# Expected value for p

1. Sep 7, 2015

### bznm

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

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?

2. Sep 7, 2015

### blue_leaf77

Is that integral too lengthy to compute?

3. Sep 7, 2015

### 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: $| \psi(x,0)>=\frac{1}{\sqrt2} (|1>+e^{i \gamma} |2>$?

4. Sep 7, 2015

### 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.

5. Sep 7, 2015

### bznm

mmh.. for example?

6. Sep 7, 2015

### 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 $\phi_n$.