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Proving That Even Potential Leads to Even or Odd Wavefunction

  1. Sep 20, 2013 #1
    1. The problem statement, all variables and given/known data

    This problem comes from the Griffiths QM book and is stated as "show that if V(x) is an even function then the solution to the time-independent Schroedinger equation can be taken to be either even or odd."

    Now, I have seen the solution to this, but am not thoroughly convinced. Specifically, I do not see how the fact that psi(x) is a solution leads to psi(-x) to be a solution. It appears to be some variable manipulation, but I am wondering if there is simpler intuitive way of thinking about this.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 21, 2013 #2

    mfb

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    You can out psi(-x) in the Schrödinger equation, simplify and see that it is a solution if and only if psi(x) is a solution.

    Maybe more intuitive: the potential is symmetric, so a mirrored wavefunction of a possible state satisfies the equation, too.
     
  4. Sep 21, 2013 #3
    I convinced myself of that using the mathematics. Continuing the same problem, suppose I show that any function ψ can be written as a linear combination of both odd and even solutions. How does that lead to the ability to assume ψ to be odd or even? After all, if a function is a linear combination of both odd and even functions, we can conclude nothing about its parity.

    This is Griffiths Problem 2.1(c), if anyone is curious.
     
  5. Sep 21, 2013 #4

    mfb

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    Consider an arbitrary function psi(x) which is neither even nor odd. Let's assume this is a solution to the time-independent Schrödinger equation with a symmetric potential. Then psi(-x) is another solution, with the same energy.

    As the Schrödinger equation is linear, c(psi(x)+psi(-x)) and c'(psi(x)-psi(-x)) with some normalization factors c, c' are solutions as well. The first one is even, the second one is odd.
    This also allows to write psi(x) as superposition of those two functions.

    Therefore, all solutions can be written as superposition of odd and even eigenfunctions. It is sufficient to find the (anti)symmetric solutions to get all solutions.
     
  6. Sep 21, 2013 #5
    I believe I understand what you wrote. However, the problem states that "the psi(x) can be taken to be either even or odd." Clearly, if you wanted to make a linear superposition of (anti)symmetric wavefunctions as you wrote them and form psi(x), we need to use both odd and even components. I understand that if you take an arbitrary solution psi(x), then psi(-x) is also a solution due to V(x)=V(-x). Then, we make up (anti)symmetric forms and claim that we can get the original function back by superposing the former? Seems like this is much ado about nothing.

    It is possible that I am simply misunderstanding what the problem is asking to do (by no means is this a novelty with Griffiths books).
     
  7. Sep 21, 2013 #6
    Or is the idea simply that if we had some arbitrary solution psi(x) and an even potential, then, if needed, we can always make up a symmetric or antisymmetric solution out of that wavefunction? Then, that superposition is either even or odd and still remains a solution.
     
  8. Sep 22, 2013 #7

    mfb

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    Sure, but it is sufficient to consider the odd and even solutions. All other solutions (if they exist at all) follow from that. That makes the analysis easier.
     
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