Given a quantum system, a mixed state can be represented as "density matrix operator" in the form:

$$\hat{\rho} = \sum_{i} p_i |\psi_i\rangle \langle\psi_i|$$

where $|\psi_i\rangle$ is a Hilbert basis of pure states. My first question is this:

  1. Is it possible to select the states $|\psi_i\rangle$, as stationary states (i.e. eigenvectores of the Hamiltonanian operator, $\hat{H}|\psi_i\rangle = E_i|\psi_i\rangle$

if the answer is yes, then the density matrix operator is constant, because: $$i \hbar \frac{\partial \hat{\rho}}{\partial t} = [\hat{H}, \hat{\rho}] = \sum_i p_i(E_i |\psi_i\rangle \langle\psi_i|-p_i |\psi_i\rangle \langle\psi_i|E_i) = 0$$

then the Von Neumann entropy remains constant for a such a mixed state:

$$S = - \operatorname{tr}(\hat{\rho} \ln \hat{\rho})$$

If the answer to my first question is YES, then it seems that in the quantum context entropy really is constant, and this seems wrong to me. Then the second question comes to mind:

  1. Can quantum entropy be constant? Does the conservation of quantum information mean that there is never really a loss of information? Does this imply that there is also no increase in entropy?

If the answer to my first question is NO, then:

  1. Under what circumstances is it not possible to express a matrix density in terms of steady states?
  • 1
    $\begingroup$ I think you're confusing kinematic and dynamic aspects here. How a density matrix can be decomposed is completely independent from the underlying dynamics, possibly specified by a Hamiltonian. A density matrix can always be decomposed as a convex combination of pure states. Generally in infinitely many ways, see eg quantumcomputing.stackexchange.com/q/5259/55. The Hamiltonian bears no relevance with this. $\endgroup$
    – glS
    Commented Jun 20, 2022 at 8:39

1 Answer 1


You appear to be asking if all state matrices are diagonal in the eigenbasis of an arbitrary Hamiltonian, and the answer is trivially no, that is a subset.

For instance, a qubit under the Hamiltonian $$H=\hbar\omega|1\rangle\langle 1|$$ has an energy eigenbasis of $$|0\rangle=\begin{bmatrix}1\\0\end{bmatrix},|1\rangle=\begin{bmatrix}0\\1\end{bmatrix}$$.

But under this Hamiltonian, most of the mixed states formed by the Hilbert basis $$|{+}\rangle=\sqrt{1\over2}\begin{bmatrix}1\\1\end{bmatrix}, |{-}\rangle=\sqrt{1\over2}\begin{bmatrix}1\\-1\end{bmatrix}$$(sometimes called the Hadamard basis) aren't going to be diagonal. Indeed let's just write them out, $$ \frac12 p \begin{bmatrix}1&1\\1&1\end{bmatrix} + \frac12 (1-p) \begin{bmatrix}1&{-1}\\{-1}&1\end{bmatrix}\\ =\frac12 \begin{bmatrix}1&2p-1\\2p-1&1\end{bmatrix}, $$ and we see that this is only a diagonal matrix if the mixture is exactly half and half. It turns out that in fact $$\frac12 |{+}\rangle\langle{+}|+\frac12 |{-}\rangle\langle{-}|=\frac12 |{0}\rangle\langle{0}|+\frac12 |{1}\rangle\langle{1}| $$in this special case, and in fact any 50/50 mixture is going to be the same state for a qubit...

In terms of the Bloch sphere, the pure states are precisely the points on the sphere, the mixed states are precisely the points inside the sphere, and you get them by identifying two opposite points on the sphere, which are Hilbert-orthogonal, and then the mixed states are the line connecting those two opposite points. So you really need the freedom to choose a basis for your mixed states that is not your eigenbasis prescribed by your Hamiltonian! Otherwise you're only going to get this one line through the Bloch sphere but not all of the points inside of it. And that's pretty typical, not a quirk of qubits. So you are right that there is some sort of flow map and the mixed states for the Hamiltonian eigenbasis are all fixed points of that flow map, which means that those states do not have varying entropy (or anything else)... but, that is always a subspace of the fuller space of mixed states.


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