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?
 A: In addition to the physical explanation given above, I'd like to point out the mathematical difference. A Hilbert space is a complex vector space equipped with an inner product that is also complete. A Minkowski space is a real vector space equipped with a symmetric bilinear form. A symmetric bilinear form is not the same as an inner product, that's why the 'norm' can be negative in Minkowski space. In fact, an inner product is defined as a symmetric bilinear form that's positive definite.
A: 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.
