Schrodinger's equation for H atom Some questions

  • #1
Hello,

I just finished learning how to solve the Schrodinger equation for the H atom and a few things trouble me.

1) The relative particle, CoM particle treatment: From what I understand, we are solving this for the case where the nucleus is taken to have mass m1+m2 and the electron has mass m1m2/(m1+m2), where m1 is the actual mass of the nucleus and m2 the actual mass of the electron. Am I correct in saying this? And isn't this going to cause an error?

2) Ignoring the nucleus: The wavefunction for the nucleus turns out to be a plane wave. Well, what I don't understand is why we ignore it. Shouldn't we add on that plane wave wavefunction at the end and multiply it to the solution of the wavefunction we get for the electron?

3) Suppose I could not make the first assumption. Let's say I had a system where m1 and m2 were comparable. Then what? How should I proceed?

Thank you for your help.
 

Answers and Replies

  • #2
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Just like in classical orbital mechanics, the two-body problem can be viewed in a center-of-mass frame where the problem reduced to an effective one-body problem. This can be done with particles of any mass.

You cannot describe the nucleus by means of a coulomb potential and schrodinger's equation, but relies on a different fundamental quantum field theory for it's description.
 
  • #3
dextercioby
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1) The relative particle, CoM particle treatment: From what I understand, we are solving this for the case where the nucleus is taken to have mass m1+m2 and the electron has mass m1m2/(m1+m2), where m1 is the actual mass of the nucleus and m2 the actual mass of the electron. Am I correct in saying this? And isn't this going to cause an error?
This is normally not different than the Kepler problem in gravitostatics: the nucleus and the electron are replaced by 2 <fictive> particles, one of which is situated in the COM frame.

McLaren Rulez said:
2) Ignoring the nucleus: The wavefunction for the nucleus turns out to be a plane wave.
Not for the nucleus, but for the <fictive> particle situated in the COM frame.

McLaren Rulez said:
Well, what I don't understand is why we ignore it.
Out of convenience and because we're really interested in finding the discrete spectrum of the atom. A discrete energy spectrum is the trademark of quantization. The observables with completely continuous spectrum are known as <unquantized>.

McLaren Rulez said:
Shouldn't we add on that plane wave wavefunction at the end and multiply it to the solution of the wavefunction we get for the electron?
Indeed, the full solution is a product of solutions (the H-space is a product space anyways).

McLaren Rulez said:
3) Suppose I could not make the first assumption. Let's say I had a system where m1 and m2 were comparable. Then what? How should I proceed?
There's no difference. In this case, the assumption M_nucleus ---> infinity cannot be done, so the full solution must be found.
 
  • #4
Thank you, dextercioby.

So now that we have the solutions for the two fictive particles, how do we get the solution for the actual two particles from there?
 
  • #5
dextercioby
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From a mathematical perspective, solving the problem for the 2 fictive particles is enough, as the 2 Hilbert spaces (real particles vs. fictive particles) are isomorphic.
 
  • #6
Thank you, dextercioby.

So now that we have the solutions for the two fictive particles, how do we get the solution for the actual two particles from there?
What dextercioby said above. However, if you want an explicit solution:

The kind of systems you're interested in have states that can be factored in an 'internal' state that describes the relative movement of the two components and an 'external' state that describes the system as a whole:

[tex]
\hat\rho = \hat{\rho}_\mathrm{ext} \otimes \hat{\rho}_\mathrm{int}.
[/tex]

For a pure state this can be expressed as the product of two wavefunctions:

[tex]\psi(\mathbf{x}_c,\mathbf{x}_r) = \psi_\mathrm{ext}(\mathbf{x}_c)\psi_\mathrm{int}(\mathbf{x}_r)[/tex],

where xc and xr signify the 'centre of mass' and the 'relative' translational degrees of freedom, respectively.

These can be expressed as a function of the translational degrees of freedom of the two component particles m1 and m2:

[tex]\mathbf{x}_1 = \mathbf{x}_c - \frac{\mu}{m}_1\mathbf{x}_r,[/tex]
[tex]\mathbf{x}_2 = \mathbf{x}_c + \frac{\mu}{m}_2\mathbf{x}_r.[/tex]

Where [tex]\mu[/tex] is the reduced mass and xr is defined as x2 - x1.

This allows us to express the wavefunctions as

[tex]\psi_\textrm{ext}(\mathbf{x}_c) \equiv \psi_\textrm{ext}(\mathbf{x}_1,\mathbf{x}_2),[/tex]

[tex]\psi_\textrm{int}(\mathbf{x}_r) \equiv \psi_\textrm{int}(\mathbf{x}_1,\mathbf{x}_2).[/tex]

The Jacobian of this transformation is unity.

The density for particle 'i' will be given by

[tex]
\rho_i(\mathbf{x}) = \int \mathrm{d}^3\mathbf{x}_1 \mathrm{d}^3\mathbf{x}_2 \delta(\mathbf{x}-\mathbf{x}_i) \left|\psi_\mathrm{ext}(\mathbf{x}_1,\mathbf{x}_2)\right|^2 \left|\psi_\mathrm{int}(\mathbf{x}_1,\mathbf{x}_2)\right|^2,
[/tex]

and its {charge, mass} density by {[tex]q_i\rho_i(\mathbf{x})[/tex], [tex]m_i\rho_i(\mathbf{x})[/tex]}.
The total {charge, mass} density will be given by

[tex]\rho_m(\mathbf{x}) = m_1\rho_1(\mathbf{x}) + m_2\rho_2(\mathbf{x}),[/tex]
[tex]\rho_q(\mathbf{x}) = q_1\rho_1(\mathbf{x}) + q_2\rho_2(\mathbf{x}).[/tex]
 
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