In quantum mechanics, I have been going through basics of the subject. In general the space of quantum states is Hilbert space (which is Euclidean - I presume). Being just curious, are there any instances where Non-Euclidean spaces are used in quantum mechanics ? I have heard that there is something called Topological QFT, is it associated to my question ?

I know my question is a little ridiculous, but it just out of sheer curiosity.

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    $\begingroup$ What do you mean by "Euclidean" in this context? $\endgroup$ – Valter Moretti Feb 4 '14 at 6:34
  • $\begingroup$ Perhaps the OP are referring to non positively defined scalar products. Krein-Hilbert spaces. Gupta-Bleuler formalism to deal with quantized EM field has something to do with them. But I think it is too technical to discuss as an example for the OP. $\endgroup$ – Valter Moretti Feb 4 '14 at 8:26
  • $\begingroup$ @V.Moretti : I am sorry, my knowledge is quite limited. The only kind of Non-Euclidean space that I learnt so far is the Riemannian metric space whose metric is not a kronecker delta. $\endgroup$ – user38249 Feb 5 '14 at 3:31

The Hilbert space of physical states of any physical system is a positively definite complex vector space i.e. the squared proper length may be computed as $$ ds^2 = |da_1|^2 + |da_2|^2 + \dots $$ We may also split the complex coordinates ("amplitudes") $a_i$ to the real parts and imaginary parts which turns the $N$-dimensional complex space to a $2N$-dimensional real Euclidean space. Under this transition, we may also define the angles between two vectors, via $$\cos(\alpha) = \frac{|\langle u| v \rangle|}{|u|\cdot |v|} $$ There are several basic generalizations of this "Euclidean" space to a non-Euclidean one.

First, some of the terms $|da_i|^2$ in the formula for the $ds^2$ could be given negative coefficients; more generally, the bilinear form could be indefinite. If it is negatively definite, we obtain an isomorphic Hilbert space and we should just flip the overall sign of $ds^2$ to achieve the usual, positively definite convention. But if $ds^2$ is really indefinite, i.e. allowing both signs, we have a problem with the interpretation of quantum mechanics because $ds^2$ is interpreted as a probability by quantum mechanics and probabilities just can't be negative (can't have both signs).

So the indefinite Hilbert spaces are not possible for a theory that may be physically interpreted. However, indefinite Hilbert spaces actually often appear in modern theories as an intermediate step – in theories with gauge symmetries, bad ghosts, and good ghosts (BRST quantization). For example, it is natural to have a Hilbert space with 4 polarizations of a photon for each allowed vector $k^\mu$; the signature of this 4-complex-dimensional space is $3+1$, just like for the spacetime. But the gauge symmetry and the related Gauss' law render $1+1$ polarizations unphysical, leaving just 2 physical polarizations (say $x,y$) with a positively definite norm. This is a master example for all analogous situations.

Another possible non-Euclidean generalization could be a curved, Riemannian space. The Hilbert space can't be deformed to a curved one because of the superposition principle – in quantum mechanics, the linear combination of two allowed vectors simply has to be allowed, too. No known consistent nonlinear deformation of quantum mechanics is known and it probably cannot exist. After all, the freedom to consider linear superpositions is just the complex counterpart of the freedom to consider linear superpositions of probability distributions.

One could also try to discuss noncommutative generalizations of the Hilbert space. But due to the irrelevance of the overall scaling of a vector, $|\psi\rangle \to a |\psi\rangle$, there should be no parameter analogous to the "noncommutativity parameter" on the Hilbert space, either. Noncommutative spaces are useful in quantum mechanics but they generalize classical phase spaces (because $p,x$ don't commute in quantum mechanics), not Hilbert spaces.


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