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QED Corrections for Hydrogen

  1. Aug 2, 2013 #1
    Binding energy equations can be derived for the main spherical orbitals of hydrogen. If derived correctly, the energies computed by the equations are negative in value, for example, the well known approximate -13.6 eV for hydrogen's spherical ground state. In QED, corrections are added to the "basic" Dirac binding energy equation. In total, do all of the several QED corrections make the basic Dirac spherical main orbital energies "more negative" ("stronger" binding energy) or "less negative" ("weaker" binding energy)?

    Thanks,
    sb635
     
  2. jcsd
  3. Aug 6, 2013 #2
    The QED corrections are responsible for the Lamb shift, which is positive for the H atom ground state (relative to the relativistic Dirac ground state) by 8173 MHz.

    See http://dspace.ubvu.vu.nl/bitstream/handle/1871/27194/224355.pdf?sequence=1 for detailed discussion of how the Lamb shift affects other levels (it can be slightly negative in some cases). An introductory discussion on the Lamb shift (for n=2 levels) is also found in R.E. Moss (Advanced Molecular Quantum Mechanics, section 11.9).
     
  4. Aug 6, 2013 #3
    Thanks for the reference. In the paper, in eq. (1), is the Dirac-Coulomb energy E_DC(n,J) to be interpreted as a positive number, or a negative number? For example, the E_n = -R/n^2 described in the second paragraph of the Introduction is obvioulsy negative in value (Rydberg R is positive). If E_DC in eq. (1) is negative, then the positive electron self energy listed in Table 1 for 1S and 2S when added in, would make their orbital binding energies "less negative" and hence "weaken" the binding energy. Is that the correct interpretation?
     
  5. Aug 7, 2013 #4
    I am not a theoretician but...

    (a) My reading of Eq. 1 is that it simply lists the sources of the energy contributions. It is a matter of choice whether you regard the Dirac energy as positive (if factor mc^2 is included) or negative (if you ignore the mc^2 part).

    (b) Yes, the Lamb shifts should be added as they stand to this term, thus weakening the binding. A diagram in Moss confirms this interpretation.
     
  6. Aug 7, 2013 #5
    I wish I had the Moss textbook. The texts I do have show my interpretation to be incorrect, which is the reason why I'm asking these questions. The texts I have show the Lamb shift makes binding energies "more negative," i.e., stronger. The paper you provided a reference to, shows that some of the QED corrections are positive and some are negative. Hence, when adding these in, it seems extremely important to know whether the "base" Dirac energy is positive or negative. In the physics I know, it has to be negative, like why the simple Bohr energies are negative in value. But I have seen many texts and papers that do not include the "-mc^2" part in the "Dirac equation," which to me, is physically incorrect. And it's the "minus mc^2" (i.e., -mc^2) part that's missing (see the Cohen-Tannoudji reference below). The physically incorrect equations that do not subtract mc^2 yield positive values. It's when the mc^2 part is subtracted, then the resultant energy is physically correct with a negative value. And then a positive QED correction makes the energy less negative, in complete contradiction to the fact that, for example, the nonspherical 2P_1/2, l = 1 level "lies under" (i,e., "is more negative" than) the spherical 2S_1/2, l = 0 level.

    The text "Quantum Mechanics" Vol 2 by Cohen-Tannoudji, Diu and LaLoe, on p. 1226, provides an "exact solution of the Dirac equation" which cannot be physically correct because it always yields positive values. This is the same equation listed in many texts/papers which do not subtract mc^2 (i.e., add -mc^2) in the equations. How they get to this result, I do not know. If you start off with total orbital energy equals kinetic energy plus potential energy where potential energy is always negative but larger than the kinetic energy in absolute value, which is the physical truth for bound motion, you get physically correct total orbital energy equations which yield negative values when evaluated, like the simple Bohr energy equation. An example of a paper that does add in -mc^2, which produces a physically correct Dirac equation, is at the link

    http://www.nist.gov/data/PDFfiles/jpcrd100.pdf [Broken].

    Eq. (2.4) of the above paper is what I am calling a "physcially correct" Dirac equation. Inspection of (2.4) shows it will always yield negative total orbital energy values.

    I originally said the Lamb shift, according to the texts I have, makes a binding energy more negative. But apparently that's not even a correct interpretation. The Lamb shift doesn't make the 2S_1/2, l = 0 level "more negative." It would seem that electron self energy actually creates an entire separate level, e.g., the 2P_1/2, l = 1 level. Logically, to me, electron self energy is always present. If so, we should never see indications of electrons existing in the 2S_1/2, l = 0 level, or any spherical orbital/level for that matter. But I had always thought actual hydrogen spectroscopic data show these spherical-to-spherical orbital transitions do occur.
     
    Last edited by a moderator: May 6, 2017
  7. Aug 7, 2013 #6

    king vitamin

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    Experiments don't measure absolute energies, only changes in energies. Whether or not one includes mc^2 in the energy won't change the spacings between levels in a bound state (the spectra of the atom).
     
  8. Aug 7, 2013 #7
    For shell trasnitions, yes, but the equations presented are for single shell/orbital/level energies. Strictly speaking, a single level bound energy equation (not a difference) that evaluates to a positive value is physically incorrect. But let's say the -mc2 is not included, so that the "base" Dirac energy to which QED corrections are added, is itself positive. Consider the ground state, and then the total QED correction contribution to the ground state (including electron self energy) is listed as a positive 8172.802 MHz. Convert this to a positive energy and add it to the positive "base" Dirac energy (without -mc2 in it). You have a bigger positive number on hand. However you think to convert this to an actual physical negative value, please do so. Now place this negative value on a graph with actual physical negative binding energies (like about -13.6 eV for n = 1) on the y-axis, with y = 0 at the upper left and minus values below it, like you see in many textbooks. Where does your now negative ground state energy value go, above (less negative) or below (more negative) a line/level made without the QED corrections?
     
  9. Aug 7, 2013 #8
    I am also having trouble interpreting some NIST hydrogen data I downloaded. Here is the file at

    http://sb635.qwestoffice.net/nist.txt

    The left hand column is "Observed Wavelength Vac" for which I take NIST literally. I assume this column has unbiased, spectroscopically obtained, observed transition wavelengths (in vacuo, in nm). I assume these data have all of nature's effects "in them" without bias. I cut and pasted a line at the bottom of the file for the n = 2 to n = 1 transition. If I am interpreting things correctly, the observed wavelgth at the left (121.5673123130200) is for the transition between the 1S1/2, l = 0 level and the 2S1/2, l = 0 level. This is apparently the 2S "metastable" level, which confuses me, because I thought the jump to ground state from there was prohibited.
     
  10. Aug 7, 2013 #9

    king vitamin

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    The bound state simply has to have less energy than the potential at infinity (so the particle can't escape). If your potential is 1/r, then your bound state energy must be negative. If your potential is mc^2-1/r, then your bound state energy must be less than mc^2.

    If there is a positive energy shift, as you say, the state is then "less bound," or less negative in your language. Looking at my reference (the last chapter of Weinberg), QED gives the 1s and 2s have a positive energy shift and the 2p_{1/2} has a negative shift (my book ignores recoil effects).

    Although the transition 2s -> 1s should be highly forbidden, "forbidden" in this context just means that the lifetime of the 2s state is very long (very low probability of decay). I'll admit that I know nothing about this particular transition, but a famous example of a forbidden transition is the 1s -> 1s hyperfine transition induced by the splitting between the two 1s levels from interaction between the magnetic moments of the electron and proton. The hyperfine transition has a half-life of ~10^7 years.

    http://en.wikipedia.org/wiki/21_cm_line
     
  11. Aug 7, 2013 #10
    It seems that Weinberg, Moss and the reference I posted are all in agreement. Consider the possibility that your text has a typo in it, or is using a different convention to describe the energy (for example, "binding energies" are always reported as positive quantities in photoelectron spectroscopy).

    Unfortunately I cannot say any more as my knowledge of QM stops soon after the Dirac equation, and well before QED.
     
    Last edited: Aug 7, 2013
  12. Aug 7, 2013 #11
    Ok, so a positive QED correction makes the binding energy less negative, and weakens the binding. Then why is the 2P1/2, l = 0 level always displayed as more negative than the 2S1/2, l = 0 level? For example, see

    http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/lamb.html

    The above says the Lamb shift (electron self energy appears to dominate) is a shift to a more negative level from the 2S1/2, l = 0 level. Is not the 4.372 x 10-6 eV value simply the sum of the QED corrections for the 2S1/2, l = 0 level? If I were to compute the actual negative value shown for the 2P1/2, l = 1 level, don't I start with the "basic" non-QED Dirac value for the 2S1/2, l = 0 level, and add in its QED corrections? After that, am I suppose to compute the 2P1/2, l = 1 level's QED corrections and then add these in also? It doesn't look like that's correct. It looks like the correct value for the 2P1/2, l = 1 level is simply the 2S1/2, l = 0 level plus its QED corrections. And it looks like on a correct "negative energy" graph, the QED corrections make the 2S1/2, l = 0 "more negative" and hence not weaken, but strengthen the binding energy.
     
  13. Aug 8, 2013 #12
    Sorry, I should have written "Then why is the 2P1/2, l = 1 level always ..."
     
  14. Aug 8, 2013 #13
    In all the math I've seen, the classic Coulombic system potential used in Dirac theory is proportional to -1/r, not mc2 - 1/r. This is not just a scaling issue to me. Correct physical equations show bound orbitals have negative total orbital energies, just like the simple Bohr orbital energy equation development shows.
     
  15. Aug 8, 2013 #14
    I take it back. It looks like to me now the correct computation (which will show the need (to me anyway) for tracking the energies as negative values), for deriving a final negative value for the 2P1/2, l = 1 level is to first use an equation like the eq. (2.4) talked about before, which yields a negative value for the 2S1/2, l = 0 level. Then documents state the "Lamb shift" QED correction for the 2S1/2, l = 0 level is a positive 1045.003 MHz (which basically equals the 4.372 x 10-6 eV shown in diagrams). Compute a positive energy from this, but then subtract this energy from the negative 2S1/2, l = 0 energy, making it more negative. That agrees with the diagrams, for where the ashperical 2P1/2, l = 1 level lies in reference (below, more negative) to the spherical 2S1/2, l = 0 level. But the job is not done yet. Documents state the QED "Lamb shift" correction for the 2P1/2, l = 1 level itself is a negative 12.8357 MHz. Forget the negative sign and compute a positive energy for this small frequency shift. Since it was negative as listed, actually now add in this energy as positive to your present negative value, which makes it "less negative" weakening the final binding for the 2P1/2, l = 1 level. Then since we are dealing with n = 2, it stops. (Of course, there are hyperfine splittings, but ingore those).
     
  16. Aug 8, 2013 #15

    king vitamin

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    Unfortunately, I think you have the basic idea, but you got the order of the two levels backwards.

    Begin with the Dirac equation. We will take the energy to be zero at infinity at your behest (although I insist there isn't a measurable difference). Then the two levels are degenerate:

    $$ E(2s_{1/2}) = E(2p_{1/2}) = - 3.4 eV $$

    Now one computes the loop correction in QED (the "fun" part). One finds that all l=0 states get shifted to a larger ("more positive") energy, while for higher l, the expression is more complicated. Specifically, one computes energy shifts in the levels, and Weinberg gives:

    $$ \delta E(2s_{1/2}) = h \times 1039 MHz $$
    $$ \delta E(2p_{1/2}) = h \times -12.88 MHz $$

    Where h is Planck's constant. These are taken from an almost 20 year old textbook so I'm sure the NIST figures are more accurate. So the new energy levels are

    $$ E(2s_{1/2}) = -3.4 eV + h \times 1039 MHz $$
    $$ E(2p_{1/2}) = - 3.4 eV - h \times 12.88 MHz $$

    and the Lamb shift is

    $$ \delta E(2s_{1/2}) - \delta E(2p_{1/2}) = E(2s_{1/2}) - E(2p_{1/2}) = h \times 1052 MHz $$

    So when you referred to the 2P1/2 state as "less negative weakening the final binding for the 2P1/2, l = 1 level," you had it backwards. The Lamb shift makes the 2P1/2 state more stable, while it makes l=0 states more unstable.
     
  17. Aug 8, 2013 #16
    Thanks for the clarification. So, to compute the transition energy from 2S1/2, l = 0 to 1S1/2, l = 0, first compute 2S1/2, l = 0 as you have done, which is simply Dirac w/o QED corrections + QED corrections. If QED is positive, keep it positive in the addition. If QED is negative, add it in as negative (subtract). The total QED for 2S1/2, l = 0 is positive, so add this positive total in no matter the sign of Dirac w/o QED corrections. This shifts 2S1/2, l = 0 "higher, more positive" no matter the sign of Dirac. Now for 1S1/2, l = 0, do the same thing. The total QED for 1S1/2, l = 0 is positive, so the final energy for ground state is also "above, more positive" than Dirac for ground state. Then simply difference these two level energies to get the transition energy.

    Using NIST modern values, I computed/predicted the above theoretical transition energy and the corresponding wavelength from the equations. I used the QED corrections listed in one the reports in this thread, specifically 8172.802 MHz for ground state and 1045.003 MHz for 2S1/2, l = 0. The predicted transition wavelength turned out to be 1.21566071 x 10-7 meters. If I am reading the NIST data correctly (see the last line in the NIST spectroscopic data file (at the bottom) I gave a link to), the observed transition wavelength is 1.2156731231302 x 10-7 meters. The prediction error (observed minus predicted) is +1.24108369 x 10-12 meters. I don't know how "significant" this error is yet, but it is interesting to note that just Dirac without QED corrections actually works better. The predicted transition wavelength is then 1.21566423 x 10-7 meters, with a smaller prediction error of +8.89716857 x 10-13 meters. This was unexpected. Using non-QED corrected Dirac values actually works better for this transition.
     
  18. Aug 9, 2013 #17
    Well, I've done another prediction with results, but this editor has cost me a couple of hrs of work trying to here report the results. The "Go Advanced" option can bomb, and you lose everything. Any suggestions on how to avoid that?

    The other prediction I made produced a result as before; the basic Dirac theory without QED corrections fared better than with the QED corrections. This should not be. The observed data I am using is from NIST, assumed to be very accurate and precise. Even for a single transition, Dirac theory without QED corrections should not do better than with the corrections. This violates the belief that QED is "perfect" when it comes understanding the mechanics inside of hydrogen.
     
  19. Aug 9, 2013 #18
    Let me try and give the equations of the prediction theory, and hopefully this editor won't bomb while doing it. The "basic" non-QED-corrected Dirac equation I used was the eq. (2.4) referenced before:

    ED = μc2[1 + α2/(n - ε)2]-1/2 - μc2

    where μ is the reduced rest mass of the electron (perfectly ok to use for spherical orbitals) and

    ε = j + 1/2 - [(j + 1/2)2 - α2]1/2

    Using NIST values for the rest mass of the electron and proton, etc., this is easy to evaluate for given n and j. I then simply added in or subtraced out the QED corrections listed in documents to produce the predictions. Adding in the QED corrections should have produced better agreement with the observed data, not worse.
     
  20. Aug 9, 2013 #19
    I found a confirmation from NIST documents that I did use the correct observed value for the 1S1/2 to 2S1/2 transition. The 2002 NIST CODATA document is at

    http://physics.nist.gov/cuu/Archive/2002RMP.pdf

    Table XI lists observed transition frequencies. The value for the 1S1/2 to 2S1/2 transition (item no. A1), listed in kHz, when converted to a wavelength, equals the observed transition wavelength I used.

    The second prediction I did was for the 1S1/2 to 2P1/2 transition. In Table XI, item nos. A14.1 and 14.2 lists two observed transition wavelengths for the 2P1/2 to 2S1/2 transition. I differenced the average of the two from the transition frequency for the 1S1/2 to 2S1/2 levels, which then canceled out the 2S1/2 energy, and produced an "observed" transition value for the 1S1/2 to 2P1/2 transition. This difference using observed data cancels out all "natural" 2S1/2 energy, however nature made it. The observed transition wavelength from the difference equals 1.21567364 x 10-7 m. For the Dirac + QED correction prediction, I subtracted 12.8357 MHz from the basic degenerate 2S1/2 Dirac value. The prediction result equals 1.21566019 x 10-7 m. The prediction error is +1.34537808 x 10-12 m. I then redid the analysis without introducing the QED corrections. The prediction then became 1.21566423 x 10-7 m, with a prediction error of +9.41864974 x 10-13 m. Without QED corrections, the prediction error is once again smaller. This is not suppose to happen.
     
  21. Aug 9, 2013 #20
    There is an aspect of the comparison of the older Sommerfeld theory with the basic Dirac theory which is intriguing. It is often stated these two theories, up to the fine structure level, produce identical energy level predictions. They do, but the sense of what is spherical and what is aspherical is switched. For example, for main orbital 2, in Sommerfeld theory the upper most sublevel is a spherical orbital (nθ = 2, nr = 0). In Dirac theory, the uppermost sublevel is the aspherical 2P3/2 sublevel. And this is true for any main orbital, in terms of this "switch" in the locations of sphericity. Dirac theory then states there is a spherical sublevel which lies below this aspherical sublevel, namely the 2S1/2 sublevel. I don't know about relating "above" and "below" with distances from the proton, like can be done with the major orbital concepts (1 is closer to the proton than 2), but if so, it is strange to me that nature would construct a spherical orbital to lie below, at a closer distance, than an outer aspherical orbital. It also sounds "weird" to me that as one "approaches" a hydrogen atom, the first shell encountered is aspherical. According to Dirac theory, every hydrogen main orbital has an aspherical most outer sublevel. We are becoming able to actually image the outer orbitals of atoms now. The "fuzzy" pictures I've seen can't resolve sphericity or asphericity. Sommerfeld theory, which to me is "logical" at least in the nature of the most outer sublevel (Sommerfeld says spherical, logical) will be tested one day, at least in this aspect. Of course, any deterministic orbital theory should be questioned, and most say, abandoned.
     
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