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Homework Help: Time-independent wave function formula

  1. Mar 30, 2008 #1
    1. The problem statement, all variables and given/known data

    Construct wavefunction with given energies and probabilities of obtaining energies in a 1-D box from 0 to a

    2. Relevant equations

    3. The attempt at a solution

    I know the general form of a time-independent wavefunction but I don't know what to do with the probabilities of obtaining energies. Is there a formula for this?
    Last edited: Mar 30, 2008
  2. jcsd
  3. Mar 30, 2008 #2
    There are lots of places you can find detailed explanations of this problem - it is one of the most common quantum models out there. Try doing a Google search for "infinite square well."
  4. Mar 30, 2008 #3
    I looked but I could not find anything on probabilities of obtaining Energies. I know about psi star psi being the probability of finding a particle in a specific region but I do not have any material on probability of obtaining Energies.
  5. Mar 30, 2008 #4
    My quantum is a little shaky, but I think [tex]<\psi_{n}^{*}|\psi_{n}>[/tex] is indeed the position operator, which is also written [tex]|\psi_{n}(x)|^{2}[/tex]. You need to normalize your psi functions so that each [tex]\psi_{n}(x)[/tex] has the correct coefficient: since the particle has to be somewhere in the box, you know that the integral over the whole region must be 1. To find the probability of a certain state, you put the Hamiltonian operator into the Bra-Ket: [tex]<\psi_{n}^{*}|\hat{H}|\psi_{n}>[/tex] So once you have the correct coefficients for your set of [tex]\psi_{n}(x)[/tex] functions, you can do [tex]<\psi_{n}^{*}|\hat{H}|\psi_{n}>[/tex] for each one to find the probability of the particle being in that state, with that energy.
    Last edited: Mar 31, 2008
  6. Mar 31, 2008 #5
    I found out the Coefficient expansion theorem and constructed the following wavefunction:

    Ψ(x,0) = 1/sqrt(2)*φ1 + sqrt(2/5)*φ3 + 1/sqrt(10)*φ5
    where φn = sqrt(2/a)*sin(n*pi*x/a)

    Is this unique why or why not? I'm thinking that it has something to do with all odd Energies.
  7. Mar 31, 2008 #6
    Beyond what I wrote before I am out of my depth, but I'll try. I believe the energies are odd because the well runs from 0 to a and not -a/2 to a/2; in the latter case I think you would have even energies or cosines. As to the uniqueness question, I would guess that the wavefunction itself is a unique solution to your particular square well, but it is of course a linear combination of eigenfunctions so I don't know how that affects uniqueness. I would imagine performing the Hamiltonian on completely different wavefunctions could return the same energies, so the energy values would not necessarily be unique. Is that what you mean?
  8. Apr 1, 2008 #7
    I think that it has something to do with the fact that this wavefunction is orthonormal.
  9. Apr 1, 2008 #8
    I believe the orthonormal bit refers to the fact that each eigenfunction that makes up your wavefunction (i.e. [tex]\psi_{1} , \psi_{3} , \psi_{5} , [/tex] etc.) is linearly independent of each of the others, and therefore said to be orthonormal, in the same way that the X, Y and Z-axes are independent and orthonormal, for example.
  10. Apr 1, 2008 #9
    I appreciate all the help on this problem. I think I got this one down. Can you check out my other forum "QM wavefunctions" and see if you could help out there too.
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