Vacuum Energy in Two Dimensional Box Through the Krein Quantization

Abstract

In this work we reexamine the Casimir effect in which the vacuum expectation value of quantum fields is calculated over a so-called Krein space. This method has already been successfully applied to study Casimir effect on non-trivial topologies and also the covariance problem in the massless minimally coupled scalar field in de Sitter space-time. It is shown that within this method, no infinite term appears in the computation of the vacuum expectation value of energy-momentum tensor. We investigate the behavior of the Krein quantization for a scalar field in a box satisfying the Dirichlet boundary condition. We show that one can recover the usual theory with the exception that the vacuum energy of the free theory is zero.

Appendix

1.1 Some Technical Details on Computation of the Casimir Effect

In (22) by separating the term with n = 0, one obtains

$$ \sum\limits_{n=0}^{\infty} f(n)=\frac{m}{2}+E_{0}(a,m) $$

(38)

In a similar manner, taking into account the change of variable a k = π t, one finds

$$ {\int}_{0}^{\infty} f(t)\,dt=E_{0}(a,m)=\frac{\pi}{2a}{\int}_{0}^{\infty} dt\,(A^{2}+t^{2})^{\frac{1}{2}}, $$

(39)

where $A\equiv \frac {ma}{\pi }$. Then the Casimir energy is found from the relation (22) as follows

$$ E^{Cas}(a,m)=-\frac{m}{4}+i\frac{\pi}{2a}{\int}_{0}^{\infty}\frac{dt}{e^{2\pi t}-1}\left[G_{A}(it)-G_{A}(-it)\right]. $$

(40)

Here, the function G _A (t) is defined by

$$ G_{A}(t)\equiv(A^{2}+t^{2})^{\frac{1}{2}},\qquad A\equiv\frac{ma}{\pi}. $$

(41)

To compute G _A (i t)−G _A (−i t) with the substitution t→ε±i t and taking ε→0⁺ we find

$$\begin{array}{@{}rcl@{}} G_{A}(it)-G_{A}(-it)&=&\sqrt{A^{2}+(\varepsilon+it)^{2}}-\sqrt{A^{2}+(\varepsilon-it)^{2}}\\ &=&\sqrt{A^{2}-t^{2}+\varepsilon^{2}+2i\varepsilon t}-\sqrt{A^{2}-t^{2}+\varepsilon^{2}-2i\varepsilon t}, \end{array} $$

(42)

When t > A

$$ G_{A}(it)-G_{A}(-it)=2i\sqrt{t^{2}-A^{2}}\theta(t-A) $$

(43)

where θ(x) is the step function. Substituting (43) in (40), one arrives at

$$ E^{Cas}(a,m)=-\frac{m}{4}-\frac{\pi}{a}{\int}_{A}^{\infty}\frac{\sqrt{t^{2}-A^{2}}}{e^{2\pi t}-1}\,dt $$

(44)

where 2π t≡y and π A = m a ≡ μ (the latter parameter has the meaning of a dimensionless mass). The first contribution on the right-hand side of (44) is associated with the total energy of the boundary points. It does not depend on a and hence does not contribute to the Casimir force:

$$ E^{Cas}(a,m)=-\frac{m}{4}-\frac{1}{4\pi a}{\int}_{2\mu}^{\infty}\frac{\sqrt{y^{2}-4\mu^{2}}}{e^{y}-1}\,dy. $$

(45)

For μ = 0 (m = 0), (45) leads to (23). In the rest of this appendix we collect some details of obtaining (29). First, we apply the Abel-Plana to perform the summation over ℓ , with the result

$$\begin{array}{@{}rcl@{}} &&\bullet\sum\limits_{\ell=0}^{\infty} f(\ell)=\frac{\pi}{2}\sum\limits_{n=1}^{\infty}\frac{n}{a}+E(a,b)\\ &&\bullet{\int}_{0}^{\infty} f(t)\,dt=\frac{b}{2\pi}\sum\limits_{n=1}^{\infty}{\int}_{0}^{\infty} dk_{\ell}\,\left( k_n^2+k_{\ell}^{2}\right)^{\frac{1}{2}},\quad bk_{\ell}=\pi t~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ &&\mspace{103mu}=\frac{\pi}{2}\sum\limits_{n=1}^{\infty}{\int}_{0}^{\infty} dt\,\sqrt{\left( \frac{n}{a}\right)^2+\left( \frac{t}{b}\right)^2} \end{array} $$

$$\begin{array}{@{}rcl@{}} &&\bullet\,\frac{1}{2}f(0)=\frac{\pi}{2}\sum\limits_{n=1}^{\infty}\frac{n}{2a}\\ &&\bullet, i{\int}_0^{\infty}\frac{f(it)-f(-it)}{e^{2\pi t}-1}\,dt=i\frac{\pi}{2}\sum\limits_{n=1}^{\infty} dt\,\frac{M(it)-M(-it)}{e^{2\pi t}-1},\qquad M(t)\equiv\sqrt{\left( \frac{n}{a}\right)^2+\left( \frac{t}{b}\right)^2}\\ &&\mspace{205mu}=\frac{\pi}{2}\sum\limits_{n=1}^{\infty}{\int}_{\frac{bn}{a}}^{\infty} dt\,\frac{2i\sqrt{\left( \frac{t}{b}\right)^2-\left( \frac{n}{a}\right)^2}}{e^{2\pi t}-1} \end{array} $$

$$ E(a,b)=\frac{\pi}{2}\sum\limits_{n=1}^{\infty}\left[-\frac{n}{2a}+{\int}_{0}^{\infty} dt\,\sqrt{\left( \frac{n}{a}\right)^{2}+\left( \frac{t}{b}\right)^{2}}-2{\int}_{\frac{bn}{a}}^{\infty} dt\,\frac{2i\sqrt{\left( \frac{t}{b}\right)^{2}-\left( \frac{n}{a}\right)^{2}}}{e^{2\pi t}-1}\right] $$

(46)

Note that in the last integral on the right-hand side of (46), we have defined $u\equiv \frac {ta}{nb}$. We apply the Abel-Plana (22) to perform the first two summations over n on the right-hand side of above equation:

First summation

$$\begin{array}{@{}rcl@{}} &&\bullet{\int}_0^{\infty} f(x)\,dx=-\frac{1}{2a}{\int}_0^{\infty} t\,dt\\ &&\bullet\,i{\int}_{0}^{\infty} dt\,\frac{f(it)-f(-it)}{e^{2\pi t}-1}=-\frac{i}{2a}{\int}_{0}^{\infty} dt\,\frac{2it}{e^{2\pi t}-1}=\frac{1}{24a}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \end{array} $$

$$ {\kern50pt}-\frac{1}{2a}{\sum}_{n=1}^{\infty} n=-\frac{1}{2a}{\int}_{0}^{\infty} t\,dt+\frac{1}{24a} $$

(47)

Second summation

$$\begin{array}{@{}rcl@{}} &&\bullet\,\frac{1}{2}f(0)=\frac{1}{2b}{\int}_{0}^{\infty} t\,dt\\ &&\bullet{\int}_{0}^{\infty} f(\upsilon)\,d\upsilon={\int}_{0}^{\infty} dt{\int}_{0}^{\infty} d\upsilon\,\sqrt{\left( \frac{\upsilon}{a}\right)^2+\left( \frac{t}{b}\right)^2}\\ &&\bullet\,i{\int}_{0}^{\infty} d\upsilon\,\frac{f(i\upsilon)-f(-i\upsilon)}{e^{2\pi\upsilon}-1}=\frac{i}{b}{\int}_{0}^{\infty} dt{\int}_{0}^{\infty} d\upsilon\,\frac{M^{\prime}(i\upsilon)-M^{\prime}(-i\upsilon)}{e^{2\pi\upsilon}-1},\qquad M^{\prime}(\upsilon)\equiv\sqrt{\left( \frac{\upsilon b}{a}\right)^2+t^2}\\ &&\mspace{215mu}=-\frac{b}{8\pi^2a^2}\zeta_R(3) \end{array} $$

$$ \sum\limits_{n=1}^{\infty}{\int}_{0}^{\infty} dt\,\sqrt{\left( \frac{n}{a}\right)^{2}+\left( \frac{t}{b}\right)^{2}}=-\frac{1}{2b}{\int}_{0}^{\infty} t\,dt+{\int}_{0}^{\infty} dt{\int}_{0}^{\infty} d\upsilon\,\sqrt{\left( \frac{\upsilon}{a}\right)^{2}+\left( \frac{t}{b}\right)^{2}}-\frac{b}{8\pi^{2}a^{2}}\zeta_{R}(3) $$

(48)

Now we substitute (47) and (48) into (46).