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Suppose that Alice applies a unitary operator $U$ with real entries to her qubit in an EPR pair $|\beta\rangle=\frac{1}{\sqrt 2}(|00\rangle+|11\rangle)$. Is this the same as having Bob apply $U^T$ to his half?

I can't quite show algebraically whether the identity

$$\frac{1}{\sqrt 2}(U|0\rangle\otimes |0\rangle+U|1\rangle\otimes|1\rangle)=\frac{1}{\sqrt 2}(|0\rangle\otimes U^T|0\rangle+|1\rangle\otimes U^T|1\rangle)$$

is true or not. I see that I can get from the LHS to the RHS by applying $U^T\otimes U^T$ as $U^T=U^\dagger=U^{-1}$, but I still haven't made progress.

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Yes, application of $U$ to Alice's half of $|\beta\rangle=(|00\rangle+|11\rangle)/\sqrt{2}$ and appliaction of $U^T$ to Bob's half are equivalent. In fact the identity

$$ (A \otimes I)|\beta\rangle = (I \otimes A^T) |\beta\rangle $$

is true for any matrix (not necessarily unitary or real). Let

$$ A = \begin{pmatrix} a_{00} & a_{01} \\ a_{10} & a_{11} \end{pmatrix} $$

and compare

$$ (A \otimes I)|\beta\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} a_{00} & 0 & a_{01} & 0 \\ 0 & a_{00} & 0 & a_{01} \\ a_{10} & 0 & a_{11} & 0 \\ 0 & a_{10} & 0 & a_{11} \\ \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} a_{00} \\ a_{01} \\ a_{10} \\ a_{11} \end{pmatrix} $$

with

$$ (I \otimes A^T) |\beta\rangle = \begin{pmatrix} a_{00} & a_{10} & 0 & 0 \\ a_{01} & a_{11} & 0 & 0 \\ 0 & 0 & a_{00} & a_{10} \\ 0 & 0 & a_{01} & a_{11} \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} a_{00} \\ a_{01} \\ a_{10} \\ a_{11} \end{pmatrix}. $$

A more elegant way to prove the identity uses tensor networks, see example 9 and equation $(46)$ on page 12 in this paper.

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  • $\begingroup$ Nice. I didn't think that this could hold more generally - hence I never thought about plugging in a generic matrix. $\endgroup$ – Oilobobolus Feb 23 at 3:45
  • $\begingroup$ I was playing around with this a bit, and bar mistakes I found that the overlap between the two states should be $\text{Tr}(U)^2/2$. What's up with this? It doesn't look right. $\endgroup$ – Oilobobolus Feb 23 at 3:56
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    $\begingroup$ $((I\otimes U^T)|\beta\rangle)^\dagger (U\otimes I)|\beta\rangle = \langle\beta|U\otimes \overline{U}|\beta\rangle = \frac{1}{2}\sum_{ij}\langle i|\langle i|U\otimes \overline{U}|j\rangle|j \rangle = \frac{1}{2}\sum_{ij} u_{ij} \overline{u_{ij}} = \frac{1}{2}\sum_{ij} u_{ij}u_{ji}^\dagger = \frac{1}{2}\mathrm{tr}(UU^\dagger) = 1$ $\endgroup$ – Adam Zalcman Feb 23 at 5:34
  • $\begingroup$ Right, so this is the best way to prove this for unitary matrices. Thanks. $\endgroup$ – Oilobobolus Feb 23 at 14:29

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