What is the momentum canonically conjugate to spin in QM?

Question

In Kopec and Usadel's Phys. Rev. Lett. 78.1988, a spin glass Hamiltonian is introduced in the form:

$$ H = \frac{\Delta}{2}\sum_i \Pi^2_i - \sum_{i<j}J_{ij}\sigma_i \sigma_j, $$ where the variables $ \sigma_i (i = 1, \ldots, N) $ are associated with spin degrees of freedom [...] and canonically conjugated to the "momentum" operators $ \Pi_i $ such that $ [\sigma_i, \Pi_j] = i \delta_{ij} $.

Now, I am accustomed to writing the "kinetic" term in a transverse-field Ising-like Hamiltonian as $ \propto \sum_i \sigma^x_i $ (working in the standard basis of $ \{\sigma^z_i\} $), so this passage is raising some questions for me.

What are these $ \Pi_i $ operators? If $ \Pi_i^2 = \sigma^x_i $, like I initially believed, then they cannot be observables, for the square of a self-adjoint operator is positive semidefinite (which $ \sigma^x_i $ is not). In fact, if one restricts to the $ i $-th spin and takes $ i = j $, one can easily prove that $$ [\sigma^z, \Pi] = \sigma^z \Pi - \Pi^\dagger \sigma^z = i \mathbb 1 $$ is satisfied for $$ \Pi = \begin{pmatrix}i/2&b\\-\bar{b}&-i/2\end{pmatrix} $$ with $ b \in \mathbb C $. This squares to a multiple of the identity matrix, which seems like an odd choice for a kinetic term. I feel I am missing something here.

More generally speaking, can one even define a momentum "canonically conjugate" to $ \sigma^z $, or any other spin operator for that matter? As far as I understand, in classical mechanics the variables conjugate to physical rotations are angles, but this cannot be ported over to QM in any obvious way.

Answer 1

Even though the explicit commutator you wrote is wrong--you should not have conjugated $\Pi$ in the second term-- your conclusion is sound that you cannot possibly satisfy the Born-Heisenberg commutation relation with 2x2 matrices.

In fact, there is a general Theorem: The Heisenberg algebra does not admit faithful finite-dimensional (matrix) representations. So, whatever else they might be, your variables $\sigma, \Pi$ are not bounded operators---and so cannot be the 2x2 matrices you are considering.

This observation was first made by P Jordan, Zeits. f. Phys. 44 1 (1927).

Answer 2

First of all, there is a very elementary reason why $[\sigma_i,\Pi_j] = i \delta_{ij}$ is impossible for finite dimensional matrices. Because that identity would result in the following contradiction: $$ \text{Tr}[A,B] = 0 \neq \text{Tr} (i \cdot \mathbb{1}) = i \cdot n$$ Where $A,B$ are $n \times n$ matrices.

However, if we read the paper by Kopec and Usadel carefully, we notice a key phrase that is not mentioned in the question:

To capture the essential physics of the problem we consider a quantized spherical model on the Bethe lattice given by the Hamiltonian: $$ H = \frac{\Delta}{2} \sum_i \Pi_i^2 - \sum_{i,j} J_{ij} \sigma_i \sigma_j $$

The highlighted word "spherical model" means that we have the spherical constraint $\sum_i \sigma_i^2 = 1$ instead of the Ising constraint $\sigma_i^2 = 1$ for all $i$. So the $\sigma_i$ in this model is in fact a constrained position operator acting on an infinite dimensional Hilbert space, and a commutation relation like $[\sigma_i,\Pi_j] = i \delta_{ij}$ becomes conceivable.

However, I have one lingering confusion. When considering spherical constraints, I am used to quantizing the Dirac brackets instead of Poisson brackets (The Dirac bracket is a modification of Poisson brackets, discussed, for example, in Weinberg QFT section 7.6). Using Dirac brackets: $$ [\Pi_i, \sigma_j] \neq i \mathbb{1}$$ So I don't quite understand why the author used the standard canonical commutation relation to quantize the system. I hope someone else will clear up this point.