Does $\exp(-i \theta \sigma_m \otimes \sigma_n)$ represent a rotation operator?

Question

It is well known that $\exp(-i \sigma_k \theta)$ where $\sigma_k$ $(k=x,y,z)$ is a Pauli matrix, represents the rotation operator about $k$-th axis. What physical interpretation does $\exp(-i \theta \sigma_m \otimes \sigma_n)$ have, where $\otimes$ is the tensor product?

Answer 1

It is well known that $\exp(-i \sigma_k \theta)$ where $\sigma_k$ $(k=x,y,z)$ is a Pauli matrix, represents the rotation operator about $k$-th axis. What physical interpretation does $\exp(-i \theta \sigma_m \otimes \sigma_n)$ have, where $\otimes$ is the tensor product?

If we use the common physics terminology and say that $\vec \sigma/2$ is an angular momentum (generator of rotations), then we can also say that $\vec \sigma_n\otimes\vec \sigma_m$ is not an angular momentum. It does not generate rotations, and it does not obey the correct commutation relations for an angular momentum.

Rather, the direct product of the two "spin-1/2" angular momenta $\vec \sigma/2$ can be decomposed into a direct sum of a "spin-1" or "triplet" angular momentum and a "spin-0" or "singlet" angular momentum. The symbolic equation for this "addition of angular momenta" is:

$$ \frac{1}{2}\otimes \frac{1}{2} = 1\oplus 0\;, $$ where the symbols refer to the "spin" value. Note that the spin 1/2 representation here is a 2x2 matrix, the spin zero representation is 1x1 (trivial) matrix, and the spin 1 representation is a 3x3 matrix, so the matrix dimensions work out correctly.

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If you would like to get a better feel for what $e^{-i\theta \sigma_m\otimes\sigma_n}$ "looks like" you can expand the exponential and use $$ (\sigma_m\otimes\sigma_n)^2 = 1 $$ to see that: \begin{align} e^{-i\theta \sigma_m\otimes\sigma_n} &=1 + -i\theta \sigma_m\otimes\sigma_n + \frac{1}{2} \left(-i\theta \sigma_m\otimes\sigma_n\right)^2 + \frac{1}{3!}\left(-i\theta \sigma_m\otimes\sigma_n\right)^3+\ldots\\ &=\left(1 - \frac{\theta^2}{2} + \ldots\right) -i \sigma_m\otimes\sigma_n \left(\theta - \frac{\theta^3}{3!}+\ldots\right)\\ &=\cos(\theta) - i\sin(\theta)\sigma_m\otimes\sigma_n \end{align}

Answer 2

As emerged from comments on your question, the rotation is better to be intended as a geometrical concept, rather than physical.

The word itself, rotation, helps to understand a geometrical concept when this is isomorphic to a 3D space. But when it is not such a case, its meaning is preserved, even if you lose the ability to visualize it in your mind.

I believe that when dealing with quantum mechanics, it is particularly useful to go beyond visual reasoning. Especially when considering scenarios like the one you are asking, since, depending on $\theta$, the operator is able to entagle a two-dimensional Hilbert state.

Answer 3

What physical interpretation does $\exp(-i \theta \sigma_m \otimes \sigma_n)$ have, where $\otimes$ is the tensor product?

No physical interpretation I could think of. I can only remind you of the basic mathematical interpretation involved. The nine 4$\times$4 unitary matrices you write down are elements of SU(4).

That is, the nine hermitian traceless independent matrices $\sigma_m \otimes \sigma_n$ constitute 9 of the 15 generators in the fundamental (4D) representation of su(4), the other six independent ones being $\sigma_m \otimes {\mathbf I}$ and ${\mathbf I}\otimes \sigma_n $, each triplet of them an su(2) subalgebra of su(4), generated by the commutators of the original 9. But note the original 9 you wrote are not in a subalgebra of su(4).

If you wish to stretch the definition of rotations to SU(4) transformations, you may think of the nine group elements you wrote down as "SU(4) rotations", but I doubt one might see physical significance in them.

Answer 4

I'm not sure whether this is what you're asking, but one can more generally figure out whether $e^{itA}$ can be interpreted as a rotation, given any Hermitian operator $A$. Or more precisely, whether given some $|\psi\rangle$, the associated orbit $$\{e^{itA}|\psi\rangle:\,\,t\in\mathbb{R}\}$$ can be interpreted geometrically as a "circular motion", when the corresponding states are represented in the Bloch representation.

For a single qubit, this works fine. The reason is that any 2x2 Hermitian operator can be written as $A=\operatorname{tr}(A)I/2+\sigma$ for some traceless Hermitian operator $\sigma$. In some basis $\sigma$ will be diagonal with eigenvalues $\pm\lambda$ with $\lambda\in\mathbb{R}$. Therefore $$e^{itA}= e^{it\operatorname{tr}(A)/2} e^{it\sigma} = -\exp\left(\frac{i\pi \operatorname{tr}(A)}{\lambda}\right) I,$$ for $t=\pi/\lambda$. In other words, at this value of $t$ we have $e^{itA}|\psi\rangle=|\psi\rangle$ for any $|\psi\rangle$. Note that this equation is more formally speaking accurate when $|\psi\rangle$ are rays in the complex projective space, rather than just vectors. Which is a fancy way to say that I'm taking into account the fact that states are defined up the global phase.

However, such an interpretation stops working in higher dimensions, in general. First of all, the Bloch representation of higher-dimensional states does not give a (hyper)sphere. See e.g. this post on qc.SE, and links therein. Still, one might wonder whether we can still get closed orbits via the action of the unitaries generated by a given Hermitian operator. Again, the answer is negative. To see it, just observe that to have a closed orbit you'd need some $t>0$ such that $e^{itA}=e^{i\phi}I$ for some $\phi\in\mathbb{R}$. In terms of the eigenvalues of $A$, this amounts to asking $e^{it\lambda_k}=e^{i\phi}$ for all eigenvalues $\lambda_k$ of $A$. Or more explicitly, $$t\lambda_k = \phi + 2\pi n_k$$ for some set of $n_k\in\mathbb{Z}$. These equations do not admit a solution for the integers $n_k\in\mathbb{Z}$, in general. To see it, observe that they amount to $$t(\lambda_j-\lambda_ k) = 2\pi n_{jk}, \qquad n_{jk}\equiv n_j - n_k.$$ You can read these as saying that, for all $j,k$, we must have $\lambda_j-\lambda_k=\kappa_{jk} \alpha$ for some $\kappa_{jk}\in\mathbb{Z}$ and $\alpha\equiv2\pi/t\in\mathbb{R}$. But this would imply that the ratio between the difference of the eigenvalues is rational, which is not true in general.

It's easy to see it already in three dimensions, where you'd have $$\lambda_1 - \lambda_2 = \kappa_{12} (2\pi /t), \qquad \lambda_1-\lambda_3 = \kappa_{13} (2\pi /t),$$ which assuming $\kappa_{12}\neq0$ gives $$\lambda_1-\lambda_3 = \frac{\kappa_{13}}{\kappa_{12}} (\lambda_1-\lambda_2),$$ that is, $\lambda_1-\lambda_3=q (\lambda_1-\lambda_3)$ for some $q\in\mathbb{Q}$. Therefore if, say $\lambda_1=0$, $\lambda_2=1$, $\lambda_3=\pi$, then you don't get a closed orbit.

On the other hand, whenever the eigenvalues are in rational ratio one with the other, then you get a closed orbit, and you can picture the associated orbits as "circles" in the higher-dimensional Bloch representation. Any tensor product of Pauli matrices will be such an example, as these have spectra in $\{-1,1\}$.