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Time-independent wave function formula

  • Thread starter droedujay
  • Start date
12
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1. Homework Statement

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


2. Homework 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?
 
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Answers and Replies

183
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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."
 
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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.
 
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I know about psi star psi being the probability of finding a particle in a specific region
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.
 
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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.
 
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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?
 
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I think that it has something to do with the fact that this wavefunction is orthonormal.
 
183
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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.
 
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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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