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Home , Dirac delta function, Geometric measure theory, Heaviside step function, Product measure

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δ(x)= H′(x)= 0, x = 0, ∀ 6 and δ(0)=+∞.

We recall the definition of :

Definition 1.1 (Dirac delta function). Dirac delta function δ(x) is defined by

+∞, if x = 0; δ(x)= (1.3) 0, if x = 0.  6 Proposition 1.2 (Sampling property of the Dirac delta function). If f is any function which is continuous on a neighborhood of 0, then

+∞ f (x)δ(x)dx = f (0). (1.4) ∞ Z− In fact, we have, for a = 0, 6 +a f (x)δ(x)dx = f (0) a Z− and a +∞ δ(x)dx = H(a x)δ(x)dx = H(a). ∞ ∞ − Z− Z− Definition 1.3. Assume that f is which vanishes outside some finite interval. There corresponds a certain number which we write as H, f , given by h i +∞ +∞ H, f := f (x)H(x)dx = f (x)dx. (1.5) h i ∞ 0 Z− Z Similarly, δ, f is given by h i +∞ δ, f := f (x)δ(x)dx = f (0). (1.6) h i ∞ Z− For an ordinary function f and a fixed a R, the symbol f denotes the translation of f with ∈ a respect to a:

f (x) := f (x a). (1.7) a − Thus

+∞ +∞ δa, f = f (x)δa(x)dx = f (x)δ(x a)dx = f (a). (1.8) h i ∞ ∞ − Z− Z− 2 From the above discussion, it is easy to see that

f (x)δ(x a)= f (x)δ (x)= f (a)δ(x). (1.9) − a This fact will be used later.

Proposition 1.4. If f is any function which has a continuous derivative f ′, at least in some neighborhood of 0, then

+∞ δ′, f := f (x)δ′(x)dx = δ, f ′ = f ′(0). (1.10) ∞ − − Z−

Proof. Since

+∞ δ(x) δ(x ǫ) 1 +∞ +∞ f (x) − − dx = f (x)δ(x)dx f (x)δ(x ǫ)dx ∞ ǫ ǫ ∞ − ∞ − Z− Z− Z−  f (0) f (ǫ) f (ǫ) f (0) = − = − , ǫ − ǫ it follows that +∞ +∞ δ(x) δ(x ǫ) δ′, f = f (x)δ′(x)dx = f (x) lim − − dx (1.11) ∞ ∞ ǫ 0 ǫ Z− Z− → +∞ δ(x) δ(x ǫ) = lim f (x) − − dx (1.12) ǫ 0 ∞ ǫ → Z− f (ǫ) f (0) = lim − = f ′(0)= δ, f ′ . (1.13) ǫ 0 − ǫ − − →

This completes the proof.

Proposition 1.5. The n-th derivative of the Dirac delta function, denoted by δ(n), is defined by the follow- ing:

δ(n), f =( 1)n f (n)(0), (1.14) − D E where n N and f is any function with continuous at least up to the n-th order in some ∈ + neighborhood of 0.

From these, we see that

δ′ , f = δ , f ′ = f ′(a), (1.15) a − a − ( ) δ n , f = ( 1)n δ , f (n) =( 1)n f (n)(a). (1.16) a − a − D E D E Suppose that g(t) increases monotonely over the closed interval [a, b]: suppose there is t (a, b) 0 ∈ such that g(t0)= 0. We have known that

dg 1(x) 1 − = . dx dg(t) dt

3 From this, we see that, via x = g(t), t [a, b], ∈ b g(b) 1 1 f (t)δ(g(t))dt = f (g− (x))δ(x)d(g− (x)) (1.17) Za Zg(a) g(b) 1 1 dg− (x) = f (g− (x))δ(x) dx (1.18) x Zg(a) d 1 1 dg− (x) f (t0) = f (g− (0)) x=g(t ) = . (1.19) dx 0 g′(t0)

Proposition 1.6. If g(t) is monotone, with g(a)= 0 and g (a) = 0, then ′ 6 δ (t) δ(g(t)) = a . (1.20) g (a) | ′ | From this, we see that 1 b δ(kx + b)= δ x + , k = 0. k k 6 | |   More generally, the delta distribution may be composed with a smooth function g(x) in such a way that the familiar change of variables formula holds, that

f (g(x))δ(g(x)) g′(x) dx = f (t)δ(t)dt (1.21) R R Z Zg( )

provided that g is a continuously with g′ nowhere zero. That is, there is a unique way to assign meaning to the distribution δ g so that this identity holds for all compactly ◦ supported test functions f . Therefore, the domain must be broken up to exclude the g′(x) = 0 . This distribution satisfies δ(g(x)) = 0 if g is nowhere zero, and otherwise if g has a real root at x0, then δ(x x ) δ(g(x)) = − 0 . (1.22) g (x ) | ′ 0 | It is natural to define the composition δ(g(x)) for continuously differentiable functions g by

δ(x xj) δ(g(x)) = ∑ − (1.23) j g′(xj) where the sum extends over all roots of g(x), which are assumed to be simple. Thus, for example

1 δ x2 a2 = (δ(x + a)+ δ(x a)) . (1.24) − 2 a −  | | In the form the generalized scaling property may be written as

+∞ f (xj) f (x)δ(g(x))dx = ∑ . (1.25) ∞ g (xj) Z− j ′

4 Example 1.7 ([1]). If T is an n n positive definite matrix and r R+, α > 0, then × ∈ πnΓ(α + 1) rn+α (r u T u )α [du]= , (1.26) Γ(n + α + ) (T) ZBn(T,r) − h | | i 1 det where B (T, r) := u Cn u T u < r and [du]= ∏n du for [dz]= d (Re(z)) d (Im(z)). n { ∈ | h | | i } j=1 j Indeed, let 1 1 v = r− 2 T 2 u.

1 1 Then [dv]= det(r− T)[du] or [du]= det(rT− )[dv]. Thus

rn+α (r u T u )α [du]= (1 v v )α [dv], (1.27) (T) 1 ZBn(T,r) − h | | i det ZBn( n,1) −h | i where B (1 , 1)= v Cn v v < 1 . Now n n { ∈ |h | i } 1 Bn( n, 1)= γ [0,1)Sn(γ), ∪ ∈ where S (γ)= v Cn v = γ . n { ∈ | k k2 } 1 (1 v v )α [dv] = dγ δ(γ v ) (1 v v )α [dv] (1.28) 1 2 ZBn( n,1) −h | i Z0 Z − k k −h | i 1 2 α = dγ(1 γ ) δ(γ v 2)[dv] (1.29) Z0 − · Z − k k 1 2 α = dγ(1 γ ) vol(Sn(γ)), (1.30) Z0 − where n 2π 2n 1 vol(S (γ)) = δ(γ v )[dv]= γ − . n 2 Γ(n) Z − k k Therefore, we obtain that

n n+α 1 α 2π r 2 α 2n 1 (r u T u ) [du] = (1 γ ) γ − dγ (1.31) Γ(n) (T) ZBn(T,r) − h | | i det Z0 − n n+α 1 π r 2 α 2n 2 2 = (1 γ ) γ − d(γ ), (1.32) Γ(n) (T) det Z0 − where

1 1 2 α 2n 2 2 α n 1 Γ(α + 1)Γ(n) (1 γ ) γ − d(γ )= (1 x) x − dx = B(α + 1, n)= . Γ(n + α + ) Z0 − Z0 − 1 This completes the proof.

Definition 1.8 (). An operation on functions, called convolution and denoted by the symbol , is defined by: ∗ f g(x)= f (x t)g(t)dt = f (t)g(x t)dt. (1.33) R R ∗ Z − Z − 5 It is easily seen that some properties of convolution:

(i) δ δ = δ . a ∗ b a+b (ii) The delta function as a convolution unit: δ f = f δ = f . ∗ ∗ (iii) Convolution as the translation: δ f = f δ = f , where f (x) := f (x a). a ∗ ∗ a a a − (iv) δ(n) f = f δ(n) = f (n). ∗ ∗ Definition 1.9 (). Let f be a complex-valued function of the real variable t which is absolutely integrable over the whole real axis R. That is,

f1(x) < +∞ and f2(x) < +∞, R R Z | | Z | | where f = f + √ 1 f . We define the Fourier transform of f to be a new function 1 − 2 iωt f (ω) := ( f )(ω)= e− f (t)dt. (1.34) R F Z Next, we consider the Fourierb integral representation of Dirac delta function which is very powerful in applications. We use the following standard result:

+∞ eiωx PV dx = iπ (1.35) ∞ x Z− where the symbol PV denotes the of the integral and ω > 0 a constant. That is,

+∞ sin(ωx) +∞ cos(ωx) dx = π and dx = 0. (1.36) ∞ x ∞ x Z− Z− Replacing ω by ω simply changes the sign of the first of these two real and leaves the − other unaltered. That is, if ω > 0

+∞ e iωx PV − dx = iπ (1.37) ∞ x − Z− Hence, if we replace ω by the usual symbol t for the independent real variable we can write

+∞ eitx PV dx = iπ sign(t), (1.38) ∞ x Z− i.e.,

1 +∞ eitx 1 1 , t > 0 dx = sign(t)= 2 (1.39) π ∞ x  1 2 Z i 2 , t < 0 − − 2 A formal differentiation of this with respect to t then yields the following result:

6 Proposition 1.10. It holds that

1 +∞ δ(t)= eitxdx. (1.40) 2π ∞ Z− This amounts to say 1(1)(t)= δ(t). Replacing t by t a, we have F − − +∞ 1 i(t a)x e − dx = δ(t a)= δa(t). (1.41) 2π ∞ − Z− This amounts to say 1(e iax)(t) = δ (t) or (δ (t))(x) = e iax. The integral on the left- F − − a F a − hand side of (1.40) is, of course, divergent, and it is clear that this equation must be understood symbolically. That is to say, for all sufficiently well-behaved functions f , we should interpret (1.40) to mean that

+∞ 1 +∞ +∞ f (t) eitωdω dt = f (t)δ(t)dt = f (0) (1.42) ∞ 2π ∞ ∞ Z−  Z−  Z− or, more generally, that

+∞ +∞ +∞ 1 i(t x)ω f (x) e − dω dx = f (x)δ(t x)dx = f (t). (1.43) ∞ 2π ∞ ∞ − Z−  Z−  Z− We can rewrite this result in the form

+∞ +∞ 1 i(t x)ω f (t) = f (x) e − dω dx (1.44) ∞ 2π ∞ Z−  Z−  +∞ +∞ +∞ 1 itω ixω 1 itω = e f (x)e− dx dω = e f (ω)dω (1.45) 2π ∞ ∞ 2π ∞ Z− Z−  Z− Proposition 1.11 (Fourier Inversion). Let f be a (real or complex valued) functionb of a single real variable which is absolutely integrable over the interval ( ∞, +∞) and which also satisfies the Dirichlet − conditions over every finite interval. If f (ω) denotes the Fourier transform of f , then at each point t we have b 1 +∞ 1 eitω f (ω)dω = [ f (t+)+ f (t )] , (1.46) 2π ∞ 2 − Z− b where f (t ) := lims t f (s). ± → ± There are several important properties of the Fourier transform which merit explicit mention.

(i) The Fourier transform of the convolution of two functions is equal to the product of their individual transforms: ( f g)= ( f ) (g). F ∗ F F (ii) ( f g) = 1 ( f ) (g). F 2π F ∗ F (iii) The Fourier transformation is linear: (λ f + λ f )= λ ( f )+ λ ( f ). F 1 1 2 2 1F 1 2F 2

7 (iv) ( f (x a))(ω)= e iωa ( f (x))(ω). F − − F (v) ( f (x)e ax )(ω)= ( f (x))(a + iω). F − F (vi) ( f )(ω)= iω ( f )(ω). F ′ F (vii) ( f (ax))(ω) = 1 ( f (x)) ω . F a F a  For example, the proofs of (iv) and (vi) are given. Indeed, f (x a) = f (x) = f δ (x), thus − a ∗ a ( f (x a))(ω) = ( f δ (x))(ω) = ( f ) (δ )(ω), that is, ( f (x a))(ω) = e iωa ( f ), F − F ∗ a F F a F − − F hence (iv). Since f = f δ , it follows that ( f )= ( f δ )= ( f ) (δ ). In what follows, we ′ ∗ ′ F ′ F ∗ ′ F F ′ calculate (δ ). By definition of Fourier transform, F ′ iωt iωt de− (δ′)(ω)= e− δ′(t)dt = = iω. (1.47) R t t=0 F Z − d

Thus ( f )(ω) = iω ( f )(ω), hence (vi). This property can be generalized: ( f (n))(ω) = F ′ F F (iω)n ( f )(ω). Indeed, F ( f (n))(ω)= ( f δ(n))(ω)= ( f )(ω) (δ(n))(ω)=(iω)n ( f )(ω). (1.48) F F ∗ F F F We can apply the sampling property of the delta function to the Fourier inversion integral:

1 +∞ 1 eixω δ(ω α)dω = eixα (1.49) 2π ∞ − 2π Z− and similarly

+∞ 1 ixω 1 ixα e δ(ω + α)dω = e− . (1.50) 2π ∞ 2π Z− Thus, recalling that the Fourier transform is defined in general for complex-valued functions, these results suggest that we can give the following definitions for the Fourier transforms of complex exponentials such as

iαx iαx (e )(ω)= 2πδ(ω α); (e− )(ω)= 2πδ(ω + α). F − F Both equations immediately yield the following definitions for the Fourier transforms of the real functions cos(αx) and sin(αx):

(cos(αx))(ω)= π (δ(ω α)+ δ(ω + α)) , (1.51) F − (sin(αx))(ω)= iπ (δ(ω α) δ(ω + α)) . (1.52) F − − − In particular, taking α = 0, we find that the generalized Fourier transform of the constant function f (t) 1 is simply 2πδ(ω). This in turn allows us to offer a definition of the Fourier transform ≡ of the unit .

8 1 1 Proposition 1.12. The Fourier transform of the H(x) = 2 + 2 sign(x), where x sign(x)= x , is given by | | 1 H(ω)= πδ(ω)+ . (1.53) iω Proof. Now b

1 +∞ eixω 1 1 , x > 0 dω = sign(x)= 2 (1.54) π ∞ ω  1 2 Z i 2 , x < 0 − − 2 2  Then we know that iω is a suitable choice for the Fourier transform of the function sign(x) in the sense that

+∞ 1 ixω 2 1 2 sign(x)= e dω = − (x). (1.55) 2π ∞ iω F iω Z−   This amounts to say that (sign(x))(ω)= 2 . Hence for Heaviside step function F iω 1 1 H(x)= + sign(x), (1.56) 2 2 the Fourier transform of it is given by

1 1 1 1 (H(x))(ω)= + sign(x) = (1)(ω)+ (sign(x))(ω), (1.57) F F 2 2 2 F 2F   i.e. 1 1 (H(x))(ω)= πδ(ω)+ H(ω)= πδ(ω)+ . (1.58) F iω ⇐⇒ iω This completes the proof. b

We can get some important properties of Dirac delta function which are listed below:

(i) The delta function is an even distribution: δ(x)= δ( x). − (ii) The delta function satisfies the following scaling property for a non-zero scalar: δ(ax) = 1 R a δ(x) for a 0 . | | ∈ \{ } (iii) The distributional product of δ(x) and x is equal to zero: xδ(x)= 0.

(iv) If x f (x) = xg(x), where f and g are distributions, then f (x) = g(x)+ cδ(x) for some constant c.

9 Previous two facts can be checked as follows: Note that ∞ ∞ 1 itx 1 itx δ( x) = e− dt = e− d( t) − 2π ∞ −2π ∞ − Z− Z− 1 ∞ 1 ∞ = − eisxds = eisxds = δ(x). −2π +∞ 2π ∞ Z Z− This is (i). For the proof of (ii), since a = 0, we observe that δ(ax) = δ( ax) by (i), hence 6 − δ(ax)= δ( a x), it follows that | | +∞ +∞ 1 it a x 1 i a t x δ(ax)= δ( a x)= e ·| | dt = e | | · dt. | | 2π ∞ 2π ∞ Z− Z− Let s = a t. Then ds = a dt, thus | | | | +∞ 1 1 is x 1 δ(ax)= e · ds = δ(x). (1.59) a 2π ∞ a | | Z− | | That is, in the sense of distribution, 1 δ(ax)= δ(x), a R 0 . (1.60) a ∈ \{ } | | 2 Dirac delta function of vector argument

Definition 2.1 (Dirac delta function of real-vector arguments). The real-vector delta function can be defined in n-dimensional Euclidean Rn as the such that

f (x)δ(x)[dx]= f (0) (2.1) Rn Z for every compactly supported continuous function f . As a measure, the n-dimensional delta function is the of the 1-dimensional delta functions in each variable separately. Thus, formally, with

n δ(x)= ∏ δ(xj), (2.2) j=1 where x = [x ,..., x ]T Rn. 1 n ∈ The delta function in an n-dimensional space satisfies the following scaling property instead:

n δ(ax)= a − δ(x), a R 0 . (2.3) | | ∈ \{ }

T T Indeed, ax = [ax1,..., axn] for x = [x1,..., xn] , thus

n n n 1 n n δ(ax)= ∏ δ(axj)= ∏ a − δ(xj)= a − ∏ δ(xj)= a − δ(x). j=1 j=1 | | | | j=1 | |

10 This indicates that δ is a homogeneous distribtion of degree ( n). As in the one-variable case, it − is possible to define the compositon of δ with a bi-Lipschitz function g : Rn Rn uniquely so → that the identity

f (g(x))δ(g(x)) det g′(x) [dx]= f (u)δ(u)[du] (2.4) Rn Rn Z Zg( ) for all compactly supported functions f . Using the coarea formula from , one can also define the composition of the delta function with a from one to another one of different dimension; the result is a type of . In the special case of a continuously differentiable function g : Rn R such that the of g is nowhere zero, the following identity holds1 → f (x) f (x)δ(g(x))[dx]= dσ(x) (2.5) Rn g 1(0) g(x) Z Z − |∇ | where the integral on the right is over g 1(0), the (n 1)-dimensional defined by g(x)= 0 − − with respect to the measure. That is known as a simple layer integral.

Proposition 2.2. It holds that

1 i t,x n δ(x)= e h i[dt] (x R ). (2.6) ( π)n Rn 2 Z ∈ Proof. Let x = [x ,..., x ]T Rn. Then by Fourier transform of Dirac delta function: 1 n ∈ n n 1 itj xj δ(x) = ∏ δ(xj)= ∏ e dtj (2.7) 2π R j=1 j=1 Z n 1 n i ∑j=1 tjxj = e ∏ dtj (2.8) (2π)n Rn Z j=1

1 i t,x n = e h i[dt] (x R ), (2.9) ( π)n Rn 2 Z ∈ n where [dt] := ∏j=1 dtj.

Proposition 2.3. For a full-ranked real matrix A Rn n, it holds that ∈ × 1 δ(Ax)= δ(x), x Rn. (2.10) det(A) ∈ | | In particular, under any reflection or rotation R, the delta function is invariant:

δ(Rx)= δ(x). (2.11) 1See https://en.wikipedia.org/wiki/Dirac_delta_function

11 The first proof. By using Fourier transform of Dirac delta function, it follows that

T 1 i t,Ax 1 i A t,x n δ(Ax)= e h i[dt]= e [dt] (x R ). (2.12) ( π)n Rn ( π)n Rn h i 2 Z 2 Z ∈ Let s = ATt. Then [ds]= det(AT) [dt]= det(A) [dt]. From this, we see that | | | |

1 1 i s,x 1 n δ(Ax)= det− (A) e h i[ds]= det− (A) δ(x) (x R ). (2.13) (2π)n Rn ∈ Z

Thus δ(Rx)= δ(x) since det(R )= 1 for reflection or any rotation R. We are done. ± The second proof. By SVD, we have two orthogonal matrices L, R and diagonal matrix

Λ = diag(λ1,..., λn) with positive diagonal entries such that A = LΛRT. Then, via y := RT x (hence δ(y)= δ(x)),

n n Λ T Λ 1 δ(Ax)= δ(L R x)= δ( y)= ∏ δ(λjyj)= ∏ λ−j δ(yj), j=1 j=1 that is, 1 n 1 1 δ(Ax)= ∏ δ(yj)= δ(y)= δ(x). ∏n ∏n ∏n j=1 λj j=1 j=1 λj j=1 λj Now det(A)= det(LΛRT x)= det(L) det(Λ) det(RT), it follows that

n T det(A) = det(L) det(Λ) det(R ) = ∏ λj. | | | | | | | | j=1

1 Therefore we get the desired identity: δ(Ax)= det− (A) δ(x). This completes the proof.

1 Clearly, letting A = a n in the above gives Eq. (2.3).

3 Dirac delta function of matrix argument

Definition 3.1 (Dirac delta function of real-matrix argument). (i) For an m n real matrix X = × [x ] Rm n, the matrix delta function δ(X) is defined as ij ∈ × m n δ(X) := ∏ ∏ δ(xij). (3.1) i=1 j=1

In particular, the vector delta function is just a special case where n = 1 in the matrix case. (ii) For an m m symmetric real matrix X = [x ], the matrix delta function δ(X) is defined as × ij

δ(X) := ∏ δ(xij). (3.2) i6j

12 From the above definition, we see that the matrix delta function of a complex matrix is equal to the product of one-dimensional delta functions over the independent real and imaginary parts of this complex matrix. In view of this observation, we see that δ(X)= δ(vec(X)), where vec(X) is the vectorization of the matrix X. It is easily checked for a rectangular matrix. For the symmetric

x11 x12 case, for example, take 2 2 symmetric real matrix X = with x12 = x21, then × " x21 x22 # T T T T vec(X) = [x11, x21, x12, x22] = [x11, x12, x12, x22] , thus vec(X) = x11[1, 0, 0, 0] + x12[0, 1, 1, 0] + x [0, 0, 0, 1]T , i.e., there are three independent variables x , x , x in the vector vec(X) just 22 { 11 12 22} like in the matrix X, thus

δ(X)= δ(x11)δ(x12)δ(x22)= δ(vec(X)).

Proposition 3.2. For an m m symmetric matrix X, we have × m m(m+1) i Tr(TX) δ(X)= 2− π− 2 e [dT], (3.3) Z where T = [t ] is also an m m real symmetric matrix, and [dT] := ∏ dt . ij × i6j ij Proof. Since m m Tr (TX) = ∑ tjj xjj + ∑ tij xij = ∑ tjj xjj + 2 ∑ tij xij j=1 i=j j=1 i

Proposition 3.3. For A Rm m, B Rn n and X Rm n, we have ∈ × ∈ × ∈ × n m δ(AXB)= det− (A) det− (B) δ(X). (3.4)

Proof. We have already known that δ(AXB )= δ(vec(AXB)) . Since vec(AXB)=(A BT) vec(X) ⊗ [11], it follows that

δ(AXB) = δ(vec(AXB)) = δ ((A BT) vec(X)) (3.5) ⊗ 1 T = det− (A B ) δ(vec(X)) (3.6) ⊗ n m = det− (A) det− (B) δ(X). (3.7)

This completes the proof.

13 Proposition 3.4. For A Rn n and X = XT Rn n, we have ∈ × ∈ × T (n+1) δ(AXA )= det(A) − δ(X). (3.8) | | Proof. By using Eq. (3.3), it follows that

T T n n(n+1) i Tr(TAXA ) δ(AXA ) = 2− π− 2 e [dT]

Z T n n(n+1) i Tr(A TAX) = 2− π− 2 e [dT]. Z Let S = ATTA. Then we have [dS]= det(A) n+1 [dT] (see Proposition 2.8 in [11]). From this, we | | see that

T (n+1) n n(n+1) i Tr(SX) (n+1) δ(AXA )= det(A) − 2− π− 2 e [dS]= det(A) − δ(X). (3.9) | | Z | | This completes the proof.

Note that Dirac delta function for complex number is defined by δ(z) := δ(Re(z))δ(Im(z)), where z = Re(z)+ √ 1Im(z) for Re(z), Im(z) R. The complex number z can be realized as a − ∈ 2-dimensional real vector Re(z) z z := . 7→ " Im(z) # Thus b

Re(z) δ(z)= δ(z)= δ . (3.10) " Im(z) #!

Then for c C, cz is represented as b ∈ Re(c)Re(z) Im(c)Im(z) Re(c) Im(c) Re(z) cz − = − , 7→ " Im(c)Re(z)+ Re(c)Im(z) # " Im(c) Re(c) #" Im(z) # we have

Re(c) Im(c) Re(z) δ(cz) = δ − " Im(c) Re(c) #" Im(z) #!

1 Re(c) Im(c) Re(z) 2 = det− − δ = c − δ(z). " Im(c) Re(c) #! " Im(z) #! | |

Therefore we have

2 δ(cz)= c − δ(z). (3.11) | | Furthermore, if z Cn, then ∈ 2n δ(cz)= c − δ(z). (3.12) | |

14 Proposition 3.5. For a full-ranked complex matrix A Cn n, it holds that ∈ × 1 2 n δ(Az)= δ(z)= det(A) − δ(z), z C . (3.13) det(AA ) | | ∈ | ∗ | In particular, for any unitary matrix U U(n), we have ∈ δ(Uz)= δ(z). (3.14)

Proof. Since Az can be represented as

Re(A) Im(A) Re(z) Az = − " Im(A) Re(A) #" Im(z) # it follows that bb Re(A) Im(A) Re(z) δ(Az) = δ − " Im(A) Re(A) #" Im(z) #!

1 Re(A) Im(A) Re(z) = det− − δ " Im(A) Re(A) #! " Im(z) #!

1 = det− (AA∗) δ(z),

Re(A) Im( A) where det − = det(AA∗) can be found in [8, 11]. Therefore we have " Im(A) Re(A) #! | |

2 n δ(Az)= det(A) − δ(z), z C . (3.15) | | ∈ If A = U is a unitary matrix, then det(U) = 1. The desired result is obtained. | | Re(A) Im(A) Re(z) The second proof. Let A = − and z = . Then by Proposition 2.3, " Im(A) Re(A) # " Im(z) #

b 1 b 1 δ(Az)= δ(Az)= det− (A) δ(z)= det− (AA∗) δ(z). (3.16)

This completes the proof. bb b b Definition 3.6 (Dirac delta function of complex-matrix argument). (i) For an m n complex × matrix Z = [z ] Cm n, the matrix delta function δ(Z) is defined as ij ∈ × m n δ(Z) := ∏ ∏ δ Re(zij) δ Im(zij) . (3.17) i=1 j=1   In particular, the vector delta function is just a special case where n = 1 in the matrix case. (ii) For an m m Hermitian complex matrix X = [x ] Cm m, the matrix delta function δ(X) is × ij ∈ × defined as

δ(X) := ∏ δ(xjj) ∏ δ Re(xij) δ Im(xij) . (3.18) j i

Proposition 3.7. For an m m Hermitian complex matrix X Cm m, we have × ∈ × 1 δ(X)= ei Tr(TX)[dT], (3.19) m m2 2 π Z where T = [t ] is also an m m Hermitian complex matrix, and [dT] := ∏ dt ∏ < dRe(t )dIm(t ). ij × j jj i j ij ij Proof. Indeed, we know that

m m Tr (TX) = ∑ tjj xjj + ∑ t¯ij xij = ∑ Re(tjj)Re(xjj)+ ∑ t¯ij xij + tij x¯ij j=1 i=j j=1 16i

m i Tr(TX) e [dT] = ∏ exp itjj xjj dtjj Z j=1 Z  ∏ exp iRe(tij) 2Re(xij) dRe(tij) × 16i

Therefore we get the desired identity.

Remark 3.8. Indeed, since

off off Tr T X = ∑ 2(Re(tij)Re(xij)+ Im(tij)Im(xij)) <   i j and

off [dT ]= ∏ dRe(tij)dIm(tij), i

16 it follows that

[dToff] exp i Tr ToffXoff Z    = ∏ dRe(tij) exp iRe(tij)(2Re(xij)) dIm(tij) exp iIm(tij)(2Im(xij)) i

From the above discussion, we see that (3.19) can be separated into two identities below:

1 diag diag δ(Xdiag) = [dTdiag]ei Tr(T X ), (3.20) ( π)m 2 Z 1 off off δ(Xoff) = [dToff]ei Tr(T X ). (3.21) m(m 1) π − Z Note that the identity in Proposition 3.7 is used in deriving the joint distribution of diagonal part of Wishart matrix ensemble [12], and it is also used in obtaining derivative principle for unitarily invariant ensemble in [9]. More generally, the fact can be found in [2] that the derivative principle for invariant measure is used to investigate the joint distribution of eigenvalues of local states from the same multipartite pure states.

Proposition 3.9. For A Cm m and B Cn n, let Z Cm n. Then we have ∈ × ∈ × ∈ × n m δ(AZB)= det− (AA∗) det− (BB∗)δ(Z). (3.22)

Proof. Now AZB can be represented as, via XY = XY,

[ Re(A) Im(A) dRe(Zb)b Im(Z) Re(B) Im(B) AZB = AZB = − − − (3.23) " Im(A) Re(A) #" Im(Z) Re(Z) #" Im(B) Re(B) # bb b Then from Eq. (3.4), we see that

[ n m δ(AZB) = δ(AZB)= δ(AZB)= det− (A) det− (B) δ(Z) (3.24) n m = det− (AA∗) det− (BB ∗)δ(Z). (3.25) bb b b b b The result is proven.

Proposition 3.10. For A Cm m, and m m Hermitian complex matrix X Cm m, we have ∈ × × ∈ × m δ(AXA∗)= det(AA∗) − δ(X). (3.26) | |

17 Proof. By using the Fourier transform of the matrix delta function (see Eq. (3.19)) 1 1 δ(AXA )= ei Tr(TAXA∗)[dT]= ei Tr(A∗TAX)[dT], (3.27) ∗ m m2 m m2 2 π Z 2 π Z where T = [t ] is also an m m Hermitian complex matrix, and [dT] := ∏ dt ∏ < dRe(t )dIm(t ). ij × j jj i j ij ij Let H = A∗TA. Then we have (see Proposition 3.4 in [11]):

m [dH]= det(AA∗) [dT]. (3.28) | | Thus

m 1 i Tr(HX) m δ(AXA )= det(AA ) − e [dH]= det(AA ) − δ(X), (3.29) ∗ ∗ m m2 ∗ | | 2 π Z | | We are done.

4 Applications

4.1 Joint distribution of eigenvalues of Wishart matrix ensemble

The first application is to calculate the joint distribution of Wishart matrix ensemble [6]. Note that the so-called Wishart matrix ensemble is the set of all complex matrices of the form W = ZZ† with Z an m n(m 6 n) complex Gaussian matrix, i.e., a matrix with all entries being standard × complex Gaussian random variables, 1 ϕ(Z)= exp Tr ZZ† , πmn −    the distribution density of W is given by

P(W) = δ W ZZ† ϕ(Z)[dZ] (4.1) Z − 1   = δ W ZZ† exp Tr ZZ† [dZ]. (4.2) πmn − − Z      That is,

1 Tr(W) † P(W) = e− δ W ZZ [dZ]. (4.3) πmn − Z   Let Z = √WY. Then ZZ† = √WYY†√W and [dZ]= det(W)n[dY]. By Proposition 3.10, we get 1 δ W ZZ† = δ √W 1 YY† √W = δ 1 YY† . − − det(W)m −         Therefore

1 n m Tr(W) † n m Tr(W) P(W) = det − (W)e− δ 1 YY [dY] ∝ det − (W)e− , (4.4) πmn m − Z   where δ 1 YY† [dY] := C is a constant, independent of W. − R  18 Proposition 4.1. It holds that

1 m(2n m+1) 1 † π 2 − δ m YY [dY]= m . (4.5) − ∏k=1(n k)! Z   − In particular, (i) for m = 1, πn πn δ (1 y, y ) [dy]= = . (4.6) −h i (n 1)! Γ(n) Z − (ii) for m = n, we have

1 n(n+1) π 2 δ 1 YY† [dY]= = 2 n vol(U(n)). (4.7) n n Γ − − ∏k=1 (k) · Z   Proof. Since

C n m Tr(W) 1 = P(W)[dW]= det − (W)e− [dW], (4.8) πmn > Z ZW 0 where via W = UwU† for w = diag(w ,..., w ) where w > > w 1 m 1 · · · m [dW]= ∆(w)2[dw][U†dU].

It follows that C Tm ∆ 2 n m Tr(w) 1 = mn vol(U(m)/ ) (w) det − (w)e− [dw]. (4.9) π w > >w Z 1 ··· m By using Selberg integral formula, we see that

m 2 n m Tr(w) ∆(w) det − (w)e− [dw]= ∏ k!(n k)!. (4.10) Z k=1 − Thus m 2 n m Tr(w) 1 ∆(w) det − (w)e− [dw]= ∏ k!(n k)!. (4.11) w > >w m! − Z 1 ··· m k=1 We conclude that Tm mn 1 † vol( )m!π δ YY [dY]= m . (4.12) − vol(U(m)) ∏k=1 k!(n k)! Z   − Now m m(m+1) 2 π 2 Tm m vol(U(m)) = m and vol( )=(2π) . ∏k=1 Γ(k) We obtain that

1 m(2n m+1) 1 † π 2 − δ YY [dY]= m . (4.13) − ∏k=1(n k)! Z   − This completes the proof.

19 Corollary 4.2. It holds that

1 m(2n m+1) † π 2 − n m δ W ZZ [dZ]= m det(W) − . (4.14) − ∏k=1(n k)! Z   − Proof. It is easily seen that

† n m † δ W ZZ [dZ] = det(W) − δ 1m YY [dY] (4.15) Z − Z −   1 m(2n m+1)   π 2 − n m = det(W) − . (4.16) ∏m (n k)! k=1 − Here we used the result in Proposition 4.1.

Remark 4.3. Denote U(m, n) := Z Cm n : ZZ† = 1 (m 6 n). Note that ∈ × m  m 1 m(2n m+1) 1 † † 2 π 2 − vol(U(m, n)) = δ m ZZ [Z dZ]= m . − ∏k=1(n k)! Z   − This indicates that

δ 1 ZZ† [Z†dZ]= 2m δ 1 ZZ† [dZ]. m − m − Z   Z   In addition, the delta integral can be reformulated in terms of another form:

δ W ZZ† [dZ]= [dZ]. − ZZ†=W Z   Z Finally, we have seen that det(W)n me Tr(W) P(W)= − − . (m) m π 2 ∏ (n k)! k=1 − 4.2 Joint distribution of eigenvalues of induced random ensemble

Any mixed state ρ (i.e., nonnegative complex matrix of trace-one) acting on , may be purified Hm by finding a pure state X in the composite , such that ρ is given by the | i Hm ⊗Hm partial tracing over the auxiliary subsystem,

X ρ = Tr ( X X ) . (4.17) | i −→ 2 | ih | In a loose sense, the purification corresponds to treating any density matrix of size m as a vector of size m2. Consider a bipartite m n composite quantum system. Pure states of this system X may ⊗ | i be represented by a rectangular complex matrix X. The partial tracing with respect to the n- dimensional subspace gives a reduced density matrix of size m: ρ = Tr ( X X ). The natural n | ih | measure in the space of mn-dimensional pure states induces the measure Pm,n(ρ) in the space of the reduced density matrices of size m.

20 Without loss of generality, we assume that m 6 n, then ρ is generically positive definite (here something is generic means it holds with probability one). In any case, we are only interested in the distribution of the positive eigenvalues. Let us call the corresponding positive reduced density matrix again ρ = XX†, where X is a m n matrix. First we calculate the distribution of × matrix elements

P(ρ) ∝ [dX]δ ρ XX† δ 1 Tr XX† (4.18) − − Z      where the first delta function is a delta function of a Hermitian matrix and in the second delta function Tr XX† may be substituted by Tr (ρ). Since ρ is positive definite we can make a transformation 

X = √ρX, (4.19) it follows that [dX]= (det ρ)n [dX] [11]. The matrixe delta function may now be written as

† m † δ √ρ 1 eXX √ρ =(det ρ)− δ 1 XX . (4.20) − −       As the result the distribution of matrixe e elements is given by e e

n m P(ρ) ∝ θ(ρ)δ (1 Tr (ρ))(det ρ) − (4.21) − where the assures that ρ is positive definite. It is then easy to show by the methods of random matrix theory that the joint density of eigenvalues Λ = λ ,..., λ of ρ is given by { 1 m} m m ∝ n m 2 Pm,n(λ1,..., λm) δ 1 ∑ λj ∏ λj − θ(λj) ∏ (λi λj) . (4.22) − j=1 ! j=1 16i

This result is firstly obtained by Zyczkowski˙ [13] in 2001.

4.3 Joint distribution of eigenvalues of two Wishart matrices

The second application is to calculate the distribution of the sum of a finite number of complex Wishart matrices taken from the same Wishart ensemble [6]. The distribution of the sum of two Wishart matrices is considered in [7]. Let us consider two independent complex matrices A and B of dimensions m n and m n taken, respectively, from the distributions × A × B 1 P (A) = exp Tr Σ 1 AA† , (4.23) A mn nA −A π A det (ΣA) − 1    P (B) = exp Tr Σ 1BB† . (4.24) B mn nB −B π B det (ΣB) −   

21 Here ΣA, ΣB are the covariance matrices. Since the domains of A and B remain invariant under unitary rotation, without loss of generality, we may take ΣA = diag(σA1,..., σAm) and ΣB = diag(σB1,..., σBm). We assume that m 6 nA, nB. And we have

[dA]PA(A)= 1 and [dB]PB(B)= 1. Z Z The matrix AA† and BB† are then n-variate complex-Wishart-distributed, i.e., AA† WC(n , Σ ) ∼ m A A and BB† WC(n , Σ ). ∼ m B B Ones are interested in the of the ensemble of m m Hermitian matrices × W = AA† + BB†.

The distribution of W can be obtained as

P (W)= [dA] [dB]δ W AA† BB† P (A)P (B). (4.25) W − − A B Z Z   In what follows, our method will be different from Kumar’s in [7]. Let A = √W A and B = √WB. Then [dA]= det(W)nA [dA] and [dB]= det(W)nB [dB]. We also have e e † † m † † δ We AA BB = det(W)e− δ 1 AA BB . − − − −     Then e e ee n +n m det A B− (W) PW(W) = n n [dA] [dB] m(nA+nB) A Σ B Σ π det ( A) det ( B) Z Z 1 † 1 † † † Tr(√WΣ− √WAA ) Tr(√WΣ− √WBB ) δ 1 AA BB e− A e − e B . × − −   e e ee That is, e e ee

nA+nB m PW(W) ∝ det − (W) 1 † 1 † † † Tr(√WΣ− √WAA +√WΣ− √WBB ) [dA] [dB]δ 1 AA BB e− A B . (4.26) × − − Z Z   e e ee e e e e√ Σee1√ † √ Σ 1√ † 1 Now consider the reduction of the sum Tr W −A W AA + W −B WBB . Since = AA† + BB†, it follows that   e e ee √ Σ 1√ † √ Σ 1√ † e e ee Tr W −A WAA + Tr W −B WBB (4.27)  1  1 1 † = Tr WΣ− + Tr √W(Σ− Σ− )√WBB (4.28) A e e B − A ee  1  1 1 † = Tr WΣ− + Tr √W(Σ− Σ− )√W AA . (4.29) B A − B ee     Hence e e n +n m det A B− (W) PW(W) ∝ Q(W), (4.30) Tr WΣ 1 e ( −A )

22 where

1 1 † † † Tr(√W(Σ− Σ− )√WBB ) Q(W) := [dA] [dB]δ 1 AA BB e A − B − − Z Z ee  1 1  † Tr(√W(Σ− Σ− )√W(1 AA )) = e [deA]e e eA − Bee − † 0

nA+nB m det − (W) Tr √W(Σ 1 Σ 1)√WAA† ∝ ( −B −A ) PW(W) 1 [dA]e − . (4.31) Σ † Tr(W −B ) 0

nA+nB m det − (W) Tr √W(Σ 1 Σ 1)√WBB† ∝ ( −A −B ) PW(W) 1 [dB]e − . (4.32) Σ † Tr(W −A ) 0

1 √ Σ 1 Σ 1 √ † Tr( W( −B −A ) WAA ) 1 [dA]e − Σ † Tr(W −B ) 0

m nA m † [dA]= 2− det − (X)[dX][dU1U ]. (4.33)

This means that e 1 † m Tr(ΛAA ) m nA m Tr(ΛX) † [dA]e = 2− [dX] det − (X)e [dU1U ] (4.34) † 0

1m 1m 1 nA m Tr(ΛX) 1 nB m Tr(ΛX) 1 [dX] det − (X)e = 1 [dX] det − (X)e− . (4.39) Tr(WΣ− ) Tr(WΣ− ) e B Z0 e A Z0

In particular, if Λ = 0 (i.e., ΣA = ΣB := Σ), then

n +n m det A B− (W) ∝ PW(W) Σ 1 . (4.40) eTr(W − )

† In the following, we calculate, via X = Udiag(x)U where diag(x)= diag(x1,..., xm),

1 1 1 m m Λ † nA m Tr(ΛX) ∝ ∆ 2 nA m Tr( Udiag(x)U ) [dX] det − (X)e (x) ∏ xj − dxj dµHaar(U)e (4.41) Z0 Z0 · · · Z0 j=1 Z

W.l.o.g, we assume that Λ is in the diagonal form, i.e. Λ = diag(λ ,..., λ ) where each λ R. 1 m j ∈ By Harish-Chandra-Itzykson-Zuber integral formula [3, 5],

m Tr(AUBU†) det exp(aibj) dµHaar(U)e = ∏ Γ(j) , (4.42) ∆(a)∆(b) ZU(m) j=1 !   it follows that

m Tr(ΛUdiag(x)U†) det exp(λi xj) dµHaar(U)e = ∏ Γ(j) . (4.43) ∆(λ)∆(x) Z j=1 !  

Therefore

1m 1 1 m nA m Tr(ΛX) 1 nA m [dX] det − (X)e ∝ ∆(x) det exp(λixj) ∏ x − dxj (4.44) ∆(λ) j Z0 Z0 · · · Z0 j=1   1 1 1 m λj xj nA m ∝ ∆(x) ∏ e x − dxj, (4.45) ∆(λ) j Z0 · · · Z0 | | j=1 implying that

1 m 1 m 1 1 m nA m Tr(ΛX) nA m λjxj [dX] det − (X)e ∝ ∏ ∂ − ∆(x) ∏ e dxj. (4.46) ∆(λ) λj Z0 j=1 ! Z0 · · · Z0 | | j=1

24 Remark 4.5. Kumar in [7] have presented analytical formula for distribution of the sum of two Wishart matrices in terms of the confluent hypergeometric function of matrix argument [8], which is defined by

1 Γ p p(c) a p c a p Tr(ΛX) 1F1(a; c; Λ)= det(X) − det(1 X) − − e− [dX], (4.47) − Γ (a)Γ (c a) 0 − p p − Z p(p 1) where Γ (a) := π 2− Γ(a)Γ(a 1) Γ(a p + 1). Kummer’s formula for the confluent hyper- p − · · · − geometric function 1F1 is given by

Tr(Λ) F (a; c; Λ)= e− F (c a; c; Λ). (4.48) 1 1 − · 1 1 − C Remark 4.6. We can present a simple approach to the similar result. Denote Wm (n, Σ) Wishart matrix ensemble for which each matrix is of the form W = AA , where A Cm n(m 6 n). Now ∗ ∈ × W = W + W , where W , W WC(n, Σ), can be rewritten as 1 2 1 2 ∈ m † m 2n W = ZZ , Z = [A, B] C × . ∈ Then W WC(2n, Σ). Thus ∼ m 1 2n m Tr(Σ− W) PW(W) ∝ det − (W)e− .

The distribution of sum of an arbitrary finite number of Wishart matrices can be derived as:

1 kn m Tr(Σ− W) PW(W) ∝ det − (W)e− , where W = ∑k W for W WC(n, Σ). j=1 j j ∼ m

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26