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Invariance of Schrödinger's equation

  1. Aug 10, 2009 #1
    I thought i had a basic to intermediate understanding of quantum physics and group theory, but when reading hamermesh's "group theory and it's application to physical problems" there's something in the introduction i don't understand.

    first of all, i know the parity (or space inversion) operator and it's eigenfunctions. so from this point of view the example in the introduction is quit easy but i don't get hamermesh's argumentation.

    he start's with the Schrödinger equation in one dimension:
    "[tex]u''+[\lambda - V(x)]u = 0[/tex]
    where [tex]\lambda[/tex] is the eigenvalue of u.
    one dimension => necessarily not degenerate." why?

    "We assume that the potential is an even function of x. ([tex]V(x)=V(-x)[/tex])"

    "replacing x by -x, we see, that if u(x) is a solution, so is u(-x)."

    what does he mean? is this a variable substitution [tex]x \rightarrow -x[/tex]
    [tex]u''(-x)+[\lambda - V(-x)]u(-x) = 0 = u''(-x)+[\lambda - V(x)]u(-x)[/tex] ???
    is such a substitution allowed?

    sorry if the answer to my question is obvious but i don't get it at the moment.

    thanks and greetings.
  2. jcsd
  3. Aug 10, 2009 #2


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    Greetings tommy01! :smile:

    (have a lambda: λ :wink:)

    Don't use x -> -x, it's too confusing.

    Use x -> y, where y = -x.

    Then d2u/dx2 = d2u/dy2,

    so d2u/dx2 + (λ - V(x))u

    = d2u/dy2 + (λ - V(-y))u :wink:
  4. Aug 10, 2009 #3
    For understanding why the states ara not degenerate I suggest Volume III of L.D. Landau and Lifgarbagez chapter III search for "general properties of motion in one dimmension"
  5. Aug 11, 2009 #4
    hi all.

    thanks for your quick answer.
    i'm going to consult landau and lifgarbagez.
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