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Expectation of position and momentum at time t, pictures

  1. Dec 3, 2015 #1
    1. The problem statement, all variables and given/known data

    Consider a particle, with mass m, charge q, moving in a uniform e-field with magnitude E and direction X_1.

    The Hamiltonian is (where X, P, and X_1 are operators):
    upload_2015-12-3_13-26-9.png

    The initial expectation of position and momentum are <X(0)> = 0 and <P(0)>=0

    Calculate the expectation of position and momentum oprator in the Schrodinger picture is X (t) = (X_1, X_2, X,3)

    2. The attempt at a solution

    I know the Hamiltonian and the initial condition. I am doing this in the schrodinger picture, and that with the time evolution operator I can represent the ket as

    |phi(t)> = U (t,0) |phi(0)>

    Is H is time-independent? As P is squared and is the magnitude of the P? If so the time evolution operator can be represented as

    U (t,0) = exp (-i*t*H/hbar)

    Then

    I can represent phi(t) in terms of those terms and phi(0) and proceed but I am not sure if that is correct.

    I know that P = mv = m dx/dt so I think I can use the derivative of the expectation value of X when I get that, correct. According to the equation:

    d/dt <O>_t = i/h <[H,O]>_t + <dO/dt>_t

    but I have to find the expectation of the position first.

    Any help would be much appreciated!
     
  2. jcsd
  3. Dec 3, 2015 #2

    DrClaude

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    You don't need to know φ(t).

    That should not be necessary.

    Start with that equation, and see what you get for P and X.
     
  4. Dec 3, 2015 #3
    Okay, plugging X into that equation (get rid of _t for legibility) I get

    d/dt <X> = i/h <[H,X]> + <dX/dt>
    = i/h <[(P^2/2m - E*q*X_1), X]>+<dX/dt>
    = i/h (-E*q) <X,X1>+<dX/dt>

    and I'm not sure where to go from here...
     
  5. Dec 3, 2015 #4
    What if I just did:

    U (t,0) = exp (-i*t*H/hbar)
    = exp (-i*t*(P^2/2m - E*q*X_1)/hbar)

    |phi (t)> = exp (-i*t*(P^2/2m - E*q*X_1)/hbar) |phi(0)>

    <phi (t)|X|phi(t)> = <phi(0) exp (i*t*(P^2/2m - E*q*X_1/hbar) | X | exp (-i*t*(P^2/2m - E*q*X_1)/hbar) |phi(0)>

    but I don't know where I would go next, could I just multiply it in? But they are operators.
     
  6. Dec 4, 2015 #5

    DrClaude

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    What is the commutator of X and P? What can you say about the time-dependence of an operator (not explicitly time dependent) in the Schrödinger picture?

    You can't go anywhere with that. Note also that you do not know what |φ(0)>, only that <φ(0)|X|φ(0)> = <φ(0)|P|φ(0)> = 0.
     
  7. Dec 4, 2015 #6
    I know that the commutator of x and p is i*hbar and that in the Schrödinger picture the operators are time independent
     
  8. Dec 4, 2015 #7

    DrClaude

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    X1 is only one of the spatial dimensions (if I understand correctly "X (t) = (X_1, X_2, X,3)"), and p2 is simply pp, so
    $$
    \begin{align*}
    [p^2,x] &= p^2x-xp^2 \\&= ppx - xpp \\
    &= ppx - xpp - pxp + pxp \\
    &= ppx - pxp + pxp - xpp \\
    &= p[p,x] + [p,x]p
    \end{align*}
    $$
     
  9. Dec 4, 2015 #8
    Ah ok so then, assuming that the commutator of x and x_1 is 0 I would get:

    d/dt <X> = i/h <[H,X]> + <dX/dt>
    = i/h <[(P^2/2m - E*q*X_1), X]>+<dX/dt>
    = i/h <[X*P^2/2m -X*P^2/2m]>+<dX/dt>
    = i/(2mh) <[X,P^2]>+<dX/dt>
    = i/(2mh) <2*i*h*p>+<dX/dt>
    = i/(2mh)*(2*i*h) <p>+<dX/dt>
    = -1/m<p> +<dX/dt>

    ==>
    <p> = m (d/dt <X> - <dX/dt>)

    But that still has a lot of unknows for a final answer as I have to find out <X> as well
     
  10. Dec 4, 2015 #9
    If I did the same procedure for P then it would be


    d/dt <P> = i/h <[H,P]> + <dP/dt>
    = i/h <[(P^2/2m - E*q*X_1), P]>+<dP/dt>
    = i/h <[P*-E*q*X_1 -E*q*X_1*P]>+<dP/dt>
    = i(-E*q)/h <[P,X_1]>+<dP/dt>
    = i/(2mh) <i*h>+<dP/dt>

    But then if [P,X_1] = ih then this doesn't really make sense...
     
  11. Dec 5, 2015 #10

    blue_leaf77

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    Sorry for butting in, just a short intervention to remind the OP that dX/dt = dp/dt = 0.
     
  12. Dec 5, 2015 #11
    Ah, thank you.

    So then

    <p> = m (d/dt <X>)

    and

    d/dt <P>= i(-E*q)/h <[P,X_1]>

    Is that the final answer?
    The second portion of the question is to, with the same Hamiltonian, knowing that in the Heisenberg Picture , X_H = (X_1H,X_2H,X_3H), find the commutator

    [X_aH (t), X_bH (0)]

    where all the X are operators

    I know that in the Heisenberg picture the operators evolve with time as U^+ (t,0) O U(t,0)
     
    Last edited: Dec 5, 2015
  13. Dec 6, 2015 #12
    So then for the heisenberg picture, because e^(0) = 1 and thus for X_bH (0), U would equal 1. I would get (where U_d represents U dagger)

    U_d X_aH U X_b - X_b U_d X_aH U

    then I don't know what to do
     
  14. Dec 6, 2015 #13

    blue_leaf77

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    No, you still have to solve those two coupled differential equations.
     
  15. Dec 6, 2015 #14
    ah, okay. Plus I still need to find <[P,X_1]> right. Could you help me with that commutator?

    Thanks
     
  16. Dec 6, 2015 #15

    blue_leaf77

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    Is P actually a vector in your notation? Is it actually ##\mathbf{P} = P_1\hat{i} + P_2\hat{j} + P_3\hat{k}##?
     
  17. Dec 6, 2015 #16
    It doesn't specify but I would assume it is a vector
     
  18. Dec 6, 2015 #17

    blue_leaf77

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    In that case, it will be ##[P_1\hat{i} + P_2\hat{j} + P_3\hat{k},X_1] = [P_1,X_1]\hat{i} + [P_2,X_1]\hat{j} + [P_3,X_1]\hat{k}##. You know the value of each term, don't you?
     
  19. Dec 6, 2015 #18
    Yes so then it would just be [P,X_1] = ih ?

    Then the equation would be, as I put in post 9

    d/t <P> = = i/(2mh) <i*h>
     
  20. Dec 6, 2015 #19

    blue_leaf77

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    Don't forget the unit vector, ##[P,X_1] = i\hbar\hat{i}##.
    That equation will then disperse into three equations, each for the momentum components and it should be easily solvable.
     
  21. Dec 6, 2015 #20
    As ih is a constant wouldn't <ih> just be equal to ih since the exponentials of the U term would become 1 and the phi(0) would as well.

    And then I could sub in the first equation to get the value for <x>?
     
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