17 Quantization of Electromagnetic Fields 17 A Gauge

17. Quantization of Electromagnetic Fields 17 A. Gauge Choice and Energy Coulomb Gauge • We want to quantize electromagnetic fields • It is unfortunately insufficient to work with E and B – We must work with A and U instead • Given E and B, the choice of A and U is ambiguous – We can perform a gauge transformation • To avoid this ambiguity, pick a gauge • We will use Coulomb gauge: • Can you always do this? Suppose • Want to pick to cancel this • We want • This is Poisson’s equation • Solution is well known:

Working With Coulomb Gauge • In Coulomb gauge, we can find the electric potential U(r, t) • This is Poisson equation again • In the absence of sources, U = 0 Energy Density and Energy • Energy density of EM fields: • Use the fact that c 2 = 1/( 0 0): • Integrate to get the total energy:

17 B. Fourier Modes and Polarization Vectors Finite Volume and Fourier Modes • Although it isn’t obvious, this is just a lot of coupled Harmonic oscillators – The curl terms just couple A(r) to “adjacent” values of r • As an aid to thinking about things, we will pretend the universe has volume V = L 3, with periodic boundary conditions – Later we will take L • Periodic boundary conditions means that any function repeats if you go a distance L in any direction • To decouple modes, write A(r, t) in terms of Fourier modes • Because of periodic boundary conditions, k is restricted to certain values:

Writing the Energy in Fourier Modes: • We now substitute this into the expression for E: • Then for example, the x-integral yields • Integral is only non-zero if k + k' = 0 • This lets us do integrals and one sum:

Some Restrictions on A: • We must make sure A(r, t) is real: • Effectively, there are only half as many Ak’s as you think there are • We are also working in Coulomb gauge: • Use these expressions to simplify E: • We have some redundancies in the sum – Because Ak and A–k are related • To avoid redundancy, arbitrarily pick half of the k’s to be positive • Rewrite sums over just half of the k’s

Polarizations: • Ak actually has only two components • Choose two unit vectors k for = 1, 2, the polarization vectors, such that: • Usually pick them real, but you don’t have to • Then any vector Ak perpendicular to k can be written: • Our energy is now:

17 C. Quantizing the Electromagnetic Fields Comparison to Complex Harmonic Oscillator • • Compare to the complex harmonic oscillator: This is just a sum of complex harmonic oscillators! We need the analogous relations: Old formulas: • New formulas:

Simplifying the Notation • The a’s have commutation relations: – All other commutators vanish • Keep in mind, we have restricted k’s to half of the values, just k > 0 • Unnecessarily complicated, since we have k label and label • Easier to combine this to a single notation • Now commutation relations: – All others vanish • The expressions above then simplify, slightly

Vector Potential as an Operator • We want to write out A in terms of a’s: • • Not manifestly Hermitian Can be fixed by redoing second term in sum with k –k Also, some complicated details of how we specified modes demands We therefore have:

Electric and Magnetic Field Operators • To get magnetic field, just use • For electric field, we use • As before, on last term, let k –k • Note all fields are functions of position r, but no longer time t

Comments on Field Operators • Note that time dependence is now gone • All fields are now operators Compare to ordinary quantum: • Position and momentum are classically functions of time, x(t) and p(t) • When you quantize it, they become operators X and P • Any time dependence will be in the state vector | (t) • Just like any other operator, field operators A(r), etc. , will be uncertain • You can measure things like E(r) and B(r) – You can’t measure A(r) because it’s not gauge invariant • We have an infinite number of operators – Which makes describing quantum states | (t) hard

17 D. Eigenstates of the Electromagnetic Field Subtracting Infinity from the Energy • • Our Hamiltonian is a sum of harmonic oscillators Ground state we will call |0 ; it has the property for all k and Problem: the energy of this state is infinite: This infinity persists even if you divide by the volume V of the universe • Solution – you can always add a constant to the Hamiltonian • Energy for all states is now finite • These infinities come up repeatedly in quantum field theory • They are generally dealt with by a process called renormalization – This one is the easiest to deal with

Labeling General Eigenstates • Normally give eigenvalues of all number operators • This would be an infinitely long list • Most of these states would have infinite energy To avoid this problem: • List only states that are non-zero: – You have n 1 photons with wave number k 1 and spin 1, etc. • Any terms with ni = 0 is irrelevant • Since it’s just a list, order doesn’t matter • Energy of this state is: • State can be constructed from the vacuum state as

Abbreviations and Notation • Sometimes understood that you only have one photon of a given type. Drop ni: • Do we ever write wave functions? No! You will go insane! Need to get good at how raising and lowering operators affect things: • Lowering operator decreases number of photons present if that photon is present to begin with: • Raising operator creates a photon whether one was present or not: • Whenever possible, try to keep expressions simple; lower first

Sample Problem An electromagnetic system at t = 0 is in the state and. (a) Find the state vector at all times (b) Find the expectation value of the electric field at all times • General solution to Schrödinger’s equation is: • The two portions of (0) given are both eigenvectors of H • Each of these terms will just pick up a phase • Electric field operator is: • So we want: , where

Sample Problem (2) (b) Find the expectation value of the electric field at all times • We must either increase the number of photons by one, or decrease it by one • The only terms that matter are therefore:

Sample Problem (3) (b) Find the expectation value of the electric field • Use some simple identities: at all times

17 E. Momentum of Quantum States Begin the Calculation • • • We have been talking as if we have particles (“photons”) with energy To cement this concept, let’s find the momentum We first need the momentum density from E & M: Integrate it over space to get the momentum: Recall our expressions for E and B: • Substitute them into Pem:

Integrals and One Sum are Easy • The integrals we have seen before: • Substitute this in. Note that the k’s automatically will cancel • Do the sum on k' • Expand the triple products. Use that k is perpendicular to k:

More Stuff Goes Away On first and last term: • Note that if you take k –k, and exchange ' the expressions would be identical • Because all k is summed over, and ' are summed over, this means there is an identical cancelling sum somewhere in the expression • Recall our polarization vectors are orthonormal • This makes the ' sum easy: • Commutation relations allow us to rewrite • Our penultimate form:

Answer and Interpretation: • On the last term, positive values of k will be cancelled by negative values • Eigenstates of H will have momentum: • Clearly, each photon has momentum p = k • Since the energy was , and = ck, this tells us E = cp – Correct for relativistic particles

17 F. Taking the Infinite Volume Limit • We said we were going to take L 3 = V , but how do we do this? • First step: Consider the following integral • We know how to do this in both finite and infinite volume • So we have: • • • Second step: Consider an integral over k in one dimension Recall how this is defined Remember that our k’s are discrete: So we have Therefore, in 1 D: In three dimensions this becomes:

17 G. The Nature of the Vacuum Things Are Not As Simple As They Seem • What is the value of the electric or magnetic field for the vacuum state |0 ? • These fields are operators, so like other operators, expect them to have a distribution – With an expectation value and an uncertainty • Let’s find expectation value and uncertainty • Let them act on the vacuum state • The annihilation terms vanish • The creation terms create one-particle states

So We Find: • It is trivial to see that: • On the other hand, the expectation value of the square of E is: • Take limit V : • Similar computation for magnetic field:

Interpretation • It is easy to see that both of these integrals diverge • Implies there should be infinite random electric fields, even for empty space – How come we don’t notice this? • All probes of electric or magnetic field are in fact finite in extent • Even though we think we are measuring E(r), we are actually measuring • The “aperture function” f(s) tells how the measurement is spread over space • Normalized to 1 • The function f smooths out E, suppressing high momentum modes • Effectively cuts off integrals over k • This will lead to finite values:

Sample Problem The electric field of the vacuum is measured using the aperture function: Find the expectation values for Ef(r) and Ef 2(r). • We first find • So we need

Sample Problem (2) The electric field of the vacuum is measured using the aperture function: Find the expectation values for Ef(r) and Ef 2(r). • It is obvious that as usual: • For Ef 2(r): • Take limit V , then let a 2 k 2 = x

17 H. The Casimir Effect Can We Really Ignore the Infinite Energy? • We found the expectation value of E 2 and B 2: • Energy density expectation value is then • This is the same infinity as the infinity we discarded in H • As Sidney Coleman of Harvard used to say: “Just because something is infinite doesn’t mean it’s zero!” • Though it is unlikely any measurement will ever yield infinity, maybe this will contribute if we change the Hamiltonian

Modes in a Capacitor • Consider a parallel plate capacitor – Area L 2, separation a – Assume L >> a • E|| must vanish at the conducting surface • We can get the modes to vanish on the boundaries z = 0 and z = a if we use modes – nx = 0, 1, 2, … – nz = 0, 1, 2, … • We still need k Ak = 0 • This means there will be two polarization modes • We can also get modes with E|| = 0 by choosing • We still need k Ak = 0 • To get it E|| = 0 on the boundary • Which we can arrange by picking Ak in the z-direction • Which means there is only one mode L L a

Energy in a Capacitor • Each mode contributes an energy • Total energy for separation a is therefore • For kx and ky, we can take the limit L using the usual rule • The energy density in the absence of the capacitor is • So in the energy in this same volume would be • Maybe we can find the difference in energy • Problem – this is infinity minus infinity L L a

Regulating the Sum • These sums/integrals are both infinite • Physically, at very high frequency, all conductors become transparent anyway • Add a regulator function that cuts off integrals at infinity • Can show the form of the regulator doesn’t make much difference • We will pick the cutoff function e– k L L a

Do Some Sums • Sum on polarizations, and substitute for kz: • Some sum theorems: • We therefore have: • Write in terms of • Cutoff frequency will be high, so w will be small

Take the Limit and Interpret • Expand in the limit of small w – Let Maple can do it for us • Note that the energy is negative – Capacitor plates prefer to be near each other • Take derivative to get a force • Note this is force per unit area, or pressure – Negative pressure, since it attracts • Experimentally observed L L a