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Can the wave function be a constant?

  1. Dec 1, 2009 #1
    I'm wondering if the wave function can be a constant in some special cases?

    Now I understand that if we have a one dimensional wave function describing the location of a particle (say, along the x-axis), then the wave function can not be a constant. If it was, then it wouldn't be normalizable.

    But what if the particle is free to move in a circle? The we have [tex]\psi(\theta) = \psi(\theta + 2 * \pi)[/tex]
  2. jcsd
  3. Dec 1, 2009 #2
    Yes exactly. The particle on a ring has a constant wavefunction for the k=0 momentum eigenstate. The s-orbital in an atom is a bit like a constant wave function too, except it decays with radius. Similarly, the wavefunction for the particle on a ring isn't constant if you take into account the space outside the ring.
  4. Dec 1, 2009 #3
    And it's the bondary conditions that allow n to be 0? I suppose so since it were the boundary conditions that dissaloved n to be 0 in the case of the infinite square well. Thanks for your answer :)

  5. Dec 1, 2009 #4


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    If the wave function is constant in position space, then it's a delta function in momentum space. I don't see how that could be a valid solution to the S.E.
  6. Dec 1, 2009 #5
    alxm, you have a slight confusion I believe.

    'Can the wavefunction be constant?' is not the same as whether it solves the time-independent SE.

    A valid wavefunction in the position representation is one which is continuous.
    If we are going to have valid wave function which is constant, it has to exist in a space with boundary conditions which allow it to be constant: so it has to exist in a periodic (compact) space.

    For a general Hamiltonian on such a space, such a wave function will evolve in time according to the TDSE and no longer be constant in space.

    For the special case of a free particle hamiltonian on a periodic space, H = p^2/2m, the constant wavefunction will also be an eigenstate with 0 energy, so it will also solve the TISE.
    Last edited: Dec 1, 2009
  7. Dec 1, 2009 #6
    If there were no constant wavefunctions, can there be stationary states?
  8. Dec 1, 2009 #7


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  9. Dec 1, 2009 #8
    I think the original question was concerned with 'constant in space' wave functions. In fact any momentum mode for a free particle on a ring has a probability density which is constant in space, but only the k=0 mode has no spatial dependence for the wavefunction.

    Constancy in time only makes sense for the probability density, since an arbitrary shift of the definition of energy can alter the time dependence of all states.

    Eigenstates of the hamiltonian are 'stationary' in the sense that the expectation value of any observable is time independent. It is certainly not the case that eigenstates of the hamiltonian have no time dependence.
  10. Dec 1, 2009 #9
    mmmm ..I think you're right, but are you suggesting that the wavefunction that describes hyrdogen electron in its ground state keep changing?
    Also i realized that my comment is irrelevant to the original post.
  11. Dec 1, 2009 #10


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    Well I didn't really say that, although I'm still not sure what you were getting at with the s-orbital analogy. (which certainly is a solution to the TISE, and does not have an definite momentum) So I got a bit confused on whether you were talking about a spherically symmetric system, or what.

    I see what you're talking about now. Yes, if you've got a 1-dimensional 'particle in a box' where you give it periodic boundary conditions, then the plane wave solutions [tex]\psi_n = \frac{1}{\sqrt{2\pi}}e^{inx}[/tex] are normalizable and valid, and the constant wave function [tex]\psi_0 = \frac{1}{\sqrt{2\pi}}[/tex] is a solution, with definite energy and momentum. (zero and zero in that case)
  12. Dec 1, 2009 #11
    What I'm getting at with the s-orbital analogy is that there are a number of spaces on which the 'constant function' is a solution to laplace's equation: one such space is the sphere; another is a torus; another is a line with periodic boundary conditions. All these spaces are compact, which allows the constant function to be normalised to unity.

    The free particle on a sphere problem (rigid rotator) has energy eigenstates which are also eigenstates of angular momentum. The l=0 spherical harmonic is a constant function.

    The free particle on a torus has energy eigenstates which are also eigenstates of momentum. The kx=ky=0 state is a constant function.

    The other point I was getting at is that the constant function is a valid state for single particle QM on such spaces, even if it is not an eigenstate of the Hamiltonian. In this case, if you prepare the system such that the particle wave function is constant, in a few ns, the wavefunction will have evolved according to the TISE and will no longer be constant.
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