= Gupta-Bleuler null-state quotient
{c}
{title2=$\mathcal H_{\mathrm{phys}}=\ker C/\operatorname{rad}(\ker C)$}
Choose $C_{\mathbf p}=a^0_{\mathbf p}-a^3_{\mathbf p}$ with the standard temporal and longitudinal <polarization vectors>. The physical pre-space is the common kernel of all $C_{\mathbf p}$. In each regulated mode, $a^0$ acts on creator <polynomials> as $-\partial_{a^{0\dagger}}$ and $a^3$ as $\partial_{a^{3\dagger}}$, so the constraint kernel consists of transverse creator <polynomials> and <polynomials> in $C^\dagger=a^{0\dagger}-a^{3\dagger}$. Since $[C,C^\dagger]=0$, these latter excitations remain constrained. They are orthogonal to every constrained state because $\langle\Phi|C^\dagger=\langle C\Phi|=0$. Quotienting this <radical of a Hermitian form> leaves only the transverse <bosonic Fock space> with a positive <inner product>. The constraint alone gives a <positive semidefinite Hermitian form>; quotienting removes its null directions. If starting from finite-particle creator <polynomials>, take the <Hilbert space completion> of this positive quotient to obtain the physical <Hilbert space>.
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