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What happens to the wave function after an operator transformation?

  1. Sep 6, 2009 #1
    for example, if the hamiltonian of a system is transformed this way:
    H(x) --> H(x+a)
    i understand that the tranformation can be represented by a unitary operator U=exp(iap/[tex]\hbar[/tex])

    but what happens to the wave function? how is it transformed?
  2. jcsd
  3. Sep 6, 2009 #2
    If the operator is transformed, then the wavefunction or state is left alone.
  4. Sep 6, 2009 #3
    what i mean to ask is how does the new wave function look like.
    the harmonic oscillator for example, H=p[tex]^{2}[/tex]/2m+m[tex]\omega^{2}x^{2}[/tex]/2
    if H(x) --> H(x+a) , then the eigenstates represented in the position basis must go through some sort of transformation as well.
    how do i represent that transformation?
    Last edited: Sep 6, 2009
  5. Sep 6, 2009 #4


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    RedX answered that question. There are two formalisms in quantum mechanics. You can think of a transformation as affecting the wave function only, not the operator, or as affecting the operator only, not the wave function. since you are transforming the Hamiltonian, you are using the second formalism. The wavefunction is not changed in any way. Of course, the eigenvalues and eigenvectors will change because now they are eigenvalues and eigenvectors of this new operator.
  6. Sep 6, 2009 #5
    thanks for the response, i think i partially understand my confusion now.

    had another question, but just figured it out, so thanks again.
    Last edited: Sep 6, 2009
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