$\newcommand{\ket}[1]{|#1\rangle}\newcommand{\bra}[1]{\langle#1|}$There is a known expression for evolution of density operator in time: $$\rho(t) = U(t,t_0)\rho(t_0)U^{\dagger}(t,t_0)$$ Let's denote $\rho(t)$ as $\ket{f} \bra{f}$ and $\rho(t_0)$ as $\ket{i} \bra{i}$, thus, we simply describe the evolution of bra and ket by our expression. $\ket{i}$ is some state in the basis $\{\ket{n}\}$ But what meaning does it have if we apply this train of operators consequently to some $\ket{\psi}$? As $U^{\dagger}(t,t_0)$ is a time reversed operator we should take $\ket{\psi(t)}$ at $t$. It looks like a projector operator with $\ket{\psi(t_0)}$ and final state $\ket{f}$. It can be written as $$\rho(t)\ket{\psi(t)} = \ket{f} \bra{\psi(t_0)}$$ where $\ket{f}$ is a state at the moment $t$. Does it have sense for functions in different time moments?

  • 3
    $\begingroup$ The equation $\rho(t)|\psi(t)\rangle = |f\rangle \langle\psi(t_0)|$ is impossible -- it has a state on the left-hand side and a projector on the right. As currently written the question reads mostly like symbol salad and it's not really possible to parse it. You should explain in more detail what you mean. $\endgroup$ Commented May 14, 2021 at 15:56
  • $\begingroup$ I want to understand the interpretation of the sequence of transformations while applying the operators from right to left to some ket state $\endgroup$ Commented May 14, 2021 at 16:00
  • 1
    $\begingroup$ But as @Emilio Pisanty has pointed out, your last question does not make sense at all. In your notation, there should be something like $\rho |\psi\rangle = |f\rangle \langle i|U^\dagger |\psi\rangle$ if you defined $\rho \equiv |f\rangle \langle f|$ with $|f\rangle \equiv U |i\rangle$, no? $\endgroup$ Commented May 14, 2021 at 16:03
  • $\begingroup$ Yes, I define $|f \rangle$ like this. Your expression for $\rho | \psi \rangle $ is what I mean. But it is still not clear for me why $|f \rangle$ is defined at time $t$ and $\langle i | U^{\dagger} | \psi \rangle$ at $t_0$. How can it be consistent? $\endgroup$ Commented May 14, 2021 at 16:14
  • $\begingroup$ I wanted to change perspective on the states evolution, i.e if we consider $U^{\dagger} | \psi \rangle$ state. Is it defined at $t_0$? If it is and $\langle i |$ corresponds to $t_0$ as well. Then we have a value defined at $t_0$, which we multiply by $| f \rangle$ at $t$ $\endgroup$ Commented May 14, 2021 at 16:26

1 Answer 1


Did you confuse yourself trying the following ? $$\begin{aligned} \hat \rho(t)|\psi_t\rangle &= \hat U(t,t_0)|\psi_{t_0}\rangle \langle \psi_{t_0}|\hat U^\dagger (t,t_0)|\psi_t\rangle \\ \hat \rho(t)|\psi_t\rangle &= \hat U(t,t_0)|\psi_{t_0}\rangle \langle \psi_{t_0}|\hat U^\dagger (t,t_0)\hat U(t,t_0)|\psi_{t_0}\rangle \\ \hat \rho(t)|\psi_t\rangle &= \hat U(t,t_0)|\psi_{t_0}\rangle \langle \psi_{t_0}|\hat 1|\psi_{t_0}\rangle \\ \hat \rho(t)|\psi_t\rangle &= \hat U(t,t_0)|\psi_{t_0}\rangle \langle \psi_{t_0}|\psi_{t_0}\rangle \\ \hat \rho(t)|\psi_t\rangle &= \hat U(t,t_0)|\psi_{t_0}\rangle 1 \\ \hat \rho(t)|\psi_t\rangle &= |\psi_{t}\rangle \\ \end{aligned}$$ with $|\psi_t\rangle =\hat U(t,t_0)|\psi_{t_0}\rangle $ and $\hat U^\dagger(t,t_0)\hat U(t,t_0)= \hat 1 $ and $\langle \psi_t|\psi_t\rangle =1 =\langle \psi_{t_0}|\psi_{t_0}\rangle$

The same is achived by simply doing, $$\begin{aligned} \hat \rho(t)|\psi_t\rangle &= \hat U(t,t_0)|\psi_{t_0}\rangle \langle \psi_{t_0}|U^\dagger (t,t_0)\ |\psi_t\rangle \\ &= |\psi_{t}\rangle \langle \psi_{t}| \ \psi_t\rangle \\ &= |\psi_{t}\rangle \\ \end{aligned}$$


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.