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I read the Schmidt basis is unique "up to a phase" or as stated here in answer given by Norbert Schuch "modulo degeneracies".

If I choose a Bell state

$|\psi\rangle = \frac{1}{\sqrt{2}}\left(|0\rangle\otimes|0\rangle + |1\rangle\otimes|1\rangle \right)$

and then define a second basis on each of the two subspaces by

$ |0\rangle \equiv \frac{1}{\sqrt{2}}(|\tilde{0}\rangle +|\tilde{1}\rangle) \quad \quad |1\rangle \equiv \frac{1}{\sqrt{2}}(|\tilde{0}\rangle -|\tilde{1}\rangle) $

then $|\psi\rangle$ becomes

$|\psi\rangle = \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}(|\tilde{0}\rangle +|\tilde{1}\rangle) \otimes\frac{1}{\sqrt{2}}(|\tilde{0}\rangle +|\tilde{1}\rangle) + \frac{1}{\sqrt{2}}(|\tilde{0}\rangle -|\tilde{1}\rangle) \otimes\frac{1}{\sqrt{2}}(|\tilde{0}\rangle -|\tilde{1}\rangle) \right) =$

$\frac{1}{2 \sqrt{2}}( |\tilde{0}\rangle\otimes|\tilde{0}\rangle + |\tilde{0}\rangle\otimes|\tilde{1}\rangle + |\tilde{1}\rangle\otimes|\tilde{0}\rangle + |\tilde{1}\rangle\otimes|\tilde{1}\rangle + |\tilde{0}\rangle\otimes|\tilde{0}\rangle - |\tilde{0}\rangle\otimes|\tilde{1}\rangle - |\tilde{1}\rangle\otimes|\tilde{0}\rangle + |\tilde{1}\rangle\otimes|\tilde{1}\rangle )$

finally

$|\psi\rangle = \frac{1}{\sqrt{2}}\left(|\tilde{0}\rangle\otimes|\tilde{0}\rangle + |\tilde{1}\rangle\otimes|\tilde{1}\rangle \right)$

which again is a Schmidt decomposition with the same coefficients as above but with different bases. What is my mistake?

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  • $\begingroup$ Why do you think you made a mistake? Also, "the first answer" might change. It is better to link to the answer. $\endgroup$ – Norbert Schuch Feb 5 at 2:19
  • $\begingroup$ It was already linked. I edited the link title and hope it is more unique now, not so degenerate ;-) $\endgroup$ – Harald Rieder Feb 5 at 7:31
  • $\begingroup$ I doubt it is linked. If I go to the "share" button on my answer, I get a different link (with a different number identifying the post). $\endgroup$ – Norbert Schuch Feb 5 at 12:28
  • $\begingroup$ Than you for the tip. I updated the link. $\endgroup$ – Harald Rieder Feb 6 at 6:30
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You're not doing anything wrong. The decomposition essentially comes from the singular value decomposition of your tensor product element when written as a matrix.

In this case a scalar matrix (multiple of the identity), so it has every vector as an eigenvector, i.e. we are in a degenerate case and every unitary change of basis will give a valid decomposition. The coefficients will still be uniquely determined.

EDIT: Remarks on unicity

In general the Schmidt decomposition is formulated for the tensor product of any two (finite dimensional) state spaces, not necessarily even of the same dimension, and even if they are equal, the (orthonormal) basis elements of both factors can generally not be chosen equal. We have unicity of the components if we additionally require them to be real and nonnegative.

When the matrix is Hermitian (or more generally, normal), it can be diagonalized by a single unitary transformation, meaning that we can write the general tensor element as

$$\sum a_i|\psi_i\rangle\otimes |\psi_i\rangle$$

where the $a_i$ are general complex numbers. To convert this into a combination of the form

$$\sum A_i|\phi_i\rangle\otimes |\psi_i\rangle$$

where the $A_i$ are nonnegative real numbers, we can simply move the phase into one of the bases or the other without changing unitarity (so now we have bases $|\phi_i\rangle$ and $|\psi_i\rangle$). These $A_i$ are unique, but the bases are only unique up to a phase (if all eigenvalues are different, i.e. no degeneracy). In the first representation the $a_i$ themselves are only unique up to a phase.

In the general case (still equal dimensions of the factors) we cannot unitarily diagonalize and we must use different bases. Here again the $a_i$ are unique when restricting to non-negative real numbers, while we can move phase factors from left to right to change the bases.

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  • $\begingroup$ Can we say the Schmidt decomposition is "unique in the non-degenerate case up to a phase"? $\endgroup$ – Harald Rieder Feb 5 at 7:27
  • $\begingroup$ @HaraldRieder I edited to say something about unicity. I wrote it a bit fast, but I think it is correct $\endgroup$ – doetoe Feb 5 at 8:22

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