Discussion of Notes 331(1,2)

OK thanks again. The change of sign does not affect Note 331(3) except for the fact that the signs of energy levels are reversed. The correct computer generated signs will of course be used in the final paper. Your point is well taken, the angular momentum operator can be written as (h bar / i) partial / partial phi, and acts on the spherical harmonics to give eigenvalues m h bar. The spherical harmonics are part of the hydrogenic wavefunctions. I will look at this tomorrow to try to assess the validity of the approximation. This is an interesting problem. The <T L sub Z> can be worked out with computer algebra and compared with <T> <L sub Z>.

Subj: Re: Computer Check of notes 331(1,2)

In (12) and (18) you wrote (1 – …) but I obtained (1 + …).

I think the operator product is an approximation. In general we have for two operators A and B:

<A*B> not equal <A> * <B>.

It is true that L_Z has the known eigenvalues but it is used here in combination with nabla squared which makes a difference. L and nabla squared operate on the same coordinates and cannot be separated as a product of functions.


Am 29.10.2015 um 13:46 schrieb EMyrone:

Many thanks again for these computer checks, with which I agree and which will be used in the final paper. The final result, Eq. (18), is the same, and was used in Note 331(3) in the discovery of relativistic Zeeman splitting, another of many effects on the list of new experiments, bearing in mind that this is the simplest possible theory (of atomic H). In Eq. (8), I used <L sub Z> = h bar m sub L, so the result in Eq. (8) follows, i. e. h bar m sub L can be taken outside the integral, i.e. integral psi* T L sub Zpsi d tau = L sub Z integral psi* T psi d tau. = L sub Z < T > . Finally L sub Z psi = h bar m sub L psi.

Sent: 29/10/2015 11:15:26 GMT Standard Time
Subj: notes 331(1,2)

Note 1 describes an interesting extension of spectra induced by relativistic magnetic effects.
Note 2: Eq.(9) is the Rydberg energy which is from Wikipedia:
E_n = -\frac{m e^4}{8 \varepsilon_0^2 h^2}\cdot \frac{1}{n^2}

This is different from eq.(9). There is no c^2 and m appears in the numerator. The correct result and all constants are listed in the attachment. The additional term in (12) is positive. The modulus of <H1> is enlarged.

How did you come to eq.(8)? It seems that you equated the expectation value of an operator product with the product of the single expectation values.

You can find all constants in the attachment. The numerical result in (18) is correct.



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