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Free particle in three dimensions (angular momentum)

  1. Apr 28, 2015 #1
    1. The problem statement, all variables and given/known data
    A particle with mass m can move freely in three dimensions. Explain why the stationary states of the particle are determinate states for angular momentum ([itex]L_z[/itex] and [itex]L^2[/itex])

    2. Relevant equations
    [itex]L^2 = L_x^2 + L_y^2 + L_z^2[/itex]
    [itex]L = r \times p[/itex]
    [itex]\hat{H} = -\frac{\hbar^2}{2m}\bigtriangledown^2[/itex]
    3. The attempt at a solution
    I am quite certain this has to do with the fact that the hamiltonian commutes with both [itex]L_z[/itex] and [itex]L^2[/itex]. However, I am not certain how to make the leap from there to every solution to the Schrodinger equation of the free particle being an eigenfunction of the two other operators. I know they have the same eigenbasis. However, how can I use this to prove that every eigenfunction of the hamiltonian is also an eigenfunction for the angular momentum operators? The way the problem is formulated, I get the impression that I do not have to write down the solution to the Schrodinger equation.

    So, how should I proceed?
     
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  3. Apr 28, 2015 #2

    DrClaude

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    Unless something is escaping me right now, I don't fully agree with this statement. So lets start by considering [itex]L^2[/itex] only.

    Hint: spherical coordinates.
     
  4. Apr 28, 2015 #3
    Are you saying I should do this by solving the Schrodinger equation in spherical coordinates? Wouldn't that, for the free particle, be easier with Cartesian coordinates? Of course, that gives me the wrong answer:
    [tex]\hat{L}_z \psi = \hbar (xk_y- yk_x)\psi[/tex]
    Since:
    [tex]\psi = Ae^{i(xk_x + yk_y + zk_z)}[/tex]
    Of course, this is not a normalizable solution. Will doing this in spherical coordinates show me a different solution?

    Edit:
    Oh, wait... I just noticed that [itex]\hat{L}^2[/itex] looks just like the angular part of the Schrodinger equation after separation of variables.
     
    Last edited: Apr 28, 2015
  5. Apr 28, 2015 #4

    DrClaude

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    Exactly! So what can you conclude?
     
  6. Apr 28, 2015 #5
    The angular part of the solution will be an eigenfunction to [itex]\hat{L}^2[/itex]. Since the [itex]\hat{L}^2[/itex] does not affect the radial part of the solution, we get:

    [itex]\hat{L}^2\psi = \hat{L}^2R(r)Y(\phi,\theta) = R(r)\hat{L}^2 Y(\phi,\theta) = aR(r)Y(\phi,\theta) = a\psi[/itex]
    Here a is the eigenvalue in [itex]\hat{L}^2 Y(\phi,\theta)= aY(\phi,\theta)[/itex].

    So, now I just have to figure out why the claim above is also true for [itex]\hat{L}_z =-i\hbar\frac{\partial}{\partial \phi}[/itex].

    Edit. I can modify this to also solve for [itex]\hat{L}_z[/itex]. [itex]Y(\phi,\theta) = \Phi (\phi)\Theta(\theta)[/itex]

    The solutions to the phi part of the solution always take the form [itex]\Phi (\phi) = Ae^{im\phi}[/itex]. So:

    [itex]\hat{L}_z \Phi =-i\hbar\frac{\partial}{\partial \phi} \Phi = m\hbar \Phi[/itex]

    Using this result, I can use the same logic as above to show that [itex]\psi[/itex] is an eigenfunction for [itex]\hat{L}_z[/itex].

    Thanks for the help!
     
    Last edited: Apr 28, 2015
  7. Apr 29, 2015 #6

    DrClaude

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    You have to be careful here. While ##\hat{L}^2## appears explicitly in the Hamiltonian, ##\hat{L}_z## does not. You could use the same argument you have for ##\hat{L}_x## and ##\hat{L}_y##. This is why I said at the beginning that I disagree with the problem statement. While a stationary state has to be an eigenstate of ##\hat{L}^2##, it does not need to be an eigenstate of ##\hat{L}_z##. It can be, since ##\hat{L}_z## commutes with both ##\hat{L}^2## and ##\hat{H}##, but that is an arbitrary choice.
     
  8. Apr 29, 2015 #7

    TSny

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    Maybe I'm not seeing something here. A stationary state (i.e., energy eigenstate) of the free particle Hamiltonian is not necessarily an eigenstate of ##\hat{L}^2##.
     
  9. Apr 29, 2015 #8
    Yes, that is true. However, [itex]\psi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi)[/itex]. So;

    [itex]\hat{L}_z \psi(r,\theta,\phi) = \hat{L}_z R(r)\Theta(\theta)\Phi(\phi) = R(r)\Theta(\theta) \hat{L}_z\Phi(\phi) = R(r)\Theta(\theta) m\hbar\Phi(\phi) [/itex]

    [itex]= m\hbar R(r)\Theta(\theta) \Phi(\phi) = m\hbar\psi[/itex]

    The reason, I think, this is accurate is because [itex]\Phi(\phi) = e^{im\phi} \textrm{ with } m \in \mathbb{Z}[/itex]. At least this is the case when Griffiths solves the Schrodinger's equation in spherical coordinates (the potential in his case is a function of r, and our potential is obviously a special case of that).
     
  10. Apr 29, 2015 #9

    DrClaude

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    The Hamiltonian can be written in spherical coordinates as
    $$
    \hat{H} = - \frac{\hbar^2}{2m} \nabla^2_r + \frac{ \hat{L}^2}{2m}
    $$
    That means that the wave function is separable, ##\Psi(x,y,z) = \psi(r) Y_{l,m} (\theta, \phi)## and that eigenstates of ##\hat{H}## are also eigenstates of ##\hat{L}^2##. Another way to see it is that conservation of angular momentum requires that a stationary state be an eigenstate of ##\hat{L}^2##.
     
  11. Apr 29, 2015 #10

    DrClaude

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    Again, that choice is arbitrary.

    Exercise: show that, for ##\psi(r)## and eigenfunction of ##- \frac{\hbar^2}{2m} \nabla^2_r##, then ##\Psi = \frac{1}{\sqrt{2}}\psi(r)\left[ Y_{1,1}(\theta,\phi) + Y_{1,0}(\theta,\phi) \right]## is a stationary state of a free particle, i.e., an eigenstate of ##\hat{H}##, and simultaneously an eigenstate of ##\hat{L}^2##, but not an eigenstate of ##\hat{L}_z##.
     
  12. Apr 29, 2015 #11

    TSny

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    The wavefunction ##\psi(x,y,z) = e^{ikx}## is an energy eigenstate for a free particle. But it is not an eigenstate of ##\hat{L}^2##.

    There exist free-particle solutions of the form ##\psi(r) Y_{l,m} (\theta, \phi)## which have different values of ##l## but have the same energy. If you take a superposition of two such states, then you can produce a state that is still an eigenfunction of ##\hat{H}## but is not an eigenfunction of ##\hat{L}^2##.
     
  13. Apr 29, 2015 #12
    Okay, I see now why that is wrong.
     
  14. Apr 30, 2015 #13

    DrClaude

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    Yes, but that wave function is not a proper state as it is not square integrable.

    Do you have an example? You raised interesting points, which have been bothering me all day, so I would like to settle it.

    And just to be sure, we should agree on the definition of stationary state of a free particle. I take it to mean that not only is it an eigenstate of the Hamiltonian, but also that the probability density is independent of time.
     
  15. Apr 30, 2015 #14

    TSny

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    I don't think there are any energy eigenstates of a free particle that are normalizable.

    If you solve for the free particle states in spherical coordinates, you get energy eigenstates of the form ##j_l(kr)Y_{lm}(\theta, \phi)## where the ##j_l##'s are spherical Bessel functions and ##k## is related to the energy as ##E=\frac{\hbar^2 k^2}{2m}##. These states are not normalizable. Once you fix the energy, ##k## is fixed but ##l## and ##m## can be picked arbitrarily. So, there are many states of the same energy with different values of ##l##. A superposition of two such states with the same value of ##k## but different values of ##l## will give an energy eigenstate that is not an eigenstate of ##\hat{L}^2##.

    If you take an energy eigenstate in the form of a plane wave, then you can expand it as shown here: http://en.wikipedia.org/wiki/Plane_wave_expansion. This gives a superposition of spherical harmonics of different ##l##.

    I think that any eigenstate of the Hamiltonian will have a probability density that is independent of time. That is, any eigenstate ##\psi## of the Hamiltonian with eigenvalue E will correspond to a time dependent state of the from ##\psi e^{-i Et/\hbar}##. So, the probability density will be time independent.
     
  16. Apr 30, 2015 #15

    BruceW

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    Interesting. So, was the question statement the wrong way around? If we reverse it, we get: "Explain why determinate states for angular momentum are stationary states of the particle". But after reading your post, I guess this is not true either. We could have a state which is made up of superpositions of states with the same ##l## and ##m## values, but different ##k## values.

    So what was the question statement trying to ask for? Maybe something like: "explain why each stationary state of the particle can be associated with at least one determinate state for angular momentum"
     
  17. Apr 30, 2015 #16

    TSny

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    Yes, that's my understanding.

    I'm also not sure of the intent of the question. There are stationary states which do not have definite angular momentum. But you can say that for any energy, there exist stationary states with definite angular momentum.
     
    Last edited: Apr 30, 2015
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