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I have this state of two qubits here: $$ |\psi_{AB}\rangle = \frac{1}{2}(|0\rangle_A |0\rangle_B + |1\rangle_A |1\rangle_B + |1\rangle_A |0\rangle_B - |0\rangle_A |1\rangle_B) $$ Which means that the density matrix (with order $|00\rangle$, $|01\rangle$, $|10\rangle$, $|11\rangle$) is: $$ \rho_{AB} = \frac{1}{4}\begin{pmatrix}1&-1&1&1\\-1&1&-1&-1\\1&-1&1&1\\1&-1&1&1\end{pmatrix} $$ I have to calculate the Schmidt decomposition WITHOUT USING THE SVD. What I did was calculating the reduced density matrix which at the end are: $$ \rho_A = \text{Tr}_B(\rho_{AB}) = \frac{1}{2}I \qquad\qquad \rho_B = \text{Tr}_A(\rho_{AB}) = \frac{1}{2}I $$ With this I get the Schmidt coefficients which are $\left\{\frac{1}{2}, \frac{1}{2}\right\}$ with eigenstates $\left\{|0\rangle_A, |1\rangle_A\right\}$ for $\rho_A$ and $\left\{|0\rangle_B, |1\rangle_B\right\}$ for $\rho_B$.

So my decomposition is (in the computational basis) $$ |\psi\rangle_{AB} = \frac{1}{2} (|0\rangle_A |0\rangle_B + |1\rangle_A |1\rangle_B) $$ But this is different than the first state! I would expect the correct decomposition to be: $$ |\psi\rangle_{AB} = \frac{1}{2} (|+\rangle_A |0\rangle_B - |-\rangle_A |1\rangle_B) $$ Is it wrong to assume that the eigenstates of the reduced density matrix are part of the Schmidt basis? How can I calculate the Schmidt basis in a correct way?

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    $\begingroup$ Since the Schmidt decomposition is just the SVD, it will be impossible to obtain it without at the same time doing (or obtaining) an SVD. $\endgroup$ Commented May 23 at 17:31

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Why do you need to calculate the Schmidt decomposition without SVD? Doing it in other ways will just lead to the same result.

That aside, the problem with the calculation here is that knowing the reduced states doesn't uniquely identify the full state. The full state must be some purification of the reduced ones, but you can't know which one of those. In other words, studying the eigenvalues of the reduced states correctly gives you the Schmidt coefficients, but does not necessarily tell you the associated states in the decomposition. More precisely, knowing that a reduced state is $\rho_A$ only tells you that the original bipartite state has the form $\operatorname{vec}(\sqrt{\rho_A}V^\dagger)$ for some isometry $V$.

The correct approach to get the Schmidt vectors would be to compute the eigendecomposition of $\Psi\Psi^\dagger$ and $\Psi^\dagger\Psi$, with $\Psi$ the original state thought of as a matrix (that is, the linear operator such that $|\Psi\rangle=\operatorname{vec}(\Psi)$). But of course, this is just the same thing as doing the SVD of $|\Psi\rangle$.

It might be interesting to note that the approach of computing the eigendecomposition of reduced states does work, sometimes. More precisely, you get the correct Schmidt vectors iff the Schmidt coefficients are non-degenerate (equivalently, iff the SVD of the original state are non-degenerate). To see it, write the SVD of the original state as $$|\Psi\rangle = \sum_k \sqrt{p_k} |u_k\rangle\otimes|v_k\rangle$$ for some pair of orthonormal vectors $\{|u_k\rangle\}_k$ and $\{|v_k\rangle\}_k$. Then the reduced states are $$\rho_A = \sum_k p_k |u_k\rangle\!\langle u_k|.$$ This relation might lead you to believe that the eigendecomposition of $\rho_A$ gives you back the vectors $|u_k\rangle$. After all, the above equation is an eigendecomposition for $\rho_A$. However, when the singular values $(p_k)$ are not all distinct, it is not the only such decomposition. For exactly the same reason why the eigenvectors of a linear operator are not uniquely defined when there are degenerate eigenvalues. Whatever method you use to compute the eigenvectors will necessarily make some choice, and won't generally give you back $|u_k\rangle$, but rather some (unitary) linear combination of them.

You can also see it directly yourself with your example: try adding an arbitrarily small perturbation and apply the same procedure to the state $$|+,0\rangle+(1+\epsilon)|-,1\rangle$$ and you should get the correct results.

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  • $\begingroup$ Perfect! Thank you a lot! I knew it had to do with the degeneracy of the eigenvalues. So at the end I could just get another eigenbasis of the degenerate eigenvalues for system A to then yield the correct result! I will go ahead and try with the SVD now, thank you again $\endgroup$ Commented May 21 at 22:37
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    $\begingroup$ But you can always construct a Schmidt decomposition (SD) by computing the reduced density matrices (RDMs), no? Fix any orthonormal (eigen) basis (ONB) of $\rho_A$, and put it in the generic SD form for some ONB of $\rho_B$. This ONB is then determined by computing suitable inner products with the given bipartite pure state $\psi$. That this always is possible should be a consequence of the SD itself; i.e. start from its existence for some bases, then perform unitary transformations in the (common) degenerate subspaces of $\rho_A$ and $\rho_B$. This also shows that the SD is not unique. $\endgroup$
    – Jakob
    Commented May 22 at 13:26
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    $\begingroup$ @Jakob I'm not sure I'm following entirely, but it sounds like that procedure is selecting an eigenbasis for the reduced states using knowledge of the full state. In which case, yea, I'm sure it should work, but then you're just doing the SVD in a convoluted way. And in particular, you're not computing the Schmidt decomposition from the local states only. The point is that local states alone don't give you enough information to characterise the full state, in general. $\endgroup$
    – glS
    Commented May 22 at 14:17
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    $\begingroup$ @glS Thanks for the response. I fully agree with you: The reduced states determine the bipartite state if and only if one of these is pure; then the bipartite state is just a (possibly mixed) product state. So yes, one needs the full bipartite state, and I guess you are right that this procedure is just the SVD in disguise (although one does not need to understand this concept in order to perform this "procedure"). $\endgroup$
    – Jakob
    Commented May 22 at 16:45
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When a question says "without using the SVD" it may be implying that there are easier ways, at least in this specific case. It's always worth just having a stare at the state for a minute (as I suspect you've done given that you have spotted the answer). I generally start just grouping things based on the state of the first qubit: $$ \frac12(|0\rangle(|0\rangle+|1\rangle)+|1\rangle(|0\rangle-|1\rangle)) $$ which is the same as $$ \frac{1}{\sqrt{2}}(|0+\rangle+|1-\rangle) $$ (make sure you're careful with normalisation). Emphasise that this is the Schmidt decomposition because of the orthonormal basis on each subsystem, and so you can just read off what the Schmidt coefficients are.

(I realise this doesn't answer your question about why your method was failing, but @glS did that already)

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  • $\begingroup$ I think this is a general thing, you can always get the Schmidt coefficients and one of the basis by calculating the reduced state of one of the subsystems and then you "fill out" the rest by rewriting the original state... $\endgroup$ Commented May 23 at 12:40
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    $\begingroup$ @AlessandroRomancino Sure you can. I was just point out that in this case, you don't even have to do that, and was suggesting that the "without using the SVD" might have been a hint that it was that simple. $\endgroup$
    – DaftWullie
    Commented May 24 at 7:17

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