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Dirac Delta Function of Matrix Argument

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Dirac Delta Function of Matrix Argument Dirac Delta Function of Matrix Argument ∗ Lin Zhang Institute of Mathematics, Hangzhou Dianzi University, Hangzhou 310018, PR China Abstract Dirac delta function of matrix argument is employed frequently in the development of di- verse fields such as Random Matrix Theory, Quantum Information Theory, etc. The purpose of the article is pedagogical, it begins by recalling detailed knowledge about Heaviside unit step function and Dirac delta function. Then its extensions of Dirac delta function to vector spaces and matrix spaces are discussed systematically, respectively. The detailed and elemen- tary proofs of these results are provided. Though we have not seen these results formulated in the literature, there certainly are predecessors. Applications are also mentioned. 1 Heaviside unit step function H and Dirac delta function δ The materials in this section are essential from Hoskins’ Book [4]. There are also no new results in this section. In order to be in a systematic way, it is reproduced here. The Heaviside unit step function H is defined as 1, x > 0, H(x) := (1.1) < 0, x 0. arXiv:1607.02871v2 [quant-ph] 31 Oct 2018 That is, this function is equal to 1 over (0, +∞) and equal to 0 over ( ∞, 0). This function can − equally well have been defined in terms of a specific expression, for instance 1 x H(x)= 1 + . (1.2) 2 x | | The value H(0) is left undefined here. For all x = 0, 6 dH(x) H′(x)= = 0 dx ∗ E-mail: [email protected]; [email protected] 1 corresponding to the obvious fact that the graph of the function y = H(x) has zero slope for all x = 0. Naturally we describe the slope as "infinite" at origin. We shall denote by δ(x) the 6 derivative of H(x): δ(x)= H′(x)= 0, x = 0, ∀ 6 and δ(0)=+∞. We recall the definition of Dirac delta function: 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 continuous function 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 derivatives 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 differentiable function 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 point. 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 integral 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 (Convolution). 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 (Fourier transform).
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