What do the commutators of the Hamiltonian with the spin operators mean precession-wise?

I proved that $[H, S_z] = 0$, while $H$ and $S_x,S_y$ do not commute.

I showed this using matrix representations, now I am to comment on my results with respect to spin precession and I need help for that - how exactly do commutators represent something about precession and rotation along the 3 axes?

The commutator of an observable like $S_x$ or $P_y$ (the spin in the x direction and the y component of the momentum respectively) with the hamiltonian will tell you about the time evolution of that observable. It tells you how that observable changes with time.

This is given through the heisenberg equation:

$\frac{dA}{dt} = \frac{i}{\hbar}[H, A]$

where A is an operator (like $S_x$ or $P_y$) and I have assumed that A does not depend explicitly on time, i.e $A = A(x(t),p(t))$ and not $A = A(x(t), p(t), t)$.

Now from the equation above, we can see that if A commutes with the hamiltonian, $[H,A] = 0$, then $\frac{dA}{dt} = 0$ and thus A is constant in time, it is conserved.

So when you showed that $S_z$ commutes with H while $S_x$ and $S_y$ do not, you showed that the z component of a particles spin is constant in time, while the x and y components are not constant, they precess.

• so this is true provided that the H is time independent ? – dumpy Nov 16 '16 at 15:04
• H can be time dependent or independent in this formalism. – CStarAlgebra Nov 16 '16 at 15:11
• so that means the same can be shown in another way using expectation values of Sx, Sy, Sz, yes? also thanks alot! – dumpy Nov 16 '16 at 15:27