# Expectation Value of Momentum Squared

1. Jan 23, 2010

### VariableX

1. The problem statement, all variables and given/known data
A particle of mass m is in the state

Psi(x,t) = Ae^(-a[(mx^2)+it])

where A and a are positive real constants.

a) Find A

b) For what potential energy function V(x) does Psi satisfy the Shrodinger equation?

c) Calculate the expectation values of x, x^2, p, and p^2

d) Find sigma_x and sigma_p. Is their product consistent with the uncertainty principle?

2. Relevant equations
Intetgral of p(x)dx = 1 (from negative infinity to infinity)
Shrodinger equation
<x> = Integral of Psi* (x) Psi dx
<p> = Integral of Psi* (hbar/i partial derivative with respect to position) Psi dx
sigma_x times sigma_p > or = hbar/2

3. The attempt at a solution
I managed to get solutions for parts a and b, and most of part c. Using the equations listed above, I got 0 for <x> and <p> and 1/4 hbar/(am). The problem I had was figuring out how to solve <p^2>. I tried squaring hbar/i and the partial derivative term but that lead to some problems.

The partial derivative is A^2 (-2amx/hbar) e^(-amx^2/hbar - iat).

When I squared only the (-2amx/hbar) part, I managed to get a solution of hbar*am, and when I solved for the standard deviations and put them into the Heisenberg Uncertainty Principle I got exactly hbar/2. However, a friend of mine informed me that e^(-amx^x/hbar - iat) is also part of the derivative, and when I tried squaring that as well I ran into problems with simplification.

Is hbar*am the correct answer? And if so, why don't I have to square e^(-amx^2/hbar - iat)? Or if I'm completely wrong, what am I supposed to do to calculate <p^2>?

-Pie

2. Jan 23, 2010

### vela

Staff Emeritus
Your friend is correct, and you could just grind it out.

Here's a trick that might help you. The time-independent Schrodinger equation is

$$\frac{p^2}{2m}\psi(x)+V(x)\psi(x) = E\psi(x)$$

Since you already found V(x) earlier, you can solve for $p^2\psi(x)$ without having to differentiate twice, and use the result when evaluating the integral.

3. Jan 24, 2010

### Altabeh

Hi

Could you please show me how did the OP get $$<p>= 1/4 \hbar/(am)$$? I'm a little bit new to this stuff.

AB

4. Jan 24, 2010

### vela

Staff Emeritus
I think the OP made a typo and meant that result was $\langle x^2\rangle$ and that both $\langle x\rangle$ and $\langle p\rangle$ were 0.

5. Nov 13, 2011

### nana2416

Hello,

I'm really new to this whole thing, would you be able to explain how you got those answers, i'm getting confussed when i try it myself