# Bell State, if Bob applies a Pauli Gate?

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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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.
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).
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