Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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 :


    My lecturer has a small section titled :

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


    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


    User Avatar

    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


    User Avatar

    Staff: Mentor

    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

  8. Apr 4, 2015 #7


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    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).##
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook