Inspired by this question: Are these two quantum systems distinguishable? and discussion therein.

Given an ensemble of states, the randomness of a measurement outcome can be due to classical reasons (classical probability distribution of states in ensemble) and quantum reasons (an individual state can have a superposition of states). Because a classical system cannot be in a superposition of states, and in principle the state can be directly measured, the probability distribution is directly measurable. So any differing probability distributions are distinguishable. However in quantum mechanics, an infinite number of different ensembles can have the same density matrix.

What assumptions are necessary to show that if two ensembles initially have the same density matrix, that there is no way to apply the same procedure to both ensembles and achieve different density matrices? (ie. that the 'redundant' information regarding what part of Hilbert space is represented in the ensemble is never retrievable even in principle)

To relate to the referenced question, for example if we could generate an interaction that evolved:

1) an ensemble of states $|0\rangle + e^{i\theta}|1\rangle$ with a uniform distribution in $\theta$


2) an ensemble of states $|0\rangle + e^{i\phi}|1\rangle$ with a non-uniform distribution in $\phi$

such an mapping of vectors in Hilbert space can be 1-to-1. But it doesn't appear it can be done with a linear operator.

So it hints that we can probably prove an answer to the question using only the assumption that states are vectors in a Hilbert space, and the evolution is a linear operator.

Can someone list a simple proof showing that two ensembles with initially the same density matrix, can never evolve to two different density matrices? Please be explicit with what assumptions you make.

Update: I guess to prove they are indistinguishable, we'd also need to show that non-unitary evolution like the projection from a measurement, can't eventually allow one to distinguish the underlying ensemble either. Such as perhaps using correlation between multiple measurements or possibly instead of asking something with only two answers, asking something with more that two so that finally the distribution of answers needs more than just the expectation value to characterize the results.

  • $\begingroup$ Hah! I addressed your update in my answer before I even saw it. $\endgroup$ Commented Apr 6, 2011 at 1:05

2 Answers 2


You only need to assume

  1. the Schrödinger equation (yes, the same old linear Schrödinger equation, so the proof doesn't work for weird nonlinear quantum-mechanics-like theories)
  2. the standard assumptions about projective measurements (i.e. the Born rule and the assumption that after you measure a system it gets projected into the eigenspace corresponding to the eigenvalue you measured)

Then it's easy to show that the evolution of a quantum system depends only on its density matrix, so "different" ensembles with the same density matrix are not actually distinguishable.

First, you can derive from the Schrödinger equation a time evolution equation for the density matrix. This shows that if two ensembles have the same density matrix and they're just evolving unitarily, not being measured, then they will continue to have the same density matrix at all future times. The equation is $$\frac{d\rho}{dt} = \frac{1}{i\hbar} \left[ H, \rho \right]$$

Second, when you perform a measurement on an ensemble, the probability distribution of the measurment results depends only on the density matrix, and the density matrix after the measurement (of the whole ensemble, or of any sub-ensemble for which the measurement result was some specific value) only depends on the density matrix before the measurement.

Specifically, consider a general observable (assumed to have discrete spectrum for simplicity) represented by a hermitian operator $A$. Let the diagonalization of $A$ be $$A = \sum_i a_i P_i$$ where $P_i$ is the projection operator in to the eigenspace corresponding to eigenvalue (measurement outcome) $a_i$. Then the probability that the measurement outcome is $a_i$ is $$p(a_i) = \operatorname{Tr}(\rho P_i)$$ This gives the complete probability distribution of $A$.

The density matrix of the full ensemble after the measurment is $$\rho' = \sum_i P_i \rho P_i$$ and the density matrix of the sub-ensemble for which the measurment value turned out to be $a_i$ is $$\rho'_i = \frac{P_i \rho P_i}{\operatorname{Tr}(\rho P_i)}$$

Since none of these equations depend on any property of the ensemble other than its density matrix (e.g. the pure states and probabilities of which the mixed state is "composed"), the density matrix is a full and complete description of the quantum state of the ensemble.

  • $\begingroup$ Oh, and for the case of an observable $A$ with a continuous spectrum, it works basically the same way. For mathematicians it might get more hairy, but as a physicist I have no problem just saying "replace all the summation signs with integrals". $\endgroup$ Commented Apr 6, 2011 at 0:59
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    $\begingroup$ You don't even need to assume Schrödinger equation, but only the fact that the evolution of a quantum state is unitary. $\endgroup$ Commented Apr 10, 2011 at 19:14
  • $\begingroup$ Can you please justify why just by weighted sum of density matrices for sub ensembles gives the thing for whole ensemble? In other words, what's difference between density matrix for a single system, as opposed to an ensemble? I understand former quite well as the "state" of system. $\endgroup$ Commented May 11, 2018 at 11:49

Density matrices are an alternative description of quantum mechanics. Consequently, if two ensembles have the same density matrix, they are not distinguishable.

Example, consider the unpolarized spin-1/2 density matrix which can be modeled as a system that is half pure states in the +x direction and half in the -x direction, or alternatively, as half pure states in the +z direction (i.e. spin up) and half in the -z direction (i.e. spin down):
$$\begin{pmatrix}0.5&0\\0&0.5\end{pmatrix} = 0.5\rho_{+x}+0.5\rho_{-x} = 0.5\rho_{+z}+0.5\rho_{-z}$$ Now compute the average value of an operator $H$ with respect to these ensembles. Let $$H = \begin{pmatrix}h_{11}&h_{12}\\h_{21}&h_{22}\end{pmatrix}$$ then the averages for the four states involved are: $$\begin{array}{rcl} \langle H\rangle_{+x} &=& 0.5(h_{11}+h_{12}+h_{21}+h_{22})\\ \langle H\rangle_{-x} &=& 0.5(h_{11}-h_{12}-h_{21}+h_{22})\\ \langle H\rangle_{+z} &=& h_{11}\\ \langle H\rangle_{-z} &=& h_{22} \end{array}$$ From the above, it's clear that taking the average over $\pm x$ will give the same result as taking the average over $\pm z$, that is, in both cases the ensemble will give an average of $$\langle H\rangle = 0.5(h_{11}+h_{22})$$

Any preparation of the system amounts to an operator acting on the states and so $H$ can stand for a general operation. Therefore there is no way of distinguishing an unpolarized mixture of +- x from an unpolarized mixture of +-z.

The argument for general density matrices is similar, but I think this gets the point across.

  • $\begingroup$ Are you saying instead of representing a state as a vector in Hilbert space, it is sufficient to represent a state as a density matrix? It seems like this view would change the counting of physical states and would have an effect in statistical mechanics or thermodynamics of a system. It almost seems like you would be reducing the entropy by mixing two ensembles. $\endgroup$
    – Ginsberg
    Commented Apr 6, 2011 at 0:32
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    $\begingroup$ Either way, the whole point of the question was to see a concrete mathematical proof. Instead of just saying it is so, can you please show how it is so, such that I can learn more? $\endgroup$
    – Ginsberg
    Commented Apr 6, 2011 at 0:34
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    $\begingroup$ @Ginsberg; Yes, a density matrix is equivalent to a collection of pure states (presumably represented by state vectors) along with a probability density for the pure states. I've not found the reference I was looking for so I'll type up an outline of a proof and edit it in. $\endgroup$ Commented Apr 6, 2011 at 0:45

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