2
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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! $\endgroup$ Nov 25, 2022 at 20:28
  • $\begingroup$ 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? $\endgroup$
    – ric.san
    Nov 25, 2022 at 20:33

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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 $p_H(t) > p_H(0)=1/2$ $\endgroup$
    – ric.san
    Nov 25, 2022 at 20:43
  • 2
    $\begingroup$ @ric.san Quantum mechanics does not model the flip of a classical coin. I do not understand your comment. $\endgroup$
    – ACuriousMind
    Nov 25, 2022 at 20:45
  • $\begingroup$ 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 $\endgroup$
    – ric.san
    Nov 25, 2022 at 20:54
  • 1
    $\begingroup$ @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. $\endgroup$
    – ACuriousMind
    Nov 25, 2022 at 21:04

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.