1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Two particles in a potential (wave equation and harmonic oscillators)

  1. Oct 6, 2009 #1
    The problem statement, all variables and given/known data

    Please bear with me, I'm not that good with LaTeX.

    Consider the harmonic oscillator problem. Define [tex]\Phi[/tex]n(x) as the n-th wave function for one particle, with coordinate x and energy (n+1/2) [tex]\overline{h}[/tex][tex]\omega[/tex], where n=0, 1,… Now, let’s consider a system consisting of two particles which have the same mass; each particle experiences the same potential energy function and therefore each has the same angular frequency [tex]\omega[/tex]. Consider a situation where the total energy of this two-particle system is Etotal = 3 [tex]\overline{h}[/tex][tex]\omega[/tex]. We write the wave functions [tex]\Psi[/tex](xA,xB) for the system of two particles, where xA and xB are the positions of the two particles.
    You will be asked below whether the following state is (or is not) allowed; it is a product of two n=1 wave functions (one for each particle):
    [tex]\Psi[/tex]maybe(xA,xB) =[tex]\Phi[/tex]1(xA)[tex]\Phi[/tex]1(xB)

    (a) Suppose first that the particles are of different species, called A and B. Is the state [tex]\Psi[/tex]maybe(xA,xB) an allowed state of the two-particle system, or not?

    (b) For this “different species case”, what is the degeneracy g of the 2-particle system with total energy Etotal=3 [tex]\overline{h}[/tex][tex]\omega[/tex]?

    (c) Now, consider instead the case when both particles are identical bosons. Is the state labeled [tex]\Psi[/tex]maybe(xA,xB) possible, or not?



    2. Relevant equations

    From what I know, the time independent schrodinger wave equation.

    H[tex]\Psi[/tex] = (-h/2m)([tex]\partial[/tex]2/[tex]\partial[/tex]x2) + V[tex]\Psi[/tex] = E[tex]\Psi[/tex]

    Right Moving particle solution
    [tex]\Psi[/tex](x) = Aexp(ipx)/[tex]\overline{h}[/tex])

    E = p2/2m + C


    3. The attempt at a solution

    I am pretty confused with what this problem is asking, so I don't have an attempt at a solution to offer. I figured that if you have [tex]\Psi[/tex]maybe(xa,xb) could be looked at as:
    [tex]\Psi[/tex](xa) * [tex]\Psi[/tex](xb)

    Unfortunately from here, I am really not sure where to go on. I've been reading this problem for a week with no such luck.
     
  2. jcsd
  3. Oct 6, 2009 #2

    MathematicalPhysicist

    User Avatar
    Gold Member

    For b), you should use the fact that E=3hw=(n+1/2)hw+(m+1/2)hw=(m+n+1)hw, so if I am not mistaken degenrate state is one which for two different eigenvalues we have the same energy, i.e check for values of m and n which satisfy n+m+1=3 and partition the ways to choose them, for example, n=0 and m=2 and n=2 and m=0 are two non degenrate states, the degenrate state is n=m=1, and thus g=?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Two particles in a potential (wave equation and harmonic oscillators)
Loading...