If two quantum states have the same Schmidt bases at all times, are they equal?

Question

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.

Answer 1

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 $|\psi'\rangle\propto|\phi'\rangle$.

Answer 2

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.