Quantum Mechanics – Are Two Quantum States with the Same Schmidt Bases at All Times Equal?

linear algebraquantum mechanicsquantum-entanglementquantum-information

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.

Best Answer

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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