# Clarification about stationary quantum states of a system

1. Mar 15, 2015

### deep838

Okay, here goes... Our teacher set a question in the last test which asked us to show that if a system initially be in a stationary state, it will remain in a stationary state even if the system evolves according to the time dependent Schrodinger equation. What I did was show that the expectation value of the operator will not change using
∂<O>/∂t = 0
But now that I think about it, I find it really stupid!! Why shouldn't the expectation value change with time? It's a quantum system after all... it's supposed to be unpredictable every instant! If I know what it is now, I shouldn't know what the system will become 2 mins later,am I right?
Anyway, I tacitly assumed that ∂ψ/∂t = 0 and ended up with that result...
What the teacher wanted was <O(t)> = <O(t0)>
Please help me get out of my own mess!! Let me know if I need to clarify anything.

2. Mar 15, 2015

### ShayanJ

The fact that the expectation value of observables in stationary states is constant, doesn't mean the actual value of that observable is constant and you can predict it. It just means the probability distribution for that observable isn't changing. But measurements separated by finite amounts of time, still give different values for the same observable.
Also, $\frac{\partial \psi}{\partial t}=0$ is in general not correct for a stationary state. An stationary state is of the form $\Psi(\vec r,t)=\psi(\vec r) e^{-i \omega t}$. So we have:
$\langle O \rangle_{\Psi}=\langle \Psi |O|\Psi \rangle= \langle \psi |e^{i \omega t}Oe^{-i \omega t}|\psi \rangle=\langle \psi |e^{i \omega t}e^{-i \omega t}O|\psi \rangle=\langle \psi |O|\psi \rangle= \langle O \rangle_{\psi}$ which means $\langle O \rangle_{\Psi}$ is independent of time.

3. Mar 15, 2015

### deep838

I see... thank you for replying so early... Yes I kind of knew that $\frac{\partial \psi}{\partial t}=0$ is wrong... but in that short period of time I didn't even try to think... please don't start criticizing me for that...