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It's easy to see that in computational basis, Pauli matrices could be represented in the outer product form:

$$ X=|0\rangle\langle1|+|1\rangle\langle0|\\ Y=-i|0\rangle\langle1|+i|1\rangle\langle0|\\ Z=|0\rangle\langle0|-|1\rangle\langle1| $$

If we want to represent the outer products in $X$ basis $|+\rangle$ and $|-\rangle$, one way I can think of is to use the identities $$ |0\rangle=\frac{1}{\sqrt{2}}(|+\rangle+|-\rangle)\\ |1\rangle=\frac{1}{\sqrt{2}}(|+\rangle-|-\rangle) $$ and plug them in the first three equations. I'm wondering is there a simpler / more direct way we can do that? Thanks!

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By far the most direct way is to realise that you'd be writing (for example) $$ Y=a_{++}|+\rangle\langle +|+a_{+-}|+\rangle\langle -|+a_{-+}|-\rangle\langle +|+a_{--}|-\rangle\langle-|. $$ You can then evaluate $$ a_{+-}=\langle +|Y|-\rangle. $$

You might shave a little bit off your calculations by realising that (i) the representation will be a Hermitian matrix, so $a_{+-}=a_{-+}^\star$, (ii) all Pauli matrices have trace 0 so $a_{++}=-a_{--}$.

Perhaps the easier computation comes from first representing each of the Pauli matrices in the standard basis. For example, $$ Y\equiv\left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right). $$ Then you realise that all you're trying to do is a basis transformation $$ U=|0\rangle\langle+|+|1\rangle\langle -|\equiv\frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right), $$ so to make the transformation you calculate $$ UYU^\dagger\equiv\left(\begin{array}{cc} 0 & i \\ -i & 0 \end{array}\right). $$ Hence, $$ Y=i|+\rangle\langle -|-i|-\rangle\langle +|. $$

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  • $\begingroup$ Thanks so much for the answer! For the first method, can I say that (i) the representation is hermitian since it represents a quantum gate? (ii) The trace is 0 because the representation matrix is the Hadamard gate, which is a combination of Pauli matrices? Thanks!! $\endgroup$ – ZR- Feb 18 at 7:59
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    $\begingroup$ The representation is hermitian because the operator is self adjoint. Effectively, if it's Hermitian in one representation, it's Hermitian in all representations! (ii) The trace is invariant under representations. Because we're performing a basis transformation $U\sigma U^\dagger$, then $\text{Tr}(U\sigma U^\dagger)=\text{Tr}(\sigma U^\dagger U)=\text{Tr}(\sigma)=0$. $\endgroup$ – DaftWullie Feb 18 at 8:52

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