Can anything be said about the inner product of a bipartite entangled state with itself but with the Schmidt vectors swapped? That is, if the Schmidt decomposition of a state is given by $$\vert \psi \rangle = \sum_{i=1}^d\sqrt{\lambda_i} \vert i_A \rangle \vert i_B \rangle,$$ then can anything be said about the inner product $$\langle \psi \vert \psi' \rangle = \sum_{i,j=1}^d \sqrt{\lambda_i\lambda_j}\langle i_A \vert \langle i_B \vert \vert j_B \rangle \vert j_A \rangle = \sum_{i,j=1}^d \sqrt{\lambda_i\lambda_j} \langle i_A \vert j_B \rangle \cdot \langle i_B \vert j_A \rangle?$$ Note that the inner products are well-defined because in the definition of the Schmidt decomposition, the underlying Hilbert space of systems $A$ and $B$ is $\mathbb{C}^d$. Because of this paper, I guess the answer is (Lemma 20) $$\frac{1}{2}\left(1 + \sum_{i=1}^d \lambda_i^2 \right),$$ but they get this result as a corollary of a more difficult theorem. I would be interested in seeing a direct calculation, since it seems like a very natural result that I can't seem to prove.

  • $\begingroup$ I feel like if anything can be said, then it might only be in the non-degenerate case because of the non-uniqueness of the Schmidt decomposition. But I would love to be proven wrong $\endgroup$
    – SescoMath
    Nov 25, 2022 at 3:19
  • $\begingroup$ If the Schmidt bases are the same on the two systems, doesn't that imply that the inner product is 1? That would seem to be incompatible with your guess. $\endgroup$
    – DaftWullie
    Nov 25, 2022 at 7:42
  • 2
    $\begingroup$ I don't think this is always well defined. It is when you assume that they are bipartite spaces in $\mathbb{C}^d\otimes\mathbb{C}^d$ for some $d$, which needs not be the case. That aside, assuming equal dimensions, I'd guess the result will change greatly with the states. You can also get $2\lambda_1\lambda_2$ for states such as $\lambda_1|01\rangle+\lambda_2|10\rangle$. $\endgroup$
    – glS
    Nov 25, 2022 at 9:16

1 Answer 1


TL;DR: In general, such "inner product" is not well-defined. This can be remedied by appropriate choice of $A$ and $B$. Still, the inner product depends on the global phase and thus cannot be observed. Its absolute value - which can be observed in principle - can take any value in $[0,1]$, even for product states (in contradiction to the proposed formula).


Suppose that $A$ is a spin-$\frac12$ particle and $B$ a particle on a line and consider the state $$ |\psi\rangle=\frac{1}{\sqrt2}\left(|{\uparrow}\rangle|p_+\rangle+|{\downarrow}\rangle|p_-\rangle\right)\tag1 $$ where $|{\uparrow}\rangle$ and $|{\downarrow}\rangle$ represent the spin-up and spin-down states of $A$ and $|p_+\rangle$ and $|p_-\rangle$ represent the states of $B$ with absolute linear momentum $p$ propagating to the right and to the left, respectively. Clearly, $|{\uparrow}\rangle$ and $|p_+\rangle$ live in different Hilbert spaces, so the "inner product" $\langle{\uparrow}|p_+\rangle$ is not defined.

Sensitivity to global phase

That said, we might choose $A$ and $B$ so that their Hilbert spaces coincide. Note that this is an additional assumption which is not needed for or implied by Schmidt decomposition$^1$. With the assumption, the inner product is defined, but only up to the unobservable global phase. Thus, we loose no generality by limiting our consideration to its absolute value.

Any value in $[0,1]$ is possible

By Cauchy-Schwarz inequality the absolute value of the inner product lies in $[0, 1]$. It is sufficient to consider product states to see that no further constraints on the inner product can be obtained. Indeed, fix $a\in[0,1]$ and let $|\phi\rangle=\sqrt{a}|0\rangle+\sqrt{1-a}|1\rangle$ and $|\psi\rangle=|0\rangle|\phi\rangle$. Then $$ \langle\psi|\psi'\rangle=\langle 0|\phi\rangle\langle\phi|0\rangle=a\tag2. $$ Thus, we can make the inner product equal to any desired number $a\in[0,1]$. In particular, the proposed formula predicts $\langle\psi|\psi'\rangle=1$ for all product states so it can't be correct.

$^1$ In particular, Schmidt decomposition is well-defined when $A$ and $B$ have Hilbert spaces of different dimension. In this case, $d$ is the smaller one.


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