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In brief: if two quantum states can be Schmidt decomposed using the same sets of joint basis at all times, no matter the evolution they go through, are the quantum states equal?

In detail:

Consider two quantum systems A and B with a joint Hilbert space $H= H_A \otimes H_B$. Consider two quantum states $|\psi\rangle$ and $|\phi\rangle$ of the joint system such that:

  1. They may be Schmidt decomposed using the same basis elements: \begin{align} |\psi\rangle&= \Sigma_i \alpha_i |A_i\rangle |B_i\rangle \\ |\phi\rangle&= \Sigma_i \gamma_i |A_i\rangle |B_i\rangle \end{align} where $\{|A_i\rangle\}$ and $\{|B_i\rangle\}$ are bases of, respectively, $H_A$ and $H_B$.

  2. For all unitary operators $U$, the evolved quantum states may be Schmidt decomposed using the same basis elements (though, in general, it may be different from before): \begin{align} U|\psi\rangle&= |\psi^\prime\rangle = \Sigma_i \beta_i |A_i^\prime\rangle |B_i ^\prime\rangle \\ U|\phi\rangle&= |\phi^\prime\rangle = \Sigma_i \omega_i |A_i^\prime\rangle |B_i ^\prime\rangle \end{align} This second condition may be re-expressed as follows: \begin{align} \forall U, \exists & \{|A_i^\prime\rangle\}, \{|B_i^\prime \rangle\}: \\ U|\psi\rangle&= |\psi^\prime\rangle = \Sigma_i \beta_i |A_i^\prime\rangle |B_i ^\prime\rangle \\ U|\phi\rangle&= |\phi^\prime\rangle = \Sigma_i \omega_i |A_i^\prime\rangle |B_i ^\prime\rangle \end{align} where $\{|A_i^\prime\rangle\}$ and $\{|B_i ^\prime\rangle\}$ are bases of, respectively, $H_A$ and $H_B$

Do these two conditions together imply $|\psi\rangle=|\phi\rangle$?

Note: strictly speaking, the first condition is redundant, since, if the quantum states are expressible using the same Schmidt decomposition basis elements at all times, then they will be at one time. As shown in the asnwer, only the second condition is necessary to prove the equality of the quantum states.

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  • $\begingroup$ Note that condition 1 is actually unnecessary: You can always find one basis where phi and psi have this property. (e.g. where one state is |00> and the other a*|00>+b*|11>, with the correct scalar product). $\endgroup$ Oct 8, 2023 at 10:45
  • $\begingroup$ @Norbert That is surprising. Aren't the bases, generically, fully fixed by the states? $\endgroup$ Oct 8, 2023 at 11:06
  • $\begingroup$ @EmilioPisanty True. I guess I wasn't precise enough. What I meant to say is that the OP considers a situation where they look at the full unitary orbit of two states phi and psi. Then, they consider the situation where "same Schmidt basis in one point of the orbit => same Schmidt basis in all points of the orbit". My point is that there is always some point in the orbit where the two states have the same Schmidt basis and thus, this is no extra condition. It might indeed not be a specific basis where the OP defines phi and psi, but I can define them in any basis I want (as one considers ... $\endgroup$ Oct 8, 2023 at 11:26
  • $\begingroup$ ... the orbit) without changing the problem, and thus, condition 1 adds no extra constraint. (You can also see this by looking at my 2nd answer: It is immediate to come up with a basis where the two states have the same Schmidt basis.) $\endgroup$ Oct 8, 2023 at 11:27
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    $\begingroup$ @Pol I agree that it makes it easier to parse, I would leave it like that (though one could add a note that condition (1) is, in fact, not required). $\endgroup$ Oct 9, 2023 at 19:02

2 Answers 2

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First, note that all what is fixed about $|\psi'\rangle=U|\psi\rangle$ and $|\phi'\rangle=U|\phi\rangle$ is their scalar product -- any other property can be changed by choosing a suitable $U$.

We are thus considering arbitrary $|\psi'\rangle$ and $|\phi'\rangle$ with a fixed scalar product $\langle \psi'|\phi'\rangle=\alpha$, which satisfy property 2., and we want to prove they must be equal.

To this end, choose $|\psi'\rangle=|0\rangle|0\rangle$, and $|\phi'\rangle=\alpha|0\rangle|0\rangle+\sqrt{1-|\alpha|^2}|0\rangle|1\rangle$. Then, $\langle \psi'|\phi'\rangle=\alpha$, yet, $|\phi'\rangle$ does not have the same Schmidt basis, in contradiction with property 2 -- unless $|\alpha|=1$, which implies that $|\psi'\rangle\propto|\phi'\rangle$.

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  • $\begingroup$ Thank you very much for your excellent reply. I have just a couple of questions to ensure I have understood. (1) Am I correct to understand that your answer relies on the following claim: For any two pairs of states $(|\psi\rangle, |\phi\rangle)$ and $(|\psi^\prime\rangle, |\phi^\prime\rangle)$ such that $\langle \psi | \phi \rangle = \langle \psi^\prime | \phi^\prime \rangle$ there is a unitary such that $U|\psi\rangle= |\psi^\prime\rangle$ and $U|\phi\rangle= |\phi^\prime\rangle$ $\endgroup$
    – Pol
    Oct 9, 2023 at 10:57
  • $\begingroup$ (2) It seems to me that your answer is formulated in terms of two qubits but can be easily generalised. I was wondering if you agree. Consider the bases $\{ |A_i\rangle\}$, $\{ |B_i\rangle \}$ of, respectively $H_A$ and $H_B$. Then choose a unitary such that $|\psi^\prime\rangle = |A_0 \rangle |B_0\rangle$ and $|\phi^\prime\rangle = \alpha |A_0 \rangle |B_0\rangle\ + \sqrt{1-|\alpha|^2} |A_0 \rangle |B_1\rangle$. The reasoning is then the same. Am I correct? $\endgroup$
    – Pol
    Oct 9, 2023 at 11:03
  • $\begingroup$ I think both (1) and (2) are correct, so I have accepted the answer. But please do let me know if I am wrong. $\endgroup$
    – Pol
    Oct 9, 2023 at 11:18
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    $\begingroup$ @Pol Regarding (1), yes, this is a more-or-less trivial (and certainly standard) linear algebra statement. If you want, expand the system in a basis spanned by |phi> and |psi> -- then you just have to choose the first two columns of U accordingly. Regarding (2), I disagree this is about qubits -- my point is that I can rotate this to any vectors I want, which I happen to call |0>|0> and |0>|1> (and assume them to be of this form). You are free to call them |A0>|B0> and |A0>|B1>, but this is really just a name, not a different statement. $\endgroup$ Oct 9, 2023 at 19:01
  • $\begingroup$ True, I was being too literal. Thanks! $\endgroup$
    – Pol
    Oct 10, 2023 at 8:25
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Let us first prove it for qubits. W.l.o.g., we can call the Schmidt basis $|0\rangle|0\rangle$ and $|1\rangle|1\rangle$. Then, \begin{align} |\phi\rangle &= a|0\rangle|0\rangle+b|1\rangle|1\rangle\ ,\\ |\psi\rangle &= a'|0\rangle|0\rangle+b'|1\rangle|1\rangle\ . \end{align}

Let us assume that $|\phi\rangle\ne|\psi\rangle$. Then, we can make linear comibinations \begin{align} |\alpha\rangle &:= x|\phi\rangle+y|\psi\rangle = |0\rangle|0\rangle\ ,\\ |\beta\rangle &:= v|\phi\rangle+w|\psi\rangle = |1\rangle|1\rangle\ . \end{align} By linearity, your condition 2 now implies it must thus hold that for all $U$, $U|\alpha\rangle$ and $|\beta\rangle$ have the same Schmidt basis. But this is clearly not true, because we just have to choose a $U$ whos first and last column (i.e., $U|00\rangle$ and $U|11\rangle$) have different Schmidt decompositions (this is easy to see -- the only condition the columns have to satisfy is that they are orthogonal, so take e.g. $|00\rangle+|11\rangle$ and $|01\rangle+|10\rangle$).

By contradiction, this implies that $|\phi\rangle=|\psi\rangle$.


The general case is left for later (or as a proof to the reader) -- I was hoping that one could restrict to a 2D subspace where the Schmidt coefficients differ, but it is not entirely obvious that/how this works.

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  • $\begingroup$ Thank you very much for your reply. Your proof in the case of qubits is very clear and very helpful. I haven’t managed to generalise it to more dimensions either, but hopefully someone will post a general answer. $\endgroup$
    – Pol
    Oct 7, 2023 at 12:29

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