Quantization of the E-M Field 2.1 Lamb Shift Revisited

2.1. LAMB SHIFT REVISITED

April 10, 2015 Lecture XXVIII Quantization of the E-M ﬁeld

2.1 Lamb Shift revisited

We discussed the shift in the energy levels of bound states due to the vacuum fluctuations of the E-M fields. Our picture was that the fluctuating vacuum fields pushed the electron back and forth over a region of space defined by the strength of the fields and the electron mass. We estimate the uncertainty in the electron position due to the fields. The strength of the Coulomb potential that binds the electron to the proton depends of course on the position of the electron. In order to account for fluctuating position we average the potential over the position uncertainty. Far from the proton the difference of the potential averaged over this finite range of electron positions is zero. But at the origin of the potential, there is a finite contribution that will tend to reduce the effective coupling. At any rate we estimate the effect and found that because the vacuum fluctuations occur at all wavelengths, the position fluctuations diverge. We used a cutoff, excluding photons with energy greater than the electron rest energy. Now we consider the effect in the language of transitions. Suppose that we have an electron in a bound state A. There is some amplitude for the electron to emit a photon with energy ¯hck and absorb it again. In the interim, energy is not necessarily conserved. There may be an intermediate bound state, but maybe not. The electron lives in a virtual state as does the photon. We write

X i(EI −EA)t/h¯ i¯hc˙I = HIAcAe photon

X X i(EA−EI )t/h¯ i¯hc˙A = HAI cI e (2.1) photon I

Here HIA corresponds to emission and HAI to absorption. These are the usual equations for the amplitudes to be in state I where at t = 0, cA(0) = 1 and cI6=A(0) = 0. We sum over all photon energies including those wherehω ¯ 6= EI − EA. Since we are looking for the shift in the energy of −i∆EAt)/h¯ EA we try cA = e so that

−i(EA+∆EA)t/h¯ ψA(t) = | uAie

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The plan now is to substitute our guess for cA into the Equations 2.1, and solve for ∆E. Since we are looking for an energy shift and not necessarily a transition rate, we will want eventually to integrate to t → ∞.

Z t 1 X 0 c = H ei(EI −EA−∆EA+¯hω)t /h¯ I i¯h IA photons 0 X ei(EI −EA−∆EA+¯hω)t/h¯ − 1 cI = HIA (2.2) (−EI + EA + ∆EA − ¯hω) photons

Next substitute Equation 2.2 and our guess for cA into the second of 2.1

ei(−∆EA+¯hω)t/h¯ − ei(EA−EI )t/h¯ −i(∆EAt/h¯) X X X −iωt ∆EAe = HAI HIAe (−EI + EA + ∆EA − ¯hω) photon I photons X X X 1 − ei(EA−EI +∆EA−hω¯ )t/h¯ ∆EA = HAI HIA (−EI + EA + ∆EA − ¯hω) photon I photons

To lowest order in ∆EA (drop the terms on the right).

X X 1 − ei(EA−EI −hω¯ )t/h¯ ∆EA = HAI HIA (2.3) (−EI + EA − ¯hω) photon I

We would like to evaluate that last equation for ∆EA as t → ∞ but it appears to oscillate. We note that

ixt ∞ 1 − e Z 0 lim = − lim i ei(x+i)t dt0 t→∞ x →0 0 1 = lim →0 x + i x i = lim − →0 x2 + 2 x2 + 2 1 = − iπδ(x) x Therefore, Equation 2.3 becomes

X X 2 1 <∆EA = |HIA| (2.4) (−EI + EA − ¯hω) photon I X X 2 Im∆EA = −π |HIA| δ(EA − EI − ¯hω) (2.5) photon I

The real part of the energy shift is just that, a shift in the energy of the bound state A. It has contributions from all photon energies and intermediate states. The intermediate states are unrestricted. That is, EI > EA is allowed. The electron is interacting with virtual photons in the vacuum. The imaginary part corresponds to the spontaneous decay rate from I to A. The delta

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function enforces conservation of energy. Only photons with energy EI −EA will contribute. Indeed there is an imaginary part only if there is a lower energy state available. Therefore 2 1 ΓA − Im[∆EA] = = ¯h τA ¯h Then −i(EA+Re[∆EA])t/h¯ −ΓAt/2¯h ψ(t) = | uAie e The probability of ﬁnding the state A |ψ(t)|2 ∼ e−Γt/h¯ Now consider the real part of the energy shift. We integrate over all of the photon phase space to account for the sum over photons.

X X 2 1 <∆EA = |HIA| (−EI + EA − ¯hω) photon I 2 2 2 Z 3 λ 2 c ¯h e X d kV 1 X |(p · )IA| = 3 V mc (2π) 2ω EA − EI − ¯hω I λ The sum over polarizations and the angular integration is the same as for our calculation of Rayleigh R P λ 2 2 scattering, namely, dΩ λ |(p · )IA| = (8π/3)|(p)IA| . Then the real part 2 2 Z 2 2 e 1 X Eγ |(p)IA| dEγ ∆EA = 2 3π 4π¯hc (mc) EA − EI − ¯hω I

2 The integral diverges. If we choose a cutoff, Emax = mc , (certainly our non-relativstic wave functions) will not be valid for energies greater than the electron rest mass, then the energy shift is equivalent to our earlier calculation of the Lamb shift. This time we take it a step further and consider the energy shift of a free electron. After all, a free electron can similarly interact with virtual photons. The self-energy of the free electron manifests itself as a shift in the mass of the free electron. We are looking only for the change in the bound state energy due the real potential. So we may be double counting. For the free electron, the difference is simply that there are no bound states. We integrate over all photon energies. Returning to equation 2.4 for the real part of the energy shift D e E H = p0, k | p · A | p IA mc r Z −ip0·x/h¯ ip·x/h¯ ¯hc e e −ik·x λ e 3 HIA = − √ e (p · ) √ d x 2kV mc V V where the initial and final states of the electron are plane waves (momentum eigenstates) with p and p0 respectively. The photon has energyhc ¯ k. Note that here we do not make the dipole approximation. By including the spatial dependence e−ik·x· in the integration d3x we end up with the momentum conserving delta function. For the energy shift of the bound state we did assume the dipole approximation. Are these results compatible? Then r ¯h e 1 H = − p · λδ(p0 − p + ¯hk). IA 2ω m V 1/2

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2 2 The denominator EA − EI − ¯hω → p /2m − (p − ¯hk) /2m − ¯hω ∼ −¯hω. Then as before

X X 2 1 <∆EA = |HIA| (−EI + EA − ¯hω) photon I 2 Z 3 ¯he λ 2 d kV 1 = 2 |p · | 3 2ωV m (2π) ¯hω Ea − EI − ¯hω

λ 2 8π 2 We average |p · | over initial state polarization and sum over ﬁnal → 3 p . ¯h e2 p2 V R k2dkdΩ ∆E = − A 2ω m2 V (2π)3¯hω 8π ¯h e2 p2 Z d3kV = − 3 2ω m2 V (2π)3¯hω 2 e2 1 Z = − p2 ¯hdω 3π 4π¯hc (mc)2 2 e2 p2 Z = − dE 3π 4π¯hc (mc)2 γ = Cp2

Insofar as 1 ∂E = 2m ∂p2 we ﬁnd that the observed mass 1 1 = + C → mobs = mbare(1 − 2mbareC) 2mobs 2mbare

2 assuming C is small which it will be if we choose Emax = mc . Our calculation of energy of state A assumed the observed rather than the bare mass. We need to correct. Let’s subtract the contribution for the free particle with momentum A. Note that 2 P 2 (p )AA = I |(p)IA|

2 Z 2 ! observed 2 e 1 X Eγ |(p)IA| 2 ∆EA = 2 + (p )AA dEγ 3π 4π¯hc (mc) EA − EI − ¯hω I 2 Z 2 ! 2 e 1 X |(p)IA| (EA − EI ) = 2 dEγ 3π 4π¯hc (mc) EA − EI − ¯hω I 2 max ! 2 e 1 X 2 Eγ = 2 |(p)IA| (EI − EA) log 3π 4π¯hc (mc) EA − EI I

The expression still diverges, but now logarithmically rather than linearly so much less sensitive to the cutoﬀ. With some eﬀort it is possible to show that X 1 Z |(p) |2(E − E ) = − ¯h |ψ |2∇2V d3x IA A I 2 A I

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p2 [Let’s try to do that. For the Hamiltonian H0 = 2m + V ,

pH0 − H0p = −i¯h∇V

hI | pH0 − H0p | Ai = hI | −i¯h∇V | Ai

pIAEA − EI pIA = −i¯h(∇V )IA

Multiply by pAI and sum and note that the result must be real

X 2 X |pIA| (EA − EI ) = −i¯h pAI · (∇V )IA I X = i¯h (∇V )AI · pIA ¯h X = −i [p, (∇V )] 2 AA ¯h2 = − (∇2V ) 2 AA ] which for the hydrogen atom becomes ∇2V = e2δ(x), so

X 1 |(p) |2(E − E ) = − e2¯h2|ψ (0)|2 IA A I 2 A I The energy shift becomes

2 max observed 2 e 1 1 2 2 2 Eγ ∆EA = 2 e ¯h |ψA(0)| log 3π 4π¯hc (mc) 2 hEA − EI iave 3 3 max 1 2 ¯h c 2 Eγ = α 4π 2 2 |ψA(0)| log 3π (mc ) hEA − EI iave 3 3 max 1 2 ¯h c 1 Eγ = α 4π 2 2 3 log 3π (mc ) πa0 hEA − EI iave 3 3 3 max 1 2 ¯h c (αmc) Eγ = α 4π 2 2 3 log 3π (mc ) π¯h hEA − EI iave 4 Emax = α5mc2 log γ 3π hEA − EI iave max 8 3 Eγ = α Eryd log 3π hEA − EI iave

max 2 If Eγ = mc then 8 ∆Eobserved = 4.9 α3E A 3π ryd