Considering a Scalar field (massless) in a $D$ dimensional space, with Dirichlet boundary condition in $x$ direction, \begin{eqnarray} \phi(0,x_2,x_3,....x_D,t)=\phi(a,x_2,x_3,...x_D,t) \end{eqnarray} So, the momenta could be defined as, \begin{eqnarray} \vec{p}=\{ \frac{n \pi}{a},p_2,p_3,,.......p_D \} =\{ \frac{n \pi}{a}, \vec{p_1}\} \end{eqnarray} where, $n\in Z^+ $. Now the Hamiltonian, \begin{eqnarray} H=\sum_n \int \frac{d^{D-1} \vec{p_1} }{(2\pi)^{D-1} } \omega_{\vec{p}} (a_{\vec{p}} ^{\dagger}a_{\vec{p}} +\frac{1}{2}) \end{eqnarray} Therefore, \begin{eqnarray} E_{vac}=\left\langle 0 \middle| H \middle| 0 \right\rangle \sim \sum_n \int \frac{d^{D-1} \vec{p_1} }{(2\pi)^{D-1} } \frac{\omega_{\vec{p}}}{2} \end{eqnarray} We will use Zeta function regularization technique to regularize it, Defining, \begin{eqnarray} E(s)=\frac{1}{2}\sum_n \int \frac{d^{D-1} \vec{p_1} }{(2\pi)^{D-1} } \omega_{\vec{p}} (l\omega_{\vec{p}} )^{-s} \end{eqnarray} The finite part of vacuum energy, \begin{eqnarray} E_{vac}'=\lim_{s\rightarrow 0} E(s) \end{eqnarray} Doing some algebra, we find out \begin{eqnarray} E(s)=l^{-s} \frac{V}{a ({4\pi})^{\frac{D-1}{2}}}(\frac{\pi}{a})^{D-s} \frac{ \Gamma (\frac{s-D}{2})}{\Gamma( \frac{s-1}{2}) }\zeta(s-D) \end{eqnarray} Here, $\zeta(s)$ is the Zeta function. In the limit $s\rightarrow 0$, $D$ being even $D=2p$, with $p>0$ and $p$ is an integer, \begin{eqnarray} E'_{vac}\propto \zeta(-2p)=0. \end{eqnarray} My question: Can you please elaborate the reason, why vacuum energy is zero in even dimension for scalar fields for such geometry where one direction is finite and the rest $D-1$ direction is infinite? May be picking $D=2$, which will result in a infinite slab like geometry.
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