Apparently, the decomposition of a state into an ensemble of pure states is not unique. I can't understand why, as if I understood correctly a "pure state ensemble decomposition" is just the diagonalization of the density operator

$$\rho=\sum_{k=0}^rp_k|\psi_k\rangle\langle\psi_k| $$

where $r$ is the rank of $\rho$ and $|\psi_k\rangle$ are its eigenvectors with associated eigenvalues $p_k$.

Such a diagonalization is unique up to permutations of the $|\psi_k\rangle$, i.e. there exists a unique basis where $\rho$ is diagonal. How can the pure ensemble decomposition not be unique? What have I misunderstood?

  • $\begingroup$ Just wanted to point out a recent paper where it was shown that distinguishing two "preparations" for a mixed state is impossible in quantum theory (and leads to signaling): arxiv.org/abs/1904.05722 $\endgroup$ Commented Jul 1, 2021 at 10:42

3 Answers 3


Suppose that we're talking about $n\times n$ density operators, so that the rank will never exceed $n$. Now suppose that you choose $N$ to be much larger than $n$, and then arbitrarily pick a probability vector $(p_1,\ldots,p_N)$ and pure states $|\psi_1\rangle,\ldots,|\psi_N\rangle$. Assuming that more than $n$ of the entries in the probability vector are nonzero, and the vectors $|\psi_1\rangle,\ldots,|\psi_N\rangle$ represent distinct states, the decomposition $$ \rho = \sum_{k=1}^N p_k |\psi_k\rangle \langle \psi_k | $$ for the operator $\rho$ you have selected through this process will surely be different from the spectral decomposition of $\rho$.

Note that for a given $\rho$, it will always be possible to do something along these lines. If the rank of $\rho$ is 1, it will be trivial (meaning that every $|\psi_k\rangle \langle \psi_k |$ will have to be equal to $\rho$, assuming $p_k >0$), but as long as the rank of $\rho$ is at least 2, there will be a continuum of different inequivalent ways to do this.

In particular, for any choice of $\varepsilon > 0$ and $|\psi\rangle$ for which $$ \rho - \varepsilon |\psi\rangle \langle \psi| $$ is positive semidefinite, you could write $$ \rho = \varepsilon |\psi\rangle \langle \psi| + (1-\varepsilon) \sigma $$ for $$ \sigma = \frac{\rho - \varepsilon |\psi\rangle \langle \psi|}{1-\varepsilon} $$ and then recurse on $\sigma$. For any choice of $|\psi\rangle$ in the image of $\rho$, not just eigenvectors of $\rho$, this will be possible so long as $\varepsilon$ is small enough. And, as long you take $\varepsilon$ small enough each time, you could recurse as long as you like, and in general you will get different decompositions of $\rho$.

  • 1
    $\begingroup$ Thanks for the answer, I think I get it! The states $|\psi_1\rangle \cdot |\psi_N\rangle$ are in the same Hilbert space as the eigenvectors of $\rho$? So they need not be linearly independent, correct? For example I can arbitrarily construct $\rho=p_0|0\rangle\langle 0| +p_1|1\rangle\langle 1| +p_2|+\rangle\langle +|$, which is a pure state decomposition but clearly not the eigendecomposition of $\rho$, which I can find by diagonalizing it, do I understand correctly? $\endgroup$ Commented Jan 24, 2019 at 19:10
  • $\begingroup$ Everything in your comment appears to be correct, so I think you have it. $\endgroup$ Commented Jan 24, 2019 at 19:45

For any matrix $\rho$, there are (typically) many different choices of $\{p_i\}$ and $\{|\psi_i\rangle\}$ such that $$ \rho=\sum_ip_i|\psi_i\rangle\langle\psi_i|. $$ The simplest way to do it, with the fewest number of terms, is always the eigenvalue decomposition, but there are other ways as well. You do not have to use the eigenvector decomposition.

We should start by noting that not even the eigenvector decomposition is unique if you have repeated eigenvalues. For example, the maximally mixed state can be written as $$ \rho=\frac{1}{d}\mathbb{I}=\frac{1}{d}\sum_{i=1}^d|u_i\rangle\langle u_i| $$ where $\{|u_i\rangle\}$ is any orthonormal basis of dimension $d$.

Now we can use this to show how there can be many different ways of decomposing something even if there aren't repeated eigenvalues. Take a one-qubit state $$\rho=p_0|v_0\rangle\langle v_0|+(1-p_0)|v_1\rangle\langle v_1|$$ where $p_0>\frac12$ and $\langle v_0|v_1\rangle=0$. We could write this as $$ \left((p_0-q)|v_0\rangle\langle v_0|+(1-p_0-q)|v_1\rangle\langle v_1|\right)+q\mathbb{I} $$ for any $q\leq 1-p_0$, and we can decompose $\mathbb{I}$ in terms of any one (or more) orthonormal bases that we want to. You have a larger ensemble, but it still describes the same $\rho$. As an example, look at $$ \rho=\left(\begin{array}{cc} \frac45 & 0 \\ 0 & \frac15 \end{array}\right) $$ I'm going to rewrite this as $$ \rho=\left(\begin{array}{cc} \frac35 & 0 \\ 0 & 0 \end{array}\right)+\left(\begin{array}{cc} \frac15 & 0 \\ 0 & \frac15 \end{array}\right) $$ You can easily check that this is the same as, for example, $$ \rho=\frac35|0\rangle\langle 0|+\frac{1}{5}(|+\rangle\langle +|+|-\rangle\langle -|). $$ This means that $\rho$, as well as being described by the ensemble $\{(\frac45,|0\rangle),(\frac15,|1\rangle)\}$, we can also use the ensemble $\{(\frac35,|0\rangle),(\frac15,|+\rangle),(\frac15,|-\rangle)\}$.

  • $\begingroup$ Hi, thank you for the answer, I'm not sure I understand your example, should the last line be equal to $\rho$? The indices of the basis $|u_i\rangle$ start from 1 in $\rho$, so what is $|u_0\rangle$? $\endgroup$ Commented Jan 24, 2019 at 12:26
  • $\begingroup$ @user2723984 Sorry, there was a typo $\endgroup$
    – DaftWullie
    Commented Jan 24, 2019 at 12:34
  • $\begingroup$ So now the last line is $$ (p_0-q)|v_0\rangle\langle v_0| - p_0|v_1\rangle\langle v_1| + q\sum_{i=1}^d |u_i\rangle\langle u_i|$$ how is this equal to $\rho$? Is there another typo in $(1-p_0-1)$? $\endgroup$ Commented Jan 24, 2019 at 12:39
  • $\begingroup$ Sorry, I must be missing something obvious, I don't really understand this answer :( I don't see how the operator you exhibited at the end represents the same $\rho$, or even if you meant it to represent $\rho$ or $p_0|v_0\rangle\langle v_0| + (1-p_0)|v_1\rangle\langle v_1|$ $\endgroup$ Commented Jan 24, 2019 at 15:23
  • $\begingroup$ @user2723984 you only need to remember that $\mathbb I =\lvert v_0\rangle\!\langle v_0\rvert + \lvert v_1\rangle\!\langle v_1\rvert$ $\endgroup$
    – glS
    Commented Jan 25, 2019 at 2:50

Just to include the corresponding general statement in the discussion:

Given a density matrix $\rho$ in an $n$-dimensional Hilbert space, and a probability vector $\mathbf p\equiv(p_1,...,p_n)$, you can find a collection of pure states $\{|\psi_k\rangle\}_{k=1}^n$ such that $$\rho = \sum_{k=1}^n p_k \, |\psi_k\rangle\!\langle\psi_k|\tag1$$ iff $\mathbf p\preceq \lambda(\rho)$, that is, iff the probability vector is majorised by the vector of eigenvalues of $\rho$.

So for example, if $\rho$ is pure, then $\lambda(\rho)=(1,0,...,0)$, which majorises any probability vector $\mathbf p$, and thus $\rho$ can be decomposed as in (1) for any choice of probability vector (and suitable choices of states $|\psi_k\rangle$). On the opposite side of the spectrum, if $\rho$ is maximally mixed, then $\lambda(\rho)=(\frac1{n},...,\frac1{n})$, which is a probability vector majorised by any other probability vector, and thus there's no way to decompose $\rho$ as in (1) with a difference choice of coefficients.

See chapter 4, and in particular Theorem 4.34, of Watrous' book for more details (Link to pdf).

A related question is Why can we write $\rho=\sum_\mu q_\mu|\varphi_\mu\rangle\!\langle\varphi_\mu|$ iff $q\preceq \mathrm{spec}(\rho)$?.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.