# Spin Quantum Number

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1. Oct 25, 2015

### Sheldon Cooper

Hello everyone,
In case of hydrogen atom, when we say spin up or spin down we refer to the z component of the spin. Why are we interested only in the z component of spin and not in the x and y components?

2. Oct 26, 2015

### Simon Bridge

The "z component" is the component that is in the direction of the applied magnetic field, see "Stern-Gerlach experiment".
The x-y components are those that are perpendicular to the z component. We are interested in them, they are just in superposition.

3. Oct 26, 2015

### Staff: Mentor

I suspect the reason we call it the "z" component, and not "x" or "y" is that when we deal with orbital angular momentum we normally express the Schrödinger wave function $\psi$ in spherical coordinates, which uses the z-axis as the axis of the coordinate system. The math describing atoms in magnetic fields is simpler if we align the axis of the coordinate system with the magnetic field.

We carry this convention for the z-axis over to spin angular momentum for consistency, which makes it easier to add spin and orbital angular momenta.

4. Oct 26, 2015

### HomogenousCow

When you solve the SE equation for a spherically symmetric Hamiltonian you want the eigenstates to be simultaneous eigenstates of L squared and L. You can only pick one of the Ls since they don't commute, we call this arbitrary direction the z direction. I'm pretty sure once you sum all the degenerate states you get a state with no bias towards any direction, correct me if I'm wrong.

5. Oct 26, 2015

### Chandra Prayaga

It is correct that in a quantum system, L2 and only one component of the vector L, can be measured simultaneously. We arbitrarily choose the z-component. But once we choose the z-component as the measurable quantity, Lx and Ly can no longer be specified exactly. I am not sure why you would want to add the degenerate states. Each one of them is a perfectly valid, measurable state. Of course, once you add a number of degenerate states with the same L2, but different Lz, the new state no longer has a definite value of Lz.

6. Oct 26, 2015

### HomogenousCow

What I mean is, in general an eigenstate in an eigenspace has no particular connection to any direction of angular momentum, it's only that when we solve the SE for rotationally invariant problems we pick an "eigenbasis" which are also Lz eigenstates.