0
$\begingroup$

I'm working on the derivation of the Casimir energy from quantum field theory. From the K-G equation (with $c=1$ and $\hbar=1)$ I found the vacuum energy:

$$\langle 0|H|0\rangle=E_{vac}=V\int_{-\infty}^{+\infty} d^3p\ \frac{1}{2}E_p$$

where $V$ is an arbitrary large volume (hence $E_{vac}$ is infinite). However, when trying to calculate the vaccum energy for a region between plates separated by a distance $a$, I had trouble finding a solution to the K-G equation with boundary conditions. We have the K-G equation to be

$$(\partial^2_t-\nabla^2+m^2)\psi=0 \, .$$

Which has the general solution, in terms of the creation and anhiquilation operators:

$$\psi (x,t)=\int d^3p\ \frac{1}{\sqrt{2E_p}}(\hat{a}e^{-ip\cdot x}+\hat{a}^\dagger e^{+ip\cdot x}) \, .$$

For simplicity, let's suppose the two plates are situation in $x_1=0$ and $x_1=L$, so our field must satisfy $\psi(x_1=0)=\psi(x_1=L)=0$. This leads to \begin{align} \psi(x=0,t) &= \int d^3p\ \frac{1}{\sqrt{2E_p}}(\hat{a}e^{-ip_\| \cdot x_\|}+\hat{a}^\dagger e^{+ip_\|\cdot x_\|})=0 \\ &= \int d^3p\ \frac{1}{\sqrt{2E_p}}(\hat{a}e^{-ip_\| \cdot x_\|}e^{-ip_1L}+\hat{a}^\dagger e^{+ip_\|\cdot x_\|}e^{ip_1L})=0 \, . \end{align}

Where I introduced the notation $p_\|\cdot x_\|=E_pt-p_2x_2-p_3x_3$ to denote the directions parallel to the plates (not sure if this helps). However, I cannot find a way to arrange the coefficients so I can get an expression (out of intuition I think some sinus should appear, but I can't justify it). How can I do that (this is the massless case)?

$\endgroup$
1
+50
$\begingroup$

The problem with your attempt is that you are trying to impose the Dirichlet boundary conditions on a solution of the KG equation which already satisfies periodic boundary conditions (i.e. infinite volume solution) hence the contradictions you run into.

The standard procedure to properly obtain the mode expansion of the scalar field is the following: you first have to find the positive and negative monochromatic solutions of the Klein-Gordon equation and then you put them together to find the most general solution, i.e. the field. In classical terms you basically build a wave packet from the plain wave solutions.

Let's call the positive and negative frequency monochromatic states $f_{\pm}(t,x_1,x_2,x_3)$ such that $$\left(\Box+m^2 \right)f_{\pm}(t,x_1,x_2,x_3)=0$$ This equation can be separated beween the "parallel" $x_\| =(t,x_2,x_3)$ and "orthogonal" $x_1$ coordinates. Therefore we can write $$\left(\Box+m^2 \right)f_{\pm}(x_\|)g(x_1)=0$$ where $g(x_1)$ obeys the equation $g''(x_1)+m^2g(x_1)=0$ and must satisfy the Dirichlet boundary condition $g(0)=g(L)=0$. This has solution $g(x_1) = A \sin(\frac{n\pi}{L}x_1)+B \cos(\frac{n\pi}{L}x_1)$ with $B=0$. The $f_{\pm}(x_\|)$ are the usual infinite volume functions.

The general form of $f_{\pm}(t,x_1,x_2,x_3)$ is then $$f_{\pm}(t,x_1,x_2,x_3)=\frac{1}{\sqrt{2E_p L}}e^{\mp i p_\| \cdot x_\|}\sin(\frac{n\pi}{L}x_1)$$ Where we have used the standard "covariant" normalization of the one particle states and where the solution in the $x_1$ direction is the same for positive and negative frquencies.

So now we can build the field as a linear combination (which is still a solution!) of the negative and positive frequencies with some coefficients that in the quantum theory become the creation and annhilation operators: $$\psi(x_\|,x_1)=\sum_{n=1}^{\infty}\int \frac{d^2p_\|}{2\pi}\frac{1}{\sqrt{2E_p L}}\sin(\frac{n\pi}{L}x_1)\left(a_n(p_\|)e^{- i p_\| \cdot x_\|}+a_n^{\dagger}(p_\|)e^{\ i p_\| \cdot x_\|}\right)$$

The dispersion relation now reads $E_p=\sqrt{p^2_{\|}+m^2+\left(\frac{n\pi}{L}\right)^2}$. With some labor the vacuum energy can be found to be $$\langle0\rvert H\rvert0\rangle=\frac{A}{(2\pi)^2}\sum_{n=1}^{\infty}\int d^2p_\|\ \sqrt{p^2_{\|}+m^2+\left(\frac{n\pi}{L}\right)^2}$$ where $A$ is the area of the plates. Obviously, to gain a meaningful result the expression will have to be regularized.

$\endgroup$
  • $\begingroup$ Thank you. I see my main mistake now, I was acting on an already established solution so that's why I was falling into contradictions. In general, I see how this scheme can be applied to different problems, so this helped me. I'll work on the last step, but I have a better idea now on how to proceed. $\endgroup$ – Charlie Sep 25 '18 at 22:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.