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I need clarification on the accepted answer in this thread. I don't have reputation to leave comments there.

The vacuum energy for a free field is the ground state energy of each field oscillator, ${1\over 2} \omega$, summed over all the modes. For a cubic periodic box of side-length $L$, you get $$ \sum_k {1\over 2} \sqrt{k^2+m^2} $$ Where the sum is all $k$'s in an infinite size 3D cubic lattice where each $k$ component is an integer multiple of $2\pi\over L$...

How do we get to this sum?
What is $m$? Is it mass? If it is — is it the intrinsic mass?
Why do we square, add and root $k$ and $m$?

Chris
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K. Kirilov
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1 Answers1

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1) What is $m$ and why do we square, add and root $\vec{k}$ and $m$

Remember that free scalar fields in Minkowski spacetime obey the Klein-Gordon equation:

$$\left(\eta^{\mu\nu}\partial_\mu\partial_\nu + m^2\right)\phi = 0, \quad \text{(1)}$$

where $\eta_{\mu\nu}=\text{diag}(1,-1,-1,-1)$ is the Minkowski metric.

One can show that plane waves of positive energy (positive frequency) form a complete set of solutions of $\text{(1)}$:

$$g_\vec{k}(\vec{x},t)=\left(2\pi\right)^{-3/2}\left(2\omega_\vec{k}\right)^{-1/2}\exp{\left[i\left(\vec{k}\cdot \vec{x} -\omega_\vec{k}\,t\right)\right]}, \quad \text{(2)}$$

where

$$\omega_\vec{k}=\left(\vec{k}^2+m^2\right)^{1/2}. \quad \text{(3)}$$

Here, $k^\mu = (\omega_\vec{k},\vec{k})$ is the wave 4-vector associated with the field mode. As DanielSank mentionend, momentum and wave vector are related. The mass $m$ is the mass of the scalar field. As far as I know you can view it as the mass of the particle described by the field.

You can obtain eq. $\text{(3)}$ plugin $\text{(2)}$ into $\text{(1)}$. Also, note that I'm using natural units, i.e., $c=1$ and $\hbar$=1$

2) How do you get a sum over $\omega$

The answer to this question is rather long. I suggest you look for the derivation of the Casimir effect. If you know/are learning QFT you should find it easy to understand where this sum comes from.

Lsheep
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