17. Quantization of Electromagnetic Fields 17A. 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 c2 = 1/(00): Integrate to get the total energy:

17B. 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 = L3, 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 condi- tions, 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:

17C. 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

17D. 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 n1 photons with wave number k1 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 , where 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:

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 at all times Use some simple identities:

17E. 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  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

17F. Taking the Infinite Volume Limit We said we were going to take L3 = 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 1D: In three dimensions this becomes:

17G. 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 Ef2(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 Ef2(r). It is obvious that as usual: For Ef2(r): Take limit V  , then let a2k2 = x

Can We Really Ignore the Infinite Energy? 17H. The Casimir Effect Can We Really Ignore the Infinite Energy? We found the expectation value of E2 and B2: 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 L2, 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 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

Energy in a Capacitor Each mode contributes an energy L 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 L 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