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Free particle in spherical box

  1. Aug 5, 2009 #1
    1. The problem statement, all variables and given/known data

    Hello everybody:

    I have a problem with the Schrödinger equation in 3D in spherical coordinates, since I'm trying to calculate the discrete set of possible energies of a particle inside a spherical box of radius "a" where inside the sphere the potential energy is zero and out the sphere is infinite.

    First problem was when I was looking at wikipedia webpage for Schrödinger equation and I have to confess that I don't understand why when we deal with spherical coordinates the first thing that appears is the spherical harmonics. I tell you this because in my case I don't see any dependency on the angular variables, theta and phi, since there is not any privilegiated direction in the free particle inside the spherical box. Therefore I should recover the same result that the one showed in the wikipedia webpage where the Schrödinger's equation in spherical coordinates is dealt with when there is rotational symmetry (I'd say!)

    What I was trying to solve was the Schrödinger equation in spherical coordinates assuming the function psi doesn't depend on the angular variables, theta and phi, since the problem has spherical symmetry (by definition of the problem). Then I got only derivatives in the radial coordinate:

    2. Relevant equations

    View attachment eq1.bmp

    Making a change of variable

    [tex]\Psi=U(r)/r[/tex]

    I found the energy of the free particle inside the spherical box:

    View attachment eq3.bmp

    The problem arises when I compare this result to that one of a free particle inside a cubic of side l in cartesian coordinates, since I expected something similar, but the problem is that I found:

    View attachment eq2.bmp

    Then I see a difference, because in the cartesian case for the cube I have one result and in the sphere I have another one. It can be said that it is not shocking because the systems are different, but I'm wondering something....

    I was studying statistical physics (perhaps I should include this now in other thread, I don't know...) and the problem arose when I saw that in the calculation of the partition function for the "particle-in-a-box" problem the partition function reads:

    [tex]\ q_{trans}=\sum_{n_{x}}\sum_{n_{y}}\sum_{n_{z}}exp(-\beta\frac{h^2}{8m l^2}(n_{x}^2+n_{y}^2+n_{z}^2))=(\sum_{n}exp(-\beta\frac{h^2 n^2}{8m l^2}))^3 [/tex]

    and if you compare it to the partition function for the case that I've solved for the particle in the spherical box, I don't find the same, since the factor 3 in the exponent is missed.

    Now my question is, why???? Is there any energy that I didn't take into account and it should be taken into account? The energy related to the angular momentum of the particle? I don't see it, even in this case that energy doesn't recover the case of the "particle-in-a-box" in cartesian coordinates.

    So, either I have made a mistake all along the steps described in the previous discussion which recovers almost the good solution without the factor 3, or there isn't anything wrong but something missed.

    3. The attempt at a solution

    In my opinion, as the partition function has to take into account all the states, even if there is degeneracy, I have the intuition that there is some kind of degeneracy that gives me the same result in both cases, but I really don't see how. Because the free particle in the sphere doesn't have (I'd say) any degeneracy, since the eigensolutions of the free particle in the spherical box is (inside the sphere)

    [tex]\Psi(r)=\frac{2i sin(\alpha r)}{\sqrt{8\pi a}r}[/tex]

    and the boundary conditions establish that [tex]\alpha a=n\pi[/tex] for n integer. And [tex]\alpha[/tex] is related to the energy through

    [tex]\alpha^2=\frac{2mE}{\hbar^2}[/tex]

    Thank you very much and sorry for the, perhaps, too technical thread.
     
    Last edited: Aug 5, 2009
  2. jcsd
  3. Aug 5, 2009 #2

    kuruman

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    Spherical harmonics appear in the 3-D Schrodinger equation because, when you separate variables (I assume a potential of spherical symmetry here), they are the solutions to the angular part of the equation. Typically, one solves the radial equation for whatever potential one has and then tacks on the angular part as spherical harmonics. The degeneracy involved is related to the spherical harmonics.

    So if your ultimate goal is to write down the partition function, you are there because you already have the energies and all you need to calculate is the degeneracies. The ground state (n=1) is spherically symmetric, therefore non-degenerate. The spherical harmonic that goes with it is Y00.

    The first excited state (n=2), can have angular momentum. (Imagine the particle swirling around inside the spherical box.) What are the spherical harmonics that go with it? How are L and M of the spherical harmonics related to your quantum number n? I suggest that you solve the 3-D Schrodinger Equation in spherical coordinates formally for your potential. "Formally" means use separation of variables for the 3-D equation and keep track of your separation variables. A good guide is to see how it is done for the hydrogen atom (if you haven't already seen that.) Any Introductory Quantum Mechanics textbook shows that. Because the spherical harmonics are eigenstates of the Hamiltonian in a spherically symmetric potential, the rectangular 3-D box is not much help as an analogue here.
     
  4. Aug 5, 2009 #3

    kuruman

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    I forgot to comment on your "missing" factor of three. The correct substitution is

    [tex]n^{2}_{x}+n^{2}_{y}+n^{2}_{z}=n^{2}[/tex]

    Imagine three mutually perpendicular axes labeled nx, ny and nz. A quantum state in this "space" is represented as a point for a given choice of n values. Any other state at the same "distance" from the origin will have the same energy because the sum of the three n's squared will be the same. This defines a sphere of radius n as given by the above expression on the surface of which the energy is the same. Do you see now what happens to the degeneracy as n increases?
     
  5. Aug 5, 2009 #4
    Thank you for your quick reply.

    I know that the spherical harmonics are the natural consequence of the solution of the angular part of the Schrödinger equation in 3D in spherical coordinates for a general potential energy and the Bessel functions are the corresponding to the radial part.

    However in my case I was assuming that the wave function (psi) had not any dependency on the angular variables (by hypothesis since the potential energy where the particle is placed is zero). Thus, I'd say (please, correct me if I'm wrong) that in the separation of variables, the angular functions would be constants, i.e.:

    [tex]\Psi(r)=R(r) \Theta(\theta) \Phi(\varphi)=C R(r)[/tex]

    Then the angular part of the Schrödinger equation would dissapear in the Laplacian operator since the first derivative of the wave function with respect to theta and to phi would vanish.

    In that case the equation that remains is the one showed in my first message. Sorry for the inconvenience of the link to the picture which refers to that equation but I wrote it from my computer and I exported it as picture, and I should have written it in LaTex, but I'm still not too good with LaTex.

    On the other hand I had to ask you for your forgiveness since I didn't explain in detail the step that connects from the partition function with three sums to that one with a single one. But I'd say that it appears because we can consider the energy of translation in each direction as independent on the other directions, thus the partition function can be splitted in three multiplication corresponding to the contribution to the partition function of each direction. As it doesn't matter on how it's called each quantum number, either nx, ny, nz or n, we multiply three times the same sum.

    I know that the degeneracy implies a symmetry on the system, and this is why I agree that there must be some other quantum numbers in the spherical box problem (as three in the cubic box), but I don't see it, because there is only a single quantum number since there is not dependency on the angular variables.

    Therefore my doubt is: why should I consider the spherical harmonics to take into account the degeneracy and the other quantum numbers (l and m) to reproduce the same partition function than that of the cubic box if I assume that there is not angular dependency on the wave function? Moreover, I saw the most general energy distribution for spherical coordinates (in Wikipedia the problem that I'm looking for is even solved with potential energy set to zero) and I think the result of the energy in the most general case in the spherical coordinates doesn't recover the result for the cubic box either. Unless the derivation is not straightforward and I think the result is different.
     
  6. Aug 5, 2009 #5

    kuruman

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    "However in my case I was assuming that the wave function (psi) had not any dependency on the angular variables (by hypothesis since the potential energy where the particle is placed is zero). Thus, I'd say (please, correct me if I'm wrong) that in the separation of variables, the angular functions would be constants..."

    The particle can still have angular momentum despite the fact that the potential is zero. This is the quantum analogue of a marble rolling around inside a hollow shell (no gravity). When you throw out the angular functions, you are essentially eliminating this possibility for the particle in the spherical box. You are throwing out the baby with the bath water as it were.
     
  7. Aug 5, 2009 #6
    Yes, I see what you mean.

    I'll try to solve the most general problem (a little longer, considering all the calculations with the spherical harmonics that need to be done to get the total energy of the free particle) and to compare it to that one of the particle inside the cubic box to try to obtain the same partition function (which was my goal).

    But I'm not sure whether in the moment to calculate the partition function corresponding to the translation energy, the contribution to the energy due to the angular momentum should be taken into account to compare it to the case of the free particle inside the cubic box. Therefore, I'm not sure what I'll get compared to the cubic box case, but in anycase I'll let you know what I get.

    Thank you very much.
     
  8. Aug 5, 2009 #7

    kuruman

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    The energy En is determined strictly by n. All states described by 3-D wavefunctions that have the same n will have the same energy regardless of what their angular wavefunction looks like. You need to find out how many these states you have for a particular n (the degeneracy number) and multiply the corresponding Exp[-β En] in your partition function by it.
     
  9. Aug 7, 2009 #8

    Avodyne

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    Sorry, no. In general (and in this case in particular), the energy eigenvalues depend on the total angular momentum quantum number [itex]\ell[/itex]. For certain very special cases (hydrogen atom, harmonic oscillator), there in some extra symmetry that removes the [itex]\ell[/itex] dependence from the energy eigenvalues.

    See post #14 in this thread https://www.physicsforums.com/showthread.php?t=325144 for how to separate variables for the analog of this problem in two dimensions.
     
  10. Aug 10, 2009 #9

    kuruman

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    Oops, I stand corrected. Avodyne's clarification is to the point.
     
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