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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:

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

The way I know to evaluate this is by using the identity

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

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

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Why is tedious? I see that using properties of Pauli matrices makes it quit easy. – TMS Feb 11 '13 at 18:13
Well, you still have 4 terms with 1-3 pauli matrices each. I was just wondering whether there's an easier way to remember the result. – lucas clemente Feb 11 '13 at 18:16
Rewrite the expression with the explicit expressions of Pauli matrices, sum and then play with $\sin,\cos$, you should get a nice result. – TMS Feb 11 '13 at 18:22

2 Answers 2

up vote 3 down vote accepted

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 $$ e^{-i\vec{\sigma}\cdot\hat{n}\theta/2} \, \vec{a}\cdot\vec{\sigma} \, e^{i\vec{\sigma}\cdot\hat{n}\theta/2} = \left[ \hat{n}(\hat{n}\cdot \vec a)+ \cos(\theta)(\vec{a}-\hat n(\hat n\cdot \vec a)) +\sin(\theta)\hat{n}\times\vec{a} \right]\cdot\vec{\sigma}. $$ Here the vector in square brackets is the rotated one: $\vec{a}$ rotated by angle $\theta$ around unit $\hat{n}$. (To see what each term does, simply take $\hat n$ along the $z$ axis; the cosine and sine terms then describe the on- and off-diagonal terms of the $x,y$ submatrix.)

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.

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For people who like mathiness:

The 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 $$ \mathrm{Ad}_{e^X} = e^{\mathrm{ad}_X} $$

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I've seen this identity called Hadamard's lemma, Baker-Campbell-Hausdorff lemma and Lie series! – auxsvr Jun 5 '14 at 18:23

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