= Solution
In a large periodic quantization volume $\mathcal V$, one <wavevector> state occupies $(2\pi)^3/\mathcal V$ in <wavevector> space. There are two transverse <photon polarizations>, and each oscillator contributes $\hbar ck/2$. A lower <wavelength> cutoff $\lambda_{\min}$ corresponds to $k_{\max}=2\pi/\lambda_{\min}$, so the <energy density> is
$$
u_{\rm vac}
=2\int_{|\mathbf k|\leq k_{\max}}\frac{\mathrm d^3k}{(2\pi)^3}
\frac{\hbar ck}{2}
=\frac{\hbar c}{2\pi^2}\int_0^{k_{\max}}k^3\,\mathrm dk
=\frac{\hbar c\,k_{\max}^4}{8\pi^2}
=\frac{2\pi^2\hbar c}{\lambda_{\min}^4}.
$$
For $\lambda_{\min}=10^{-10}\,\mathrm m$,
$$
\boxed{u_{\rm vac}\simeq6.24\times10^{15}\,\mathrm{J\,m^{-3}}.}
$$
The factor two is <polarization>, not an extra independent count of positive and negative frequencies. The continuum <wavevector> integral already counts independent travelling-wave modes in the usual field quantization convention. The quartic cutoff dependence also shows why an assumed shortest <wavelength> is crucial.
Back to article page
