In Quantum Mechanis (QM), the dynamical variables are the (quantized) coordinates $x_j$ and their canonical conjugate $p_j = -i\partial_j$ with the commutation relation $[x_j,p_k]=i\delta_{jk}$ acting as operators on the quantum state space.

What exactly happens to that state space when we change the underlying geometry or topology of "physical" space - the (spacetime) manifold that serves as the background for the quantum system? How does this change in geometry/topology reflect on the Hilbert space?

In Quantum Field Theory (QFT), the dynamical objects are the (quantized) fields $\phi (x^\mu)$ and the coordinates $x_i$ are demoted to mere labels. What happens in this case? How does a change in geometry/topology alter the resulting Fock space?

I am new to this area, so what I need would be a basic explanation (for QM and QFT) how to make the connection between the two concepts geometry/topology of physical space and resulting properties of the quantum state space - if such a wish makes sense at all.

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    $\begingroup$ To be precise - with "geometry/topology of physical space" you mean the structure of the classical phase space, which in turn is the cotangent bundle of the configuration space (any manifold, basically) charted by the coordinates $x_j$? And we are doing some variant of geometric quantization? (In the context of structure of the phase space, one must recall Darboux' theorem, heuristically indicating that no "local" effects can be seen since, locally, there is just one symplectic geometry) $\endgroup$
    – ACuriousMind
    Feb 10, 2015 at 15:46
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    $\begingroup$ changing coordinates doesn't change the geometry/topology of a manifold, so it is not entirely clear to me what you're asking here. As to whether there are changes between configuration spaces of different dimensions then the answer is that in general there is no difference because of von Neumann's uniqueness theorem. $\endgroup$
    – Phoenix87
    Feb 10, 2015 at 15:55
  • $\begingroup$ With "physical space" I just mean the (spacetime) manifold in which I place my quantum system. I guess one would call it semi-classical as it is QM/QFT on a fixed background. For instance, I could analyse a particle in a curved space. How would such a change of geometry affect the quantum state space? $\endgroup$ Feb 10, 2015 at 15:56
  • $\begingroup$ @ACuriousMind By changing the physical space, for example in the case of a scalar, he means changing $g_{\mu\nu}$ in $\mathcal L \sim g_{\mu\nu}\partial_\mu \phi \partial_\nu \phi + \dots$ That is, the underlying spacetime, or in string theoretic language, the target space. $\endgroup$
    – JamalS
    Feb 10, 2015 at 16:03
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    $\begingroup$ OP, are you perhaps looking for QFT in curved spacetime? $\endgroup$
    – ACuriousMind
    Feb 10, 2015 at 16:13

2 Answers 2


OP comments as an example of what the question is about:

Let us consider the case of an electron confined to a curved surface. Does the geometry of the background have any consequences for the state space?

A simple answer can be given for ordinary QM: A (scalar, i.e. spin-0) particle moving in one-dimension has state space $L^2(\mathbb{R})$, a particle in three dimensions has state space $L^2(\mathbb{R}^3)$. The state space of a particle moving on a submanifold $\mathcal{M} \subset \mathbb{R}^3$, e.g. a particle moving on any smooth surface, is then by analogy simply $L^2(\mathcal{M})$ i.e. the functions whose square-integral over $\mathcal{M}$ exists.

Note that Fourier (i.e. relating the position and momentum representations) transforms on manifolds that are not $\mathbb{R}^n$ are somewhat complicated, cf. this math.SE post.

  • $\begingroup$ OK, thank you! So to keep it simple: an electron confined to 1D (thin wire) is described by exactly the same state space for every possible bending of the wire (since the functions on it stay the same) - no quantum states change at all? No shift in energy levels or anything like that? $\endgroup$ Feb 10, 2015 at 17:08
  • $\begingroup$ @quantumorsch: I believe that is the case - one dimension is one dimension, and ordinary QM has no way to "see" the curvature. $\endgroup$
    – ACuriousMind
    Feb 10, 2015 at 17:12
  • $\begingroup$ That is exactly what I wanted to know! The Hilbert space cannot see the curvature/geometry/topology of the underlying manifold (background space). I find it quite remarkable that all those systems in arbitrarily complicated geometries (but with the same dimension) are exactly the same and there are no consequences for the quantum states and operators at all! $\endgroup$ Feb 10, 2015 at 17:17
  • $\begingroup$ @quantumorsch: The spacetime (or space) is not really the "underlying manifold" of the Hilbert space of a quantum system. The Hilbert space of the quantum version of a classical system is obtained from the classical phase space of the system, not the spacetime it lives in. $\endgroup$
    – ACuriousMind
    Feb 10, 2015 at 17:26
  • $\begingroup$ Sure, thanks. Of course the wave function lives on configuration/phase space, not space - but the system itself does. So I still find it quite confusing that a change in the geometry of that space does not affect quantum states... $\endgroup$ Feb 10, 2015 at 17:49

Comments to the question (v3):

  1. The possible quantum states depend on the topology of spacetime. For instance, the momentum in the extra (fifth) direction of a 5-dimensional Kaluza-Klein theory is quantized/discrete variable if the extra (fifth) dimension is a compact circle $S^1$, but a continuous variable if the extra (fifth) dimension instead is a non-compact real line $\mathbb{R}$.

  2. Given a classical system with certain geometric data, say, a Hamiltonian system defined on a symplectic manifold $(M,\omega)$, it is a huge research area in physics to try to device a general recipe of how to quantize the theory, and determine the corresponding Hilbert space of physical states and observables. See e.g. the topic of geometric quantization, cf. Refs. 1-3.


  1. Wikipedia.

  2. nLab.

  3. John Baez.


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