Real Hydrogen Atom

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I've heard that the hydrogen atom that we originally learn about in QM that deals with the Coulomb force is an incomplete description. I'm having trouble understanding all of these effects.

When describing the electron energy of the real Hydrogen atom, how do things like the zeeman effect, stark effect, fine structure, hyperfine splitting, etc, affect the energy levels? What changes the energy the most and how do they?
 

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A. Neumaier
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I've heard that the hydrogen atom that we originally learn about in QM that deals with the Coulomb force is an incomplete description. I'm having trouble understanding all of these effects.

When describing the electron energy of the real Hydrogen atom, how do things like the zeeman effect, stark effect, fine structure, hyperfine splitting, etc, affect the energy levels? What changes the energy the most and how do they?
If you already have trouble to understand the textbook version of the hydrogen atom, there is little point in explaining what happens with the real hydrogen atom. All the effects persist, but their discussion gets far more complicated since relativistic effects and effects of field quantization must be taken into account, too. (The hyperfine splitting is an additional effect, though, not present in case of a Dirac electron in an external Coulomb potential.)
 
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dextercioby
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I've heard that the hydrogen atom that we originally learn about in QM that deals with the Coulomb force is an incomplete description. I'm having trouble understanding all of these effects.

When describing the electron energy of the real Hydrogen atom, how do things like the zeeman effect, stark effect, fine structure, hyperfine splitting, etc, affect the energy levels? What changes the energy the most and how do they?
I'm not as drastic as Arnold. I would say the complete theory of a hydrogen atom can only be explored through approximate methods, the so-called perturbation theory. Surely that in its most simple form, the Hydrogen atom based on the Schroedinger's original equation of 1926 is completely solvable and the full solution for the discrete spectrum of the Hamiltonian is displayed in every book.

Then you wish to include the theory of special relativity into the 1926 original picture and write down and solve the Dirac equation. This was done by the physicists Gordon and Darwin in 1928. But even Dirac's theory is not the whole picture. It should be generalized to include the relative motion of the proton wrt the electron (which is not so simple) and the efect of the proton's magnetic momentum (the so-called hyperfine splitting).

And then of course, you've got the fact that quantum electrodynamical corrections (Lamb shift) must be taken into consideration as well.

Eventually, the Balmer's 1885 formula for the energetic spectrum of the H atom would be so changed, that you wouldn't recognize it was about the H atom, unless you know the whole picture behind it.
 
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Pengwuino
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As people are alluding to, the hydrogen atom shown in intro textbooks is quite simple. The proton is allowed to be stationary, spins are ignored sometimes, the magnetic moment is ignored, special relativistic and frame of reference effects are ignored, the intrinsic structure of the proton is ignored, the fields are treated classically, all sortsa stuff!
 
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alxm
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The usual description with the Schrödinger equation and a stationary nucleus gives you accuracy to about 1E-4.

That's the largest error, assuming a stationary nucleus. (the first-order correction is quite easy to do, though) After higher-order corrections for that, you have the second-largest error, which are relativistic mass corrections, at around 1E-7. From then on, you can't really use the Schrödinger equation anymore (since it's non-relativistic) other than for a rough approximation of these errors.

After that come various other S.R. effects, such as the Breit-Pauli interactions and Darwin term. After that, you have QED effects, such as the anomalous magnetic moment and Lamb shift.

After that, you would have the effect of the finite-sized nucleus, but by this point, you're up to about 13-14 digits of accuracy, which is about as far as you can go. Because we just don't know the fundamental constants or hydrogen levels to higher accuracy than that.
 
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dextercioby
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After that, you would have the effect of the finite-sized nucleus, but by this point, you're up to about 13-14 digits of accuracy, which is about as far as you can go. Because we just don't know the fundamental constants or hydrogen levels to higher accuracy than that.
Excellent point, the Rydberg constant is not that accurate.
 

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