3

Linear field theories

Linear field theories form the classical counterparts to many important QFT's in condensed matter physics, modeling a wide range of materials, from the mundane (semiconductors), to the exotic (topological insulators and superconductors). Linear field theories are exactly soluble, and, when quantized, form great first approximations to complicated QFT's.

Non-interacting particles from a Quadratic Form

Suppose I have a Hamiltonian given by a Hermitian quadratic form on the phase space $(\mathbb C^N,S)$, where $S$ is a symplectic form. This represents a field theory on a one-dimensional lattice $\mathbb Z/N$. The Hamiltonian, explicitly, is

$$H=\sum_{ij}A^{ij}\bar z_iz_j\tag{1}$$

Bosonic statistics

(1) provides a model for a system of non-interacting, identical bosons, because, if I apply geometric quantization to this theory, while using the polarization indicated by the complex structure, we get the following QFT: $$\hat H=\sum_{ij}A^{ij}b_i^\dagger b_j$$ The field operators here are bosons, because Geometric Quantization replaces classical Poisson brackets with commutators: $$[b_i^\dagger,b_j]=\{\bar z_i,z_j\}=\delta_{ij}.$$ If all I get is commutators, then how do I apply geometric quantization to get fermions? It seems to be a mystery.

David Roberts
  • 987
  • Geometric quantization doesn't merely "replace Poisson brackets with commutators", that's canonical quantization (although geometrical quantization in essence wants to reproduce the simple replacement procedure more rigorously, avoiding the no-go theorems of Groenewold and van Hove). – ACuriousMind Apr 09 '16 at 15:34
  • How do I apply geometric quantization to get fermions? – David Roberts Apr 09 '16 at 18:10

1 Answers1

0

If you are interested In obtaining half-integer spin particles, you will need a classical phase space of the form $\mathbb{R^3}\times S^2$. You can find this in Souriau's book. Or more generally, Geometric Quantization obtains spin in quantum mechanics via the quantization of the sphere. This is strongly related to the coadjoint orbit method. In case you want to obtain not only the Hilbert space of spin, you need to consider a classical phase space $\mathbb{R}^3 \times S^2$.

DanielC
  • 4,333
ProphetX
  • 644
  • Could you perhaps say a bit more about the "quantization of the sphere"? – flippiefanus Aug 23 '23 at 08:54
  • Of course. Let us consider the set \begin{equation} S^{2}:={(x,y,z) \in \mathbb{R}^{3}| x^2+y^2+z^2=1 }.\end{equation} We equip it with the subset topology of $\mathbb{R}^3$ and using the implicit function theorem, we obtain a smooth structure, thus $S^2$ becomes a smooth manifold. We now can equip with with a volume form $$ds=x dy \wedge dz + ydz \wedge dx+ z dx \wedge dy.$$ We claim that $\omega=-c ds$ for some $c \in \mathbb{R} \setminus {0}$ is a symplectic form on $S^2$, which makes the pair $(S^2,\omega)$ into a symplectic manifold. – ProphetX Sep 14 '23 at 14:42
  • A symplectic manifold $(M,\omega)$ is said to be quantizable if and only if we can build a complex line bundle $\pi:L \to M$ with connection $\nabla$ over it, such that the $Curv(L,\nabla)=\omega$. One can show that the Hopf Fibration $S^3 \to S^2$ with connection 1-form $\alpha=-in\hbar z^{*} dz$ is such a line bundle and for the condition on the curvature to be satisfied, it must hold that $c=-\frac{n \hbar}{2}$, which gives rise to 'spin'. – ProphetX Sep 14 '23 at 14:52
  • An alternative way to see the quantization of the sphere is to realize it as coadjoint orbits of $SU(2)$. Then, as coadjoint orbits are symplectic manifolds (cf. Kirillov), they can be quantized if quantizability condition is met. Then, this quantization condition will essentially boil down to what we had before. We can equip $S^2$ with a Kahler structure, thus giving a Kahler polarization. We now determine the polarization preserving functions, and assign to each the Kostant-Souriau prequantum operator. – ProphetX Sep 14 '23 at 14:59
  • These will form a lie algebra, which can be represented on the space of square integrable polarized sections of the prequantum bundle. Lifting this Lie algebra representation to the Lie group will give rise to all unitary representations of $SU(2)$. I have reference for both approaches, so in case you are interested in more details or step by step computations, let me know. – ProphetX Sep 14 '23 at 15:00
  • Perhaps you can add all this to your answer instead of putting it in the comments. – flippiefanus Sep 15 '23 at 04:26