On time evolution of density operator (matrix) in quantum mechanics

Suppose I have at time $$t=0$$ a statistical ensemble of quantum states $$\{|\psi_n\rangle\langle\psi_n|\}_{n=1}^N$$ The probability of finding the result $$a$$ for an observable $$A$$ in this ensemble is given by the Born rule (assuming discrete spectrum, without degeneracy):

$$\sum_{n=1}^N p_n \langle \psi_n | \Pi_a|\psi_n \rangle = \operatorname{Tr}\left[\rho\,\Pi_a\right]$$

where $$\Pi_a$$ is the projector in the $$a$$-eigenspace of $$A$$, and

$$\rho:= \sum_{n=1}^N p_n |\psi_n\rangle\langle\psi_n|$$ is the density operator or density matrix.

At time $$t>0$$, this density $$\rho$$ takes the form

$$\rho(t):= \sum_{n=1}^N p_n |\psi_n(t)\rangle\langle\psi_n(t)|$$

what puzzles me the most is that those $$p_n$$ stay the same. Why should I expect that? Why couldn't the probabilities change as each state $$|\psi_n(t)\rangle\langle\psi_n(t)|$$ evolves with time?

• I don't understand. If your ensemble is "created" (e.g. you have some probabilistic preparation scheme) in the sense that at $t_0$ your system is associated a wave function $|\psi_n\rangle$ with probability $p_n$, then the probability that your system at time $t$ is in $|\psi_n(t)\rangle$ is still $p_n$ - simply because the time evolution is unitary/deterministic! Nov 25, 2022 at 20:28
• From what I've understood, $p_n$ is the probability of finding the eigenvalue $a$ in the $n$-th state $|\psi_n\rangle\langle\psi_n|$ at time $t=0$. But... if the $n$-th state evolves, together with all the other $N-1$, who tells me that $p_n$ couldn't also change? Nov 25, 2022 at 20:33

The idea is really just that you start out with a statistical ensemble, i.e. the quantum system is in the state $$\lvert \psi_n\rangle$$ with probability $$p_n$$ at the start. The density matrix $$\sum_n p_n\lvert \psi_n\rangle\langle \psi_n\rvert$$ is just a way to summarize this information into a single object. The notion of density matrices does not add any new "mechanics" to quantum mechanics.
Each of these states evolves in time like quantum states normally do, i.e. $$\lvert \psi_n(t)\rangle = U(t)\lvert \psi_n\rangle$$ with $$U(t)$$ the ordinary time evolution operator. So if the system started in the state $$\lvert \psi_n(0)\rangle$$, of course it is in the state $$\lvert \psi_n(t)\rangle$$ at time $$t$$. Since it had probability $$p_n$$ to start in the state $$\lvert \psi_n(0)\rangle$$, it still has probability $$p_n$$ to be in $$\lvert \psi_n(t)\rangle$$ at time $$t$$, and the density matrix at time $$t$$ is $$\sum_n p_n\lvert \psi_n(t)\rangle\langle \psi_n(t)\rvert$$. There isn't anything more to this, the probabilities here just act like classical probabilities.
• I might explain my doubt with an example I made up, hoping not to make things worse. Suppose at time $t=0$ I have a coin. Its density matrix is $\rho= 1/2 |H \rangle\langle H| + 1/2|T \rangle\langle T|$. But suppose that this coin has a hidden engine at time $t>0$ that makes heads come out more frequently than tails. Evidently $p_H(t) > p_H(0)=1/2$ Nov 25, 2022 at 20:43
• The classical analogy was just a pretext to say I can't see a good reason on assuming $p_n(t)=p_n$. But it's clearly my fault, since this expression for $\rho(t)$ leads to the quantum Liouville equation, so it must be right of course Nov 25, 2022 at 20:54
• @ric.san We're not "assuming $p_n(t) = p_n$". We're just saying "we start with a system which can be in a bunch of different states with probability $p_n$ at $t=0$" and see what we get when we apply ordinary QM time evolution to that", that's the point of my answer. Nov 25, 2022 at 21:04