1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Expectation value of the square of Momentum

  1. Mar 4, 2015 #1
    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. jcsd
  3. Mar 4, 2015 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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##?
     
  4. Mar 4, 2015 #3
    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.
     
  5. Mar 4, 2015 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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
  6. Mar 5, 2015 #5
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted