In my class it was told that ensemble decompositions of a density operator $\rho$ are not unique, but that the ones that exist are related by a unitary operator. I'm trying to prove this, but I get stuck somewhere along the way.

Lets begin by assuming two different decompositions of density operator $\rho$: $\rho = \sum_{j=1}^n{p_j|\psi_j\rangle\langle\psi_j|} = \sum_{k=1}^m{q_k|\phi_k\rangle\langle\phi_k|}$

Now, these two decompositions live in a Hilbert space $\mathcal A$. We can then define a purification of both, using a system described by a Hilbert space $\mathcal B$ of dimension $k=\max (n,m)$, so that we get $|\Psi_1\rangle_{\mathcal A\mathcal B} = \sum_{j=1}^n{\sqrt{p_j}|\psi_j\rangle \otimes |b_j\rangle}$ and $|\Psi_2\rangle_{\mathcal A\mathcal B} = \sum_{k=1}^m{\sqrt{q_k}|\phi_k\rangle \otimes |b_k\rangle}$.

Now, here we can use that as these pure states are purifications of the same density operator, there must be a unitary $U$ connecting them: $(1_A \otimes U_B)|\Psi_1\rangle_{\mathcal A\mathcal B} = |\Psi_2\rangle_{\mathcal A\mathcal B}$.

Here is where I get stuck. I should be able to use this to prove the unitary relation between the $\psi$ and the $\phi$, but it is not obvious to me how I should do this.

Update: after reviewing the comments to the first question, I should have written that the $\psi$ and $\phi$ states do NOT have to be orthonormal, per se.

  • $\begingroup$ I once wrote a blog post about this: marozols.wordpress.com/2012/05/09/… $\endgroup$ Jan 5, 2015 at 22:11
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    $\begingroup$ I might be mistaken, but your blog post is about the unitary equivalence of the purification, not the ensamble decomposition, right? $\endgroup$
    – user129412
    Jan 6, 2015 at 11:46
  • $\begingroup$ Your are right, I overlooked that $\psi$s and $\phi$s need not be orthonormal in your question. Norbert's answer then is what you are looking for! $\endgroup$ Jan 7, 2015 at 0:12

2 Answers 2


Result for one orthogonal and one non-orthogonal decomposition

We will prove the following Theorem:

Let $\rho=\sum_{i=1}^Np_i\lvert\phi_i\rangle\langle\phi_i\rvert$ be a eigenvalue decomposition and $\sigma=\sum_{i=1}^Mq_i\lvert\psi_i\rangle\langle\psi_i\rvert$ ($M\ge N$). Then, $\rho=\sigma$ if and only if \begin{equation} \sqrt{q_j}\lvert\psi_j\rangle=\sum_i v_{ij}\sqrt{p_i}\lvert\phi_i\rangle\ , \end{equation} with $\sum_j v_{ij}v_{i'j}^*=\sum_k \delta_{ii'}$, i.e., $V\equiv(v_{ij})$ is an isometry.


The "if" direction is straightforward: \begin{align} \rho&=\sum_{j}q_j\lvert\psi_j\rangle\langle\psi_j\rvert\\ &=\sum_{i,i',j} v_{ij} v_{i'j}^* \sqrt{p_i}\sqrt{p_{i'}} \lvert\phi_i\rangle \langle\phi_{i'}\rvert \\ &=\sum_i p_i\lvert\phi_i\rangle \langle\phi_{i}\rvert = \sigma\ , \end{align} where in the last step we have used that $\sum_j v_{ij}v_{i'j}^*=\delta_{ii'}$.

To prove the converse, let $$ v_{ij} := \langle\phi_i\lvert\psi_j\rangle\,\sqrt{q_j/p_i}\ .$$ Then, $$ \sum_i v_{ij} \sqrt{p_i}\lvert\phi_i\rangle = \sum_i \lvert\phi_i\rangle\langle \phi_i\lvert\psi_j\rangle\sqrt{q_j} = \sqrt{q_j}\lvert\psi_j\rangle\ ,$$ i.e., $v_{ij}$ is the desired basis transformation. Further, $$ \sum_{ii'}\underbrace{\delta_{ii'}\sqrt{p_ip_{i'}}}_{=:a_{ii'}}\lvert\phi_i\rangle\langle\phi_i\rvert = \sum_ip_i\lvert\phi_i\rangle\langle\phi_i\rvert = \sum_jq_j\lvert\psi_j\rangle\langle\psi_j\rvert = \sum_{ii'}\underbrace{\sum_jv_{ij}v^*_{i'j}\sqrt{p_ip_i'}}_{=:b_{ii'}}\lvert\phi_i\rangle\langle\phi_{i'}\rvert\ . $$ Now, since the $\lvert\phi_i\rangle$ are orthogonal (as they form an eigenbasis), the $\lvert\phi_i\rangle\langle\phi_{i'}\rvert$ are linearly independent, and thus, $a_{ii'}=b_{ii'}$, which implies $\sum_jv_{ij}v^*_{i'j}=\delta_{ii'}$.

Extension to two non-orthogonal decompositions

If the $\lvert\phi_i\rangle$ do not form an orthonormal basis, we can generalize the theorem by going through an orthonormal basis $\rho=\sum r_k\lvert\chi_k\rangle\langle\chi_k\rvert$: Then, \begin{align} \sqrt{p_i}\lvert\phi_i\rangle&=\sum_k u_{ki}\sqrt{r_k}\lvert\chi_k\rangle\\ \sqrt{q_j}\lvert\psi_j\rangle&=\sum_k w_{kj}\sqrt{r_k}\lvert\chi_k\rangle \end{align} with $\sum_i u_{ki}u_{k'i}^*=\delta_{kk'}$ and $\sum_j w_{kj}w_{k'j}=\delta_{kk'}$. The second equation then yields $\sqrt{r_k}\lvert\chi_k\rangle= \sum_j w_{kj}^*\sqrt{q_j}\lvert\psi_j\rangle$. After inserting this in the first equation, we obtain $$ \sqrt{q_j}\lvert\psi_j\rangle=\sum_i v_{ij}\sqrt{p_i}\lvert\phi_i\rangle $$ with $v_{ij} = \sum_k u_{ki}w_{kj}^*$, i.e., $V=UW^\dagger$ is a partial isometry.


You certainly want to have $\langle\psi_i|\psi_j\rangle=\delta_{ij}$ and $\langle\phi_k|\phi_l\rangle=\delta_{kl}$, or some equivalent condition ensuring that both ensemble decompositions are optimal. Otherwise, it is easy to come up with many different totally unrelated decompositions.

Because of the optimality condition, both decompositions are eigendecompositions of $\rho$. Hence $n=m$ and $p_i=q_i$, if we order $p_j$ and $q_k$ according to decreasing magnitude. The $|\psi_j\rangle$ and $|\phi_k\rangle$ for a specific eigenvalue are an orthonormal basis basis for the corresponding eigenspace, hence they are related by an unitary operator. For the global unitary operator, extend $|\psi_j\rangle$ and $|\phi_k\rangle$ to an orthonormal basis, and then just map $|\psi_i\rangle$ to $|\phi_i\rangle$. This works even if the $p_i$ and $q_k$ are not ordered, just because of $\langle\psi_i|\psi_j\rangle=\delta_{ij}$ and $\langle\phi_k|\phi_l\rangle=\delta_{kl}$.

So maybe the real task here is to show that being an optimal decomposition implies $\langle\psi_i|\psi_j\rangle=\delta_{ij}$. This is related to properties of the singular value decomposition, which gives a succinct description of the optimal approximations with respect to the Frobenius norm and the spectral norm. Note: The condition $\langle\psi_j|\psi_j\rangle=1$ and $\sum_{j=1}^n{p_j}=1$ is not sufficient to describe this sort of optimality, even so I initially claimed this. So this question turn out to be a bit advanced, because describing the sense in which the decomposition is optimal is non-trivial, if you don't just reference the singular value decomposition for that part.

Here is a simple counterexample to the comment by Norbert Schuch that "two decompositions are related by isometries is true regardless of whether the vectors are orthonormal": $\begin{bmatrix}1 & 0 \\ 0 & \epsilon^2\end{bmatrix} = \frac{1}{2}\begin{bmatrix}1 \\ \epsilon\end{bmatrix}\begin{bmatrix}1 & \epsilon\end{bmatrix} + \frac{1}{2}\begin{bmatrix}1 \\ -\epsilon\end{bmatrix}\begin{bmatrix}1 & -\epsilon\end{bmatrix}=\begin{bmatrix}1 \\ 0\end{bmatrix}\begin{bmatrix}1 & 0\end{bmatrix}+\epsilon^2\begin{bmatrix}0 \\ 1\end{bmatrix}\begin{bmatrix}0 & 1\end{bmatrix}$

No isometry can map $(\begin{bmatrix}1 \\ \epsilon\end{bmatrix},\begin{bmatrix}1 \\ -\epsilon\end{bmatrix})$ to $(\begin{bmatrix}1 \\ 0\end{bmatrix},\begin{bmatrix}0 \\ 1\end{bmatrix})$, not even with rescaling, because isometries preserve angles. The first pair of vectors is nearly parallel, while the second pair of vectors is orthogonal.

  • $\begingroup$ You are right, I forgot to include the fact that both the $\psi$ and $\phi$ form an orthonormal basis and that the $p$ and $q$ sum to one. Do you perhaps have a link or something of the sorts, for the fact that the orthonormal basis are related by a unitary operator? $\endgroup$
    – user129412
    Jan 5, 2015 at 20:09
  • $\begingroup$ The unitary you are looking for is constructed simply by mapping $\psi_j$ to $\phi_j$ for any $j$ (under optimality you have the same number of orthonormal vectors, say $n$, appearing in the decomposition). At this stage you only have a partial isometry, which then you extend to a unitary on the whole Hilbert space (using, e.g., Graham-Schmidt) $\endgroup$
    – Phoenix87
    Jan 5, 2015 at 20:22
  • $\begingroup$ The statement that any two decompositions are related by isometries is true regardless of whether the vectors are orthonormal: The isometry is chosen such that $\sqrt{q_j}\lvert\phi_j\rangle = \sum V_{ij} \sqrt{p_i}\lvert\phi_i\rangle$. (The case where both decompositions are eigenvalue decompositions is a very special case.) $\endgroup$ Jan 5, 2015 at 22:46
  • $\begingroup$ @NorbertSchuch The problem with your proposal is that $V_{ij}$ is in general not an isometry. Remember that an isometry is not just a linear map, but also preserves distances, and hence angles. I added a counterexample to my answer, because my answer indeed makes the claim that orthogonality is important, without explicitly proving it. $\endgroup$ Jan 5, 2015 at 23:39
  • $\begingroup$ @ThomasKlimpel : You are thinking of an isometry which acts on the Hilbert space itself. This is not what I said: The isometries act on the $\lvert\phi_i\rangle$, i.e., they express the $\lvert\psi_j\rangle$ as linear combinations of the $\lvert\phi_i\rangle$. In your case, with the (unnormalized) $\lvert\phi_{0,1}\rangle=(1,\pm\epsilon)$, $\lvert\psi_0\rangle=(1,0)$, $\lvert\psi_1\rangle=(0,\epsilon)$, the isometry is given by $\left(\begin{smallmatrix}1&1\\1&-1\end{smallmatrix}\right)/2$. That's certainly the way I know the statement about the relation of different ensemble decompositions. $\endgroup$ Jan 6, 2015 at 0:19

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