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After Alice and Bob share a Bell state, Bob applies a Pauli gate to his qubit. What will be the situation of the Bell state? What happens?

Then Alice applies the same gate to her qubit – again, what happens?

I would be thankful if you could explain it slowly, as I am a beginner in this field.

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up vote 3 down vote accepted

Pauli gates applied to one qubit will just transform one Bell state into another one. Take, for example, the antisymmetric singlet state: $$|\psi_{AB}^-\rangle = \frac{1}{\sqrt{2}}(|\uparrow_A\rangle\otimes|\downarrow_B\rangle - |\downarrow_A\rangle\otimes|\uparrow_B\rangle). $$ Alice applies the Pauli X gate, defined as $\sigma^x|\uparrow\rangle = |\downarrow\rangle$, $\sigma^x|\downarrow\rangle = |\uparrow\rangle$. The gate is local, so it acts as the identity on Bob's qubit (it does nothing to Bob's qubit). This is written as \begin{eqnarray}(\sigma^x_A\otimes 1_B) |\psi_{AB}^-\rangle &=& \frac{1}{\sqrt{2}}(\sigma^x_A|\uparrow_A\rangle\otimes 1_B|\downarrow_B\rangle - \sigma^x_A|\downarrow_A\rangle\otimes 1_B|\uparrow_B\rangle) \\ &=&\frac{1}{\sqrt{2}}(|\downarrow_A\rangle\otimes|\downarrow_B\rangle - |\uparrow_A\rangle\otimes|\uparrow_B\rangle) \\ &=& |\phi_{AB}^-\rangle,\end{eqnarray} which is just a different Bell state. This demonstrates the local unitary equivalence of Bell states.

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Thank you for your help, that means if apply any Pauli to a qubit of bell state, it will only affect the local qubit but not the other, right ? – Semih Korkmaz Dec 9 '12 at 13:54
if you apply any unitary to just one qubit (this is called a local unitary operation), it doesn't affect the other. – Mark Mitchison Dec 9 '12 at 18:39

In your question, you were also wondering what would happen if both Alice and Bob apply the same Pauli gate on their respective qubit. To complement Mark's answer, I would like to expand on this point. Interestingly, the answer is that nothing happens, the action of their individual gates is equivalent to the identity map! This is a consequence of the more general result that for any maximally entangled pure state $|\psi\rangle$ and any unitary $U$, it holds that $$U\otimes U^T|\psi\rangle=|\psi\rangle$$ where $U^T$ is just the transpose of $U$. All the Bell states are maximally entangled and the Pauli operators satisfy $\sigma_x^T=\sigma_x$, $\sigma_y^T=-\sigma_y$, $\sigma_z^T=\sigma_z$ so that if both Alice and Bob apply the same Pauli gate, from the above result the state is unchanged (up to a global phase).

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Nice, I noticed this when I was writing my answer, but didn't know why it seemed to be true for all the examples I could think of. Does this hold for arbitrary dimensions? – Mark Mitchison Dec 10 '12 at 2:52
It does work in arbitrary dimensions. Specifically, let $|\psi\rangle = \frac{1}{\sqrt{N}}\sum_i |i\rangle \otimes |i\rangle$ and let $U$ be any unitary. Then $U \otimes U^* |\psi\rangle = |\psi\rangle$ where the star denotes complex conjugate. – Dan Stahlke Dec 10 '12 at 17:31
Are you sure it's a complex conjugate and not a transpose? – Juan Miguel Arrazola Dec 10 '12 at 21:59

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