I have started to learn about the mathematics behind ebits and I have a question. Assume $\color{red}{\text{Alice}}$ and $\color{blue}{\text{Bob}}$ share the following ebit: $\begin{align}\vert\Phi^+ \rangle= \dfrac{\vert\color{red}{0}\color{blue}{0}\rangle^{\color{red}{A}\color{blue}{B}} + \vert\color{red}{1}\color{blue}{1}\rangle^{\color{red}{A}\color{blue}{B}}}{\sqrt{2}} .\end{align}$

What would happen if $\color{red}{\text{Alice}}$ performs Pauli-$X$ gate on their portion of the ebit?

Here is my answer but I am not sure if it is correct:

The $\color{red}{0}$ and $\color{red}{1}$ will swap and end up with: $\color{red}{\text{Alice}}$ and $\color{blue}{\text{Bob}}$ share the following ebit: $\begin{align}\vert\Phi^+ \rangle= \dfrac{\vert\color{red}{1}\color{blue}{0}\rangle^{\color{red}{A}\color{blue}{B}} + \vert\color{red}{0}\color{blue}{1}\rangle^{\color{red}{A}\color{blue}{B}}}{\sqrt{2}} .\end{align}$

Any help is appreciated!

  • 1
    $\begingroup$ question of the type "is this calculation correct?" aren't really a good fit for SE. I'd suggest reframing the post as asking a specific question. For example, you can focus on a specific aspect, and ask about how to show X, including your current reasoning, that you are however not completely sure about. Also, you should ask a single, laser-focused question per post. You can open different posts to ask different questions $\endgroup$
    – glS
    Commented Aug 20, 2021 at 22:31
  • $\begingroup$ +1 but one question per post please. For question 1, I think you've applied the X-gate correctly. $\endgroup$ Commented Aug 21, 2021 at 4:08

1 Answer 1


Your reasoning is correct, this is indeed the resulting state they would get. More generally, you can think about it as applying a $X\otimes I$ gate on the whole system, where $I$ is the identity gate.

However, do note that if Alice and Bob do not communicate, there is nothing Alice can do to influence the measurement she will get, since its qubit is in the maximally mixed state: no matter the gate she applies, she will always get 50% chance of measuring $|0\rangle$ and an equal probability of measuring $|1\rangle$.

  • $\begingroup$ Very clear! Thank you for the answer! $\endgroup$ Commented Aug 23, 2021 at 2:09
  • $\begingroup$ in the above case, Bob and Alice have not communicated, and Bob measures his qubit, and say it comes as 0. Now, he expects Alice's qubit to be 0 .But Alice's qubit is 1 now. This is assuming Bob is not aware of X gate at Alice's side. Is this correct? $\endgroup$
    – Kittu A
    Commented Nov 26, 2021 at 14:12
  • $\begingroup$ @Kittu A That's correct yes! $\endgroup$
    – Tristan Nemoz
    Commented Nov 26, 2021 at 15:48

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