When I evaluate the following equation using $U = ZX$ and $a = 0, b = 0$ [for Bell state], I am getting LHS not equal to RHS.

Before describing the protocol, let us first review the teleportation protocol and introduce the notation used in the paper. In general, Bell states are of the following form

$$|B_{xy}\rangle = \frac{1}{\sqrt 2}(|0,x\rangle + (-1)^y|1, x\oplus 1\rangle),\tag{1}$$

where $x, y \in \{0, 1\}$ and $\oplus$ represents addition modulo $2$. The relationship between Bell states and classical bits can be defined as

$$|B_{xy}\rangle \leftrightarrow xy, x, y \in \{0,1\} \tag{2}$$

For any qubit $|\phi\rangle$ and any single-qubit unitary operation $U$, a general teleportation equation, based on an initial Bell state $|B_{ab}\rangle$, $a, b \in \{0, 1\}$, shared between the two users, can be written as

$$\boxed{|\varphi\rangle_1 \otimes (I\otimes U)|B_{ab}\rangle_{2,3} = \frac{1}{2}\sum_{\mathrm{x\in\{0,1\}}}\sum_{\mathrm{x\in\{0,1\}}}(-1)^{b.x}|B_{x,y}\rangle_{1,2}\otimes UZ^{y\oplus b}X^{x\oplus a}|\varphi\rangle} \tag{3}$$

where $X = (|0\rangle\langle 1| + |1\rangle\langle 0|)$, $Z = (|0\rangle\langle 0| - |1\rangle\langle 1|)$ and the subscripts denote different systems.

Paper Reference


1 Answer 1


There's a mistake. It's incorrect even for $a=0,b=0,U=I$. In this case the correct formula is $$ |\phi\rangle \otimes |B_{00}\rangle = \frac{1}{2} \bigg(|B_{00}\rangle \otimes |\phi\rangle + |B_{01}\rangle \otimes Z|\phi\rangle + |B_{10}\rangle \otimes X|\phi\rangle + |B_{11}\rangle \otimes XZ|\phi\rangle\bigg) $$ but their formula swaps $X$ and $Z$ in the last term. It looks like if we swap $X$ and $Z$ terms in their formula then it will be correct (I checked on the computer).


The correct formula is with the swapped $X$ and $Z$ terms: $$ |\phi\rangle \otimes (I \otimes U)|B_{ab}\rangle = \frac{1}{2} \bigg(\sum_{x \in \{0,1\}}\sum_{y \in \{0,1\}} (-1)^{b\cdot x}|B_{xy}\rangle \otimes UX^{x\oplus a}Z^{y\oplus b}|\phi\rangle\bigg) $$

I've used Julia programming language to check this. Note that both sides of equality are linear on $|\phi\rangle$, so we can check it only for $\phi = 0$ and $\phi = 1$. Also both sides are linear on $U$, so we can check equality only for $I, X, Z$ and $XZ$.

using LinearAlgebra

const ⊗ = kron
const ⊕(x,y) = (x+y) % 2

const Z = zeros(2,2) + [1 0; 0 -1]
const X = zeros(2,2) + [0 1; 1 0]
const Id = zeros(2,2) + I

function basis(x)
  return Id[:,x+1]

function bell(x,y)
  ret = zeros(4)    
  ret += basis(0) ⊗ basis(x)
  ret += (-1)^y * basis(1) ⊗ basis(x ⊕ 1)
  ret /= sqrt(2)
  return ret

function check()
  for a=0:1, b=0:1, ϕ=0:1, U in [Id, X, Z, X*Z]
    lhs = basis(ϕ) ⊗ ( (Id ⊗ U) * bell(a,b) )
    rhs = zeros(8)
    for x=0:1, y=0:1
      rhs += (-1)^(b*x) * bell(x,y) ⊗ (U * X^(x⊕a) * Z^(y⊕b) * basis(ϕ))
    rhs /= 2
    err = norm(lhs - rhs)    


Note that this check is numerical, though one can also try to use symbolic math packages, like SymPy.

  • $\begingroup$ could you please share the related code used for checking the equation... $\endgroup$ Sep 24, 2019 at 13:32
  • $\begingroup$ yes, updated the answer $\endgroup$
    – Danylo Y
    Sep 24, 2019 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.