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Separation of variables to solve Schrodinger equation

  1. Feb 12, 2015 #1
    How do we know that separable solutions of Schrodinger equation (in 3d) form a complete basis? I understand that the SE is a linear PDE and therefore every linear combination of the separable solutions will also be a solution , but how do we know that the converse, i.e 'every solution can be written as a linear combination of separable solutions', is true? If we can separate out the variables to get ordinary differential equations, can we take it for granted that the solutions we'll get will be complete?
    ( I guess my question is not limited to the schrodinger equation alone. I have seen this done to laplace equation also in Electrostatics and I got the same doubt)
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  3. Feb 12, 2015 #2


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    The answer is, we do not know it. We must be told that there is some symmetry that forces it to be true. For example, if the system is spherically symmetric or rotationally symmetric, then we can separate out the appropriate angle. Or if the system is invariant in time then we can separate out time.
  4. Feb 12, 2015 #3
    That doesn't seem right to me. I am not talking about the case where we can eliminate variables using symmetry.
    Consider the case of the 3d rectangular potential well. We find the wave function by assuming it to be a product of three separate functions X(x), Y(y) and Z(z) depending on x, y and z respectively. Plugging it into the schrodinger equation gives us three ordinary differential equations that X, Y and Z satisfy. solving these equations, we find the functions X,Y,Z which turn out to be sinusoidal functions wth arguments that contain nx, ny, nz that can take any positive integer value.
    Then we write the general wave function as Ψ= ∑X(x)Y(y)Z(z) over all values of nx, ny, nz
    This is surely 'a' solution, but how do we know that this is THE GENERAL solution.
    Last edited: Feb 12, 2015
  5. Feb 12, 2015 #4


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    We do not know that this is the general solution. It is not the general solution. There is no necessity that it is even a solution at all.

    In your example we were told that it is a "3d potential well." And then you skipped over a bunch of things where the characteristics of this potential well were omitted. Things like, the well is presumably rectangular. Meaning it has very particular symmetry. And symmetry means that certain things about the wave function must be true.

    Now, since you omitted the characteristics of the well, what if it was assumed it was spherically symmetric instead? Would splitting the SE into x, y, and z factors make sense then? Probably not. What about r, theta, and phi? Yes it would.

    What if it was a general 3d potential well with arbitrary shaped walls. Maybe the particle is stuck inside a lumpy pyramid or a half deflated crumpled soccer ball. Could we separate the variables then? No, we could not.

    Symmetry. It is one of the more powerful tools in our toolbox, particularly in quantum mechanics.
  6. Feb 12, 2015 #5
    OK. I meant rectangular potential well. Sorry about that.
    I agree that you cannot always separate out variables. Sometimes you can. For a rectangular well, we can if we assume the wave function, like i said in the original post, to be product of 3 functions.
    I meant that in cases where we CAN separate them, after solving the resulting Ordinary differential eqns , we find the individual functions and multiply the together to get an infinite family of functions. Let us say that we are able to find some linear combination of them that satisfies the boundary conditions.Do you claim that this solutions NEED NOT be the general solution, i.e other solutions may exist to the same problem?
    This is causing me some confusion, as every textbook I've read has conveniently left this out. Not just QM, but even in Electrodynamics the solutions to laplace equations subject to different boundary conditions have been done in the same way. I will quote a statement from electrodynamics by Griffiths on separation of variables to solve laplace equation:
    We look for solutions that are products of functions, each of which depends only on one coordinate... On the face of it, this is an absurd restriction- the overwhelming majority of solutions to Laplace's equation do not have such a form.... obviously, we're only going to get a tiny subset of all possible solutions, and it would be a miracle if one of them happened to fit the boundary conditions of our problem... But hang on, because the solutions we do get are very special, and it turns out that by pasting them together, we can construct the general solution.
    I don't know what to make of this.
    (i'm sorry for writing everything as sentences. I know i could explain better if I could write some equations, but i am just starting to learn LATEX and it will take days for me to write the code)
  7. Feb 12, 2015 #6


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    I think you're confusing two different issues.
    1. Is there a energy eigenstate of the form: [itex]\psi = X(x)Y(y)Z(z)[/itex]?
    2. Can every square-integrable wave function be written in the form: [itex]\sum_{i,j,k} C_{i,j,k} X_i(x) Y_j(y) Z_k(z)[/itex]?
    Unless there is rectangular symmetry, the answer to 1. is "no", but the answer to 2. is still "yes".

    Completeness is about the second issue, and whether the potential is spherically symmetric doesn't affect the answer to question number 2 (as far as I know).
  8. Feb 12, 2015 #7
    <<Moderator note: Reply to deleted post removed.>>

    Thank you for taking the time to understand my question. The question 2 that you posted was exactly my point.
    Last edited by a moderator: Feb 13, 2015
  9. Feb 13, 2015 #8


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    The completeness of the solutions goes back to the Sturm-Liouville theorem, which states that a Sturm-Liouville operator (with appropriate boundary conditions) has a complete set of eigenfunctions. The reasoning goes something like this (assuming your operator is minus Laplace):

    1. Expand the Laplace operator in your coordinates.
    2. If in one of the coordinates you have a SL operator, expand the solution in its eigenfunctions. The coefficients of this expansion will generally be functions of the remaining coordinates.
    3. Insert your expansion in the original PDE.
    4. Repeat step 2 and 3 until you have separated everything and have a solution on the form of a series expansion containing products of functions of the individual coordinates.

    Now, it is not certain that this will work out in any coordinate system. The assumption is essentially that your differential operator is separable in your chosen coordinates, the boundaries of your domain are coordinate surfaces, and that the separation gives you SL operators. For example, the Laplace operator does not allow separation in an arbitrary coordinate system.
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