What is the definition of "precession"? How is it applicable to abstract objects such as Spinors? I understand the mathematics, but don't understand what one means by "precession angle" etc when it applies to abstract objects such as two spins?

In my (naive?) view, precession is defined as the rotation of a vector $\vec{v}_1$ about another vector $v_{2}$ in 3D space, with the following constraints:

  • The norm of $\vec{v}_1$ is an invariant.
  • The projection angle of $\vec{v}_1$ on $\vec{v}_2$ is an invariant.

I am stuck after this, because I know that there must be something else that has to do with the trajectory in the plane that is always orthogonal to both vectors (the cross product part of the equation).

Once you fix the definition, how to generalize it to quantum spin-systems? I am having a huge conceptual problem because in my view Spinors are not vectors and I don't know how they can be treated as such when we combine them with a magnetic field vector?


1 Answer 1


Your definition of precession sounds fine to me: it's basically rotation of one vector around another. In many cases, precession is assumed to be at constant angular speed, but even if the angular speed is not constant it would still be called precession.

To apply this to quantum spin systems (for spin-1/2 particles, which is the most common case), you have to get around the fact that the different components of spin are not simultaneously measurable. This means that if you're going to write the spin as a vector, that vector needs to have elements which don't commute. That rules out using a vector of numbers, but you can still use a vector of matrices (actually operators), specifically the Pauli matrices

$$\begin{align}\sigma_x &= \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix} & \sigma_y &= \begin{pmatrix}0 & -i \\ i & 0\end{pmatrix} & \sigma_z &= \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\end{align}$$

With these, you can construct a vector spin operator $\mathbf{S} = \frac{\hbar}{2}(\sigma_x \hat{x} + \sigma_y \hat{y} + \sigma_z \hat{z})$. This operator doesn't describe a specific spin state; it's just an operator that represents the general quantum mechanical concept of spin. But you can apply it to a specific state in all the ways that one normally uses quantum operators; in particular, you can take the expectation values $\langle\chi|S_x|\chi\rangle$, $\langle\chi|S_y|\chi\rangle$, $\langle\chi|S_z|\chi\rangle$ (where $|\chi\rangle$ is a spinor). Then you can make a vector out of these components, which is the closest you can get to expressing a spinor as a vector.

If you calculate the expectation values of the spin components for a spin-1/2 particle in a magnetic field and construct the resulting vector $\langle\mathbf{S}\rangle$, you will find that $\langle\mathbf{S}\rangle$ precesses around $\mathbf{B}$. (This comes about due to the spin-field coupling in the Hamiltonian, $H = -\gamma\mathbf{B}\cdot\mathbf{S}$.) But of course this is just an expectation value. If you actually construct such a quantum system and measure it, your actual results will follow some distribution.

I'm not sure if that covers what you wanted to know... if not, maybe you can clarify what it is you're confused about and I can try to edit this accordingly.

  • $\begingroup$ Thank you, that clarified quite a bit. There are a few more things I am unclear about, but I need to organize my thoughts properly. I will edit my question once I do so. $\endgroup$ Commented Feb 16, 2012 at 22:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.