Pauli matrix rotations

Question

When doing physics with two-level systems and introducing rotations, a term that appears quite often is the rotation of a pauli matrix by another one:

$e^{- i \sigma_i \theta/2} \sigma_k e^{i \sigma_i \theta/2}$

The way I know to evaluate this is by using

$\exp(i \sigma_i \theta/2) = I \cos \frac{\theta}{2} + i \sigma_i \sin \frac{\theta}{2}$

and expanding the previous equation. If $i = j$ it's trivial, but in the other cases it get's quite tedious. Is there an easier way to do that (or some shortcut)?

Answer 1

As TMS mentioned, if you play around with the Pauli matrix properties and the double-angle trig formulae, you should get a nice result. (Of course, what "nice" means may depend on what you're going to use it for.)

I find it useful, however, to step back and give the geometrical nature of the object you're dealing with take the reins. How would I do this? It's often the particulars of the formulation that bog you down and it's best to generalize the setting a bit. Thus, in particular, consider the frame-of-reference-free expression $$ e^{-i\vec{\sigma}\cdot\hat{n}\theta/2} \, \vec{a}\cdot\vec{\sigma} \, e^{i\vec{\sigma}\cdot\hat{n}\theta/2} $$ where $\hat{n}\cdot\hat{n}=1$. You need to apply the identity you mention and the property that $$ (\vec{a}\cdot\vec{\sigma})(\hat{n}\cdot\vec{\sigma})=\vec{a}\cdot\hat{n}+i\vec{\sigma}\cdot(\vec{a}\times\hat{n}), $$ as well as double-angle formulae and a triple vector product (i.e. nothing intimidating).

It's definitely a worthwhile exercise to work through that; the result is $$ e^{-i\vec{\sigma}\cdot\hat{n}\theta/2} \, \vec{a}\cdot\vec{\sigma} \, e^{i\vec{\sigma}\cdot\hat{n}\theta/2} = \left[ \cos(\theta)\vec{a}+\sin(\theta)\hat{n}\times\vec{a} \right]\cdot\vec{\sigma}. $$ Here the vector in square brackets is the rotated one: $\vec{a}$ rotated by angle $\theta$ around unit $\hat{n}$.

I'm not sure what kind of a shortcut you're looking for or whether this is useful in that sense. I just want to say that sometimes stepping back to a greater generality can ease the computational burden, or at least make clearer what's going on.

Answer 2

For people who like mathiness!

The Hadamard Lemma (see the section with the title Hadamard Lemma) says the following:

Let $X$ and $Y$ be square, complex matrices, then $$ e^{X} Y e^{-X} = Y + [X,Y] + \frac{1}{2!}[X,[X,Y]] + \cdots. $$

You should (in principle) be able to use this to derive whatever expression you get.

This lemma can also be written using the operations $\mathrm{Ad}$ and $\mathrm{ad}$ called "adjoint" which are important in the representation theory of Lie groups and Lie algebras (in fact all this Pauli matrix spin stuff falls under this umbrella). These operations are defined as $$ \mathrm{Ad}_g(X) = g X g^{-1}, \qquad \mathrm{ad}_X(Y) = [X,Y] $$ This notation allows us to write the lemma as $$ \mathrm{Ad}_{e^X} = e^{\mathrm{ad}_X} $$