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Solution to the 1D Free Schrodinger Equation

  1. Apr 3, 2014 #1
    So starting from the time dependent schrodinger equation I perform separation of variables and obtain a time and spatial part. The spatial part is in effect the time independent schrodinger equation.

    Since we are dealing with a free particle I can take the time independent equation, set V = 0 and solve.

    I can do this successfully to obtain :

    [itex]Ae^{+i\sqrt{{2mE}/{\hbar^{2}}}x}+Be^{-i\sqrt{{2mE}/{\hbar^{2}}}x}[/itex]

    My lecturer has a small section titled :

    So when I follow through and solve for the spatial part of his final solution I obtain :

    [itex]A'e^{+ip/\hbar}+B'e^{-ip/\hbar}[/itex]

    He seems to have conveniently ignored/left out one half of the above solution. Why ?

    I am also aware that [itex]e^{-{iEt}/{\hbar}}[/itex] is simply the solution the time dependent part of the equation so my only issue is why he has left out a certain bit.
     
  2. jcsd
  3. Apr 3, 2014 #2

    jtbell

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    Staff: Mentor

    I don't think anyone here can read his mind. Ask him. :smile:

    My guess is that he simply did not intend to be completely rigorous.
     
  4. Apr 4, 2014 #3
    He's just allowing ##p## to be either positive or negative. Since ##E## is proportional to ##p^2##, either positive or negative values will lead to the same value of energy.
     
  5. Apr 4, 2014 #4

    jtbell

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    Correct. In addition, for the same value of E, both positive and negative values of p can be present, in a superposition, so the two-term solution is a bit more general. We actually use this solution for the "particle in a box" where the boundary conditions lead to coefficients such that the two terms combine to give a (real) sine or cosine.

    Of course, that's not a free particle, but it's still V = 0.
     
  6. Apr 4, 2014 #5
    Right, but in a free particle with no boundary conditions, you are already allowed to have a superposition of different values of ##E## anyway, so the entire expression is already meant to be taken as only one term in the superposition. If that's the case, it's redundant to also go and write out the positive- and negative- momentum terms separately--you can just say "for any value of ##E##, and for any value of ##p## for which ##E = \frac{p^2}{2m}##.
     
  7. Apr 3, 2015 #6
    please somebody tell us how he cancelled out B exp(-ikx) in the spatial part and mention the necessary boundary condition

    thx
     
  8. Apr 4, 2015 #7

    vanhees71

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    He is simply not clear enough about what he wants to do! Obviously he is looking for the energy eigensolutions for a free particle, and then it's clear that you get for each non-zero spectral value of the Hamiltonian two linearly independent generalized eigenfunctions, namely the ones you found. This is called "degeneracy". The reason for a degeneracy is often (if not always) a symmetry of the problem. Here it's invariance under spatial reflections, i.e., the symmetry of the Hamiltonian under ##x \rightarrow -x##, ##p \rightarrow -p##. Thus, to completely specify the eigenstate you can choose parity as another observable, and this leads to clearly defined eigenfunctions (setting ##\hbar=1##):
    $$u^{(+)}_E=N \cos(\sqrt{2 m E} x), \quad u^{(-)}_E(x)=N \sin(\sqrt{2m E x}).$$
    Of course, also your professor's solution is correct, but it's incomplete, because for any ##E>0## there are two solutions, making up the complete (generalized) basis of the eigenspace ##\mathrm{Eig}(\hat{H},E)##.

    Another thing is the momentum eigenvalue problem. The momentum spectrum is not degenerate, and then your professor's solution is unique,
    ##u_p(x)=N \exp(\mathrm{i} p x).##
     
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