$\newcommand{\ket}[1]{|#1\rangle}$ Suppose a pure state $\ket{\psi}$ has a Schmidt decomposition given by $\ket{\psi^{SD}}$, which can be obtained via the diagonalization of the reduced density matrix of $\ket{\psi}$. Is there a unitary transform one can construct from the diagonalization process that will directly convert $\ket{\psi}$ to $\ket{\psi^{SD}}$? That is, $\ket{\psi^{SD}} = U\ket{\psi}$ for some unitary transform $U$. If so, how would one construct such a unitary transform?

  • $\begingroup$ could you clarify the relation between $|\psi\rangle$ and $|\psi^{\rm SD}\rangle$ here? A Schmidt decomposition is just a way to write the state, so what do you mean that a state "has the Schmidt decomposition of another state"? $\endgroup$
    – glS
    Commented Jul 19, 2021 at 9:39
  • $\begingroup$ $\newcommand{\ket}[1]{|#1\rangle}$ @glS Suppose the reduced density matrix can be diagonalized as $\rho_A = Tr_B (\rho) = PDP^{-1}$ for some diagonal matrix $D$ and invertible matrix $P$. Then, the diagonal entries of $D$ are the Schmidt coefficients in the decomposition, i.e. $\sqrt{D_{ii}} = \sqrt{\lambda_i}$. The column vectors of $P$ ($\ket{i_A}$) and row vectors for $P^{-1}$ ($\ket{i_B}$) form orthogonal bases for the two subsystems. Then, we can easily write the state in the Schmidt decomposed form $\ket{\psi}^{SD} = \sum_i \sqrt{\lambda_i} \ket{i_A}\ket{i_B}$. $\endgroup$ Commented Jul 19, 2021 at 19:11
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    $\begingroup$ that all sounds right, except for the last sentence. The Schmidt decomposition is not another state. The Schmidt decomposition of $|\psi\rangle$ is another way to write $|\psi\rangle$ itself. It sounds to me like what you are doing is considering another state which has the same Schmidt decomposition of the original one. Of course, you can do it, and the two would be related by a local unitary transformation (and vice versa any local unitary transformation preserves the Schmidt coefficients). But it is important to note that we are talking about different states here $\endgroup$
    – glS
    Commented Jul 19, 2021 at 19:37

1 Answer 1


The Schmidt decomposition is a singular value decomposition. It decomposes the state (re-arranged into a matrix) into a product $U \cdot S \cdot V$ where $S$ is diagonal. The unitaries you want are $U$ and $V$. (Well, one of them is transposed depending on how you ordered things.)


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