Why do purifications only differ by a local unitary?

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)

$$\newcommand{\ket}[1]{\vert#1\rangle}$$ 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.)

• 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 $|\tilde b_i\rangle$. – Danylo Y Jan 3 at 17:46
• @DanyloY Fair point! – Norbert Schuch Jan 3 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.

• 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. – StarBucK Jan 2 at 13:19
• Like if one gives you two purification of A how to prove that necesserally they are related via a unitary transformation on B – StarBucK Jan 2 at 13:20
• 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/… – Purva Thakre Jan 2 at 23:32

Another way to see this is remembering that the density matrix of a bipartite pure state is essentially its covariance matrix, in the sense that if a state $$|\psi\rangle$$ has components $$\psi_{ij}$$ and $$\rho=|\psi\rangle\!\langle\psi|$$, then $$\rho=\psi\psi^\dagger$$ (where we are thinking of $$\psi$$ as a matrix, so that $$(\psi\psi^\dagger)_{ij}=\sum_\ell \psi_{i\ell}\bar\psi_{j\ell}$$).

This means that, if $$\psi$$ and $$\phi$$ are pure states corresponding to the same density matrix $$\rho$$, as matrices they satisfy $$\psi\psi^\dagger=\phi\phi^\dagger=\rho$$. 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 can see that the above equation is equivalent to the statement about purifications differing only by local unitaries.