- #1
AStaunton
- 105
- 1
[problem is:
the wavefunction for a particle in a stationary state of a one dimensional infinite potential well is given by:
[tex]\Psi_{1}(x,t)=A\cos(\frac{\pi x}{2a})e^{-i\frac{Et}{\hbar}}[/tex] for [tex]-a\leq x\leq a[/tex]
= 0 otherwise.
using the normalised wavefunction calculate the expectation value of momentum squared:
[tex]\langle p^{2}\rangle[/tex]
My attemted solution:
i know that to find momentum p, the operator is:
[tex]\bar{p}=-\hbar\bar{\nabla}^{2}[/tex]
so we say:
[tex]\langle p\rangle=\int\Psi^{\star}\bar{p}\Psi dx[/tex]
is it simply a matter of squaring p at this stage to get p^2?
Also, it says to use the normalised wave function, have I done this already in my integration step or is there something else needs be done?
Andrew
the wavefunction for a particle in a stationary state of a one dimensional infinite potential well is given by:
[tex]\Psi_{1}(x,t)=A\cos(\frac{\pi x}{2a})e^{-i\frac{Et}{\hbar}}[/tex] for [tex]-a\leq x\leq a[/tex]
= 0 otherwise.
using the normalised wavefunction calculate the expectation value of momentum squared:
[tex]\langle p^{2}\rangle[/tex]
My attemted solution:
i know that to find momentum p, the operator is:
[tex]\bar{p}=-\hbar\bar{\nabla}^{2}[/tex]
so we say:
[tex]\langle p\rangle=\int\Psi^{\star}\bar{p}\Psi dx[/tex]
is it simply a matter of squaring p at this stage to get p^2?
Also, it says to use the normalised wave function, have I done this already in my integration step or is there something else needs be done?
Andrew