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Homework Help: Virial theorem problem

  1. Jan 29, 2013 #1
    Hi everybody

    1. The problem statement, all variables and given/known data
    I showed the following:

    [tex] \frac{1}{\hbar i} [H,xp]=x \frac {dV}{dx} - \frac {p^2}{m}[/tex]

    Now I want to use this to show that:

    [tex] \langle \frac {p^2}{2m} \rangle = \frac{1}{2} \langle x \frac {dV}{dx}\rangle[/tex]

    whereas |E> is an eigenstate to H with the eigenvalue E. H x and p are operators. Furthermore in my 2nd equation the <> are the expectation value.

    2. Relevant equations

    3. The attempt at a solution

    Can anyone help me with this? I'm learning for my final exams and I found this task in an old exam. I tried to solve <E|H|E> but that didn't lead me anywhere near it.

    Thanks for your help
  2. jcsd
  3. Jan 29, 2013 #2


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    Can you show <[H,xp]> = 0?
    I don't know if that helps:
    <[H,xp]> = <x[H,p] + [H,x]p>
  4. Jan 29, 2013 #3
    [H,xp]= 0 ??

    I had to show the first equation which I solved. [H,xp] wasnt 0 there. Or am I misunderstanding something?

    edit: oops, is just saw that you mean the expectation value. How to I calculate the expectation value of a commutator? Do I have to solve <E|[H,xp]|E> ?
  5. Jan 29, 2013 #4


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    [H,xp] does not have to be 0 everywhere, but its expectation value should vanish if the virial theorem holds.
    This directly follows from the virial theorem and your equation, by the way.
  6. Jan 30, 2013 #5
    can you show that expectation of commutator will be
    now <x.p> for stationary state is independent of time.
  7. Jan 30, 2013 #6
    I tired what mfb said to show that the expectation value is zero

    [tex] \langle E \mid [H,xp] \mid E \rangle=\langle E \mid Hxp\mid E \rangle - \langle E \mid xpH \mid E \rangle = E(\langle E \mid xp \mid E \rangle -\langle E \mid xp \mid E \rangle) [/tex]

    Is that correct? If so I showed that the expectation value of the commuator is zero, but it still don't understand how this helps me to get to the 2nd equation in my first post.

    Thanks for your help
  8. Jan 30, 2013 #7


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    @Lindsayyyy: Take the expectation value of both sides of the first equation, and use linearity of the expectation value? <a+b>=<a>+<b>
  9. Jan 30, 2013 #8
    I'm too stupid to solve it,sorry.

    I get zero for the left side as I told above but for the right side I got:

    [tex] 0=\langle E \mid x \frac {dV}{dx} \mid E \rangle -\langle E \mid \frac {p^2}{m} \mid E \rangle[/tex]

    I mean I can just multiply 1/2 on both sides to get to the solution now. But I don't think that's the point here :eek:

    thanks for your help
  10. Jan 30, 2013 #9


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    There is no <E|.|E> in the expectation value, unless you know that you consider the state E only (where did you say that?). And a multiplication with 1/2 gives you the solution, indeed.
  11. Jan 30, 2013 #10
    I thought when I want to calculate an expectation value in quantum mechanics it has always the form of:


    the task says that E is an eigenstate to H with eigenvalue E, that's why I did <E|[H,xp]|E>.
    Or how do I calculate an expectation value of [H,xp] to show it is zero?

    thanks for your help
  12. Jan 30, 2013 #11


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    But that is just one eigenstate. What happens if the particle is in a different state?
    Do you have the full problem statement somewhere?
  13. Jan 30, 2013 #12
    I try to translate it, as I'm not a native speaker and the task is not in english:

    The Hamiltonian operator
    [tex] \hat{H}= \frac {\hat{p}^2}{2m}+V(\hat{x}) [/tex]

    describes a particle in one dimension.

    (i)Show the following:

    [tex]\frac{1}{\hbar i} [\hat {H},\hat {x}\hat{p}]=\hat {x} \frac {dV}{dx} - \frac {\hat {p}^2}{m}[/tex]

    (ii) |E> is an eigenstate of H with the eigenvalue E.Deduce from the expression from task (i) that

    [tex] \langle \frac {\hat{p}^2}{2m}\rangle_E =\frac {1}{2} \langle \hat{x} \frac{dV}{dx} \rangle_E[/tex]

    whereas [tex] \langle...\rangle_E[/tex] describes the expectation value to the quantum state |E>

    I hope that helps.

    Thanks for the help and sorry for the sloppy english:shy:
  14. Jan 30, 2013 #13


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    Ok, just the expectation value for E. In that case, it is fine.
    <.>_E is just <E|.|E>.
  15. Jan 30, 2013 #14
    So my solution was right?

    But I still don't understand the factor 1/2. Why is it even there? I mean it would be the same without it, wouldn't it?
  16. Jan 30, 2013 #15


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    You can multiply both sides by 2, if you like, and the equation is still true. ##<\frac{p^2}{2m}>## is the expectation value of the kinetic energy, so usually the factor of 1/2 is included.
  17. Jan 31, 2013 #16
    let us take the right way of doing it.
    for any quantum mechanical operator following relation holds
    now with O=x.p,which does not contain time explicitly last one is zero.
    so we have
    now<x(t).p(t)>=<n|e-iEnt x.p eiEnt|n>=<x(0).p(0)> so it's independent of time.And the required form follows.
  18. Jan 31, 2013 #17
    Ok thanks for the help. I have one more task for this problem.

    V is now [tex] V(\lambda x)=\lambda^n V(x)[/tex]

    I shall show the following:

    [tex] \langle T \rangle_E= \frac {n}{2} \langle V \rangle_E[/tex]

    I tried to derive the V for lambda, but that didn't work out. I think I need (ii) from the post above, but I don't get rid of the x operator.

    Thanks for your help
  19. Feb 1, 2013 #18
  20. Feb 2, 2013 #19
  21. Feb 8, 2013 #20
    If anyone in the future will have the same problem and might find this thread, I solved the problem in a much more (in my opinion) easier way than in the links state above with the euler theorem

    we have the potential [tex] V(\lambda x)=\lambda^n V(x)[/tex]

    derive on both sides wheres I substutite lambda*x=u

    [tex] \frac {dV}{du} \frac {du}{d\lambda}= n \lambda^{(n-1)}V(x)[/tex]

    left side equals
    [tex] \frac {dV}{du} x[/tex]

    set lambda to 1 and you get

    [tex] \frac {dV}{dx} x = n V(x)[/tex]
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