# Quantization of single mode field in a cavity

I am studying quantum optics from Gerry and Knight's book. This question has to do with section 2.1 of the book, where a single mode light field in a cavity is quantized.

1. The formal equivalence of a single-mode field and a harmonic oscillator of unit mass is established just by showing that the form of the Hamiltonian is same in both cases. Is this the necessary and sufficient condition for formal equivalence? What does formal exactly equivalence mean here?

2. How does the correspondence in Hamiltonians imply correspondence in all properties of the operators (their commutation relations, their hermiticity, their physical observability)?

## 1 Answer

1. For the quantum harmonic oscillator one can show that the Hamiltonian can be expressed in terms of ladder operators as $H=\hbar \omega (\hat{a}^\dagger \hat{a} + \frac{1}{2})$ where $[\hat{a}^\dagger, \hat{a}]=1$. Subtracting the constant term from the Hamiltonian, which does not change the dynamics of the system, gives $H=\hbar \omega \hat{a}^\dagger \hat{a}$.
2. Photons are bosons. They therefore obey bosonic commutation relations.
3. The single cavity mode is assumed to be discrete.
4. We can define an operator $\hat{b}^\dagger$ that puts a photon in the mode. Because of 2 and 3 the commutation relations therefore have to be $[\hat{b}^\dagger, \hat{b}]=1$. This agrees with the harmonic oscillator.
5. We assume the mode to be at energy $\hbar \omega'$. The energy in the mode will then be the number of photons in the mode times $\hbar \omega'$. Therefore $H=\hbar \omega' \hat{b}^\dagger \hat{b}$. This agrees with the harmonic oscillator up to identification of the frequencies.
6. We have therefore shown that the Hamiltonian of the system and the commutation relations of the particle creation operators are the same. Therefore the systems will display the same dynamics, i.e. are equivalent.

What does formal exactly equivalence mean here?

That the systems behave the same dynamically if the correct identifications are made. In this case this means:

• Cavity mode = harmonic oscillator
• Cavity photon creation operator = Harmonic oscillator creation operator
• $\omega'=\omega$

Note that the main reason why this works is because photons are bosons (hence you can put infinitely many in one mode at the same energy) and the energy levels of the harmonic oscillator have constant spacing (and there are infinitely many).

The formal equivalence of a single-mode field and a harmonic oscillator of unit mass is established just by showing that the form of the Hamiltonian is same in both cases. Is this the necessary and sufficient condition for formal equivalence?

It is neither. It can not be necessary since you can easily do some operator redefinitions (like projecting onto a different basis). It is not sufficient since you also need the operators to have the same commutation relations.

But they do have the same commutation relations, so the analogy is fine.

How does the correspondence in Hamiltonians imply correspondence in all properties of the operators (their commutation relations, their hermiticity [...]

It does not.

[...] their physical observability)?

Well, now that can indeed be quite different and would also depend on how you implement your harmonic oscillator in a physical system. Hm, why not take a some photons in a cavity and look at the cool stuff they can do. :)

• Still, I was wondering why do we have to "assume" the single cavity mode to be discrete? Is it not ensured by the boundary conditions in any case? Also, why is the energy of the mode "assumed" to have that expression? May 12 '17 at 9:41
• @Navya In case of a perfect cavity it is quite easy to see by looking at how you quantize the field (see e.g. isites.harvard.edu/fs/docs/icb.topic820704.files/…). I can give you an intuitive reason too: canonical quantization is a procedure where you assign each mode of the system an operator and assign bosonic/fermionic commutation relations. The modes are just standing waves (the solutions of Maxwell equations for more complicated geometries). Those are obviously discrete if the cavity does not extend to infinity (as you say, due to boundary conditions). May 12 '17 at 11:55
• @Navya the absolute energy of these is somewhat arbitrary, since you can add arbitrary constants to the Hamiltonian. May 12 '17 at 11:56