## 2013-04-15

### Harmonic Oscillator from Fields not Potentials

I finally started writing my paper for 8.06 yesterday. Before that, though, I had asked a couple questions about the topic to my UROP supervisor, whose primary area of expertise is actually in QED and Casimir problems. I was asking him why the book The Quantum Vacuum by Peter Milonni uses the magnetic potential $\vec{A}$ instead of the electromagnetic fields $\vec{E}$ and $\vec{B}$ to expand in Fourier modes and derive the harmonic oscillator Hamiltonian. He said that is just a matter of choice, and in fact the same derivation can be done using the fields rather than the potential; moreover, he encouraged me to try this out for myself, and I did just that. Lo and behold, the right answer popped out by modifying the derivation in that book to use the fields and the restrictions of the Maxwell equations instead of the potential and the Coulomb gauge choice; follow the jump to see what it looks like. I'm going to basically write out the derivation in the book and show how at each point I modify it.

The starting point is the same and lies in the Maxwell equations without charges or currents: $\nabla \cdot \vec{E} = 0 \\ \nabla \cdot \vec{B} = 0 \\ \nabla \times \vec{E} = -\frac{1}{c} \frac{\partial \vec{B}}{\partial t} \\ \nabla \times \vec{B} = +\frac{1}{c} \frac{\partial \vec{E}}{\partial t} .$
That is all fine and dandy. Then the book decides to introduce the magnetic potential $\vec{A}$ as satisfying the wave equation $\nabla^2 \vec{A} = \frac{1}{c^2} \frac{\partial^2 \vec{A}}{\partial t^2}$ contingent on the Coulomb gauge choice $\nabla \cdot \vec{A} = 0$ along with $\phi = 0$ where $\vec{E} = -\left(\nabla\phi + \frac{1}{c} \frac{\partial \vec{A}}{\partial t}\right)$. Instead, I stick with the electric field satisfying the wave equation arising out of the Maxwell equations: $\nabla^2 \vec{E} = \frac{1}{c^2} \frac{\partial^2 \vec{E}}{\partial t^2} .$ The solutions of a given mode are separable and are generally complex, but $\vec{E}$ and $\vec{B}$ can be chosen as respectively the real and imaginary parts thereof: $\vec{E}(\vec{r}, t) = \alpha^{\star} (t) \vec{E}_0^{\star} (\vec{r}) + \alpha(t) \vec{E}_0 (\vec{r})$ from the wave equation, where $\alpha(t) = \alpha(0)e^{-i\omega t}$ solving $\ddot{\alpha} = -\omega^2 \alpha$ & $\vec{E}_0$ satisfies the Helmholtz equation $\nabla^2 \vec{E}_0 = -k^2 \vec{E}_0$ where $k = |\vec{k}|$ & $\omega = ck$, and from the Faraday law, $\vec{B} (\vec{r}, t) = \frac{ic}{\omega} \left(\alpha^{\star} (t) \nabla \times \vec{E}_0^{\star} (\vec{r}) - \alpha(t) \nabla \times \vec{E}_0 (\vec{r})\right) .$

Digressing for a moment, it is instructive to instead consider the Fourier transforms of the electric field as written and of the magnetic field from the Ampère law instead, as these are anyway the fields of a single mode: $\tilde{\vec{E}} (\vec{k}, t) = \alpha^{\star}_{\vec{k}} (t) \tilde{\vec{E}}_0^{\star} (\vec{k}) + \alpha_{\vec{k}} (t) \tilde{\vec{E}}_0 (\vec{k}) \\ \vec{e}_{k} \times \tilde{\vec{B}} (\vec{k}, t) = i\left(\alpha^{\star}_{\vec{k}} (t) \tilde{\vec{E}}_0^{\star} (\vec{k}) - \alpha_{\vec{k}} (t) \tilde{\vec{E}}_0 (\vec{k})\right)$ where $\vec{e}_{k} = \frac{\vec{k}}{k}$. This can be rewritten as $\alpha_{\vec{k}} (t) \tilde{\vec{E}}_0 (\vec{k}) = \tilde{\vec{E}} (\vec{k}, t) + i\vec{e}_{k} \times \tilde{\vec{B}} (\vec{k}, t)$ which better shows how $\alpha$ can indeed be interpreted as the complex number representation of the phase space coordinates $(\vec{E}, \vec{B})$ for a given mode with $\vec{E}$ being analogous to position and $\vec{B}$ being analogous to momentum; now I no longer have to go through contortions trying to interpret $\alpha$ in terms of $(\vec{A}, \vec{E})$ just because everyone else seems to prefer using $(\vec{A}, \vec{E})$ instead of $(\vec{E}, \vec{B})$.

Coming back to the original writing of the fields, the Gauss law without charges in conjunction with $\vec{E}_0$ solving the Helmholtz equation allows for the following simplifying integrals: $\int (\nabla \times \vec{E}_0)^2 \, d^3 r = k^2 \int \vec{E}_0^2 \, d^3 r \\ \int (\nabla \times \vec{E}_0^{\star})^2 \, d^3 r = k^2 \int \vec{E}_0^{\star 2} \, d^3 r \\ \int |\nabla \times \vec{E}_0|^2 \, d^3 r = k^2 \int |\vec{E}_0|^2 \, d^3 r$ where because $\vec{E}_0^2 \equiv \vec{E}_0 \cdot \vec{E}_0$ and $\vec{E}_0^{\star 2} \equiv \vec{E}_0^{\star} \cdot \vec{E}_0^{\star}$ then I think $|\vec{E}_0|^2 \equiv \vec{E}_0 \cdot \vec{E}_0^{\star}$. Furthermore, the following normalization can be made: $\int |\vec{E}_0|^2 \, d^3 r \equiv 2\pi\hbar\omega .$ There are two things to note from the book: the book instead uses $\int |\vec{A}_0|^2 \, d^3 r \equiv 1$, but I choose $2\pi\hbar\omega$ so that the numerical prefactors work out in replacing $\alpha$ and $\alpha^{\star}$ with $a$ and $a^{\dagger}$, and so that the dimensions of $\vec{E}$ and $\vec{E}_0$ are the same so $\alpha$ is dimensionless (as is the operator $a$ that it eventually becomes). The second is that the book says that the normalization $\int |\vec{A}_0|^2 \, d^3 r = 1$ can be done without loss of generality. The issue I have with this is that if the fields considered are for a single mode, then $\vec{E}_0 (\vec{r}) \propto e^{i\vec{k} \cdot \vec{r}} \vec{e}_{\vec{E}_0}$. The integral of this function (or rather, of the square of its norm, which would just be 1) is well-defined for a finite volume of integration but not for an infinite volume. I think the issue is that implicitly, periodic boundary conditions are being considered and that this particular mode is allowed within those boundaries; furthermore, the large-end limit of the size of the cavity whose boundaries are periodic can be taken, and while the integrals themselves may diverge, I think the differences between them are still OK. Anyway, that particular bit doesn't worry me too much; after all, the Casimir force is found to be finite despite being the difference between two infinite quantities.

Once that is done, the Hamiltonian density is $H = \int \frac{1}{8\pi} \left(\vec{E}^2 + \vec{B}^2 \right) \, d^3 r .$ Plugging in $\vec{E}$ and $\vec{B}$ yields $H = \int \frac{1}{8\pi} \left(\alpha^{\star 2} \vec{E}_0^{\star 2} + 2\alpha^{\star}\alpha \vec{E}_0^{\star} \cdot \vec{E}_0 + \alpha^2 \vec{E}_0^2 - \frac{1}{k^2} \left(\alpha^{\star 2} (\nabla \times \vec{E}_0^{\star})^2 \\ - 2\alpha\alpha^{\star} (\nabla \times \vec{E}_0^{\star}) \cdot (\nabla \times \vec{E}_0) + \alpha^2 (\nabla \times \vec{E}_0)^2 \right) \right) \, d^3 r$ as $i^2 = -1$. Noting that $\vec{E}_0^{\star} \cdot \vec{E}_0 = |\vec{E}_0|^2$ and $(\nabla \times \vec{E}_0^{\star}) \cdot (\nabla \times \vec{E}_0) = |\nabla \times \vec{E}_0|^2$, and using the above simplifications, then the first & fourth terms and third & sixth terms cancel. What is left then is $H = \int \frac{1}{4\pi} \left(\alpha^{\star}\alpha |\vec{E}_0|^2 + \alpha\alpha^{\star} |\vec{E}_0|^2 \right) \, d^3 r .$ Using the normalization condition then yields $H = \frac{\hbar\omega}{2} \left(\alpha^{\star}\alpha + \alpha\alpha^{\star} \right)$ which is exactly the harmonic oscillator Hamiltonian.

Doing this for momentum is a bit more complicated. In fact, I don't think it is possible to do it in a completely general manner for $\vec{E}_0 (\vec{r})$ and $\vec{E}_0^{\star} (\vec{r})$, and this is supported by the book implying this as well; it specifies the forms of these vectors before plugging them in to the formula for the linear momentum of the electromagnetic field. That can be done right now as well: because these fields have been specified for only a single mode, the spatial components must be plane waves of definite momentum given by $\vec{E}_0 (\vec{r}) = \epsilon e^{i\vec{k} \cdot \vec{r}} \vec{e}_{\vec{E}}$ where $\epsilon$ is a real constant with dimensions of the electric field and $\vec{e}_{\vec{E}}$ is a real-valued unit vector pointing in the direction of $\vec{E}$. Plugging this into the expression for $\vec{E}$ yields $\vec{E} (\vec{r}, t) = \epsilon \left(\alpha^{\star} (t) e^{-i\vec{k} \cdot \vec{r}} + \alpha(t) e^{i\vec{k} \cdot \vec{r}} \right) \vec{e}_{\vec{E}} .$ Furthermore, $\nabla \times (e^{i\vec{k} \cdot \vec{r}} \vec{e}_{\vec{E}}) = ie^{i\vec{k} \cdot \vec{r}} \vec{k} \times \vec{e}_{\vec{E}}$. This means that $\vec{B}$ is now $\vec{B} (\vec{r}, t) = \epsilon \left(\alpha^{\star} (t) e^{i\vec{k} \cdot \vec{r}} + \alpha(t) e^{i\vec{k} \cdot \vec{r}} \right) \vec{e}_{\vec{k}} \times \vec{e}_{\vec{E}} = \vec{e}_{\vec{k}} \times \vec{E} (\vec{r}, t)$  as expected, where $\vec{e}_{\vec{k}} = \frac{\vec{k}}{k}$. The Poynting vector is then $\vec{S} = \frac{c}{4\pi} \vec{E} \times \vec{B} = \frac{c\epsilon^2}{4\pi} \left(\alpha^{\star 2} e^{-2i\vec{k} \cdot \vec{r}} + \alpha^{\star} \alpha + \alpha\alpha^{\star} + \alpha^2 e^{2i\vec{k} \cdot \vec{r}} \right) \vec{e}_{\vec{k}}$ and the momentum is then $c^2 \vec{p} = \int \vec{S} \, d^3 r .$ My only remaining issue with this is that the first and third terms in the Poynting vector would need to have a volume integral that is zero, but I'm not entirely sure how that would happen. I think the idea is that $e^{2i\vec{k} \cdot \vec{r}} = (\cos(2k_x x) + i\sin(2k_x x))(\cos(2k_y y)$ $+ i\sin(2k_y y))(\cos(2k_z z) + i\sin(2k_z z)),$ and the integral of this over a volume would be zero because the mode $\vec{k}$ is allowed by definition and the boundary conditions are periodic. Assuming that does happen, though, then using the normalization condition that $\int \epsilon^2 \, d^3 r = 2\pi\hbar\omega$ and that $\omega = ck$ yields the desired momentum: $\vec{p} = \frac{\hbar\vec{k}}{2} \left(\alpha^{\star} \alpha + \alpha\alpha^{\star} \right) .$