Find Expected Value for p - Can You Suggest a Quicker Way?

[bznm]

I have done a lot of counts but I'm sure that there will be a quicker way.. Can you suggest me?

We have a particle in 1D that can moves only on ##[0,a]## because of the potential ##V(x)=\begin{cases}0, x\in(0,a)\\ \infty, otherwise\end{cases}##

At t=0, ##\displaystyle\psi(x,0)=\frac{\phi_1(x)+e^{i\gamma}\phi_2(x)}{\sqrt2}##

Find the expected value for ##p##.

I have followed the way ##\bar{p}=-i \hbar\int \psi(x)* \cdot \frac{\partial}{\partial x} \psi(x) \ dx##...

Can I follow a quicker way?

 

[blue_leaf77]

Is that integral too lengthy to compute?

 

[bznm]

It is, but what it asks me to compute after that is even longer.. ##\bar {p}^2##!

Can't I express the state at t=0 as: ##| \psi(x,0)>=\frac{1}{\sqrt2} (|1>+e^{i \gamma} |2>##?

 

[blue_leaf77]

How is it lengthy, it's just integral of product sinusoids, you can make use of their (anti)symmetry property to eliminate some terms without actually calculating it.

 

[bznm]

mmh.. for example?

 

[Avodyne]

##p^2## is much easier than ##p##. I just computed its expectation value in my head.

This is because ##p^2## is a special operator for this problem. It's specialness is related to the definition of ##\phi_n##.