Consider a two particle system subject to a one-dimensional harmonic oscillator potential. So, the Hamiltonian is $H=H_1+H_2$, where $H_1$ acts only on the state space of particle $1$ and $H_2$ on the state space of particle $2$.

At $t=0$, the system is in the energy eigenstate:


Then, a measurement of the total energy $H$ is performed and the result found is $2\hbar\omega$ (at $t=0$).

I am trying to calculate the mean values of position and momentum of particle $1$ at $t>0$.

My attemp is the following: Because the measurement of $H$ is made, I start with the collapsed eigenstate:


Now, the time evolution of the system for particle $1$ is:

$$|\psi(t)\rangle=\sum_{n_1}e^{-iE_{n}t/\hbar}|\psi_{n_1,n_2}(0)\rangle=\frac{1}{\sqrt{2}}(e^{-i3\omega t/2}|\psi_{1,0}\rangle+e^{-i\omega t/2}|\psi_{0,1}\rangle)$$

where I have made use of the energy eigenvalues $E_{n_1,n_2}=(n_1+n_2+1)\hbar\omega$.

In order to calculate the mean value of the position, I use the position operator in terms of creation and annihilation operators acting on the fist particle state space:

$$X|\psi(t)\rangle=\sqrt{\frac{\hbar}{2m\omega}}(a_{1}+a^{\dagger}_{1})|\psi(t)\rangle=\frac{1}{2}\sqrt{\frac{\hbar}{m\omega}}(e^{-i3\omega t/2}|\psi_{0,0}\rangle+e^{-i\omega t/2}|\psi_{1,1}\rangle).$$

It is obvious from this last equation that $\langle X\rangle(t)=0$. Similarly I obtain that $\langle P\rangle(t)=0$.

This result doesn't make sense to me because it contradicts the Ehrenfest theorem given that $X_1$ and $P_1$ do not commute with the Hamiltonian, so the expectation values can't be zero.

Also, in my calculation I made steps that I am not sure at all. For example, I made the time evolution of the state for the particle 1 only. Although, even if I do it for the complete system, I still get the same result of null mean values.

Is it possible that this results are due to the measurement done just before the time evolution?

How the time evolution of a two-particle system state should be calculated? Is it right to calculate it just for one particle, or it must always be done to the whole system?


1 Answer 1


The energies $E_n$ in the exponential must be the total energy, as you say, not just the energy of particle 1. You don't need to do any time evolution, though, because the system is left in an energy eigenstate after the measurement, and energy eigenstates are stationary states: nothing will depend on time.

And you're misreading the Ehrenfest theorem; there are mean values around the commutator too:

$$i \hbar \frac{d\langle A \rangle}{dt} = \langle [A, H] \rangle$$

Here $[X,H]\propto P$ and $[P,H]\propto X$, so you can see that $\langle X \rangle = 0$ and $\langle P \rangle = 0$ is a solution to the above equation. Even if the commutator is nonzero, its mean value can be zero, as it is here.

  • $\begingroup$ That is true, time evolution is not needed! I thought immediatly in using the evolution operator because the problem asks for the mean values at $t>0$, but you are right, it is easier than that. I Still don't understand if there is any meaning in doing the time evolution for a single particle. $\endgroup$
    – rsaavedra
    Commented Nov 8, 2016 at 21:39
  • $\begingroup$ @Saavestro: No, as far as I know it doesn't make sense to do time evolution for a single particle, except if the particles are independent and the state is a product state. You would just be looking at particle 1 and forgetting about particle 2. In presence of interaction and/or entanglement (as in your question), however, time evolution uses the full Hamiltonian. $\endgroup$
    – Javier
    Commented Nov 9, 2016 at 1:24

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