# Square of the Pauli matrices and the identity matrix

The square of any of the three Pauli Spin matrices is equal to the identity.

Is there any physical meaning to this? Would you expect it? Maybe in the context of the $SU(2)$ group?

• In fact, something stronger holds. One has $\{\sigma_a,\sigma_b\}=2\delta_{ab}$, where $\{,\}$ is the anti-commutator. This is an example of a Clifford algebra. en.wikipedia.org/wiki/Clifford_algebra Sep 8, 2014 at 23:26
• Perhaps related to the fact that measuring the square of spin should give you a single value.
– BMS
Sep 8, 2014 at 23:46

Yes, the Pauli matrix $\sigma_j$ represents (up to a proportionality factor) the spin in the $j$th direction of a spin $\frac{1}{2}$ system. Such system has only two spin states: $\uparrow$ and $\downarrow$, with opposite eigenvalues. The square $\sigma_j^2$ can no longer see the sign, so it only has one eigenvalue, cf. comment by BMS. In other words, the square $\sigma_j^2$ is proportional to the identity matrix.
• Because the eigenvalues of the matrix are $1$ and $-1$... Sep 9, 2014 at 8:59
This is because there are just two possible values to the spin in any direction, $-\frac{\hbar}{2}$ and $\frac{\hbar}{2}$, they just differ in a sign, so when you square it you get a single value $\frac{\hbar^2}{4}$. Think about this, the only possible value when you measure the square of $S_z$ is $\frac{\hbar^2}{4}$ for any state, so $$<\psi|S_z^2|\psi>=\frac{\hbar^2}{4} \quad \forall \, |\psi>$$ So it must be a multiple of the identity operator $$S_z^2=\frac{\hbar^2}{4} I$$ Remember that $S_z$ is proportional to the Pauli matrices, $S_z =\frac{\hbar}{2} \sigma_z$.