
& \text{and }\max\{\alpha_2,\beta_2\}>N+1,\\
2(\alpha_1-\beta_2) & \alpha_1>\beta_2\text{ and }\alpha_2>N+1\text{ and }\beta_2>N+1.
\end{cases} \tag*{(110)} E ( α 1 , α 2 , β 1 , β 2 , N ) = E = ⎩ ⎨ ⎧ 0 ( max { α 2 , β 2 } − N − 2 ) + ( α 2 − N ) ( β 2 − N ) α 1 − β 2 − 1 − 2 2 ( α 1 − β 2 ) α 1 < β 2 or α 2 ≤ N or β 2 ≤ N , α 1 = β 2 and one or both of α 2 or β 2 is N + 1 , α 1 = β 2 and α 2 > N + 1 and β 2 > N + 1 , α 1 > β 2 and α 2 = β 2 = N + 1 , α 1 > β 2 and min { α 2 , β 2 } = N + 1 and max { α 2 , β 2 } > N + 1 , α 1 > β 2 and α 2 > N + 1 and β 2 > N + 1. (110) When we combine (99) and (110) we need to account for the fact that if α 1 − N = α 1 − β 2 + 1 \alpha_1-N=\alpha_1-\beta_2+1 α 1 − N = α 1 − β 2 + 1 (99) and E E E α 1 − N ≠ α 1 − β 2 − r \alpha_1-N\ne\alpha_1-\beta_2-r α 1 − N = α 1 − β 2 − r − 1 ≤ r ≤ 1 -1\le r\le1 − 1 ≤ r ≤ 1 N ≠ β 2 + r N\ne\beta_2+r N = β 2 + r β 1 − N ≠ α 1 − β 2 − r \beta_1-N\ne\alpha_1-\beta_2-r β 1 − N = α 1 − β 2 − r α 1 + α 2 = β 1 + β 2 \alpha_1+\alpha_2=\beta_1+\beta_2 α 1 + α 2 = β 1 + β 2 N ≠ α 2 + r N\ne\alpha_2+r N = α 2 + r − 1 ≤ r ≤ 1 -1\le r\le1 − 1 ≤ r ≤ 1 n n n β 2 = N + 1 \beta_2=N+1 β 2 = N + 1 α 2 = N + 1 \alpha_2=N+1 α 2 = N + 1 N + 1 N+1 N + 1 (79) ). This results in the following 2-swap contribution (where the indicator function is 1 when both conditions hold and 0 otherwise)
1 β 2 > N + 1 , α 2 > N + 1 ( min { α 1 , β 1 } − N ) + × ∑ r ∈ I g r , 1 ( ∣ α 1 − β 1 ∣ − r ) + ( β 2 − ( α 1 − β 1 ) + − N − 1 + r ) + + E ( α 1 , α 2 , β 1 , β 2 , N ) . (111) \mathbf{1}_{\beta_2>N+1,\alpha_2>N+1}(\min\{\alpha_1,\beta_1\}-N)_+
\times\sum_{r\in I}g_{r,1}(|\alpha_1-\beta_1|-r)_+(\beta_2-(\alpha_1-\beta_1)_+-N-1+r)_+
+E(\alpha_1,\alpha_2,\beta_1,\beta_2,N). \tag*{(111)} 1 β 2 > N + 1 , α 2 > N + 1 ( min { α 1 , β 1 } − N ) + × r ∈ I ∑ g r , 1 ( ∣ α 1 − β 1 ∣ − r ) + ( β 2 − ( α 1 − β 1 ) + − N − 1 + r ) + + E ( α 1 , α 2 , β 1 , β 2 , N ) . (111) General trace moments
The designation of terms as zero-swap, one-swap, etc, is not just convenient book-keeping. It was shown in a series of papers [12 , 27 , 2 , 14 , 3 ] to be of critical importance when one studies the n n n n n n L L L [24 , 23 , 9 , 11 , 10 , 26 ] ). For example, if one investigates
∫ U ( N ) ∑ j 1 , … , j n F ( θ j 1 , … , θ j n ) d U (112) \int_{U(N)}\sum_{j_1,\ldots,j_n}F(\theta_{j_1},\ldots,\theta_{j_n})\,\mathrm{d}U \tag*{(112)} ∫ U ( N ) j 1 , … , j n ∑ F ( θ j 1 , … , θ j n ) d U (112) in order to probe the behaviour of n n n F F F (112) can be calculated using J ∗ J^* J ∗ Theorem 2.1 . If the Fourier transform of the test function has support only when the sum of certain Fourier variables is less than 2 q − ϵ 2q-\epsilon 2 q − ϵ ϵ \epsilon ϵ J ∗ J^* J ∗ ∣ S ∣ = ∣ T ∣ < q |S|=|T|<q ∣ S ∣ = ∣ T ∣ < q α \alpha α N + 1 N+1 N + 1 2 N + 2 2N+2 2 N + 2
In the following sections, we will address the zero- and one-swap terms for a general trace moment, one of the consequences of which is the following theorem, which approximately doubles the range of Diaconis and Shahshahani’s original result:
Theorem 7.1. Given integers α 1 , … , α K , β 1 , … , β L \alpha_1,\ldots,\alpha_K,\beta_1,\ldots,\beta_L α 1 , … , α K , β 1 , … , β L α 1 + ⋯ + α K = β 1 + ⋯ + β L < 2 N + 2 \alpha_1+\cdots+\alpha_K=\beta_1+\cdots+\beta_L<2N+2 α 1 + ⋯ + α K = β 1 + ⋯ + β L < 2 N + 2 K = { 1 , … , K } \mathcal{K}=\{1,\ldots,K\} K = { 1 , … , K } L = { 1 , … , L } \mathcal{L}=\{1,\ldots,L\} L = { 1 , … , L }
∫ U ( N ) Tr ( ( U † ) α 1 ) ⋯ Tr ( ( U † ) α K ) Tr ( U β 1 ) ⋯ Tr ( U β L ) d U = δ ( K = L ) ∑ ( K : L ) ( ∏ j = 1 L β j ) δ ( α k 1 = β ℓ 1 , α k 2 = β ℓ 2 , … , α k L = β ℓ L ) + ∑ R ⊊ K , Q ⊊ L ∣ R ∣ = ∣ Q ∣ ( ∑ ( R : Q ) ( ∏ j = 1 ∣ Q ∣ β q j ) ( δ ( α r 1 = β q 1 , … , α r ∣ Q ∣ = β q ∣ Q ∣ ) ) ) × ∑ S ⊂ R c , T ⊆ Q c ∣ S ∣ < ∣ R c ∣ 0 < ∣ T ∣ ( − 1 ) ∣ T ∣ + ∣ R c ∣ − ∣ S ∣ − 1 ( ∑ t ∈ T β t − ∑ s ∈ S α s − N ) + . (113) \begin{aligned}
\int_{U(N)} \operatorname{Tr}\left((U^\dagger)^{\alpha_1}\right)\cdots\operatorname{Tr}\left((U^\dagger)^{\alpha_K}\right)\operatorname{Tr}\left(U^{\beta_1}\right)\cdots\operatorname{Tr}\left(U^{\beta_L}\right)\,dU
={}&\delta(K=L)\sum_{(\mathcal{K}:\mathcal{L})}\left(\prod_{j=1}^{L}\beta_j\right)\delta(\alpha_{k_1}=\beta_{\ell_1},\alpha_{k_2}=\beta_{\ell_2},\ldots,\alpha_{k_L}=\beta_{\ell_L})\\
&+\sum_{\substack{R\subsetneq\mathcal{K},\,Q\subsetneq\mathcal{L}\\ |R|=|Q|}}\left(\sum_{(\mathcal{R}:\mathcal{Q})}\left(\prod_{j=1}^{|Q|}\beta_{q_j}\right)\left(\delta(\alpha_{r_1}=\beta_{q_1},\ldots,\alpha_{r_{|Q|}}=\beta_{q_{|Q|}})\right)\right)\\
&\times\sum_{\substack{S\subset R^c,T\subseteq Q^c\\ |S|<|R^c|\\ 0<|T|}}(-1)^{|T|+|R^c|-|S|-1}\left(\sum_{t\in T}\beta_t-\sum_{s\in S}\alpha_s-N\right)_+ . \tag*{(113)}
\end{aligned} ∫ U ( N ) Tr ( ( U † ) α 1 ) ⋯ Tr ( ( U † ) α K ) Tr ( U β 1 ) ⋯ Tr ( U β L ) d U = δ ( K = L ) ( K : L ) ∑ ( j = 1 ∏ L β j ) δ ( α k 1 = β ℓ 1 , α k 2 = β ℓ 2 , … , α k L = β ℓ L ) + R ⊊ K , Q ⊊ L ∣ R ∣ = ∣ Q ∣ ∑ ( R : Q ) ∑ j = 1 ∏ ∣ Q ∣ β q j ( δ ( α r 1 = β q 1 , … , α r ∣ Q ∣ = β q ∣ Q ∣ ) ) × S ⊂ R c , T ⊆ Q c ∣ S ∣ < ∣ R c ∣ 0 < ∣ T ∣ ∑ ( − 1 ) ∣ T ∣ + ∣ R c ∣ − ∣ S ∣ − 1 ( t ∈ T ∑ β t − s ∈ S ∑ α s − N ) + . (113) Here the notation ( K : L ) (\mathcal{K}:\mathcal{L}) ( K : L ) K \mathcal{K} K L \mathcal{L} L ( α k 1 , β ℓ 1 ) , ( α k 2 , β ℓ 2 ) … ( α k ∣ K ∣ , β ℓ ∣ L ∣ ) (\alpha_{k_1},\beta_{\ell_1}),(\alpha_{k_2},\beta_{\ell_2})\ldots(\alpha_{k_{|\mathcal{K}|}},\beta_{\ell_{|\mathcal{L}|}}) ( α k 1 , β ℓ 1 ) , ( α k 2 , β ℓ 2 ) … ( α k ∣ K ∣ , β ℓ ∣ L ∣ ) δ \delta δ R c R^c R c R R R K \mathcal{K} K ( x ) + (x)_+ ( x ) + x ≤ 0 x\leq0 x ≤ 0 x x x
Proof. The proof follows from the derivation of the zero- and one-swap terms in the following sections, with the additional information that two-swap and higher terms will always have α 1 + ⋯ + α K ≥ 2 N + 2 \alpha_1+\cdots+\alpha_K\geq2N+2 α 1 + ⋯ + α K ≥ 2 N + 2 (76) , where we can see from (77) and (78) that the two selected A A A B B B N + 1 N+1 N + 1 H S , T ( W ) H_{S,T}(W) H S , T ( W ) (6) , but the contribution of these to the sum of exponents on all the A A A B B B (12) and (13) . Similarly, for higher swap terms, say ℓ \ell ℓ ℓ \ell ℓ A A A ℓ \ell ℓ B B B N + 1 N+1 N + 1
Zero-swap terms
We examine the zero-swap terms of a general average of traces. This is covered by Kuperberg and Lalín [25 ] , but we include it here for completeness and to ensure consistency of notation. Write a general moment of logarithmic derivatives as
∫ U ( N ) ∏ k ∈ K ( − e − α k ) Λ U ′ Λ U ( e − α k ) ∏ ℓ ∈ L ( − e − β ℓ ) Λ U † ′ Λ U † ( e − β ℓ ) d U (114) \int_{U(N)}\prod_{k\in\mathcal{K}}(-e^{-\alpha_k})\frac{\Lambda_U'}{\Lambda_U}(e^{-\alpha_k})\prod_{\ell\in\mathcal{L}}(-e^{-\beta_\ell})\frac{\Lambda_{U^\dagger}'}{\Lambda_{U^\dagger}}(e^{-\beta_\ell})\,dU \tag*{(114)} ∫ U ( N ) k ∈ K ∏ ( − e − α k ) Λ U Λ U ′ ( e − α k ) ℓ ∈ L ∏ ( − e − β ℓ ) Λ U † Λ U † ′ ( e − β ℓ ) d U (114) with K = { 1 , … , K } \mathcal{K}=\{1,\ldots,K\} K = { 1 , … , K } L = { 1 , … , L } \mathcal{L}=\{1,\ldots,L\} L = { 1 , … , L } (114) , which we will call S 0 S_0 S 0 K \mathcal{K} K L \mathcal{L} L ∣ K ∣ = ∣ L ∣ |\mathcal{K}|=|\mathcal{L}| ∣ K ∣ = ∣ L ∣ (6) we can pair every a a a b b b
S 0 = 1 ∣ K ∣ = ∣ L ∣ ∑ ( K : L ) ( ∏ j = 1 ∣ L ∣ ( z ′ z ) ′ ( a k j + b ℓ j ) ) (115) S_0=1_{|\mathcal{K}|=|\mathcal{L}|}\sum_{(\mathcal{K}:\mathcal{L})}\left(\prod_{j=1}^{|\mathcal{L}|}\left(\frac{z'}{z}\right)'(a_{k_j}+b_{\ell_j})\right) \tag*{(115)} S 0 = 1 ∣ K ∣ = ∣ L ∣ ( K : L ) ∑ j = 1 ∏ ∣ L ∣ ( z z ′ ) ′ ( a k j + b ℓ j ) (115) where the notation ( K : L ) (\mathcal{K}:\mathcal{L}) ( K : L ) K \mathcal{K} K L \mathcal{L} L ( a k 1 , b ℓ 1 ) , ( a k 2 , b ℓ 2 ) … (a_{k_1},b_{\ell_1}),(a_{k_2},b_{\ell_2})\ldots ( a k 1 , b ℓ 1 ) , ( a k 2 , b ℓ 2 ) … ∣ K ∣ ≠ ∣ L ∣ |\mathcal{K}|\ne|\mathcal{L}| ∣ K ∣ = ∣ L ∣
With ∣ K ∣ = ∣ L ∣ |\mathcal{K}|=|\mathcal{L}| ∣ K ∣ = ∣ L ∣ A j = e − a j A_j=e^{-a_j} A j = e − a j B j = e − b j B_j=e^{-b_j} B j = e − b j ∣ A j ∣ , ∣ B j ∣ < 1 |A_j|,|B_j|<1 ∣ A j ∣ , ∣ B j ∣ < 1 j = 1 , 2 , … , ∣ L ∣ j=1,2,\ldots,|\mathcal{L}| j = 1 , 2 , … , ∣ L ∣ ∣ A ∣ L ∣ ∣ < ⋯ < ∣ A 1 ∣ |A_{|\mathcal{L}|}|<\cdots<|A_1| ∣ A ∣ L ∣ ∣ < ⋯ < ∣ A 1 ∣ ∣ B ∣ L ∣ ∣ < ⋯ < ∣ B 1 ∣ |B_{|\mathcal{L}|}|<\cdots<|B_1| ∣ B ∣ L ∣ ∣ < ⋯ < ∣ B 1 ∣
S 0 = 1 ∣ K ∣ = ∣ L ∣ ∑ ( K : L ) ( ∏ j = 1 ∣ L ∣ A k j B ℓ j ( 1 − A k j B ℓ j ) 2 ) = 1 ∣ K ∣ = ∣ L ∣ ∑ ( K : L ) ( ∏ j = 1 ∣ L ∣ ( ∑ n j = 1 ∞ n j ( A k j B ℓ j ) n j ) ) = 1 ∣ K ∣ = ∣ L ∣ ∑ ( K : L ) ∑ n 1 = 1 ∞ ⋯ ∑ n ∣ L ∣ = 1 ∞ n 1 ⋯ n ∣ L ∣ A k 1 n 1 ⋯ A k ∣ L ∣ n ∣ L ∣ B ℓ 1 n 1 ⋯ B ℓ ∣ L ∣ n ∣ L ∣ . (116) \begin{aligned}
S_0={}&\mathbf{1}_{|\mathcal{K}|=|\mathcal{L}|}\sum_{(\mathcal{K}:\mathcal{L})}\left(\prod_{j=1}^{|\mathcal{L}|}\frac{A_{k_j}B_{\ell_j}}{(1-A_{k_j}B_{\ell_j})^2}\right)\\
={}&\mathbf{1}_{|\mathcal{K}|=|\mathcal{L}|}\sum_{(\mathcal{K}:\mathcal{L})}\left(\prod_{j=1}^{|\mathcal{L}|}\left(\sum_{n_j=1}^{\infty}n_j(A_{k_j}B_{\ell_j})^{n_j}\right)\right)\\
={}&\mathbf{1}_{|\mathcal{K}|=|\mathcal{L}|}\sum_{(\mathcal{K}:\mathcal{L})}\sum_{n_1=1}^{\infty}\cdots\sum_{n_{|\mathcal{L}|}=1}^{\infty}n_1\cdots n_{|\mathcal{L}|}A_{k_1}^{n_1}\cdots A_{k_{|\mathcal{L}|}}^{n_{|\mathcal{L}|}}B_{\ell_1}^{n_1}\cdots B_{\ell_{|\mathcal{L}|}}^{n_{|\mathcal{L}|}}. \tag*{(116)}
\end{aligned} S 0 = = = 1 ∣ K ∣ = ∣ L ∣ ( K : L ) ∑ j = 1 ∏ ∣ L ∣ ( 1 − A k j B ℓ j ) 2 A k j B ℓ j 1 ∣ K ∣ = ∣ L ∣ ( K : L ) ∑ j = 1 ∏ ∣ L ∣ n j = 1 ∑ ∞ n j ( A k j B ℓ j ) n j 1 ∣ K ∣ = ∣ L ∣ ( K : L ) ∑ n 1 = 1 ∑ ∞ ⋯ n ∣ L ∣ = 1 ∑ ∞ n 1 ⋯ n ∣ L ∣ A k 1 n 1 ⋯ A k ∣ L ∣ n ∣ L ∣ B ℓ 1 n 1 ⋯ B ℓ ∣ L ∣ n ∣ L ∣ . (116) So, the contribution from zero-swap terms to the trace average
∫ U ( N ) Tr ( ( U † ) α 1 ) ⋯ Tr ( ( U † ) α K ) Tr ( U β 1 ) ⋯ Tr ( U β L ) d U (117) \int_{U(N)}\operatorname{Tr}((U^\dagger)^{\alpha_1})\cdots\operatorname{Tr}((U^\dagger)^{\alpha_K})\operatorname{Tr}(U^{\beta_1})\cdots\operatorname{Tr}(U^{\beta_L})\,dU \tag*{(117)} ∫ U ( N ) Tr (( U † ) α 1 ) ⋯ Tr (( U † ) α K ) Tr ( U β 1 ) ⋯ Tr ( U β L ) d U (117) is
1 K = L ∑ ( K : L ) β 1 ⋯ β L δ ( α k 1 = β ℓ 1 , α k 2 = β ℓ 2 , … , α k L = β ℓ L ) . (118) \mathbf{1}_{K=L}\sum_{(\mathcal{K}:\mathcal{L})}\beta_1\cdots\beta_L\delta(\alpha_{k_1}=\beta_{\ell_1},\alpha_{k_2}=\beta_{\ell_2},\ldots,\alpha_{k_L}=\beta_{\ell_L}). \tag*{(118)} 1 K = L ( K : L ) ∑ β 1 ⋯ β L δ ( α k 1 = β ℓ 1 , α k 2 = β ℓ 2 , … , α k L = β ℓ L ) . (118) Note that this must coincide with the formula of Diaconis and Shahshahani (1) if β 1 + ⋯ + β L ≤ N \beta_1+\cdots+\beta_L\leq N β 1 + ⋯ + β L ≤ N
∫ U ( N ) ∏ j = 1 l ( Tr M j ) a j ( Tr M j ) b j d U = δ a b ∏ j = 1 l j a j a j ! (119) \int_{U(N)}\prod_{j=1}^{l}(\operatorname{Tr}M^j)^{a_j}(\operatorname{Tr}M^j)^{b_j}\,dU=\delta_{ab}\prod_{j=1}^{l}j^{a_j}a_j! \tag*{(119)} ∫ U ( N ) j = 1 ∏ l ( Tr M j ) a j ( Tr M j ) b j d U = δ ab j = 1 ∏ l j a j a j ! (119) (note these a a a b b b (118) and (119) are both zero if the list of exponents on the traces of the U U U ∏ j = 1 l j a j a j ! \prod_{j=1}^{l}j^{a_j}a_j! ∏ j = 1 l j a j a j ! (119) is just the product of the list of exponents, and so corresponds to β 1 ⋯ β L \beta_1\cdots\beta_L β 1 ⋯ β L α \alpha α (118) and all the a j a_j a j (119) . But if there is a set of the α \alpha α β \beta β a a a a ! a! a ! α \alpha α β \beta β a ! a! a ! (118) that capture these arrangements. This extends naturally to any number of blocks of identical exponents, and so accounts for the a j ! a_j! a j ! (119) .
One-swap terms
The one-swap terms of (114) , which we will call S 1 S_1 S 1 [14 ] . We will use this as our starting point. Below the sets R R R Q Q Q a a a b b b H S , T ( W ) H_{S,T}(W) H S , T ( W ) (6) and the indices k k k l l l S S S T T T (5) .
S 1 = ∑ R ⊄ K , Q ⊄ L ∣ R ∣ = ∣ Q ∣ ( R : Q ) ( ∑ R 1 ⊄ R c , Q 1 ⊄ Q c k ∈ R 1 c , l ∈ Q 1 c × ∏ j = 1 ∣ Q ∣ ( z ′ z ) ′ ( a r j + b q j ) ) × ∑ R 1 ⊄ R c , Q 1 ⊄ Q c k ∈ R 1 c , l ∈ Q 1 c e − N ( a k + b l ) z ( a k + b l ) z ( − a k − b l ) × ( ∏ s ∈ R 1 z ′ z ( a s − a k ) ) ( ∏ s ∈ R 1 c ∖ { k } z ′ z ( a s + b l ) ) × ( ∏ t ∈ Q 1 z ′ z ( b t − b l ) ) ( ∏ t ∈ Q 1 c ∖ { l } − z ′ z ( b t + a k ) ) . (120) \begin{aligned}
S_1={}&\sum_{\substack{R\not\subset K,\ Q\not\subset L\\ |R|=|Q|}}
(R:Q)\left(\sum_{\substack{R_1\not\subset R^c,\ Q_1\not\subset Q^c\\ k\in R_1^c,\ l\in Q_1^c}}\right.\\
&\quad\left.\times\prod_{j=1}^{|Q|}\left(\frac{z'}{z}\right)^{\prime}(a_{r_j}+b_{q_j})\right)\\
&\quad\times\sum_{\substack{R_1\not\subset R^c,\ Q_1\not\subset Q^c\\ k\in R_1^c,\ l\in Q_1^c}}e^{-N(a_k+b_l)}z(a_k+b_l)z(-a_k-b_l)\\
&\quad\times\left(\prod_{s\in R_1}\frac{z'}{z}(a_s-a_k)\right)\left(\prod_{s\in R_1^c\setminus\{k\}}\frac{z'}{z}(a_s+b_l)\right)\\
&\quad\times\left(\prod_{t\in Q_1}\frac{z'}{z}(b_t-b_l)\right)\left(\prod_{t\in Q_1^c\setminus\{l\}}-\frac{z'}{z}(b_t+a_k)\right). \tag*{(120)}
\end{aligned} S 1 = R ⊂ K , Q ⊂ L ∣ R ∣ = ∣ Q ∣ ∑ ( R : Q ) R 1 ⊂ R c , Q 1 ⊂ Q c k ∈ R 1 c , l ∈ Q 1 c ∑ × j = 1 ∏ ∣ Q ∣ ( z z ′ ) ′ ( a r j + b q j ) × R 1 ⊂ R c , Q 1 ⊂ Q c k ∈ R 1 c , l ∈ Q 1 c ∑ e − N ( a k + b l ) z ( a k + b l ) z ( − a k − b l ) × ( s ∈ R 1 ∏ z z ′ ( a s − a k ) ) s ∈ R 1 c ∖ { k } ∏ z z ′ ( a s + b l ) × t ∈ Q 1 ∏ z z ′ ( b t − b l ) t ∈ Q 1 c ∖ { l } ∏ − z z ′ ( b t + a k ) . (120) with K = { 1 , … , K } \mathcal{K}=\{1,\ldots,K\} K = { 1 , … , K } L = { 1 , … , L } \mathcal{L}=\{1,\ldots,L\} L = { 1 , … , L } R c R^c R c R R R K \mathcal{K} K R 1 c R_1^c R 1 c R 1 R_1 R 1 R c R^c R c R 1 R_1 R 1 W = { a } W=\{a\} W = { a } (6) from which we select the first term of H S , T ( W ) H_{S,T}(W) H S , T ( W ) H H H R 1 c ∖ { k } R_1^c\setminus\{k\} R 1 c ∖ { k } W = { a } W=\{a\} W = { a } (6) from which we select the second term of H S , T ( W ) H_{S,T}(W) H S , T ( W ) k k k R 1 c R_1^c R 1 c [14 ] that we continue here. The notation on the L L L
Once again A j = e − a j A_j=e^{-a_j} A j = e − a j B j = e − b j B_j=e^{-b_j} B j = e − b j ∣ A j ∣ , ∣ B j ∣ < 1 |A_j|,|B_j|<1 ∣ A j ∣ , ∣ B j ∣ < 1 ∣ A K ∣ < ⋯ < ∣ A 1 ∣ |A_K|<\cdots<|A_1| ∣ A K ∣ < ⋯ < ∣ A 1 ∣ ∣ B L ∣ < ⋯ < ∣ B 1 ∣ |B_L|<\cdots<|B_1| ∣ B L ∣ < ⋯ < ∣ B 1 ∣
S 1 = ∑ R ⊄ K , Q ⊄ L ∣ R ∣ = ∣ Q ∣ ( R : Q ) ( ∑ R 1 ⊄ R c , Q 1 ⊄ Q c k ∈ R 1 c , l ∈ Q 1 c ∏ j = 1 ∣ Q ∣ ( ∑ n j = 0 ∞ n j A r j n j B q j n j ) ) × ∑ R 1 ⊄ R c , Q 1 ⊄ Q c k ∈ R 1 c , l ∈ Q 1 c A k N B l N ( − ∑ n = 0 ∞ n A k n B l n ) × ( ∏ s ∈ R 1 s < k ( ∑ n s = 0 ∞ A s n s A k − n s ) ) ( ∏ s ∈ R 1 s > k ( − ∑ n s = 0 ∞ A s n s + 1 A k − n s − 1 ) ) × ( ∏ s ∈ R 1 c ∖ { k } ( ∑ n s = 0 ∞ A s n s + 1 B l n s + 1 ) ) × ( ∏ t ∈ Q 1 t < l ( ∑ n t = 0 ∞ B l n t B t − n t ) ) ( ∏ t ∈ Q 1 t > l ( − ∑ n t = 0 ∞ B t n t + 1 B l − n t − 1 ) ) × ( ∏ t ∈ Q 1 c ∖ { l } ( ∑ n t = 0 ∞ B t n t + 1 A k n t + 1 ) ) . (121) \begin{aligned}
S_1={}&\sum_{\substack{R\not\subset K,\ Q\not\subset L\\ |R|=|Q|}}
(R:Q)\left(\sum_{\substack{R_1\not\subset R^c,\ Q_1\not\subset Q^c\\ k\in R_1^c,\ l\in Q_1^c}}
\prod_{j=1}^{|Q|}\left(\sum_{n_j=0}^{\infty}n_jA_{r_j}^{n_j}B_{q_j}^{n_j}\right)\right)\\
&\quad\times\sum_{\substack{R_1\not\subset R^c,\ Q_1\not\subset Q^c\\ k\in R_1^c,\ l\in Q_1^c}}A_k^NB_l^N\left(-\sum_{n=0}^{\infty}nA_k^nB_l^n\right)\\
&\quad\times\left(\prod_{\substack{s\in R_1\\s<k}}\left(\sum_{n_s=0}^{\infty}A_s^{n_s}A_k^{-n_s}\right)\right)
\left(\prod_{\substack{s\in R_1\\s>k}}\left(-\sum_{n_s=0}^{\infty}A_s^{n_s+1}A_k^{-n_s-1}\right)\right)\\
&\quad\times\left(\prod_{s\in R_1^c\setminus\{k\}}\left(\sum_{n_s=0}^{\infty}A_s^{n_s+1}B_l^{n_s+1}\right)\right)\\
&\quad\times\left(\prod_{\substack{t\in Q_1\\t<l}}\left(\sum_{n_t=0}^{\infty}B_l^{n_t}B_t^{-n_t}\right)\right)
\left(\prod_{\substack{t\in Q_1\\t>l}}\left(-\sum_{n_t=0}^{\infty}B_t^{n_t+1}B_l^{-n_t-1}\right)\right)\\
&\quad\times\left(\prod_{t\in Q_1^c\setminus\{l\}}\left(\sum_{n_t=0}^{\infty}B_t^{n_t+1}A_k^{n_t+1}\right)\right). \tag*{(121)}
\end{aligned} S 1 = R ⊂ K , Q ⊂ L ∣ R ∣ = ∣ Q ∣ ∑ ( R : Q ) R 1 ⊂ R c , Q 1 ⊂ Q c k ∈ R 1 c , l ∈ Q 1 c ∑ j = 1 ∏ ∣ Q ∣ n j = 0 ∑ ∞ n j A r j n j B q j n j × R 1 ⊂ R c , Q 1 ⊂ Q c k ∈ R 1 c , l ∈ Q 1 c ∑ A k N B l N ( − n = 0 ∑ ∞ n A k n B l n ) × s ∈ R 1 s < k ∏ ( n s = 0 ∑ ∞ A s n s A k − n s ) s ∈ R 1 s > k ∏ ( − n s = 0 ∑ ∞ A s n s + 1 A k − n s − 1 ) × s ∈ R 1 c ∖ { k } ∏ ( n s = 0 ∑ ∞ A s n s + 1 B l n s + 1 ) × t ∈ Q 1 t < l ∏ ( n t = 0 ∑ ∞ B l n t B t − n t ) t ∈ Q 1 t > l ∏ ( − n t = 0 ∑ ∞ B t n t + 1 B l − n t − 1 ) × t ∈ Q 1 c ∖ { l } ∏ ( n t = 0 ∑ ∞ B t n t + 1 A k n t + 1 ) . (121) There are several things to notice about (121) . We only see terms where the total sum of exponents on all the A A A B B B A A A B B B R \mathcal{R} R Q \mathcal{Q} Q A A A R c \mathcal{R}^c R c B B B Q c \mathcal{Q}^c Q c s ∈ R 1 s\in R_1 s ∈ R 1 s < k s<k s < k A s A_s A s (121) ) and so can’t contribute to a term of the form A 1 α 1 ⋯ A K α K B 1 β 1 ⋯ B L β L A_1^{\alpha_1}\cdots A_K^{\alpha_K}B_1^{\beta_1}\cdots B_L^{\beta_L} A 1 α 1 ⋯ A K α K B 1 β 1 ⋯ B L β L α 1 , … , α K , β 1 , … , β L ≥ 1 \alpha_1,\ldots,\alpha_K,\beta_1,\ldots,\beta_L\ge1 α 1 , … , α K , β 1 , … , β L ≥ 1 s ∈ R 1 s\in R_1 s ∈ R 1 s < k s<k s < k t ∈ Q 1 t\in Q_1 t ∈ Q 1 t < l t<l t < l (121) ). Inside the sum over R 1 R_1 R 1 Q 1 Q_1 Q 1 A 1 α 1 ⋯ A K α K B 1 β 1 ⋯ B L β L A_1^{\alpha_1}\cdots A_K^{\alpha_K}B_1^{\beta_1}\cdots B_L^{\beta_L} A 1 α 1 ⋯ A K α K B 1 β 1 ⋯ B L β L B l B_l B l β l = N + n − ∑ t ∈ Q 1 , t > l β t + ∑ s ∈ R 1 c ∖ { k } α s \beta_l=N+n-\sum_{t\in Q_1,t>l}\beta_t+\sum_{s\in R_1^c\setminus\{k\}}\alpha_s β l = N + n − ∑ t ∈ Q 1 , t > l β t + ∑ s ∈ R 1 c ∖ { k } α s n = ( β l + ∑ t ∈ Q 1 , t > l β t − ∑ s ∈ R 1 c ∖ { k } α s − N ) + n=(\beta_l+\sum_{t\in Q_1,t>l}\beta_t-\sum_{s\in R_1^c\setminus\{k\}}\alpha_s-N)_+ n = ( β l + ∑ t ∈ Q 1 , t > l β t − ∑ s ∈ R 1 c ∖ { k } α s − N ) + α \alpha α β \beta β n n n β l \beta_l β l α \alpha α β \beta β α k \alpha_k α k l l l β \beta β n n n n n n A k A_k A k α k = N + n − ∑ s ∈ R 1 , s > k α s + ∑ t ∈ Q c ∖ { l } β t \alpha_k = N + n - \sum_{s\in R_1,s>k}\alpha_s + \sum_{t\in Q_c\setminus\{l\}}\beta_t α k = N + n − ∑ s ∈ R 1 , s > k α s + ∑ t ∈ Q c ∖ { l } β t A A A R c R^c R c B B B Q c Q^c Q c
So, the contribution from one-swap terms to the trace average
∫ U ( N ) Tr ( ( U † ) α 1 ) ⋯ Tr ( ( U † ) α K ) Tr ( U β 1 ) ⋯ Tr ( U β L ) d U (122) \int_{U(N)} \operatorname{Tr}((U^\dagger)^{\alpha_1})\cdots\operatorname{Tr}((U^\dagger)^{\alpha_K})\operatorname{Tr}(U^{\beta_1})\cdots\operatorname{Tr}(U^{\beta_L})\,dU \tag*{(122)} ∫ U ( N ) Tr (( U † ) α 1 ) ⋯ Tr (( U † ) α K ) Tr ( U β 1 ) ⋯ Tr ( U β L ) d U (122) is
∑ R ⊊ K , Q ⊊ L ∣ R ∣ = ∣ Q ∣ ( ∑ ( R : Q ) ∏ j = 1 ∣ Q ∣ β q j ) ( δ ( α r 1 = β q 1 , … , α r ∣ Q ∣ = β q ∣ Q ∣ ) ) × ∑ S ⊂ R c , T ⊆ Q c ∣ S ∣ < ∣ R c ∣ 0 < ∣ T ∣ ( − 1 ) ∣ T ∣ + ∣ R c ∣ − ∣ S ∣ − 1 ( ∑ t ∈ T β t − ∑ s ∈ S α s − N ) + . (123) \sum_{\substack{R\subsetneq K,\ Q\subsetneq L\\ |R|=|Q|}}
\left(\sum_{(R:Q)}\prod_{j=1}^{|Q|}\beta_{q_j}\right)
\left(\delta(\alpha_{r_1}=\beta_{q_1},\ldots,\alpha_{r_{|Q|}}=\beta_{q_{|Q|}})\right)
\times
\sum_{\substack{S\subset R^c,\ T\subseteq Q^c\\ |S|<|R^c|\\ 0<|T|}}
(-1)^{|T|+|R^c|-|S|-1}
\left(\sum_{t\in T}\beta_t-\sum_{s\in S}\alpha_s-N\right)_+. \tag*{(123)} R ⊊ K , Q ⊊ L ∣ R ∣ = ∣ Q ∣ ∑ ( R : Q ) ∑ j = 1 ∏ ∣ Q ∣ β q j ( δ ( α r 1 = β q 1 , … , α r ∣ Q ∣ = β q ∣ Q ∣ ) ) × S ⊂ R c , T ⊆ Q c ∣ S ∣ < ∣ R c ∣ 0 < ∣ T ∣ ∑ ( − 1 ) ∣ T ∣ + ∣ R c ∣ − ∣ S ∣ − 1 ( t ∈ T ∑ β t − s ∈ S ∑ α s − N ) + . (123) The number ∣ T ∣ + ∣ R c ∣ − ∣ S ∣ + 1 |T|+|R^c|-|S|+1 ∣ T ∣ + ∣ R c ∣ − ∣ S ∣ + 1 (121) as described below. This expression agrees with all of the smaller cases in previous sections.
We arrive at (123) because the set R 1 c ∖ { k } R_1^c\setminus\{k\} R 1 c ∖ { k } n = ( β l + ∑ t ∈ Q 1 , t > l β t − ∑ s ∈ R 1 c ∖ { k } α s − N ) + n=(\beta_l+\sum_{t\in Q_1,t>l}\beta_t-\sum_{s\in R_1^c\setminus\{k\}}\alpha_s-N)_+ n = ( β l + ∑ t ∈ Q 1 , t > l β t − ∑ s ∈ R 1 c ∖ { k } α s − N ) + R c R^c R c R c R^c R c R 1 R_1 R 1 k ∈ R 1 c k\in R_1^c k ∈ R 1 c (121) , so this is represented in (123) by the sum over the subsets S ⊂ R c S\subset R^c S ⊂ R c ∣ S ∣ < ∣ R c ∣ |S|<|R^c| ∣ S ∣ < ∣ R c ∣ ( − 1 ) ∣ R c ∣ − ∣ S ∣ − 1 (-1)^{|R^c|-|S|-1} ( − 1 ) ∣ R c ∣ − ∣ S ∣ − 1 (121) ) associated with each A i A_i A i i ∈ R c i\in R^c i ∈ R c not chosen for S S S k k k β l + ∑ t ∈ Q 1 , t > l β t \beta_l+\sum_{t\in Q_1,t>l}\beta_t β l + ∑ t ∈ Q 1 , t > l β t β i \beta_i β i i ∈ Q c i\in Q^c i ∈ Q c Q 1 Q_1 Q 1 l l l (121) . This is represented by T ⊆ Q c T\subseteq Q^c T ⊆ Q c 0 < ∣ T ∣ 0<|T| 0 < ∣ T ∣ ( − 1 ) ∣ T ∣ (-1)^{|T|} ( − 1 ) ∣ T ∣ (121) ) associated with each B i B_i B i i ∈ T i\in T i ∈ T
It remains to argue that each choice of S S S T T T (121) . We could imagine, perhaps, the same set of β \beta β β \beta β β l \beta_l β l l l l β l + ∑ t ∈ Q 1 , t > l β t \beta_l+\sum_{t\in Q_1,t>l}\beta_t β l + ∑ t ∈ Q 1 , t > l β t k k k R c R^c R c R 1 c ∖ { k } R_1^c\setminus\{k\} R 1 c ∖ { k } S S S T T T R 1 R_1 R 1 Q 1 Q_1 Q 1 k k k l l l (121) .
