Vacuum energy ($E_0$)

For finitely many free physical degrees of freedom with nonnegative masses, assign positive weights to bosons and negative weights to fermions. With a common spherical ultraviolet cutoff, the unsubtracted vacuum energy density is finite precisely when the three displayed supertraces vanish. Expanding $\sqrt{p^2+m^2}=p+m^2/(2p)-m^4/(8p^3)+O(p^{-5})$ proves that they cancel the quartic, quadratic and logarithmic ultraviolet divergences. The remaining radial integral converges as ${\int }^{\infty }{p}^{-3}dp$. Equal-mass boson–fermion pairing is sufficient but stronger than the three conditions; a renormalized vacuum energy with a free counterterm need not satisfy them.

The vacuum energy density is the bulk limit ${\rho }_{\mathrm{vac}}={\lim }_{\mathcal{V}\to \infty }{E}_0(\mathcal{V})/\mathcal{V}$, with boundary effects removed and a fixed convention for the Hamiltonian. In an effective field theory at a homogeneous vacuum it is the value of the renormalized scalar potential. A global supersymmetry algebra expresses the regulated vacuum energy as a sum of supercharge norms, giving nonnegative density; the scalar potential of supergravity instead includes a negative gravitational term.

On a one-dimensional interval with Dirichlet boundary conditions, $k_n=n\pi /L$ and ${\omega }_n^2=k_n^2+m^2$. The massive-field vacuum energy density can depend on position even though its integral is the oscillator sum ${\sum }_n{\omega }_n/2$. All terms require a common ultraviolet regulator when vacuum energies in different geometries are compared.

If a free real scalar field's three-momentum ultraviolet cutoff is identified with the absolute vacuum expectation value of another scalar field, the displayed scalar potential is even and has its unique global minimum at zero. It grows as ${\phi }^4/(16{\pi }^2)$ for $|\phi |\gg m$, but as $m|\phi {|}^3/(12{\pi }^2)$ for $|\phi |\ll m$ with $m>0$. The positive derivative with respect to $|\phi |$ proves that this contribution alone does not produce spontaneous symmetry breaking. This cutoff identification is a model assumption, distinct from deriving a complete effective potential for a dynamical field.

Counting two transverse polarizations and zero-point energy $\hslash ck/2$ per mode gives $u=\hslash c{k}_{\max }^4/(8{\pi }^2)=2{\pi }^2\hslash c/{\lambda }_{\min }^4$. This is a bare cutoff-dependent mode sum, not a classical incident radiation bath or a renormalization-independent absolute observable.

A fermionic mode with canonical anticommutation relations $\{b,b^{†}\}=1$ has the symmetrically ordered Hamiltonian operator $H=\frac{E}{2}(b^{†}b-b{b}^{†})=E(b^{†}b-1/2)$. Its empty-state zero-point contribution is $-E/2$. A quantum harmonic oscillator with bosonic canonical commutation relations contributes $+E/2$ instead. For a Dirac field, sum over both particle and antiparticle modes and over spin: there are four independent oscillators per spatial momentum in four spacetime dimensions. Comparing bosonic and fermionic vacuum energies requires a common regularization in quantum field theory, matching degrees of freedom and a stated vacuum-energy subtraction convention.