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Spherically Symmetric Potentials

  1. Jul 2, 2004 #1


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    I have a question concerning the stationary states of a spherically symmetric potential (V=V(r), no angular dependence)
    By seperation of variables the eigenfunctions of the angular part of the Shrödinger equation are the spherical harmonics.
    However, (apart from Y^0_0) these are not spherically symmetrical.
    for example: Y^0_1=(3/(4Pi))^{1/2}cos(theta)

    So the probability of finding the particle in the xy-plane (so theta=1/2 Pi) is zero (regardless of r).
    But why would the probability of finding the particle at in the xy-plane differ from any plane through the origin? (which is due only because of a particular choice of the coordanation axes) How can nature prefer a specific angular direction when the potential is depends only one the distance from the origin?
    Isn`t this in violation with the isotropy of space??

    BTW: I know LateX, but how do you use it in this posting on this forum?
  2. jcsd
  3. Jul 2, 2004 #2


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    Your forgetting about the kinetic term in the Hamiltonian. Only the ground state has the spherically symmetric Y00(θ,φ) angular solution. The higher order spherical harmonics correspond to nontrivial angular momentum. It's kind of like asking why you never see the Earth's moon directly above the north pole: because the moon's angular momentum is (in the Newtonian/Keplerian scheme) conserved, so it must stay in its orbital plane. The spherical harmonics are solutions, but a particular spherical harmonic is not necessarily the solution.

    I don't know how to use the latex very well, so I just stick with html. It's a shame; the latex is so much more attractive.
  4. Jul 3, 2004 #3
    You can see how a latex equation was typed by clicking on the equation. A separate window opens. I downloaded a program called TexAide that allows you to compose the equation by drag and drop from a palette. Then you cut and paste the produced latex into your post. I like that better than typing.
    Last edited: Jul 3, 2004
  5. Jul 3, 2004 #4
    With respect to spherical harmonics, I think Turin gave you a good explanation.
    These harmonics would correspond to what in chemistry they call orbitals, and the chemistry books have pictures of them. It is an interesting question if these orbitals would have a definite orientation in space when "nobody is looking". Maybe they get into a superposition of orientations, but as soon as the atom starts interacting with its surroundings they take a definite orientation. Probably they also take some more defined orientation when the atom is placed in a magnetic field. All this is my speculation.
  6. Jul 3, 2004 #5
    I forgot that you said you know latex. In that case, to type the latex in this forum, is to type "[ tex ]" at the beginning, then type your latex code, and close it with the tag "[ / tex ]" (no quotes and no spaces within the tags).
    All the tags that are enclosed in "[ ]" are a language used in this forum called vbcode. So these vbcode tags tell the forum's engine on the server that what is between them is latex. The engine then reads the latex and produces an image for the eguation that is what you see when you read the post. I guess one of the differences with html is that html is rendered by the browser, while latex is rendered at the server and sent to the user as image.
  7. Jul 3, 2004 #6


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    This is an interesting point, often raised.
    The question can be formulated in a more abstract way in the following form: If the original problem is symmetrical under a group G (here, rotations in space), then how come that the solution is not symmetrical under that group ?

    The issue is general (in classical physics as well as in modern physics) but is more obvious in quantum theory ; but you understand that the same "paradox" appears also in classical electrodynamics for example.

    The apparent "paradox" comes from a misunderstanding of what is the "solution". Indeed, the problem has, as you say yourself, several solutions (the spherical harmonics). The vector space SPANNED BY THESE spherical harmonics is the true solution, and that SOLUTION SPACE does satisfy the symmetry of the problem, meaning that if S is an element of the solution set, then so is T = g S where g is an element of G.
    But if YOU pick out a particular solution, like Y(1,0), then YOU have BROKEN THE SYMMETRY, just by picking out a particular solution, where you have chosen the axes x,y and z just to define what you mean by Y(1,0).
    If you don't know this, then if you rotate Y(1,0), then you will always find that you can write the result as a linear combination of Y(1,-1), Y(1,0) and Y(1,1) ; in fact the coefficients vary as the coefficients of a vector in 3-space. This is because the set of spherical harmonics forms representations of the rotation group, and is in fact the underlying reason why you found them as a solution with spherical symmetry in the first place.

    So to summarize:
    a problem that is symmetrical under a group G will produce a solution SET that is invariant under the group G, meaning that elements of the solution set are transformed into (other or the same) elements of the solution set under each of the symmetry operations in G. In group theory people say that such a solution set is a REPRESENTATION of the group G.
    If that solution set only contains one single element, then of course the solution will have to be symmetrical itself (that's the case of Y(0,0) !).

  8. Jul 3, 2004 #7
    RE: "Only the ground state has the spherically symmetric Y00(θ,φ) angular solution. "

    Why would that be true? The hydrogen atom is spherically symmetric, yet there are an infinite number of states having the spherically symmetric Y00 angular solution (for example, Rydberg states).
  9. Jul 4, 2004 #8


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    Thanks for the answers everyone. It clears some of the questions I had.
    Actually I didn`t, but this term is spherically symmetrical as well. Never mind though.

    Thanks alexepascual for the info!

    Thanks for the clear explanation. I don`t know applied group theory, but I do have knowledge of groups from algebra so I understand what you mean.
    (I WILL follow lectures om group theory next year though).

    However this still doesn't clear the physical problem I see in it.
    Suppose an electron is in a stationary state of hydrogen, then we can label the state by three quantum numbers (n,l and m, or |n l m> ).
    So if we say the electron is in the state |1 1 0>, we know that there's a plane where the particle will never be found, but we don`t know which plane, because we haven`t specified our axes. So if we have a hundred (identically prepared and seperate) electrons all in the state |1 1 0>, then they might all have a different wavefunctions (but they can be transformed into one another by a rotation), but if they have different wavefunctions, they are not in the same state. Even if we know the election is in the state |1 1 0> we dont know how that plane is oriented in space.
    From a mathematical point of view I have no objections, but I don`t understand the physical implication of this.
  10. Jul 4, 2004 #9
    due to the spherical symmetry of the situation, the choice of the z-axis is completely arbitrary. In the absence of any other information, the state |1 1 0> is to be interpreted as follows: the electron is in the first excited state, it has one unit of net angular momentum, AND there exists an axis z such that the angular momentum is in an eigenstate of that axis' L operator with eigenvalue zero. I am quite certain that such an axis of definite quantization will always exist (I can show it in the case of electron spin); it is certainly unique if it does.

    However, if we make some some effort to prepare some electrons in the |1 1 0> state, say via Zeeman splitting in an external magnetic field, then it will be us who have made the choice of the z axis and somehow selected those electrons that qualify. More prosaically, since we have selected one direction in space for special treatment, this amounts to making a measurement of the angular momentum along that direction, and each wavefunction will collapse to an eigenstate of that axis' L operator.
  11. Jul 4, 2004 #10


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    That's where things go wrong: if you specify the state |1 1 0>, you HAVE to choose a preferred direction, which is induced by the way you prepared (or selected) your atoms. If you insist on not doing this, then you CANNOT prepare them in a pure state which has l different from 0, but into a statistical mixture, which would include equal contributions from |1,1,-1>, |1,1,0> and |1,1,1>.

    Let's compare to a classical problem: the gravitational potential around the sun is (to a high degree of accuracy) spherically symmetrical. This means that the solution SET of orbits has to obey spherical symmetry: indeed, for each possible orbit that is a solution, if you rotate that orbit somehow around an axis through the sun, you find another orbit that is also a solution ; that's what I wanted to point out in my previous post. But now you say: How come that a planet in a particular orbit, say with major axis one astronomical unit, is always found in a certain plane, and not outside of it ?
    So if you insist on a spherically symmetrical situation of planets around a sun, you have to have a statistical mixture ; not one single pure state (orbit) will do, except for the trivial one which is a planet that is at rest in the center of the sun. In fact, QM allows for more than just one trivial solution: all the l=0 solutions will do.

  12. Jul 4, 2004 #11


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    I'm sorry; I said "onlyground state," which was not correct. However:

    The hydrogen atom is not (necessarily) spherically symmetric; it is the coulomb potential in the Born approximation generated by the proton in the hydrogen nucleas that is spherically symmetric. I do not know what are "Rydberg states." If these are states that have a constant angular dependence (arbitrary energy but zero angular momentum), then they are states that have an angular dependence of Y00(θ,φ). There are other states in which the hydrogen atom can be. These other states with an angular dependence of Yl0(θ,φ), where l /= 0, are not spherically symmetric, yet they are not ruled out by the spherically symmetric Hamiltonian.

    To put it in a different perspective, consider the stationary state of the electron in the hydrogen atom expressed in terms of three quantum numbers: ψ = |n,l,m>. The first number in this list gives the energy or the state. The other two numbers give the (orbital) angular momentum (magnitude and direction) of the state. While the spherical symmetry does not depend on the value of n, it does require that l = m = 0. Notice that, what is spherically symmetric is an energy, V(r) in the Hamiltonian, and that the state is spherically symmetric regardless of its energy, but the angular momentum, that does not show up in the Hamiltonian as spherically symmetric, is what can make the state not symmetric. The symmetry is in the energy (the Hamiltonian), but the finer details of the state are not necessarily symmetric.
    Last edited: Jul 4, 2004
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