# Particle in a box momentum and position, expectation and uncertainty in stationary states

For stationary states solution for a particle in a box in infinite wall. i.e. $\psi=Ax(a-x)$ for $x\in[0,a]$.

$<x>=L/2$ would be a definite solution. Similarly, $<p>=0$, where $<p^2>\neq 0$.

1. Question 1, was the expectation value of stationary states solution $<x>$ and $<p>$ always independent of time?(I thought so as if write $\psi=\sum \psi_n(x)\phi_n(t)$ and then the time dependent part$\phi_n(t)^*\phi_n(t)=1$)

If so, then I understood that the uncertainty principle was still followed by each observation. However $<p>=0$ implied that the expectation of speed $<v>=0$ as through $p=mv$. The name stationary states was in deed stationary states.

1. Question 2, but what's the point of the uncertainty principle here then? As the wave function seemed to be "set". In a sense that since the system was independent of time, we could well measure the system couple of times. For instance, one time for momentum, and the other time for position. Or simply measure over 30 times and then use normal distribution.
• Have you checked that your wavefunction $\psi$ is actually stationary? Because having a constant expectation value is not sufficient to show a state is stationary. You must show that the wavefunction $\psi$ does not evolve by more than an overall phase under the Hamiltonian of the system; the eigenstates of this particular Hamiltonian are sinusoids with nodes at $0$ and $a$. Think of it this way: a pendulum bob has a constant average position and zero average velocity, but that doesn't mean it's stationary. – probably_someone Mar 20 '18 at 8:50
• @probably_someone for this specific $\psi$ as in the example, it was. I got confused by the implication of the time separation part $\phi(t)$ . – J C Mar 20 '18 at 19:57