# Homework Help: Expectation value of the square of Momentum

1. Mar 4, 2015

### Milsomonk

1. The problem statement, all variables and given/known data
The expectation value of <P^2>= -ħ∫ψ* ∂^2ψ/∂x^2 dx
For the Guassian wave-packet ψ(x)=(1/(π^1/4)(√d))e^-((x^2)/(2d^2))
Limits on all integrals are ∞ to -∞.
2. Relevant equations
<P^2>= -ħ∫ψ* ∂^2ψ/∂x^2 dx
ψ(x)=(1/(π^1/4)(√d))e^(iKx)-((x^2)/(2d^2))

3. The attempt at a solution
Ok, the previous question was to calculate <P> given <P>=-iħ∫ψ* ∂ψ/∂x dx. I did this by calculating the partial derivative of ψ and then taking the integral of ψ* ∂ψ/∂x, then multiplying by -iħ, I got the result <P>=ħK which seemed to make sense. So I tried to take the same method to this question;
Firstly working out the second derivative of ψ, since I already know the first derivative from the first question, I simply differentiated again and found it to be the following

∂^2ψ/∂x^2 = (∂ψ/∂x)(iK-(x/d^2))-(1/d^2)ψ Used Product rule, Plugged ∂ψ/∂x and ψ back in for simplicity.

Next I simplified the Integrand by factoring out ψ.
Giving the integral -iħ∫((x^2)/(d^4))-((2iKx)/(d^2))-(K^2)-(1/(d^2))ψ*ψ dx
I from the previous question that ψ*ψ is (1/((√π)(d)))e^-((x^2)/(d^2)).

So I split this into the sum of 4 seperate integrals, two had x values multipied by e^-x values, I concluded that when plugging in infinited these would each go to zero.
so left with two integrals,
<P^2>=(-ħ/((√π)(d)))(-K^2∫e^-((x^2)/(d^2))-1/(d^2) ∫e^-((x^2)/(d^2))

It is given that ∫e^-((x^2)/(d^2)) between ∞ and -∞ is (√π)(d)

So I simplify this all and get <P^2> = ħ(K^2)+(ħ/d)

I have been through my calculations a number of times and cant find and error but I can't help but think that the ħ/d term doesn't really make sense. Any thought's would be much appreciated :)

2. Mar 4, 2015

### TSny

Note that the operator $\hat{p}^2$ will have a factor of $\hbar^2$ rather than $\hbar$. In your four separate integrals, did you get one that has an integrand with a factor of $x^2$?

3. Mar 4, 2015

### Milsomonk

Thanks for the reply, this is why I distrust the answer that I calculated because it doesn't really agree with my value of <P>. But the only time ħ appears is in the equation given in the question, so there is no way to get ħ^2. The only x^2 i find is in the power of e, which is in each of the four integrals.

4. Mar 4, 2015

### TSny

If $\hat{p} = -i\hbar \frac{\partial}{\partial x}$, then $\hat{p}^2 =-\hbar^2 \frac{\partial^2}{\partial x^2}$. So, there is definitely a factor of $\hbar^2$ in $\hat{p}^2$. Maybe there was a misprint in the statement of the problem.

EDIT: If you evaluate $\frac{\partial^2}{\partial x^2} e^{-x^2}$, then you should get a term proportional to $x^2e^{-x^2}$ as well as other terms.

Last edited: Mar 4, 2015
5. Mar 5, 2015

### Milsomonk

You're absolutely right, my bad, it is ħ^2. We are given the integral for e^-(x^2)/(d^2) as (√π)d though.