Using the normalise wavefunction, calculate momentum squared

[AStaunton]

[problem is:

the wavefunction for a particle in a stationary state of a one dimensional infinite potential well is given by:

\Psi_{1}(x,t)=A\cos(\frac{\pi x}{2a})e^{-i\frac{Et}{\hbar}} for -a\leq x\leq a

= 0 otherwise.

using the normalised wavefunction calculate the expectation value of momentum squared:

\langle p^{2}\rangle


My attemted solution:

i know that to find momentum p, the operator is:

\bar{p}=-\hbar\bar{\nabla}^{2}

so we say:

\langle p\rangle=\int\Psi^{\star}\bar{p}\Psi dx

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

 

[funke]

Just square the momentum operator first, then integrate between the two Psi's (I'm completely new to this forum, so I can't write tex code yet. bear with me)

 

[AStaunton]

in terms of squaring the momentum operator, who is this done?

ie...the first part simple goes to (i^2hbar^2) but what does it mean to square the laplacian?

 

[conana]

In position representation (one dimension), the momentum operator

\hat{P}\rightarrow-i\hbar\dfrac{d}{dx}.

You want to square this. Squaring a derivative operator just turns it into a second derivative. The process is then taking

\langle\hat{P}^2\rangle=\langle\Psi_1|\hat{P}^2|\Psi_1\rangle.

As for the normalization condition, what they want you to do is first normalize the wave function, i.e. solve for A such that

\langle\Psi_1|\Psi_1\rangle=1.

Sometimes an even easier way of taking <P^2> is to let one P operate on \langle\psi| and the other on |\psi\rangle so that you only have to work with first derivatives. In this case it won't make much of a difference, but it is a nice trick to have up your sleeve.