# On the boundary conditions of the Casimir effect and quantization of the wave vector

I'm reviewing the famous Casimir effect. I'm uploading an image with the starting setup and frame of reference.

The electric field field operator is:

where $$\textbf{e}$$ is the polarization vector, $$\vec{k} = (k_x,k_y,k_z)$$ is the wave vector, $$\omega_k = c|\vec{k}|$$, V is the volume we are considering that is $$V = L^2d$$, $$\hat{a}$$ is the creation operator with its hermitian conjugate.

The quantization of the wave vector comes from imposing the right boundary conditions on the electric field. I should obtain this: $$\vec{k} = (\frac{2\pi}{L}m, \frac{2\pi}{L}n, \frac{\pi}{d}l)$$ with $$m$$ and $$n$$ integers, while $$l$$ a positive integer.
The x and y components of the wave vector come from imposing the so-called periodic boundary conditions. I'm having some problems deriving the third component. It comes from evaluating the electric field on the metal plates, so at $$z=0$$ and $$z=d$$. I did get that the component is the one I wrote before, but I did not get to the condition that $$l$$ should be a positive integer. From my passages I get that $$l$$ could also be negative, just like $$m$$ and $$n$$. I have not found a complete derivation of this passage, so I'm asking for some help.

The Fourier series of a function $$f(x)$$ with periodic boundary conditions $$f(x+L)=f(x)$$ has the form $$f(x)=\sum\limits_{n \in \mathbb{Z}} a_n \, e^{2\pi i n x/L}.$$ On the other hand, a function $$g(z)$$ with boundary conditions $$g(0)=g(d)=0$$ has the series expansion $$g(z)=\sum\limits_{\ell \in \mathbb{N}} c_\ell \, \sin(\ell \pi z/d).$$ In the first case, $$\exp (2\pi i n x/L)$$ and $$\exp(-2\pi i n x/L)$$ are linear independent (thus the summation over all integers $$n \in \mathbb{Z}$$), in the second case, you have $$\sin(-\ell \pi z/d)= -\sin(\ell \pi z/d)$$ and the summation goes only over $$\ell \in \mathbb{N}=\{1,2,3,\ldots\}$$.

The explicit form of the electric field of your problem is most easily obtained from the associated vector potential $$\vec{A}$$ in the Coulomb gauge, fulfilling $$\vec{\nabla} \cdot \vec{A} =0, \qquad \square \vec{A}=0, \qquad \vec{A}=\vec{A}^\ast \tag{1} \label{1}$$ together with the boundary conditions \begin{align} \vec{A}(t,x,y,z)&=\vec{A}(t,x+L,y,z)=\vec{A}(t,x,y+L,z) \tag{2} \label{2} \\[3pt] A_{x,y}(t,x,y,0)&=A_{x,y}(t,x,y,d)=0, \tag{3} \label{3} \\[3pt] \nabla_z A_z(t,x,y,0)&=\nabla_z A_z(t,x,y,d)=0. \tag{4} \label{4} \end{align} Eq. \eqref{2} corresponds to periodic boundary conditions in $$x$$- and $$y$$-direction, whereas \eqref{3} and \eqref{4} guarantee $$\vec{E}_{||}=0$$, $$\dot{\vec{B}}_\perp=0$$ and $$\vec{\nabla} \cdot \vec{A} =0$$ on the surface of the plates at $$z=0$$ and $$z=d$$.

The general solution of this system of equations is given by \begin{align} A_{x,y}(t,\vec{r})&= \! \sum\limits_{\vec{n}} \sum\limits_s \mathcal{N}(\vec{n}) \, \varepsilon_{x,y}(\vec{n},s) \sin\frac{n_z \pi z}{d}\left[i e^{\frac{2\pi i}{L}(n_x x +n_y y)} e^{-i \omega(\vec{n})t}a(\vec{n},s) + \text{c.c.} \right], \tag{5} \label{5}\\[5pt] A_z(t, \vec{r}) &= \! \sum\limits_{\vec{n}} \sum\limits_s \mathcal{N}(\vec{n}) \, \varepsilon_z(\vec{n},s) \cos \frac{n_z \pi z}{d} \left[e^{\frac{2\pi i}{L}(n_x x+n_y y)} e^{-i \omega(\vec{n})t} a(\vec{n},s) +\text{c.c.} \right], \tag{6} \label{6} \end{align} where $$\vec{r}=(x,y,z)$$, $$\vec{n}=(n_x,n_y,n_z) \in \mathbb{Z}\times \mathbb{Z} \times \mathbb{N}$$. Defining $$\vec{k}(\vec{n})= \left(\frac{2\pi n_x}{L}, \frac{2\pi n_y}{L},\frac{\pi n_z}{d} \right),\qquad n_{x,y} \in \mathbb{Z}, \; n_z \in \mathbb{N}, \tag{7} \label{7}$$ the angular frequency $$\omega(\vec{n})$$ is given by $$\omega(\vec{n})= |\vec k(\vec{n})| = \sqrt{ \left(\frac{2\pi}{L}\right)^{\! 2}(n_x^2+n_y^2)+\left(\frac{\pi }{d}\right)^2 n_z^2}. \tag{8} \label{8}$$ The two (normalized) polarization vectors $$\vec{\varepsilon}(\vec{n},1)$$, $$\vec{\varepsilon}(\vec{n},2)$$ (corresponding to the two possible linear polarizations for a given vector $$\vec{k}(\vec{n})$$) satisfy $$\vec{k}(\vec{n}) \cdot \vec{\varepsilon}(\vec{n},s) =0, \qquad \vec\varepsilon(\vec{n},s)\cdot \vec\varepsilon(\vec{n},s^\prime) = \delta_{s s^\prime}. \tag{9}$$ Finally, $$\mathcal{N}(\vec{n})$$ is a conveniently chosen normalization factor to be discussed later.

Taking advantage of \eqref{7}, the vector potential can be written in the slightly more compact form \begin{align}A_{x,y}(t, \vec{r})&= \sum\limits_{\vec{k}, s} \mathcal{N}(\vec{k}) \, \varepsilon_{x,y}(\vec{k}, s) \sin(k_z z) \left[i e^{i(k_x x+k_y y)} e^{-i \omega(\vec{k}) t} a(\vec{k},s) +\text{c.c.} \right], \tag{10} \label{10} \\[5pt] A_z(t, \vec{r}) &= \sum\limits_{\vec{k},s} \mathcal{N}(\vec{k}) \, \varepsilon_z(\vec{k}, s) \cos(k_z z) \left[e^{i(k_x x+k_y y)} e^{-i \omega(\vec{k}) t} a(\vec{k},s) + \text{c.c.} \right]. \tag{11} \label{11} \end{align} The correct expression for the electric field $$\vec{E}= -\dot{\vec{A}}$$ is thus given by \begin{align} E_{x,y}(t, \vec{r}) &=-\sum\limits_{\vec{k},s} \mathcal{N}(\vec{k})\, \omega(\vec{k})\, \varepsilon_{x,y}(\vec{k},s) \sin(k_z z) \left[ e^{i(k_x x+k_y y)} e^{-i \omega(\vec{k}) t} a(\vec{k},s) + \text{c.c.} \right], \tag{12} \label{12} \\[5pt] E_z(t, \vec{r}) &= i \sum\limits_{\vec{k},s} \mathcal{N}(\vec{k}) \, \omega(\vec{k}) \, \varepsilon_z(\vec{k},s) \cos(k_z z) \left[e^{i(k_x x+k_y y)} e^{-i \omega(\vec{k}) t} a(\vec{k},s) + \text{c.c.} \tag{13} \label{13} \right]. \end{align} Note the expression for $$\vec{E}$$ displayed in your question is apparently in conflict with the boundary conditions discussed above (in particular, $$\vec{E}_{||}=0$$ at the plates is not fulfilled).

The quantization of the system is now straightforward. Choosing $$\mathcal{N}(\vec{k})$$ such that the energy $$H= \frac{1}{2}\int\limits_0^L\! \!dx \int\limits_0^L \!\!dy \int\limits_0^d \!\!dz \,(\vec{E}^2 + \vec{B}^2) \tag{14}$$ takes the form $$H = \sum\limits_{\vec{k},s} \omega(\vec{k}) a^\ast(\vec{k},s) a(\vec{k},s), \tag{15}$$ the Fourier coefficients are promoted to operators acting in Fock space, obeying the commutation relations $$[a(\vec{k},s), a^\dagger(\vec{k}^\prime, s^\prime)]= \delta_{\vec{k} \, \vec{k}^\prime} \, \delta_{s \, s^\prime}, \qquad [a(\vec{k},s), a(\vec{k}^\prime,s^\prime)]=0, \tag{16}$$ describing a a system of infinitely many (uncoupled) quantum harmonic oscillators.

• I understand the point here. Still, I have some difficulties seeing the sine function arising inside the Electric field operator once I plug $k_z$... In fact, if I plug $k_z = \frac{l\pi}{d}$ the electric field should be zero, but I don't see it really... ( I'm referring to the expression of the electric field written above) Commented Feb 23 at 21:56
• @GiulianoArtale $e^{i \alpha}-e^{-i \alpha}=2 i \sin \alpha$ Commented Feb 23 at 22:03
• Yeah but in this case I got something more like $Ae^{i\alpha} - Be^{-i\alpha}$ where inside A and B are the creation and annihilation operators... Are they not playing a role? Commented Feb 23 at 22:19
• @GiulianoArtale You have to identify the independent Fourier coefficients. The most transparent way to find the result is the following: 1. Write down the expression for the vector potential as a Fourier series with the correct boundary conditions. 2. Impose the constraints following from $\vec{A}^\ast=\vec{A}$, $(\partial_t^2-\Delta)\vec{A}=0$ and $\vec{\nabla} \cdot \vec{A}=0$. 3. Compute $\vec{E}=- \partial_t \vec{A}$. Commented Feb 23 at 22:44
• @GiulianoArtale Expanding my last comment, I have now added a full discussion of the derivation of the electric field with the relevant boundary conditions in my answer. Commented Feb 26 at 13:52