# Time-varying expectation values

1. Mar 26, 2009

### Niles

Hi all.

I have a question which arose from the answer of a homework problem. A particle is in the state given by

$$\left| \psi \right\rangle = \frac{1}{{\sqrt 3 }}\left[ {\left| \psi \right\rangle _1 + \left| \psi \right\rangle _2 + \left| \psi \right\rangle _3 } \right],$$

where ${\left| \psi \right\rangle }_i$ is a stationary state of the Hamiltonian. Finding the expectation value of an operator Q in this state, it turns out that <Q> is time-dependent (in fact it oscillates). I was wondering how it is possible to have a time-dependent expectation value? If there was a time-dependent perturbation, then I would be convinced, since the state ${\left| \psi \right\rangle }$ would change over time, but in this case ${\left| \psi \right\rangle }$ doesn't change.

2. Mar 26, 2009

### naturale

the three states has different energies (eigenvalue of the hamiltonian) and thus they evolve in a different way.

3. Mar 26, 2009

### Nick89

When you have just a single state, calculating the expectation value of any operator requires you to multiply the wavefunction by it's complex conjugate. For that reason, the time dependency (the complex exponent) cancels out.
$$e^{ix} (e^{ix})^* = e^{ix}e^{-ix} = 1$$

If however you have multiple states like now, the complex exponents do not cancel out. Actually, a few terms cancel out like usual, but you are left with cross terms that do not, since the states have different energies.
$$e^{ix_1} (e^{ix_2})^* = e^{ix_1}e^{-ix_2} \neq 1$$ (in general)

4. Mar 26, 2009

### Niles

But what does this mean physically? I'm having a hard time interpreting how this can physically be possible without any external (time-varying) perturbation.

5. Mar 26, 2009

### naturale

think of every eigenstate as a wave with different frequency, in this way the modulo has different value at different times.