# What is the momentum canonically conjugate to spin in QM?

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.

• have a look at JxJy .... uncertainty relations en.wikipedia.org/wiki/… . – anna v Feb 14 '16 at 11:51
• Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files. – Qmechanic Mar 21 '18 at 17:46
• @Qmechanic Duly noted. Thank you for pointing that out. – derpy Mar 21 '18 at 20:40

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).

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.

• You didn't quote enough from the paper. The very next sentence clears up your point: ''located on the Bethe lattice with coordination z". Thus there is no spherical constraint imposed; the $\sigma_j$ just run over the lattice. Instead, it is said a few lines later, that only the mean of the spherical relation is enforced to be reproduced. – Arnold Neumaier Mar 28 '18 at 18:44
• @ArnoldNeumaier Thanks for pointing that out. I in fact did not notice this statement. However this statement makes the paper more confusing for me. If $\sigma_j$'s are discrete variables, then the commutation relation is impossible. Now, if we take $\sigma_j$'s to be continuous variables satisfying the averaged constraint, then I don't understand equation (2) in the paper, which clearly uses a delta function to impose the exact spherical constraint! In either case I don't understand why we are allowed to do canonical quantization instead of Dirac quantization. – Zhengyan Shi Mar 29 '18 at 3:09
• Yes, they don't do what they say and in fact enforce in the path integral (2) not the averaged constraint but the exact constraint. Using the integral representation of the delta-constraint introduced after (3) converts this (nonrigorously) to an integral over a path integral of the unconstrained system. This allows them to evaluate the path integral using the standard commutation rules. – Arnold Neumaier Apr 1 '18 at 17:04