[Solved]: How to apply a 1-qubit gate to a single qubit from an entangled pair?

Problem Detail: 

Reading about superdense coding I came upon a calculation I can not understand.

We have an EPR entangled pair of qubits $\frac{1}{\sqrt2}(|00\rangle + |11\rangle)$ and we want to apply a Pauli X gate to the first of the entangled qubits. On wikipedia, there is an example showing that such an operation would lead to the state $\frac{1}{\sqrt2}(|10\rangle + |01\rangle)$. I can't see an intuition behind it. I know that the Pauli X gate operates as a quantum NOT, but at the same time I do not know how to write it algebraically.

How can a single qubit from an entangled pair be represented? Is it $\frac{1}{\sqrt2}(|0\rangle + |1\rangle)$ for the pair above? And if so, why does the Pauli X gate change it at all (if we wrote the operation using matrices, the qubit would be represented as $\left[\frac{1}{\sqrt2} \frac{1}{\sqrt2}\right]$, which multiplied by the Pauli X gate matrix would not change at all...)

Asked By : 3yakuya

Answered By : 3yakuya

I came to a conclusion that we do not actually take away a single qubit from an entangled pair and operate it, but operate on both qubits with Identity operator on the one we do not want to change.

So basically to apply Pauli X gate to the first qubit from an entangled pair, we apply X tensored with $I$ to the entangled pair to get the result. That is, to operate a not on the first qubit, we evolve the system by $$ \sigma_x \otimes I = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0& 0 \end{pmatrix} $$

Intuitively it is like applying a single operation to the first qubit and identity to the second one (effectively not changing its state).

Thus if we start with the entangled state $|\psi\rangle \frac{1}{\sqrt2}(|00\rangle + |11\rangle)$ we end up with $$ (\sigma_x \otimes I)|\psi\rangle = \begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0& 0 \end{pmatrix} \cdot \frac1{\sqrt{2}} \begin{pmatrix}1 \\0 \\0 \\1 \end{pmatrix} = \frac1{\sqrt{2}} \begin{pmatrix}0 \\1 \\1 \\0 \end{pmatrix} = \frac1{\sqrt{2}}(|01\rangle+|10\rangle) $$

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Question Source : http://cs.stackexchange.com/questions/34085

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