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Schrodinger Equation and seperation of variables

  1. Jul 2, 2012 #1

    In Griffiths QM, it is claimed that to solve the Schrodinger Equation, we take the solution wavefunction [itex]\Psi(x,t)[/itex] to be of a seperable form [itex]\psi(x)\phi(t)[/itex].

    He then says that a superposition of these seperable forms can always give us the general solution. Can someone help me prove that statement? How do I know that every solution [itex]\Psi(x,t)[/itex] is a linear combination of solutions of the form [itex]\psi(x)\phi(t)[/itex]

    Thank you!
  2. jcsd
  3. Jul 2, 2012 #2


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    This is not mainly caused by the fact that these are solutions to the differential equation. It is just that, in general, you can write any suitably well-behaved function F(x,y) as a linear combination:
    [tex]F(x,y) = \sum_{i,j} c_{ij} f_i(x) g_j(y)[/tex]
    where c_ij are the coefficients and f_i/g_i are functions spanning the complete x/y function space, respectively. The solution functions of the separated problems give you such sets normally.

    The separation ansatz just gives you a set of such functions which is likely to be sensible in some regard (i.e., to not require overly many such product functions to approximate a real solution to a desired degree). This works best if the degrees of freedom you are separating into are only loosely coupled---or actually independent. And of course, any linear combination of solutions of a linear differential equation will also be a solution to the differential equation.
  4. Jul 2, 2012 #3


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    The separable form is a convenience only, textbook example. One can offer nonseparable solutions for various potentials. As for the superposition, i don't think it's true.
  5. Jul 2, 2012 #4


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    "Superposition" of solutions to get a general solution is true for any linear differential equation.
  6. Jul 3, 2012 #5
    So is there a proof for this result? That the general solution of any linear DE is a superposition of the separable solutions. Thank you!
  7. Jul 3, 2012 #6


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    You're postulating too strong of a result.

    To start, we must have results on existence and uniqueness of solutions to the initial or boundary value problem for a given differential equation. If the coefficients of the PDE are analytic functions (admit a locally-convergent power series expansion), then a major result is the Cauchy–Kowalevski theorem. This result establishes the existence and uniqueness of an analytic solution to the initial value problem.

    I'm no expert on the subject, but the online notes at http://www.math.ucdavis.edu/~hunter/pdes/pdes.html seem to be a nice treatment. The heat and Schrodinger equation (at least in one space dimension) are discussed in Ch. 5.

    Now, certain PDEs admit separable solutions. If we can establish a separable solution to an initial value problem that satisfies the restrictions of the Cauchy–Kowalevski theorem, then we can conclude that it is the unique solution.

    Next, we can ask what PDEs will admit separable solutions. The time dependent Schrodinger equation

    $$ i \partial_t \Psi(\mathbf{x},t)= \hat{H} \Psi(\mathbf{x},t)$$

    $$ \hat{H} = - \Delta + V(\mathbf{x},t)$$

    separates in space and time when the potential is independent of time, ##V(\mathbf{x},t)= V(\mathbf{x})##. Similarly, the time-independent Schrodinger equation

    $$ \hat{H} \psi(\mathbf{x})= E \psi(\mathbf{x})$$

    will admit separable solutions whenever there is some symmetry of the potential that would cause

    $$\frac{d V(\mathbf{x})}{du} =0,~~~ u = f(\mathbf{x}).$$

    This is a complicated way of saying that there is a change of variables such that the potential function is independent of at least some of the variables. In principle, there are even time-dependent potentials that are separable after a change of variables that mix space and time coordinates.
  8. Jul 3, 2012 #7


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    The key term is Sturm-Liouville. A Sturm-Liouville differential equation can be cast in terms of a Hermitian differential operator, and its solutions written in terms of functions that are eigenvectors of this operator. One then can bring to bear the full machinery of linear operators in a Hilbert space to write the general solution as a linear combination of the eigensolutions.
  9. Jul 4, 2012 #8
    Excuse me if I am wrong but if we separate variables like ψ(x)[itex]\phi[/itex](t), then we would be eliminating any possibility of motion for the particle since [itex]\phi[/itex](to), for any to, would affect equal to all the points in the space and the integral of |ψ(x)|2 for all the space must be equal to zero, so there would not be flux of probability.
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