Let's consider $\rho_A$ a density matrix. I introduce a space $B$ and an entangled state $|\Psi\rangle$ (the purification) so that:

$$\newcommand{\tr}{\operatorname{Tr}}\rho_A = \tr_B(|\Psi\rangle \langle \Psi |_{AB}) $$

Let's consider another purification of $\rho_A$: a vector $|\widetilde{\Psi}\rangle$. By definition we have:

$$ \rho_A = \tr_{B}(|\widetilde{\Psi}\rangle \langle \widetilde{\Psi} |_{AB})$$

I have read that those two states are indeed purification of $\rho_A$ if and only if:

$$ |\widetilde{\Psi}\rangle = I_A \otimes U_B |\Psi\rangle_{AB} $$

I understand the sufficient part, but I don't see why it would be necessary for $ |\widetilde{\Psi}\rangle$ to have this form. Could someone explain me how to get this result ? I found the statement but not a proof for this and I don't manage to find it (I'm probably missing something very obvious)


3 Answers 3


$\newcommand{\ket}[1]{\vert#1\rangle}$ First, write $\ket\psi$ and $\ket{\tilde\psi}$ in their Schmidt decomposition: $$ \begin{aligned} \ket\psi &= \sum \lambda_i \ket{a_i}\ket{b_i}\ , \\ \ket{\tilde\psi} & = \sum \tilde\lambda_i \ket{\tilde a_i}\ket{\tilde b_i}\ . \end{aligned} $$ Let us assume for simplicty that the $\lambda_i$ are non-degenerate. Since they both have the same reduced density operator $\rho_A$, we have that $\lambda_i=\tilde \lambda_i$ and $\ket{a_i}=\ket{\tilde a_i}$. Now construct $U$ such that $\ket{\tilde b_i}=U\ket{b_i}$ -- this is possible since both are orthogonal bases. Then, we see that $$ \ket{\tilde\psi} = (I\otimes U)\ket{\psi} $$ -- that is, given two arbitrary purifications of the same $\rho_A$, we have shown they are related by a unitary on the purifying system.

(In case some $\lambda_i$ are degenerate, the Schmidt decomposition is not unique - you want to choose it such that $\ket{a_i}=\ket{\tilde a_i}$ also for the degenerate $\lambda_i$, which is always possible, since the degenerate eigenvectors must span the same subspace.)

(Another note: In principle, the purifying system can be larger than the Schmidt rank; in that case, the action of $U$ is only fixed on the subspace spanned by the Schmidt vectors on B.)

  • $\begingroup$ Even if $\lambda_i$ are all different then $|\tilde a_i\rangle$ are not necessary equal to $|a_i\rangle$ $-$ they can differ by a phase. But we can "move" those phases to $|\tilde b_i\rangle$. $\endgroup$
    – Danylo Y
    Jan 3, 2020 at 17:46
  • $\begingroup$ @DanyloY Fair point! $\endgroup$ Jan 3, 2020 at 18:36

It is due to Schmidt decomposition. For some $|\psi \rangle_{AB} \in H_A \otimes H_B$, there exists a decomposition in terms of the orthonormal basis (Schmidt bases) of system A and B. $\lambda_i$ are the Schmidt coefficients calculated from $Tr_B(|\psi\rangle \langle\psi|_{AB})$ whose eigenvalues are $\lambda^2_i$. Given below is the Schmidt decomposition, {$|a_i\rangle$} are the eigenvectors of system A and {$|b_i\rangle$} are the eigenvectors of system B.

$|\psi \rangle_{AB} = \sum_i \lambda_i |a_i\rangle |b_i\rangle $ where $\lambda_i \geq 0$ and $\sum_i \lambda^2_i =1$

So, based on the equation you have in the question, there is a purification being done on system B while keeping system A unchanged. Thus, the eigenvectors of A and {$\lambda_i$} should remain fixed in both decompositions. Because we want to preserve {$\lambda_i$}, eigenvalues of system B should be preserved i.e a unitary applied only to system B while identity is applied to A i.e. $I_A\otimes U_B$.

In conclusion, when the partial trace on system B is done, then this change does not show up in the final purification.

  • $\begingroup$ Maybe I didnt get your point but I know that if I have a purification, only acting with a local unitary on B wont change the purification. My question is the other way around: are all purification of A all related through unitary acting locally on B. $\endgroup$ Jan 2, 2020 at 13:19
  • $\begingroup$ Like if one gives you two purification of A how to prove that necesserally they are related via a unitary transformation on B $\endgroup$ Jan 2, 2020 at 13:20
  • 1
    $\begingroup$ That is what my last statement is discussing. Any unitary acting on B as long as A is left unchanged will make two decomposition equivalent. The proofs have been solved at these links : marozols.wordpress.com/2012/05/09/…, physics.stackexchange.com/questions/156777/… $\endgroup$ Jan 2, 2020 at 23:32

If the density matrix $\rho\in\mathrm{Lin}(\mathcal H_A)$ has purification $|\psi\rangle\in\mathcal H_A\otimes\mathcal H_B$, meaning $\rho=\operatorname{Tr}_B(|\psi\rangle\!\langle\psi|)$, then we can write $\rho=\psi\psi^\dagger$ where $\psi$ denotes the "unvectorisation" of $|\psi\rangle$, that is $$\psi\in\mathrm{Lin}(\mathcal H_B,\mathcal H_A), \qquad |\psi\rangle=\operatorname{vec}(\psi), \qquad \psi_{i,j} = \langle i,j|\psi\rangle.$$

This means that, if $\psi,\phi\in\mathrm{Lin}(\mathcal H_B,\mathcal H_A)$ represent pure states purifying the same $\rho$, then $$\psi\psi^\dagger=\phi\phi^\dagger=\rho.$$ This implies that the singular value decomposition of $\psi$ and $\phi$ must have both singular values and left principal components in common, that is, we can write $$ \psi = \sum_k s_k \mathbf u_k \mathbf v_k^*, \\ \phi = \sum_k s_k \mathbf u_k \mathbf w_k^*, $$ for some $s_k\ge0$ (whose square equal the eigenvalues of $\rho$) and orthonormal sets of vectors $\{\mathbf u_k\}_k,\{\mathbf v_k\}_k,\{\mathbf w_k\}_k$, and with $\mathbf v^*$ denoting the dual of $\mathbf v$.

Going back to the notation with ket vectors (that is, considering again the vectorifications of $\psi,\phi$), we see that $$|\psi\rangle = \sum_k s_k (\mathbf u_k\otimes\bar{\mathbf v}_k), \qquad |\phi\rangle = \sum_k s_k (\mathbf u_k\otimes\bar{\mathbf w}_k),$$ meaning the purifications only differ by local unitaries.


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