# I Use the Dirac Equation to calculate transition frequencies in Hydrogen

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1. Mar 6, 2018

### neilparker62

I would just like to understand how to use the above Dirac energy equation to calculate (for example) the 1s-2s transition frequency in hydgrogen. Does one substitute n=1, j=0 for 1s energy level and n=2, j=0 for 2s energy level ?

From previous reading I understand the mass referred to in the formula will need to be the reduced electron mass.

The graphic comes from this reference.

Last edited by a moderator: Mar 6, 2018
2. Mar 7, 2018

### Staff: Mentor

You are confusing $j$ with $l$. $j$ is the total angular momentum of the electron, $\vec{\jmath} = \vec{l} + \vec{s}$. For s orbitals, you get $j=1/2$.

No, it is the rest mass of the electron. In the article cited, they give the correction with the reduced mass of the atom in eq. (23).

3. Mar 7, 2018

### neilparker62

Thanks very much - now the substitution is working perfectly for calculating energy level 1. Can I hazard a guess that for the 2p (3/2) level, j will be 3/2 ? And in general that for the 'outermost' energy level of any particular prime quantum number n, j will be n - 1/2 ?

BTW I meant that for calculating H (1s) I would need to be using the reduced electron mass.

4. Mar 7, 2018

### Staff: Mentor

That's not how addition of angular momenta works. For a state with quantum numbers $l$ and $s$, the possible values of $j$ are
$$j = \left| l - s \right |, \left| l - s \right | + 1, \ldots, l + s$$
So for $l=1$ and $s=1/2$, the possible values of $j$ are $j=1/2, 3/2$ (that's the fine-structure splitting due to spin-orbit coupling).

What do you mean by "reduced electron mass"?

5. Mar 7, 2018

6. Mar 7, 2018

### neilparker62

Thanks again.

l (max) = n-1, s=1/2. Then j = n-1-1/2 or j = n-1/2 . Is that correct ?

7. Mar 7, 2018

### Staff: Mentor

I wasn't sure what you meant because it isn't called the "reduced mass of the electron." It is the reduced mass of the hydrogen atom.

And as I said, simply replacing $m_e$ by $\mu$ is not correct. The article gives the equation to use, namely eq. (23), which is much more complicated.

I don't see why you write it in terms of $n$, since $l$ is independent of $n$. I know that you wrote l (max), but I don't see why you would concentrate only of the maximum value of $l$. Anyway, for a single electron, you indeed get
$$j = l - \frac{1}{2} \mbox{ or } l + \frac{1}{2}$$

8. Mar 7, 2018

### neilparker62

Try for starters:

This will give you the Dirac component of Hydrogen's ionization energy. You may compare your answer with this result:

You will need to use the reduced electron mass and CODATA values for all the constants!

The number I give above is from the following reference: https://research.vu.nl/ws/portalfiles/portal/2485538

Thanks once again for all your assistance - it has resolved the conundrum I had establishing equivalence between above and the Dirac energy formula.

9. Mar 7, 2018

### neilparker62

I'm not sure why you write this. In this thread https://www.physicsforums.com/threads/the-bohr-radius-and-reduced-mass.811962/#post-5096893, it's clear we must use reduced electron mass for precision in these type of calculations. Indeed I do believe it may have been yourself who - some years ago - advised the use of reduced mass. At the time I was making some first attempts at calculating transition frequencies and was worried that the answers were very imprecise. Using reduced mass improved the precision considerably.

Even Bohr himself initially did not use reduced mass and again improved the precision of his results (against the spectral measurements of the time) considerably once he did.

10. Mar 8, 2018

### neilparker62

Thanks for confirming. The reason I focus on n, l(max)=n-1 , j=l+1/2 is because in that case the term n-j-1/2 in the Dirac energy formula is zero and the Dirac energy equation simplifies considerably into the following form:

$$E_n=m_{e_r}c^2 \sqrt{1-\frac{\alpha^2}{n^2}}$$

In this case for the transition from energy level n to energy level m with n,m (n>m) being prime quantum numbers, we have:

$$\Delta E=h \Delta f = m_{e_r}c^2 \left( \sqrt{1-\frac{\alpha^2}{n^2}} - \sqrt{1-\frac{\alpha^2}{m^2}}\right)$$

Noting that for Hydrogen Z=1 in the Dirac formula - hence have not included it in the above.

Last edited: Mar 8, 2018
11. Mar 8, 2018

### Staff: Mentor

Taking the reduced mass instead of the electron mass is a good first step. But you seem to be after a high precision, so you could try instead eq. (23).

12. Mar 8, 2018

### neilparker62

A 1974 measurement of the Balmer alpha (2p 3/2 - 3d 5/2) line (H I) by laser absorption spectroscopy gave the following result:

15233.07021 cm^-1

The above formula (without any corrections) determines this line at:

15233.07093 cm^-1

I would guess the main corrections would be hyperfine structure adjustments and perhaps lamb shift.

Reference for the above measurement: http://link.aps.org/abstract/PRL/v32/p1336

13. Mar 13, 2018

### neilparker62

Measuring "Ground State Lamb Shift": The Herzberg Experiment

Quoting from the following reference:

http://rspa.royalsocietypublishing.org/content/234/1199/516

''According to modern quantum electrodynamics, the 1S level (ground state) of hydrogen and deuterium is predicted to lie $0,272 cm^{-1}$ above the energy given by the Dirac theory. In order to obtain an experimental value for this Lamb shift, the absolute wavelength of the $L_\alpha$ line of deuterium has been determined by means of a 3 m vacuum grating spectrograph in fifth order. Lines of the series $6^{1}S_{0}-n^{1}P_{1}$ and $6^{1}S_{0}-n^{3}P_{1}$ of $^{198}Hg$ observed in the same order were used as standards. The wavelengths of these standards were obtained to $\pm0.0002{\buildrel _{\circ} \over {\mathrm{A}}}$ by the combination principle from lines in the visible and near ultra-violet regions, some of which were newly measured. Both the far ultra-violet Hg lines and $L_\alpha$ of deuterium were simultaneously recorded on the plate in absorption. In this way, from six independent plates the wavelength of the $L_\alpha$ line of deuterium was found to be $1215.3378\pm0.0003 {\buildrel _{\circ} \over {\mathrm{A}}}$. This value refers to an unresolved doublet. If the relative intensity of the two doublet components (2:1)and the temperature of the absorbing column (80°K) is taken into account and the result compared with the Dirac value of $L_\alpha$, a shift of the $1^{ 2}S$ level of $0.26\pm 0.03cm^{-1}$ is obtained. The agreement with the predicted value is very satisfactory.''

Using the formula $\Delta E= m_{e_r}c^2 \left( \sqrt{1-\frac{\alpha^2}{2^2}} - \sqrt{1-\frac{\alpha^2}{1^2}}\right)$ , where $m_{e_r}$ is the reduced mass of the electron for deuterium, the wave number for the above transition is determined as $82281.939 cm^{-1}$. Deducting Herzberg's very carefully obtained experimental value of $82281.664 cm^{-1}$ a first measure of "ground state lamb shift" $0.275 cm^{-1}$ is obtained. This compares favourably against the predicted Lamb Shift value of $0.272 cm^{-1}$. Note that in this calculation a measured $L_\alpha$ wavelength of $1215.3376{\buildrel _{\circ} \over {\mathrm{A}}}$ was used based on a later revision of the $^{198}Hg$ standards.

Last edited: Mar 13, 2018
14. Apr 2, 2018

### neilparker62

Ok - I think I misunderstood you when you said using reduced mass is incorrect. Rather is it not a case of using a first order approximation when there are second and higher order terms as in Equation 23 per reference article? I tried using Eqn 23 - at a first attempt the calculated ionization energy reduced by about 24Mhz. But I'm not sure if I am using the formula correctly.