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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?

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  • $\begingroup$ 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. $\endgroup$ – Chiral Anomaly Nov 7 '19 at 3:19
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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.

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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.)

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$^1$ 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.

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