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What's the wave function for a two-particle system given distance-dependent potential

  1. Jul 13, 2007 #1
    1. The problem statement, all variables and given/known data
    I have a problem in which I have a two-atomic molecule, and I'm supposed to find the energy and wave function in the ground state, given the particles' masses [tex]m_1,m_2[/tex] and the potential [tex]V(r)=kr^2[/tex], where [tex]r[/tex] is the distance between the particles.

    I don't necessarily need this problem solved (you can change V to whatever you want). I just want to see what a solution looks like.

    2. Relevant equations
    The time-independent schrödinger equation for a two-particle system is
    [tex]-\frac{\hbar^2}{2m}\nabla_1^2\psi-\frac{\hbar^2}{2m}\nabla_2^2\psi+V\psi=E\psi.[/tex]

    If the potential is dependent only on [tex]\mathbf r=\mathbf r_1-\mathbf r_2[/tex], then the above equation can be separated into the variables [tex]\mathbf r[/tex] and [tex]\mathbf R=(m_1\mathbf r_1+m_2\mathbf r_2)/(m_1+m_2)[/tex] so that
    [tex]-\frac{\hbar^2}{2(m_1+m_2)}\nabla^2\psi_R=E_R\psi_R[/tex]
    [tex]-\frac{\hbar^2}{2\mu}\nabla^2\psi_r+V\psi_r=E_r\psi_r[/tex], [tex]\mu[/tex] is the reduced mass
    [tex]E=E_r+E_R[/tex]

    3. The attempt at a solution
    My book mentions (Introduction to Quantum Mechanics by Griffiths, in problem 5.1) that if the potential for a two-particle system is dependent only on the separation between the particles, then we can separate the time-independent schrödinger equation into two single-particle equations (see the above section), one that looks like a free particle equation (its potential is 0), and the other with the same potential as the original problem.

    The second equation is just an harmonic oscillator, but I don't know what to do with the first. The problem is that a free particle doesn't have a ground state (right?), so I don't know how the whole thing can have one.
     
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  3. Jul 13, 2007 #2

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    The ground state is the one with lowest energy, and the lowest energy a free particle can have is zero, corresponding to [itex]\psi_R=[/itex]const. This isn't normalizable, nor realizable in nature, but neither is any perfect plane wave state. That being said, I've never seen the term ground state applied to anything other than bound systems, so at best it's a little confusing.
     
  4. Jul 13, 2007 #3
    Why constant? The differential equation allows linear functions (in which case I wouldn't know how to get the coefficients).
     
  5. Jul 13, 2007 #4

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    That's a good point. The linear solutions are typically rejected. I don't have a fantastic answer as to why, the best I can say is that for a particle in a box, as the box gets larger and larger, the ground state solution approaches a constant function, and since in practice our experiments are always finite, this is the limit we should take.

    I believe it also has something to do with something called the rigged Hilbert space, which is an extension of the ordinary wavefunctions to include some that are not quite normalizable, but "close", in a certain sense that makes them still useful for many purposes. Constant functions and plane waves are close in this sense, while linear ones, which blow up at infinity, are not.

    Incidentally, it stumped me for a while why this only happens for the case k=0, until I realized that it happens in general, it's just that the linear envelope moves in time, so you need to use the time dependent schrodinger equation. That is, you can check that the following wavefunction (or something close to it) satisfies the free particle schrodinger equation:

    [tex] \Psi(x,t) = \left (a \left(x-\frac{\hbar k}{m} t\right) + b \right) e^{i (k x- \frac{\hbar k^2}{2m}t)} [/tex]
     
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