## 2013-03-20

### Nonzero Electromagnetic Fields in a Cavity

Anyway, I've been able to read some more papers on the subject, including Casimir's original paper on it as well as Lifshitz's paper going a little further with it. One of the things that confused me in those papers (and in my recitation leader's explanation, which was basically the same thing) was the following. The explanation ends with the notion that quantum electrodynamic fluctuations in a space with a given dielectric constant, say in a vacuum surrounded by two metal plates, will cause those metal plates to attract or repel in a manner dependent on their separation. This depends on the separation being comparable to the wavelength of the electromagnetic field (or something like that), because at much larger distances, the power of normal blackbody radiation (which ironically still requires quantum mechanics to be explained) does not depend on the separation of the two objects, nor does it really depend on their geometries, but only on their temperatures. The explanation of the Casimir effect starts with the notion of an electromagnetic field confined between two infinite perfectly conducting parallel plates, so the fields form standing waves like the wavefunctions of a quantum particle in an infinite square well. This is all fine and dandy...except that this presumes that there is an electromagnetic field. This confused me: why should one assume the existence of an electromagnetic field, and why couldn't it be possible to assume that there really is no field between the plates?

Then I remembered what the deal is with quantization of the electromagnetic field and photon states from 8.05 — Quantum Physics II. The derivation from that class still seems quite fascinating to me, so I'm going to repost it here. You don't need to know QED or QFT, but you do need to be familiar with Dirac notation and at least a little comfortable with the quantization of the simple harmonic oscillator.

Let us first get the classical picture straight. Consider an electromagnetic field inside a cavity of volume $\mathcal{V}$. Let us only consider the lowest-energy mode, which is when $k_x = k_y = 0$ so only $k_z > 0$, stemming from the appropriate application of boundary conditions. The energy density of the system can be given as $H = \frac{1}{8\pi} \left(\vec{E}^2 + \vec{B}^2 \right)$ and the fields that solve the dynamic Maxwell equations $\nabla \times \vec{E} = -\frac{1}{c} \frac{\partial \vec{B}}{\partial t}$ $\nabla \times \vec{B} = \frac{1}{c} \frac{\partial \vec{E}}{\partial t}$ as well as the source-free Maxwell equations $\nabla \cdot \vec{E} = \nabla \cdot \vec{B} = 0$ can be written as $\vec{E} = \sqrt{\frac{8\pi}{\mathcal{V}}} \omega Q(t) \sin(kz) \vec{e}_x$ $\vec{B} = \sqrt{\frac{8\pi}{\mathcal{V}}} P(t) \cos(kz) \vec{e}_y$ where $\vec{k} = k_z \vec{e}_z = k\vec{e}_z$ and $\omega = c|\vec{k}|$. The prefactor comes from normalization, the spatial dependence and direction come from boundary conditions, and the time dependence is somewhat arbitrary. I think this is because the spatial conditions are unaffected by time dependence if they are separable, and the Maxwell equations are linear so if a periodic function like a sinusoid or complex exponential in time satisfies Maxwell time evolution, so does any arbitrary superposition (Fourier series) thereof. That said, I'm not entirely sure about that point. Also note that $P$ and $Q$ are not entirely arbitrary, because they are restricted by the Maxwell equations. Plugging the fields into those equations yields conditions on $P$ and $Q$ given by $\dot{Q} = P$ $\dot{P} = -\omega^2 Q$ which looks suspiciously like simple harmonic motion. Indeed, plugging these electromagnetic field components into the Hamiltonian [density] yields $H = \frac{1}{2} \left(P^2 + \omega^2 Q^2 \right)$ which is the equation for a simple harmonic oscillator with $m = 1$; this is because the electromagnetic field has no mass, so there is no characteristic mass term to stick into the equation. Note that these quantities have a canonical Poisson bracket $\{Q, P\} = 1$, so $Q$ can be identified as a position and $P$ can be identified as a momentum, though they are actually neither of those things but are simply mathematical conveniences to simplify expressions involving the fields; this will become useful shortly.

Quantizing this yields turns the canonical Poisson bracket relation into the canonical commutation relation $[Q, P] = i\hbar$. This also implies that $[E_a, B_b] \neq 0$, which is huge: this means that states of the photon cannot have definite values for both the electric and magnetic fields simultaneously, just as a quantum mechanical particle state cannot have both a definite position and momentum. Now the fields themselves are operators that depend on space and time as parameters, while the states are now vectors in a Hilbert space defined for a given mode $\vec{k}$, which has been chosen in this case as $\vec{k} = k\vec{e}_z$ for some allowed value of $k$. The raising and lowering operators $a$ and $a^{\dagger}$ can be defined in the usual way but with the substitutions $m \rightarrow 1$, $x \rightarrow Q$, and $p \rightarrow P$. The Hamiltonian then becomes $H = \hbar\omega \cdot \left(a^{\dagger} a + \frac{1}{2} \right)$, where again $\omega = c|\vec{k}|$ for the given mode $\vec{k}$. This means that eigenstates of the Hamiltonian are the usual $|n\rangle$, where $n$ specifies the number of photons which have mode $\vec{k}$ and therefore frequency $\omega$; this is in contrast to the single particle harmonic oscillator eigenstate $|n\rangle$ which specifies that there is only one particle and it has energy $E_n = \hbar \omega \cdot \left(n + \frac{1}{2} \right)$. This makes sense on two counts: for one, photons are bosons, so multiple photons should be able to occupy the same mode, and for another, each photon carries energy $\hbar\omega$, so adding a photon to a mode should increase the energy of the system by a unit of the energy of that mode, and indeed it does. Also note that these number eigenstates are not eigenstates of either the electric or the magnetic fields, just as normal particle harmonic oscillator eigenstates are not eigenstates of either position or momentum. (As an aside, the reason why lasers are called coherent is because they are composed of light in coherent states of a given mode satisfying $a|\alpha\rangle = \alpha \cdot |\alpha\rangle$ where $\alpha \in \mathbb{C}$. These, as opposed to energy/number eigenstates, are physically realizable.)

So what does this have to do with quantum fluctuations in a cavity? Well, if you notice, just as with the usual quantum harmonic oscillator, this Hamiltonian has a ground state energy above the minimum of the potential given by $\frac{1}{2} \hbar\omega$ for a given mode; this corresponds to having no photons in that mode. Hence, even an electrodynamic vacuum has a nonzero ground state energy. Equally important is the fact that while the mean fields $\langle 0|\vec{E}|0\rangle = \langle 0|\vec{B}|0\rangle = \vec{0}$, the field fluctuations $\langle 0|\vec{E}^2|0\rangle \neq 0$ and $\langle 0|\vec{B}^2|0 \rangle \neq 0$; thus, the electromagnetic fields fluctuate with some nonzero variance even in the absence of photons. This relieves the confusion I was having earlier about why any analysis of the Casimir effect assumes the presence of an electromagnetic field in a cavity by way of nonzero fluctuations even when no photons are present. Just to tie up the loose ends, because the Casimir effect is introduced as having the electromagnetic field in a cavity, the allowed modes are standing waves with wavevectors given by $\vec{k} = k_x \vec{e}_x + k_y \vec{e}_y + \frac{\pi n_z}{l} \vec{e}_z$ where $n_z \in \mathbb{Z}$, assuming that the cavity bounds the fields along $\vec{e}_z$ but the other directions are left unspecified. This means that each different value of $\vec{k}$ specifies a different harmonic oscillator, and each of those different harmonic oscillators is in the ground state in the absence of photons. You'll be hearing more about this in the near future, but for now, thinking through this helped me clear up my basic misunderstandings, and I hope anyone else who was having the same misunderstandings feels more comfortable with this now.