Quantization in the Radiation Gauge

A^α(t, x), ˙A^β(t, y)



=−iη^αβδ(x− y) = iδ^αβδ(x− y) (3.164) we run into trouble if we take the divergence of both sides with respect to x^α. We find that



∂αA^α(t, x), ˙A^β(t, y)



=−i∂^βδ(x− y) = 0 (3.165) violating the∇ · A = 0 condition. The second gauge⁵²is Lorentz invariant but if we impose the commutation relation



Aⁱ(t, x), ˙A^j(t, y)



=−iη^ijδ(x− y) (3.166) The time component again creates a problem because the right hand side now has a flipped sign for the time component vis-a-vis the space com-ponent. This leads to the existence of negative norm states and related difficulties. All these can be handled but not as trivially as in the case of scalar fields. We will use the first gauge and get around the difficulty mentioned above by a different route.

3.6.1 Quantization in the Radiation Gauge

Let us consider a free electromagnetic field, with the vector potential Aⁱ= (φ, A) satisfying the gauge conditions: φ = 0,∇ · A = 0. As usual we expand A (t, x) in Fourier space as:

A(t, x) =

 d³k

(2π)³qk(t) exp ik.x =

 d³k (2π)³

ak(t) +a^∗_−k(t) e^ik.x

(3.167)

51If you write the Lagrangian as (E²− B²) and note that E = −∇φ − (∂A/∂t), B = ∇ × A, it is obvious that the Lagrangian has ˙A but no ˙φ.

So clearly the momentum conjugate to φ vanishes.

52Almost universally called the Lorentz gauge because it was discov-ered by Lorenz ; yet another example of dubious naming conventions used in this subject. This gauge was first used by L. V. Lorenz in 1867 when the more famous H.A. Lorentz was just 14 years old and wasn’t particularly concerned about gauges. We shall give the credit to Lorenz and use the correct spelling. The original reference is L.Lorenz, On the Identity of the Vibrations of Light with Electrical Currents, Philos. Mag. 34, 287-301 (1867).

53Notice that the operator a^†_kλ cre-ates a photon with a polarization state labelled by λ and momentum labelled by k which will define a direction (of propagation) in the 3-dimensional space. The polarization vectors ^α(k, λ) necessarily have to de-pend on the momentum vector k of the photon. For example, if we try to pick 3 four-vectors with compo-nents ¹_a= (0, 1, 0, 0), ²_a = (0, 0, 1, 0) and ³_a = (0, 0, 0, 1) as the bases, the Lorentz transformations will mix them up with the “time-like polarization”

vector ⁰_a= (1, 0, 0, 0).

which has exactly the same structure as in the case of the scalar field, given by Eq. (3.105). The condition∇ · A = 0, translates into

k· qk= k· ak= k· a^∗k= 0; (3.168) i.e., for every value of k, the vectorak is perpendicular to k. This allows the vectors q_k,akto have two componentsakλ(λ = 1, 2) and q_kλ(λ = 1, 2) in the plane perpendicular to k with {ak1,ak2, k} forming an orthogonal system. The electric and magnetic fields corresponding to this vector po-tential Aⁱ= (0, A) are

E=− ˙A = −

 d³k

(2π)³q˙_ke^ik·x; B=∇ × A = i

 d³k

(2π)³(k× qk) e^ik·x. (3.169) Substituting this into the action in Eq. (3.63) and using

(k× qk)· (k × q^∗k) = q^∗_k· [(k × qk)× k] = (q^∗k· qk) k² (3.170) we can again express the action as that of a sum of oscillators:

A = 1 2

 d³x

E²− B²

= 1 2



λ=1,2

 d³k (2π)³

| ˙qkλ|²− k²|qkλ|²

. (3.171)

So, once again, we find that the action for the electromagnetic field can be expressed as a sum of the actions for harmonic oscillators with each oscillator labeled by the wave vector k and a polarization. The polarization part is the only extra complication compared to the scalar field and the rest of the mathematics proceeds exactly as before. The dispersion relation now corresponds to ω_k =|k| showing that the quanta are massless, as we expect for photons.

The wave equation satisfied by A (t, x) implies that q_k(t) satisfies the harmonic oscillator equation ¨q_k+ k²q_k = 0, allowing us to take the time evolution to beak∝ exp (−ikt). With the usual normalization, this gives the mode expansion for the vector potential in the standard form

A(t, x) = 

λ=1,2

 d³k (2π)³

√1 2ω_k

akλe^−ikx+a^†kλe^ikx



(3.172)

It is convenient in manipulations to separate out the vector nature from the creation and annihilation operators and writeakλ=(k, λ)akλ where a_kλ is the annihilation operator for the photon with momentum k and polarization λ while (k, λ) is just a c-number vector carrying the vector nature of A. Since (k, 1), (k, 2) and k/|k| form an orthonormal basis, they satisfy the constraint:⁵³

2 λ=1

^α(k, λ) ^β(k, λ) +k^αk^β

k² = δ^αβ (3.173)

The creation and annihilation operators satisfy the standard commutation rules, now with an extra polarization index:



ak^λ^, a^†_kλ



= δ(k^− k)δλλ^, [ak^λ^, akλ] =



a^†_kλ, a^†_kλ



= 0 (3.174)

Everything else proceeds as before. The Hamiltonian governing the free electromagnetic field can be expressed as a sum of Hamiltonians for har-monic oscillators, with each oscillator labeled by a wave vector k and polar-ization index α. The quantum states of the system are labeled by by a set of integers|{nkα}, one for each oscillator labeled by kα. The expectation value of H in such a state will be

E = 

α=1,2

 d³k (2π)³ωk



n_kα+1 2



(3.175)

where the second term is the the energy of the system when all the oscil-lators are in the ground state — which can again be removed by normal ordering.

You might have thought all this is pretty obvious and standard. But, if you think about it, the second term in Eq. (3.172) involvinga^†_kλis pre-cisely the term about which we made a song and dance in Eq. (1.130) while discussing the scalar field. It is related to propagating negative energy an-tiphotons backward in time! The quantization of the electromagnetic field based on the separation in Eq. (3.172) describes photons and antiphotons, carrying positive frequency and negative frequency modes and propagating forward and backward in the language we have used earlier.⁵⁴ This term creates an antiphoton from the vacuum, but since the antiphoton is the same as the photon, you think of it as a creation of photons — which every light bulb does. This is the simplest example of a particle popping out of nowhere thereby making the standard description in terms of a Schrodinger equation completely inadequate. The more esoteric processes — like, for example, e⁺− e⁻ annihilation producing some exotic new particles is fun-damentally no different from a light bulb emitting a photon through the action of the second term which hasa^†_kλ in it.

Let us next take a look at the commutation relation between the fields.

Once again, we have quantized a field by mapping the relevant dynamical variables to a bunch of independent harmonic oscillators and then quan-tizing the oscillators. While discussing the scalar field, we said that this procedure is equivalent to imposing the canonical commutation relations between the dynamical variables and the canonical momenta; indeed — in the case of the scalar field — one can derive the commutation rela-tions between the creation and annihilation operators from the canonical commutation relations between the dynamical variables and the canonical momenta. But in the case of the electromagnetic field in the radiation gauge, the situation is somewhat different. We know from Eq. (3.164) and Eq. (3.165) that we cannot impose the standard commutation relations and maintain consistency with the radiation gauge condition. So, one cannot start from Eq. (3.164) and get the commutation rules for the creation and annihilation operators in a straightforward manner.

Fortunately, our procedure — of first identifying the relevant oscillators and then just quantizing them — bypasses this problem. Having quantized the system by this procedure, we can now go back and compute the relevant commutators between the field and its momentum and see what happens.

The canonical momenta⁵⁵corresponding to A^α:

π^α= ∂L

∂(∂0Aα) =−∂0A^α= E^α (3.176)

54While every textbook makes some noise about these “new features” when it quantizes the scalar field, these are not emphasized in the context of the familiar electromagnetic field. You talk about π⁺ and π⁻, electrons and positrons and their strange relation-ships but the same thing happens with photons and antiphotons contained in Eq. (3.172). It is the same maths and same physics (except for m = 0).

55Note the placement of α and the sign flips due to our signature; the components of A, E etc. have, by def-inition, a superscript index.

Exercise 3.18: Prove this.

56The way we approached the prob-lem, this should not worry you; find-ing the oscillators and quantizfind-ing them is the safest procedure, when it can be done. If, instead, you start from the commutator for fields and canoni-cal momenta, you need to first argue — somewhat unconvincingly — that we should introduce the transverse delta function for the commutators and then proceed with the quantization.

where the last equality defines the transverse delta function:

δ_αβ^⊥ (x− x^)≡ space representation) thereby making the commutation relations consis-tent with ∇ · A = 0. One can write down the explicit form of δ_⊥^αβ(x) in coordinate space by noting that it can be written, formally, as

δ^⊥_αβ(x) = By defining the inverse of the Laplacian as an integral operator, one can easily show that: have. In the radiation gauge, if we identify the correct harmonic oscillator variables and quantise them, everything works fine and this commutator has⁵⁶a value given by Eq. (3.179).

The price we have paid is the loss of manifest Lorentz invariance and, of course, loss of gauge invariance. The first problem can be avoided by using a Lorentz invariant gauge like the Lorenz gauge. The second problem is going to stay with us because we have to always fix a gauge — naively or in a more sophisticated manner — to isolate the physical variables but we expect the meaningful results to be gauge invariant. We will say more about this in the next section.

Let us next consider the commutator between the fields at arbitrary events. A straightforward calculation gives:

A^α(x), A^β(y) This can again be expressed in the coordinate space using the non-local projection operator and in terms of the function Δ(x)≡ G+(x)− G+(−x) we introduced in Eq. (1.89) but now evaluated for m = 0; it is usual to define

D(x)≡ −iΔ(x)|m=0=−1

2πθ(t)δ(x²) (3.184) in the massless case where the last equality is easy to obtain from any of the integral representations for Δ. We find

A^α(x), A^β(y)

A straightforward evaluation of the integrals reveals, at first sight, another problem: this commutator now does not vanish outside the light cone — something we have been claiming is rather sacred! But this is sacred only for observable quantities and Aⁱ is not an observable. (In fact, if we use some other gauge we will get a different result for this commutator; in the Lorenz gauge it does happen to vanish outside the light cone.) If the commutator between the electric fields, for example, does not vanish outside the light cone, you will be in real trouble. This, as you might have guessed from the fact that we are still in business, does not happen. The commutator of the electric fields is given by:

E^α(x), E^β(y)

which vanishes for spacelike separation. Further, this commutator has the same expression in all gauges, as one would have expected.