1. 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:

2. 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:

3. 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:

4. 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:

5. 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

6. 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:

7. 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:

8. 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

9. 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:

10. 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

11. 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

12. 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

13. 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

14. 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

15. 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:

16. 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:

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

18. 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:

19. 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:

20. 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:

21. 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

22. 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:

23. 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

24. 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:

25. 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:

26. 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

27. 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

28. 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

29. 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

30. 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

31. 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

32. 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

33. 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