1. Limited time only! Sign up for a free 30min personal 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 kinetic energy (QM)

  1. Jun 2, 2013 #1

    fluidistic

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    In a QM problem I must calculate the expectation value of the kinetic energy, namely ##\langle \Psi _c ,\frac{ \hat { \vec p } ^2}{2m} \Psi _c \rangle##. Where ##\Psi _c=ce^{-\alpha x^2}##.

    2. Relevant equations

    ##\int _{-\infty}^\infty \exp \{ -a x^2 +bx \} dx = \sqrt{\frac{\pi}{a}}\exp \{ -\frac{b^2}{4a} \}##.

    3. The attempt at a solution
    ##\langle \Psi _c ,\frac{ \hat { \vec p } ^2}{2m} \Psi _c \rangle = \int _{-\infty}^\infty c e^{-\alpha x^2} \frac{\hbar ^2}{2m} \frac{d^2}{dx^2}(ce^{-\alpha x^2})dx####=...=-\frac{\hbar ^2}{m}\alpha c^2 \underbrace{\int _{-\infty}^\infty e^{-2\alpha x^2}dx}_{=I}+\frac{2\hbar ^2}{m} \alpha c^2 \underbrace{\int _{\infty}^\infty x^2 e^{-2\alpha x^2}dx}_{=J}##.
    After some algebra, I reached that ##I=\sqrt{\frac{\pi}{2\alpha}}## while ##J=\frac{\sqrt \pi}{2^{5/2}\alpha ^{3/2}}##.
    From which I reached that ##\langle \Psi _c ,\frac{ \hat { \vec p } ^2}{2m} \Psi _c \rangle = \frac{\hbar ^2 c^2 \pi ^{1/2}}{2^{3/2}m} \left ( \frac{1}{\alpha ^{1/2}} -\alpha ^{1/2} \right )##.
    But I'm having a hard time to swallow the fact that this number is negative if alpha is greater than 1. What do you think? I made some mistake somewhere?! Right?
     
  2. jcsd
  3. Jun 2, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Note that the two terms inside the parentheses are inconsistent with dimensions. Check to make sure you got the correct number of factors of ##\alpha## in front of the J integral.

    EDIT: Also, don't forget that the momentum operator has a factor of ##i##.
     
    Last edited: Jun 2, 2013
  4. Jun 2, 2013 #3

    fluidistic

    User Avatar
    Gold Member

    I see. I've just checked out my 2 integrals with Wolfram Alpha and I didn't make any mistake there. So indeed, it should be in the factor(s) outside the integral(s).

    ##\hat {\vec p}^2=-\hbar ^2 \frac{d^2}{dx^2}##. So already I'm off a sign. I forgot the minus sign in my calculations. I'll recheck the rest and post tomorrow (too late here).
    Thank you very much so far.
     
  5. Jun 3, 2013 #4

    fluidistic

    User Avatar
    Gold Member

    I've redone the whole algebra but the 2 integrals and I reached that the factor in front of I is ##2c^2\hbar ^2 \alpha## and the one in front of J is ##-4c^2 \hbar ^2 \alpha ^2## which makes the result for the expectation value of the momentum: ##c^2\hbar ^2 \sqrt {\frac{\pi \alpha }{2}}## which is always positive. However I'm not sure the units make sense.
    If I'm not wrong, the units of c^2 is density of probability, i.e. 1/m (1 over meter). The units of alpha are 1/m^2 because the exponent must be adimensional. And the units of hbar squared are (J*s)^2, joule per second squared. So I get that the units of expectation of the momentum are J^2s^2/m^2 instead of J.


    Nevermind I forgot to divide all my algebra by 2m (2 masses). By doing so I get a result for the expectation value of ##\frac{c^2\hbar^2 \sqrt{\pi \alpha }}{2^{3/2}m}##. The units are joule, a unit of energy which is a good sign.
     
  6. Jun 3, 2013 #5

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I think that's the right answer.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Expectation value of kinetic energy (QM)
Loading...