# Hilbert space and the interval of special relativity?

One requirement of the Hilbert space is that it is normed. Furthermore, in the case of special relativity, the norm has the following problematic requirements:

$$||u||\geq 0 \\ ||u||= 0 \implies u=0$$

This requirements are respected in 3d space with metric $$ds^2=dx^2+dy^2+dz^2$$.

However, in special relativity, the interval is given by

$$ds^2=dx^2+dy^2+dz^2-dt^2$$

In the case of a photon traveling at the speed of light, the interval is always zero, yet u could be (3,4,0,5) such that $$3^2+4^2+0^2-5^2=0$$. In this case $$u\neq 0$$, but $$||u||=0$$. In some cases, the norm is negative such as $$u=(0,0,0,5)$$.

Consequently, the space of functions whose inner product is the interval of special relativity does not form a Hilbert space.

How does one reconcile Hilbert space with special relativity? Or are Hilbert spaces only relevant for non-relativistic quantum mechanics?

• The Hilbert space and Minkowski inner products are unrelated. An element of the Hilbert space represents the state of the whole physical system, not a 4-component vector. Quantum physics can be reconciled with special relativity using quantum field theory. QFT is (or at least can be) expressed in terms of operators on a Hilbert space, just like in non-relativistic quantum mechanics. Perturbative treatments of QFT sometimes obscure this basic fact, but any halfway-decent intro to QFT should at least make it clear for free fields, which is enough to address this question. – Chiral Anomaly Nov 7 '19 at 3:19

Analogously to indefinite metric tensors (bilinear form), famously used in relativity, one can introduce indefinite (pre-)Hilbert spaces $${\cal H}$$ with negative norm states.
This is e.g. done in the BRST operator formulation of gauge theories in QFT with a Grassmann-odd self-adjoint BRST operator $$\hat{Q}$$ on an indefinite$$^1$$ (pre-)Hilbert spaces $${\cal H}$$ that is nilpotent $$\hat{Q}^2~=~0.$$ The physical Hilbert space$$^2$$ $${\cal H}_{\rm phys}~:=~{\rm Ker}(\hat{Q})/{\rm Im}(\hat{Q})$$ is constructed such that the induced sesquilinear form $$\langle\cdot,\cdot\rangle:~ {\cal H}_{\rm phys}\times {\cal H}_{\rm phys}\to \mathbb{C}$$ in the physical sector is positive definite. (This is necessary in order for physical probabilities to be non-negative.)
$$^1$$ In fact the indefinite (pre-)Hilbert space sesquilinear form is typically a consequence of the indefinite metric of spacetime as the BRST formulations of relativistic theories are manifestly Lorentz-covariant. Moreover, the vector space is actually a super vector space.
$$^2$$ One may show that kets in the image $${\rm Im}(\hat{Q})$$ always have zero norm.