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Challenging Quantum Problem

  1. Dec 2, 2005 #1
    The question reads: Prove that for a particle ina potential V(r) the rate of change of the expectation value of the orbital angular momentum L is equal to the expectation value of the torque:
    d<L>/dt = <N>
    where N = r x (-grad V)
    Now I was thinking the key of this problem was messing with the Ehrenfest's theorem until it looked like the first equation. Obviously the r is going to come from substituting r x p for L. Any help would be appreciated, thanks.

    I simplified it to this:
    d/dt(L) = i/(hbar) ( p^2/2m + V , -i(hbar) x (grad) ) + ( dr/dt x p ) + (r x dp/dt)


    and I know dr/dt points in the same direction as p, so dr/dt x p is 0.


    So I simplified it down to this,
    d/dt(L) = i/(hbar) ( p^2/2m + V , -i(hbar) x (grad) ) + (r x dp/dt)

    and I need everything on the left except for the last term to equal zero.
     
    Last edited: Dec 2, 2005
  2. jcsd
  3. Dec 2, 2005 #2

    Galileo

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    If you see the time derivative of the expectation value of an operator, you should be reminded of the QM 'equation of motion'. For time independent operators I think it was something like:

    [tex]\frac{d}{dt}\langle Q \rangle =\frac{\hbar}{i}\langle [H,Q]\rangle[/tex]
     
  4. Dec 2, 2005 #3
    ya i know, I expanded that out where H = T + U, and Q equaled L, but what to do with that, I dont know.

    Edit: Acutally the theorem I looked up also has a "+ (dL/dt)" <---partial deriviative. So I think those high school algebra steps i took at the beginning have nothing to do with solving the problem. I think I have to go back to square 1.

    Also, there is a (b) part of the problem, that says show that d<L>/dt= 0 for any spherically symmetric potential. I haven't even looked at that part yet, not sure exactly how to solve that either..
     
    Last edited: Dec 2, 2005
  5. Dec 2, 2005 #4
    any help would really be appreciated, I put a lot of time into this problem, and haven't gotten very far.
     
  6. Dec 2, 2005 #5

    Physics Monkey

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    A useful operator identity is [tex] [AB,C] = A [B,C] + [A,C] B [/tex]. Try applying this formula to the first component of the angular momentum operator, you will find that the resulting commutators are easy to evaluate. You can then obtain similar expressions for the other components of angular momentum by cyclically permuting indices.
     
  7. Dec 2, 2005 #6
    when you say "the first component" of the angular momentum operator, are you talking about the operator in the equation I mentioned, or the sperate operators for each dimension of angular momentum?
     
  8. Dec 2, 2005 #7

    Physics Monkey

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    The angular momentum operator is a vector, so when I say the first component I mean the first component of that vector i.e. [tex] L_x = y p_z - z p_y [/tex].
     
  9. Dec 2, 2005 #8
    perhaps this seems like a stupid question but wouldn't I need to have 2 of the angular momentum components in order to apply that formulaa? For example:

    [Lx,Ly] = ihLz
     
  10. Dec 2, 2005 #9

    Physics Monkey

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    You want to calculate the time derivative of the expectation value of angular momentum. Galileo gave you a formula and so you need to calculate [tex] [L_x,H] [/tex]. What's the problem?
     
  11. Dec 2, 2005 #10
    hmm interesting, and so i am guessing h/i = A, H = B , Lx = C if I were to apply it to your formula

    Edit: no, nevermind h/i is a cosntant, so you're saying have [tex] [ y p_z - z p_y , H ][/tex] ?

    sry if these questions sound stupid, but if you knew how little our teacher taught us, and how poor this book is at explaing commuators...
     
    Last edited: Dec 2, 2005
  12. Dec 2, 2005 #11

    Physics Monkey

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    Yes, you've got it. Now first use the simple identity [tex] [A+B,C] = [A,C] + [B,C] [/tex] to separate the commutator into two terms, then you can apply the identity I told you above to each term. This is useful because the commutator of position or momentum with the Hamilitonian is easier to evaluate (I suspect you already know the results in fact).
     
  13. Dec 2, 2005 #12
    Yes I think I do, I know
    [tex] [ H , y ] = Hy-yH [/tex]
    and
    [tex] [ H , P_z ] = HP_z - P_zH [/tex]
    are those the ones you mean (along with the other permutation of the idecies)?
    Edit: I did the algebra/whatever and i got this
    (Hbar)/ i [tex]{ [ ( Hy-yH ) P_z + y ( HP_z - P_zH ) ] - [ ( Hz - zH ) P_y + z ( HP_y - P_yH ) ] } [/tex]
     
    Last edited: Dec 2, 2005
  14. Dec 3, 2005 #13

    Galileo

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    Correct, although written a bit more neatly as (omitting the constant factor hbar/i):

    [tex][H,y]p_z+y[H,p_z]-[H,z]p_y-z[H,p_y][/tex]
    Let's look at, say, the first commutator [H,y]. How would you evaluate this?

    You know that [itex]H=p^2/2m+V(r)[/itex] (are you given that V depends only on the radial distance [itex]r=\sqrt{x^2+y^2+z^2}[/itex]?)
    With the commutation relations [itex][x_i,p_j]=i\hbar\delta_{ij}[/itex] you can evaluate most terms.
    Something like [V,p_z] may require a little additional calculation.
     
  15. Dec 3, 2005 #14
    Ok so Ill expand the hamiltonian out and use [tex] [A+B,C] = [A,C] + [B,C] [/tex], where [itex] A = p^2/2m , B = V(r) [/itex] . One question, what do I do about the momentum vector p^2 do I have to break that up into its components?
     
    Last edited: Dec 3, 2005
  16. Dec 4, 2005 #15
    sry if these questions seem simple, however if you knew how little the professor teaches us, you'd understand.
     
  17. Dec 4, 2005 #16

    Galileo

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    You must do whatever you must do to solve the problem. If you think expanding out the commutator helps you to evaluate the commutator, do so!
    There doesn't seem to be much choice in this case, although you can speeden up the calculation by noting that [p^2,p_i]=0, [V,x_i]=0. I hope these identities seem obvious.
     
  18. Dec 4, 2005 #17
    ya I had tried that, and I had used those identies last night, however, I'm still getting stuck, I have no idea how im going to get from this combination of commutators to the cross product [tex] d<L>/dt = r x (- /nabla V) [/tex] where x is a cross product. Should the answer be obvious to me?

    Here is where I am currently at :

    [tex] P_z/2m[p^2_z, y] + P_z/2m [P^2_x,y] - P_y/2m[P^2_y, z] -P_y/2m[P^2_x, z] + y [V,P_z] - z[V, P_y] [/tex]
     
    Last edited: Dec 4, 2005
  19. Dec 4, 2005 #18

    Galileo

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    What we're doing is equating the components of the left and right side. So instead of using the vector equation with the cross-product we treat it as a set of 3 equations. One for each component. (There may be other ways to do it, but this is one is relatively easy and straightforward).

    You've got a lot of terms in the expression. Many of them are zero, like [p_x,y]. Do you see why?

    Also, you should be able to evaluate [V,P_z]. Let it act on a test function and see what happens.

    You're close to the answer, although it may not look like it :smile:

    BTW, there was a slight error in the motion equation I posted. The factor [itex]\hbar/i[/itex] should've been [itex]i/\hbar[/itex], so it's

    [tex]\frac{d}{dt}\langle Q \rangle =\frac{i}{\hbar}\langle [H,Q]\rangle[/tex]
     
    Last edited: Dec 4, 2005
  20. Dec 4, 2005 #19
    I am not sure why those terms, are zero, I was under the impression the only terms that could equal zero were the ones that were like [tex] [P_x, x] [/tex] or [tex] [P_y, y] [/tex] which I did set to zero in the equation in my previous post.
     
    Last edited: Dec 4, 2005
  21. Dec 4, 2005 #20

    Galileo

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    Nonono. These are the ones that are not zero! They do however cancel out in the expression, so that's a coincidence.

    Surely you've seen before that [itex][x,p_x]=i\hbar[/itex] ??

    You can demonstrate it by taking an arbitrary function f to have [x,p_x] act upon it:

    [tex][x,p_x]f=x\frac{\hbar}{i}\frac{\partial}{\partial x}f-\frac{\hbar}{i}\frac{\partial}{\partial x}(xf)=\frac{\hbar}{i}\left(x\frac{\partial}{\partial x}f-f-x\frac{\partial}{\partial x}f\right)=(i\hbar)f[/tex]
    So [itex][x,p_x]=i\hbar[/itex]. All I've used is the product rule.

    You can work out all terms this way, but I figured you've seen this before. In condensed notation the canonical ones are (see post 13):

    [tex][x_i,x_j]=[p_i,p_j]=0[/tex]

    [tex][x_i,p_j]=i\hbar \delta_{ij}[/tex]

    where i,j=x,y,z.
     
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