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QM - Hydrogenic wavefunctions - normalization

  • #1
I just want to make sure I understand this point:

The eigenfunctions of the hydrogenic Hamiltonian are

[tex]\varphi[/tex][tex]_{nlm}[/tex]=R[tex]_{nl}[/tex]Y[tex]^{m}_{l}[/tex]

If I need to find the probability of finding the electron in the nucleus (in r<R0), and I use the normalized R[tex]_{nl}[/tex], can I simply calculate the integral
integral[0 -->R0] (|R[tex]_{nl}[/tex]|^2r^2)dr
?
without calculating the whole triple integral? the constants that should be obtained from the angular part of the integral are already included in the normalized R[tex]_{nl}[/tex] function?

And another question - we analyze the hydrogen atom as a two body problem, so the total Hamiltonian eigenfunctions have the form [tex]\varphi[/tex]CM[tex]\varphi[/tex]rel.
Why do we always consider only the relative part, and not the general solution?
 

Answers and Replies

  • #2
reilly
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When in doubt, go back to first principles -- shortcuts are typically the product of experience. That is, start from scratch and you will find out whether or not your conjecture is true.

Ask yourself, what role a hydrogen nucleus will play in 1. a 2P->1S transition for a hydrogen atom in a very cold hydrogen gas, and 2. what role the H nucleus will play in a high energy collision with another H atom?
Regards,
Reilly Atkinson
 
  • #3
olgranpappy
Homework Helper
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also, it's not hard to do the whole triple intergral since the integral over solid angle is trivial.

so, suppossing you have normalizaed eigenfunctions already (which you can just look up), then
[tex]
P_{\rm in nucleus}=\int_0^{R_0}dr r^2 \int_{4\pi} d^2\Omega R_{nl}^2 Y_{lm}Y_{lm}^*
=\int_{0}^{R_0}dr r^2 R_{nl}^2\;.
[/tex]

Anyways--the short answer to ur question is "yes".
 
  • #4
olgranpappy
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And another question - we analyze the hydrogen atom as a two body problem, so the total Hamiltonian eigenfunctions have the form [tex]\varphi[/tex]CM[tex]\varphi[/tex]rel.
Why do we always consider only the relative part, and not the general solution?
Why do "we"? Well, "we" don't have to do anything we don't want to do, do we? Anyways, you (or we) can consider the CM part too: What equation does the CM part of the wavefunction obey? How do you solve it? What do the solutions look like?
 
  • #5
Ask yourself, what role a hydrogen nucleus will play in 1. a 2P->1S transition for a hydrogen atom in a very cold hydrogen gas, and 2. what role the H nucleus will play in a high energy collision with another H atom?
the first case is where the relative part should be considered and in the second the CM part is relevant?

Why do "we"? Well, "we" don't have to do anything we don't want to do, do we?
no, but everywhere I looked, in the book, on the internet, in the tutorials - only the relative part is treated, and considered as "the hydrogen eigenfunctions".

What equation does the CM part of the wavefunction obey? How do you solve it? What do the solutions look like?
the CM part obey the Schroedinger equation of a free particle.. and the solutions are plane waves in coordinate space or again, solutions of the form R(r)Y(theta,phi) in the momentum space...
 
Last edited:
  • #6
olgranpappy
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the CM part obey the Schroedinger equation of a free particle.. and the solutions are plane waves in coordinate space or again, solutions of the form R(r)Y(theta,phi) in the momentum space...
so, the CM part is not very interesting since it just describes the free motion ofthe center of mass. you already know all about that sort of thing. the interesting part is the relative motion which leads to bound states, etc. Solving the relative-coordinate equation is the hard part and that's why people discuss it at greater length than the CM part.
 
  • #7
I see. thanks.
 
  • #8
reilly
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so, the CM part is not very interesting since it just describes the free motion ofthe center of mass. you already know all about that sort of thing. the interesting part is the relative motion which leads to bound states, etc. Solving the relative-coordinate equation is the hard part and that's why people discuss it at greater length than the CM part.
Not at all, the CM part can indeed be interesting--it all depends on the physics of the situation. First, even with the neglect of recoil, the Schrodinger Eq. for an attractive 1/r potential will have scattering solutions (E>0)--sometimes called Coloumb Wave Functions -- in addition to bound states. In fact, the usual treatment of the structure of the H atom is often considerably more simple than the description of a process that involves the atom as a whole. For example:

Suppose a hydrogen atom is zapped by a X-ray, that is a photon with E>>13.5 ev. What's the probability that the atom will stay intact? What role does recoil play?
Regards,
Reilly Atkinson
 
  • #9
olgranpappy
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Not at all, the CM part can indeed be interesting--it all depends on the physics of the situation. First, even with the neglect of recoil, the Schrodinger Eq. for an attractive 1/r potential will have scattering solutions (E>0)--sometimes called Coloumb Wave Function -- in addition to bound states.
But, those are solutions to the equation in the relative coordinate, right?

In fact, the usual treatment of the structure of the H atom is often considerably more simple than the description of a process that involves the atom as a whole. For example:

Suppose a hydrogen atom is zapped by a X-ray, that is a photon with E>>13.5 ev. What's the probability that the atom will stay intact? What role does recoil play?
Regards,
Reilly Atkinson
I figured that the OP was not considering any coupling to an external field. But, again, isn't most of the interesting physics coming from (mostly dipole) matrix elements between hydrogenic (relative coordiate) wavefunctions?

I don't think about recoil too much myself, since I'm in condensed matter. But, can you explain further what you mean? For example, the energy in the photon can either excite internal states or can go into overall translational kinetic energy of the atom, right?
 
  • #10
reilly
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Right.

What's intriguing to me, among other things, is that both the proton and the electron can absorb a photon. If the H atom is zapped by a very energetic photon, then
possible final states are 1. free electron and free proton (ionized H atom) and 2. intact H atom(not necessarily in the ground state) and a radiated photon, considering only two particle final states. The kinematics will show the need to consider recoil; the interaction matrix elements are tricky. (This is the kind of problem people worked on in the 1930s -- see The Theory of Atomic Collisions, Mott and Massey,Oxford, 1933)
Regards,
Reilly
 
  • #11
olgranpappy
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thnx
 
  • #12
reilly
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You are most welcome.r
 

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