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Homework Help: 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

    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


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    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


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    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.
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