Tutorial 6 - MUB and Complex Inner Product

Transcription

1 Tutorial 6 - MUB and Complex Inner Product Mutually unbiased bases Consider first a vector space called R It is composed of all the vectors you can draw on a plane All of them are of the form: ( ) r v =, r where both r and r are reals An orthonormal basis { v, v } is a set of two vectors such that each one is of length, ie v j v j = for j =,, and they are orthogonal, ie v v = 0 (Recall standard definition of the scalar product if you have forgotten) Two orthonormal bases B = { v, v } and B = { w, w } are called mutually unbiased if each vector from one basis has equal overlap with any vector from another basis Formally, v j w k = for all j, k A set of more than two orthonormal bases is called mutually unbiased if they are all pairwise mutually unbiased (a) Draw two distinct orthonormal bases in R which are mutually unbiased Consider B = { v, v } = {( ) ( )} 0, 0 and B = { w, w } =, FIG : Mutually unbiased bases { if i = j Obviously ( v i, v j ) = ( w i, w j ) = δ i,j = 0 if i j Hence B and B are orthonormal bases in R Also v w = v w = v w = v w = = B and B are mutually unbiased orthonormal bases (b) How should one arrange vectors v and w for their matrix multiplication to agree with the scalar product formula? ( v, w) = w d v i w i = (v,, v d ) = vt w w d i=

2 (c) How many mutually unbiased bases are there in R? There are only two mutually unbiased bases in R ( ) x Suppose not Let B 3 be another distinct orthonormal basis mutually unbiased to B and B Let u = B y 3 Now ) ( ) = u v = x ( x = ± Similarly ) ( ) y 0 = u v = x 0 ( y = ± y ( ) { ( ) ( ) ( ) ( )} x Thus u = y,,, = { u, u, u 3, u 4 } But u w =, u w =, u 3 w = 0, u 4 w = 0 Hence u does not exist = B 3 does not exist There are only two mutually unbiased bases in R Note Let S be the set of mutually unbiased bases The maximal number of mutually unbiased bases in each S is the same Hence it does not matter which orthonormal basis we used in search for the maximal number of mutually unbiased bases Claim There are no mutually unbiased basis in R 3 For R 3 two orthonormal bases B = { v, v, v 3 } and B = { w, w, w 3 } are said to be mutually biased if and only if v j w k = for all j, k Without loss of generality, let B = { v, v, v 3 } = 0,, 0 be an orthonormal basis in R 3 Let u = y B, another set of orthonormal basis mutually unbiased to B Now 0 0 x z 3 = u v x = y 0 x = ± Similarly, y = z = ± and z u, 3, 3, 3 3,, 3 3,, 3 3 But none of the vectors in the set is orthogonal to each other, ie ( u i, u j ) 0 for all i j in the set, contradicting the fact that u B an orthonormal basis Therefore u does not exist and there is no pair of mutually unbiased bases in R 3

3 3 The vector space C is defined in analogy to R It is composed of the vectors: ( ) c v =, where now c and c are complex numbers The scalar product above gets replaced by the following inner product rule: where d and d are the complex components of vector w c v w c d + c d, (d) How should one arrange vectors v and w for their matrix multiplication to agree with the inner product formula? ( v, w) = i w vi w i = (v,, vd) = v w w d (e) How many mutually unbiased bases are there in C? There are three unbiased bases in C {( ) ( )} 0 First, we show that the set B = { v, v } =, is an orthonormal basis in C 0 Let ( ) ( ) ( ) a + bi 0 C v = = α + β for some a, b, c, d R and α, β C Then α = a + bi and c + di 0 ( ) ( ) 0 β = c + di This implies B spans C Obviously and are linearly independent Hence B 0 is a basis for C Since ( v, v ) = δ ij, B is an orthonormal basis for C and dim(c ) = In a similar way, we can show that the set B = { w, w } =, is an orthonormal basis of C also We have shown in (a) that S = {B, B } is a set of mutually unbiased bases [] Let B 3 be another distinct orthonormal basis mutually unbiased to B and B Consider u = B 3 where a, b, c, d R Now ( z ) = z ( ) a + bi c + di ( ) ( ) = u v = a + bi c + di 0 ( ) ( ) = u v = a + bi 0 c + di ( ) ( ) = u w = a + bi c + di ( ) ( ) = u w = a + bi c + di a + b = () c + d = () (a + c) + (b + d) = (3) (a c) + (b d) = (4) [] Since the basis vectors in {B, B } are real, the complex inner product operation reduces to the real scalar product

4 Solving for the equations () to (4) we have either { a =, b = c = 0, d = ± } Therefore u = ( z ) = z { a =, b = c = 0, d = ± } ( ) { ( ) ( ) ( ) ( )} a + bi c + di,, i i, i = { u i, u, u 3, u 4 } Clearly u = u 4, u = u 3 and ( u, u ) = 0 As a result the third set of orthonormal basis is B 3 = { u, u } =, i i 4 or and there are three mutually unbiased bases in C Note: If the dimension of a Hilbert space (C d ) d is an integer power of a prime number, then the maximal number of mutually unbiased bases is d+ Since is a prime number, the maximal number of mutually unbiased bases in C is 3 In summary, let S be the set containing all mutually unbiased orthonormal bases {( ) ( )} { ( 0 (i) In R, S = {B, B } where B =, and B 0 = 0 0 (ii) In R 3, S = {B} where B = 0,, {( ) ( )} { ( 0 (iii) In C, S = {B, B, B 3 } where B =,, B 0 =, i i Inner product between functions ) ( )}, ) ( )}, and B 3 = Mathematicians abstracted essential features of vector spaces such as those discussed above and noted that many objects behave in a similar way Here you will prove that it is possible to define inner product between (wave) functions and during the lecture it will be shown that indeed one can think about the wave functions as vectors Inner product is a generalisation of the scalar product It is any relation between two objects (vectors, functions, polynomials, etc) f and g, usually written as (f, g) that satisfies the following: Note symmetry under conjugation (f, g) = (g, f), linearity (f, αg) = α(f, g), (f, g + h) = (f, g) + (f, h), α C, positive-definiteness (f, f) 0, (f, f) = 0 f = 0 For the linearlity condition } (i) (f, αg) = α(f, g) (f, αg + h) α(f, g) + (f, h) for all α C (ii) (f, g + h) = (f, g) + (f, h) (= ) ( =) (f, αg + h) (ii) = (f, αg) + (f, h) (i) = α(f, g) + (f, h) Let α =, we have (ii) Let h = 0 we have (f, αg) α(f, g) + (f, 0) [] = α(f, g), ie part(i) [] Let α = 0 in (i) above Then (f, 0) = 0(f, g) = 0

5 5 (a) Show that the scalar product in R 3 satisfies properties above (with α R) Scalar product in R 3 is defined as ( v, w) = v T w = i v i w i R for all v, w R 3 Symmetric ( v, w) = i v i w i = i w i v i = ( w, v) Linearity ( u, α v + w) u T (α v + w) = α u T v + u T w α( u, v) + ( u, w) Positivity ( v, v) = v T v = v 0 and 0 = ( v, v) = v T v = v v = 0 v = 0 (b) Show that the inner product in C satisfies properties above Inner product in C is defined as ( v, w) = v w = i v i w i C where v is the conjugate transpose of column vector v ie v = ( v T ) = ( v ) T Symmetric Linearity ( v, w) = vi w i = i ( ) wi v i = ( w, v) ( u, α v + w) u (α v + w) = α u v + u w α( u, v) + ( u, w) i Positivity ( v, v) = v v = v 0 and 0 = ( v, v) = v v = v v = 0 v = 0 (c) Show that the following integral between wave functions f(x) and g(x) satisfies the properties of the inner product: (f, g) f (x)g(x)dx Symmetric: Linearity: (f, g) ( ( ) f (x)g(x)dx = f(x)g (x)dx) = g (x)f(x)dx (g, f) (f, αg + h) f (x) [αg(x) + h(x)] dx = [αf (x)g(x) + f (x)h(x)] dx = α f (x)g(x)dx + f (x)h(x)dx = α(f, g) + (f, h)

6 6 Positivity: Discussion (f, f) f (x)f(x)dx = f(x) dx 0 and 0 = (f, f) = f(x) dx f = 0 The mutually unbiased bases are quite extensively studied due to their applications in quantum mechanics Yet, their maximal number in vector spaces C d is unknown whenever d is not a power of a prime The simplest case being d = 6 If you see more than three mutually unbiased bases in this case, contact lecturer immediately! Application of Mutually Unbiased Bases (MUB) Quantum: Information theory, Entanglement, Cryptography, Key distribution, etc More generally, mutually unbiased bases in Hilbert space C d { f,, f d } such that e j f k = d j, k {,, d} are two orthonormal bases { e,, e d } and