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Time independent Schrodinger equation and uncertainty in x

  1. May 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Is the gaussian
    $$\sqrt{\frac{\pi}{2\alpha}}e^{-\alpha x^{2}}$$
    an eigenfunction of ## \widehat{T} = \frac{\hat{p}^{2}}{2m}## ? If so, what is the corresponding eigenvalue? If not, find a P.E. operator ##\widehat{U} = U(\hat{x}) ## which gives rise to a Hamiltonian ##\widehat{H}## for which this Gaussian is an energy eigenfunction. What physical system are we talking about?
    Find the uncertainty in ##\Delta x##.

    2. Relevant equations

    $$\hat{p} = \frac{\hbar}{i}\frac{\partial}{\partial x}$$

    time independent Schrodinger equation
    ##\{-\frac{\hbar^{2}}{2m} \frac{\partial^{2}}{\partial x^{2}}+U(x)\}\psi_n(x)=E_n\psi_n(x)##


    3. The attempt at a solution
    I have worked out the relative derivatives and determined that it isn't an eigenfunction.
    ##\widehat{T}\psi(x)= -\frac{\hbar}{2m} 2\alpha e^{-\alpha x^{2}}(2\alpha x^{2} -1)##
    Then I plug all that into the TISE and solve for ##\widehat{U}## and get

    ##\widehat{U}=\frac {\hbar^{2} + 4\alpha^{2}x^{2}}{2m}##

    Hopefully I am correct up to here.

    Now I am asked to find ##\Delta x ##

    I think I need

    ## \Delta x= \sqrt{ \langle x^{2} \rangle - \langle x \rangle^{2}}##

    where

    ## \langle x \rangle = \int_{-\infty}^\infty x|\psi|^{2} \,\mathrm{d}x##

    which gives me 0

    but I get confused here where I think I need

    ## \langle x^{2} \rangle = \int_{-\infty}^\infty x^{2}|\psi|^{2} \,\mathrm{d}x##

    is the x2 term just the square of my function? So in principal I am just multiplying the square of my function by the mod of the function squared? THen taking the integral?

    Thanks

    edit :- sorted formatting of equation
     
    Last edited: May 3, 2015
  2. jcsd
  3. May 3, 2015 #2

    Orodruin

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    If you intended to write
    ## \langle x^{2} \rangle = \int_{-\infty}^\infty x^2 |\psi|^{2} \,\mathrm{d}x##
    then yes, this is the expectation value of ##x^2##. Your wording makes it unclear what you are actually having trouble with.
     
  4. May 3, 2015 #3
    That is what the equation was meant to look like.

    My problem is do I square the function ##\psi## and multiply that by the mod squared of the function or multiply the function by x2. In short i am not sure what x2 is in the above formula

    thanks
     
  5. May 3, 2015 #4

    vela

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    When you calculated ##\langle x \rangle##, how did you treat the factor of ##x##? Why do you think it's different in the case of ##\langle x^2 \rangle## in how you deal with ##x##?
     
  6. May 8, 2015 #5
    Thanks very much I think I am getting the hang of what is going on. I don't have to treat ##\langle x^2 \rangle## any different to how I dealt with ##x## i.e.I just use ##x^{2}## when workingout ##\langle x^2\rangle ##
     
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