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Proving an eigenstate

  1. Oct 1, 2012 #1
    show that e-ikx is an eigenstate energy.

    Do I start by multiplying the hamiltonian operator by ψ(x)?

    So far I have ψ(x)(1/2m)(-i[STRIKE]h[/STRIKE]d/dx)2=e-ikx
     
  2. jcsd
  3. Oct 1, 2012 #2
    The Hamiltonian usually goes on the other side of [itex]\psi[/itex]
    Do you know schrodingers equation?
     
  4. Oct 1, 2012 #3
    -[STRIKE]h[/STRIKE]2/2m d2/dx2 eikx-iωt right?

    To prove the eigenstate, I could just have Hψ(x)=hψ(x) and break H down into it's smaller parts, right? I thought I took good enough notes on eigenvalues... Tyvm!
     
  5. Oct 1, 2012 #4
    Where H is the Hamiltonian operator
     
  6. Oct 1, 2012 #5
    There's a right hand side to schrodingers equation too :P
     
  7. Oct 2, 2012 #6

    vanhees71

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    It has not so much to do with the Schrödinger equation, but with the eigenvalue problem for the Hamiltonian [itex]\hat{H}[/itex]. In position space the Hamilton operator for a free particle is given by the differential operator
    [tex]\hat{H}=-\frac{\hbar^2}{2m} \Delta.[/tex]
    It's indeed easy to schow that the plane wave
    [tex]u_{\vec{k}}(\vec{x})=N \exp(\mathrm{i} \vec{k} \cdot \vec{x})[/tex]
    is a generalized eigenfunction of [itex]\hat{H}[/itex] (with [itex]N[/itex] an arbitrary constant). Just take the derivatives and check that it fulfills the eigenvalue equation
    [tex]\hat{H} u_{\vec{k}}(\vec{x})=E_{\vec{k}} u_{\vec{k}}(\vec{x}).[/tex]
    You'll easily find the energy eigenvalue.

    Also think about, whether this function can ever represent a state of the particle in the sense of quantum theory. To help a bit: The answer is a clear no!

    Sometimes the eigenvalue equation for the Hamilton operator is called "the time-independent Schrödinger equation". Indeed, the relation with the Schrödinger equation,
    [tex]\mathrm{i} \hbar \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x}),[/tex]
    is that the function
    [itex]\psi_{\vec{k}}(t,\vec{x})=u_{\vec{k}}(\vec{x}) \exp \left (-\mathrm{i} \frac{t E_{\vec{k}}}{\hbar} \right)[/itex]
    is a solution. The eigenfunctions of the Hamilton operator represent the "stationary solutions", because it is constant in time up to the phase factor [itex]\exp(-\mathrm{i} t E_{\vec{k}}/\hbar)[/itex].
     
  8. Oct 2, 2012 #7
    Thanks guys, I got help from my classmates and they walked me through the solution. I wish quantum mechanics (or what I've been exposed to it thus far in my Modern Physics course) was less "symbolic" and easier to apply. As an engineering student, I always seek ways to apply my knowledge, and this really messes with me haha.
     
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