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Position operators and wavefunctions

  1. May 29, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the eigenfunctions and the eigenvalues of the following Hamiltonian
    [itex]\hat{H} = \frac{1}{2m} \left ( \frac{ \hbar}{i} \vec{\nabla}-\frac{q}{c}(0, B_z x,0) \right ) ^2 = \frac{1}{2m} \left ( - \hbar^2 \vec{\nabla}^2-\frac{\hbar q}{ic}(\vec{\nabla}·(0, B_z\hat{x},0) + (0,B_z x, 0)·\vec{\nabla}) +\left(\frac{q}{c}\right)^2(0, B_z x,0)^2\right ) ^2 = [/itex]

    Assume the wavefunction is of the form [itex]\Psi(x) = e ^{i(k_y y + k_z z)}\chi(x)[/itex]
    2. Relevant equations

    The time dependent Schrodinger Equation:
    [itex]\hat{H} \psi(x) = E \psi (x)[/itex]

    3. The attempt at a solution
    Computing the scalar products we obtain:
    [itex]\hat{H} =\frac{1}{2m} \left ( - \hbar^2 \vec{\nabla}^2-\frac{q}{c}B_z x \frac{\hbar \partial}{i \partial y} + \left(\frac{q}{c}B_z x\right)^2 \right ) [/itex]

    And we can rewrite [itex]\frac{\hbar \partial}{i \partial y}[/itex] as [itex]\hat{p} _y[/itex], so:
    [itex]\hat{H} =\frac{1}{2m} \left ( - \hbar^2 \vec{\nabla}^2-\frac{q}{c}B_z x \hat{p}_y + \left(\frac{q}{c}B_z x\right)^2 \right ) [/itex]

    Next I state the time-dependent Shcrodinger Equation:
    [itex]\frac{-\hbar ^2}{2m}\vec{\nabla}^2\psi-\frac{q}{2mc}B_z\hat{x}\hat{p}_y \psi+\frac{1}{2m}\left (\frac{q}{c} B_z \hat{x}\right)^2 \psi = E \psi[/itex]

    The x that appears along in the Bz term: is it the position operator or an scalar operator with the value x? If its the position operator how does it interact with the wavefunction?
     
    Last edited: May 30, 2014
  2. jcsd
  3. May 29, 2014 #2

    strangerep

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    You seem to have any extra "##^2##" power at the end, which is carried through part of your working, though not in the eventual Schrodinger equation. So I guess that's just a cut-and-past typo.

    Are you sure you didn't miss a factor of 2 in the middle term?

    I don't think there's any need to do that here. Just leave it as a derivative.

    The position operator acts on these wavefunctions simply as multiplication by ##x##. But you don't need to add the explicit hat on ##x## for this exercise.

    You've been given an ansatz for the solution, so just plug that into the Schrodinger equation. (Leave ##x## and ##\partial/\partial y## as they were). The solution method for this type of partial differential equation is just standard separation of variables. (If necessary, look that up on Wikipedia to see a sketch of the general idea.)
     
  4. May 30, 2014 #3
    Yes, its a typo, thank you.


    I think not. Isn't this correct?

    ## \frac{1}{2m}\frac{\hbar q}{ic}(\vec{\nabla}·(0, B_z\hat{x},0) + (0,B_z x, 0)·\vec{\nabla})##
    ## \frac{1}{2m}\frac{\hbar q}{ic}((0 + \frac{\partial}{\partial y}B_z\hat{x} +0) + (0 + B_z x\frac{\partial}{\partial y} + 0))##
    ## \frac{1}{2m}\frac{\hbar q}{ic}((0 + 0 +0) + (0 + B_z x\frac{\partial}{\partial y} + 0))##

    [itex]\frac{1}{2m} \left (\frac{q}{c}B_z x \frac{\hbar \partial}{i \partial y}\right)[/itex]

    It turns out that the differential equation is not differential on the wavefunction; I can make ##\psi## disappear by multiplying by ##\psi^*## on the right on both sides (and assuming it is normalized). However I am left with derivatives of ##\chi(x)##, should I next try to solve for ##\chi(x)## so that I obtain a closed form of ##\psi^## ?

    Thanks for your help.
     
    Last edited: May 30, 2014
  5. May 30, 2014 #4

    vanhees71

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    [corrected!]

    You must be careful with operators. You always have to make sure to apply the entire expression to the wave function. It's easier to first work it out in the abstract formalism. Here we have
    [tex]\hat{H}=\frac{1}{2m} \left [\hat{\vec{p}}-\frac{q}{c} \vec{A}(\hat{\vec{x}}) \right]^2.[/tex]
    Multiplying this out gives
    [tex]\hat{H}=\frac{1}{2m} \hat{\vec{p}}^2 - \frac{q}{2mc} (\hat{\vec{p}} \cdot \hat{\vec{A}} + \hat{\vec{A}} \cdot \hat{\vec{p}}) + \frac{q^2}{2mc^2} \hat{\vec{A}}^2.[/tex]
    Now applying this to the wave function in position representation this translates into (setting [itex]\hbar=1[/itex])
    [tex]\hat{H} \psi(\vec{x})=-\frac{1}{2m} \Delta \psi +\frac{\mathrm{i} q}{2mc} (\vec{\nabla} \cdot \vec{A}) \psi+\frac{\mathrm{i} q}{mc} \vec{A} \cdot \vec{\nabla} \psi +\frac{q^2}{2mc^2} \hat{\vec{A}}^2.[/tex]
    Since you work in the Coulomb gauge for the vector potential of the magnetic field the 2nd term vanishes identically, and you are left with
    [tex]\hat{H} \psi(\vec{x})=-\frac{1}{2m} \Delta \psi +\frac{\mathrm{i} q}{mc} \vec{A} \cdot \vec{\nabla} \psi +\frac{q^2}{2mc^2} \hat{\vec{A}}^2.[/tex]
    Now simply plug the given ansatz into the eigenvalue equation of this operator.

    You should also think about the question, why one can see from the Hamiltonian, why this ansatz must work out to simplify the problem a lot. Think about symmetries (and maybe even the analogous classical problem about the motion of a spin-0 charge in an magnetic field).
     
    Last edited: May 30, 2014
  6. May 30, 2014 #5
    Where do those ##\frac{1}{2mc}## factors in the second and fourth term come from?

    Looking at the next expression you wrote I assume it was a typo and you meant:
    [tex]\hat{H}=\frac{1}{2m} \hat{\vec{p}}^2 - \frac{q}{2mc} \left(\hat{\vec{p}} \cdot \hat{\vec{A}} + \hat{\vec{A}} \cdot \hat{\vec{p}} \right )+ \frac{q^2}{2mc^2} \hat{\vec{A}}^2 .[/tex] Am I right?

    I don't understand why the third term is ##-\frac{q}{mc} \hat{\vec{A}} \cdot \hat{\vec{\nabla}}## and not ##-\frac{q}{2mc} \hat{\vec{A}} \cdot \hat{\vec{\nabla}}##. I guess its the "factor of 2" strangerep said I missed, but I can't see where it comes from.

    Also, I understand the ##\hbar## we get when going from ##\hat{\vec{p}}## to ##-\hbar i\vec{\nabla}## disappears when setting it to unity, but what about the ##i##?

    Thanks.
     
    Last edited: May 30, 2014
  7. May 30, 2014 #6

    vanhees71

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    I've corrected my typos (hopefully all). Concerning the factor 2:

    [tex]
    (\hat{p} \cdot \vec{A}+\vec{A} \cdot \hat{p}) \psi = -\mathrm{i} \left (\vec{\nabla} \cdot (\vec{A} \psi) + \vec{A} \cdot \vec{\nabla} \psi \right ) = -\mathrm{i} \left [ (\vec{\nabla} \cdot \vec{A}) \psi + \vec{A} \cdot \vec{\nabla} \psi + \vec{A} \cdot \vec{\nabla} \psi \right ]= -\mathrm{i} \left [ (\vec{\nabla} \cdot \vec{A}) \psi + 2\vec{A} \cdot \vec{\nabla} \psi \right ].[/tex]
    It's just using the product rule of differentiation on the first term.
     
  8. May 30, 2014 #7
    Oh, so the del operator acts on ##\vec{A}\psi##, and not just on ##A##. This helps me in a couple of other problems on which I was stuck. Thank you very, very much.
     
  9. May 30, 2014 #8

    vanhees71

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    Yes, this is a very common mistake. I guess nearly everybody makes it when learning the operator formalism :-). It is easily avoided by making oneself clear what an "product" of operators means: When writing [itex]\hat{A} \hat{B}[/itex] it means to first apply [itex]\hat{A}[/itex] on a Hilbert-space vector (or a wave function as its representant in the position representation) and then on the result (which is a again a Hilbert-space vector/wave function) the operator [itex]\hat{B}[/itex].

    In other words, "multiplication" in the context of operators has the meaning of "composition of mappings".
     
  10. May 30, 2014 #9

    strangerep

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    So... are you (carllacan) able to solve the PDE by separation of variables now?
     
  11. Jun 4, 2014 #10
    I thought I could, but I just realized that I made the same mistake with the ## \vec{\nabla}\vec{\nabla}## operator that I had made with the ##\vec{\nabla}\vec{A}## operator, so I 'm doing it again, more carefully.

    That is, when developing the square of the operators I did ##(\vec{\nabla}[·]+...)^2= \vec{\nabla}^2[·]+...## (I use ##[·]## as a placeholder for the function on which the ops act) when the right result would be ##(\vec{\nabla}[·]+...)^2= 2\vec{\nabla}·\vec{\nabla}[·]+... = ##, right?
     
    Last edited: Jun 4, 2014
  12. Jun 4, 2014 #11

    strangerep

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    If I understand you correctly, that placeholder shouldn't be inside the parentheses.

    Why not just write it out first in component notation like this:
    $$(\partial_k - A_k)\Big( (\partial_k - A_k) \Psi \Big) ~,~~~~~~ \text{(with implicit summation over the index k)} ~.$$I've written ##A_k## for the components of the vector ##\frac{q}{c}(0, B_z x, 0)##, (and omitted factors of ##\hbar/i## which you can put back in). Then expand it using the Leibniz product rule carefully. Some things should disappear because only ##A_y## is nonzero but depends only on ##x##, hence terms like ##(\partial_k A_k)\Psi## disappear. (Do you see why?)
     
    Last edited: Jun 4, 2014
  13. Jun 5, 2014 #12
    Yep, you're right.
    That's more or less what I'm doing. Though it looks that I also got the ##vec{A}## wrong (its expression in terms of B comes from another excercise which I had doen wrong).

    Thank you very much for your help, I'll keep working on this.
     
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