I am studying spin recently, and textbook gives some identities of Pauli matrices, one said that

for any two unit vectors $\bf m$ and $\bf n$, $[\bf m \cdot \bf{\sigma},\bf {n \cdot \sigma}]= 2i(m\times n)\cdot\sigma$

I know how to derive it, but is there any physical meaning of this identity?

  • $\begingroup$ possible duplicate of Rotation in configuration space $\endgroup$
    – ACuriousMind
    Feb 8, 2015 at 13:44
  • $\begingroup$ not exactly a duplicate, since this question deals with the physical meaning, while the linked question deals more with the maths and answers this question only implicitly. $\endgroup$
    – Martin
    Feb 9, 2015 at 13:15

2 Answers 2


Yes. This commutation relation is that of the Lie algebra $\mathfrak{so}(3)$ corresponding to the rotation group in three dimensions. Thus the commutation relation states that the Pauli matrices generate rotations.

To understand why this is the commutation relation of $\mathfrak{so}(3)$, one can draw a diagram showing that the commutator of two infinitesimal rotations is an infinitesimal rotation around the perpendicular axis. It is of course also possible to reason without the picture. To work that out, you need that the effect of an infinitesimal rotation around $\mathbf n$ on $\mathbf v$ is $$\mathbf v \mapsto \mathbf v + \varepsilon \mathbf n \times \mathbf v.$$


Just to clarify to Robin Ekman's answer, superpositions of the Pauli matrices exponentiate to $SU(2)$, not $SO(3)$, but both these Lie groups have $\mathfrak{so}(3)$ as their Lie algebra - but I am sure you already know this.

Also, there is another way to look at the problem that you might find helpful, even though it is a mathematical insight rather than a physical one. I've seen $\mathfrak{so}(3)\cong\mathfrak{su}(2)$ called the "cross product algebra" in some texts, and the following is why. Your identity is a particular case of a general identity for Lie algebra homomorphisms for the special case of the adjoint representation of the cross product Lie algebra.

Let's begin with the cross product operating on three dimensional vectors and work backwards in the opposite direction to your question. The cross product with 3D vectors is an abstract Lie algebra over the reals: it is billinear, skew-symmetric and fulfils the Jacobi identity.

There is a fiddly theorem called Ado's theorem that essentially says that every abstract Lie algebra over a field of characteristic nought can be realised as a matrix Lie algebra. The proof of Ado's theorem is essentially a very messy algorithm for deducing that matrix Lie algebra from a set of abstract commutation relations, in principle.

In this particular case, though, the alogorithm of Ado's theorem is very simple, because the Lie algebra is centreless: no member of the algebra commutes with everything else.

What this means is that, by dint of the identity:

$$\mathrm{ad}([X,\,Y]) = \mathrm{ad}(X)\,\mathrm{ad}(X)-\mathrm{ad}(Y)\,\mathrm{ad}(X) = [\mathrm{ad}(X),\,\mathrm{ad}(Y)]\tag{1}$$

(which happens to be the Jacobi identity in disguise and is also your exact identity in disguise see footnote) the image of the "cross product algebra" under the adjoint representation is the same Lie algebra. In more everyday language, when you write down the matrices for the linear map $Y\mapsto X\times Y$, that is:

$$Y=\left(\begin{array}{c}y_1\\y_2\\y_3\end{array}\right) \mapsto X\times Y = \left(\begin{array}{ccc}0&-x_3&x_2\\x_3&0&-x_1\\-x_2&x_1&0\end{array}\right) \left(\begin{array}{c}y_1\\y_2\\y_3\end{array}\right)\tag{2}$$

the vector space spanned by the matrices of the "basis" linear operations of the form $U\mapsto \hat{X}\times U$, $U\mapsto \hat{Y}\times U$, $U\mapsto \hat{Z}\times U$ form the same Lie algebra as the cross product algebra, when one uses the wonted matrix commutator bracket as the Lie product (here $\hat{X},\,\hat{Y},\,\hat{Z}$ are the normal Euclidean unit basis vectors). This new algebra, namely the image of the cross product algebra under the adjoint representation, is of course none other than $\mathfrak{so}(3)$, and the Ado theorem procedure here realises the algebra as the $3\times 3$ real matrices rather than the "quaternion" Pauli matrices algebra $\mathfrak{su}(2)\cong\mathfrak{so}(3)$. As we've already noted, though, these two Lie algebras are isomorphic.

In summary, in your identity, the left hand side of (1) plays the role of the right hand side of your identity. The cross product is the Lie bracket in the cross product with 3D vectors algebra. The left side of your identity corresponds to the right hand side of (1). We can rewrite (1) as:

$$\mathrm{ad}(X \times Y) = [\mathrm{ad}(X),\,\mathrm{ad}(Y)]$$

and the factor $i$ is added to make the Hermitian Pauli matrices into the skew-Hermitian quaternion units (the latter are the Lie algebra).


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