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In the Dür, 2000 paper, he gave a statement that

(...) Any state $\vert\psi\rangle$ can be obtained from Bell State with certainty

From his paper too, it's known that we can transform $\vert\psi\rangle$ and $\vert\phi\rangle$ into each other with certainty by means of LOCC iff they are related by local unitaries (LU). So, I can write the transformation down such that (correct me if I'm wrong)

$U_1 \otimes U_2 \vert\Phi^+\rangle = \alpha_{00}\vert00\rangle + \alpha_{01}\vert01\rangle + \alpha_{10}\vert10\rangle + \alpha_{11}\vert11\rangle \\ \left(U_1 \otimes U_2\right)\left(\frac{1}{\sqrt{2}}\left(\vert00\rangle + \vert11\rangle\right)\right) = \alpha_{00}\vert00\rangle + \alpha_{01}\vert01\rangle + \alpha_{10}\vert10\rangle + \alpha_{11}\vert11\rangle$

From this step, I find it difficult how to obtain the LU transformation.

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  • $\begingroup$ LU and LOCC are two different things. $\endgroup$ Jan 10, 2023 at 20:57
  • $\begingroup$ But it stated that it can be obtained with 'certainty' by means of LOCC, and this coincides with LU. Am I wrong to represent that statement? $\endgroup$
    – Steve J.
    Jan 11, 2023 at 5:44
  • $\begingroup$ related: quantumcomputing.stackexchange.com/q/29575/55 $\endgroup$
    – glS
    Jan 16, 2023 at 11:33

1 Answer 1

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What you want is the theory of the Schmidt decomposition. Let's say you have a two-qubit state $|\psi\rangle$ which is written $$ |\psi\rangle=\alpha_{00}|00\rangle+\alpha_{01}|01\rangle+\alpha_{10}|10\rangle+\alpha_{11}|11\rangle, $$ then it's helpful to construct a matrix $$ C=\left(\begin{array}{cc} \alpha_{00} & \alpha_{01} \\ \alpha_{10} & \alpha_{11} \end{array}\right). $$ This matrix $C$ has a singular value decomposition, meaning there exist unitaries $U$ and $V$ such that $$ C=UDV^\dagger, $$ where $D$ is diagonal with positive semi-definite entries.

You will find that $$ (U^\dagger \otimes V^\dagger )|\psi\rangle=|\phi\rangle $$ where $|\phi\rangle$ is the Bell state. Hence, really what you want is to know how to do the singular value decomposition. If you're not familiar with it, it can help to relate it to the more famliar eigenvalues: $$ CC^\dagger=UD^2U^\dagger,\qquad C^\dagger C=VD^2V^\dagger. $$

Note that $|\phi\rangle$, the Bell state, already has $U=V=I$ when written in this way.

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  • $\begingroup$ I don't get it when you states that $\left(U^\dagger \otimes V^\dagger\right)\vert\psi\rangle = \vert\phi\rangle$. How can you find that, meanwhile in earlier, we're talking about the coefficient matrix $C$? $\endgroup$
    – Steve J.
    Jan 10, 2023 at 12:53
  • $\begingroup$ You build the coefficient matrix $C$ and use that to find what $U$ and $V$ are. $\endgroup$
    – DaftWullie
    Jan 10, 2023 at 13:06
  • $\begingroup$ But $U$ and $V$ are the unitary matrices that build state $\vert\psi\rangle$ via SVD. How can those matrices can also be LU transformation matrices to $\vert\phi\rangle$? $\endgroup$
    – Steve J.
    Jan 10, 2023 at 13:15
  • $\begingroup$ I would like to see a more clarifying comment, as it is not evident that U and V from C can be used to transform the generic state $|\psi\rangle$ to the Bell state. So far, I could only find that two pure bipartite states with the same Schmidt coefficients can be turned one into the other by means of local unitary transformations (see N&C exercise 2.80). $\endgroup$ Jan 12, 2023 at 6:54
  • $\begingroup$ @MicheleAmoretti That's exactly what's going on. The $C$ matrix and its diagonalisation are how (or at least one method) you find the Schmidt coefficients and the Schmidt basis. They way that you convert between the two states is to convert between their Schmidt bases. $\endgroup$
    – DaftWullie
    Jan 12, 2023 at 16:15

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