What happens to spin after measurement, does it evolve and randomise like position evolves after measurement?

Question

When a particle's position is measured, if one considers the wavefunction to collapse then one can assume it collapses into a delta function peaked at the measured position, and then after some finite time the wavefunciton 'evolves' outward away from this position.

Does a similar thing happen to spin?

From reading, it seems that instead if one measures a particle to be 'spin up' in the z-direction, and then wait some finite time, so long as you measure the particle in the z-direction again, it will always measure spin up. This would suggest that the particle's spin doesn't 'evolve into randomness after measurement'?

Is this correct?

One idea I had was that if we assume the wavefunctions only evolve due to the Hamiltonian in the TISE, if we assume a free particle Hamiltonian, this would indeed affect the spatial part of the wavefunciton but not the spin part.

Answer 1

Cliff notes: a spin up particle won't in general stay spin up, but in certain situations it will (for example, if it's free).

The standard way to describe a physical quantity (observable) whose value stays the same is to say that "it's conserved". Your question can be generalized as follows:

Which observables are conserved?

The answer is: whether or not any particular observable is conserved doesn't just depend on the observable itself, but on the physical situation, i.e. the Hamiltonian.

As an easy example, in classical mechanics for a particle in free space the position won't be conserved but the momentum will. For a particle in a gravitational field neither the position nor the momentum are conserved.

In quantum mechanics, it can be shown that an observable is conserved if and only if it commutes with the Hamiltonian.

With regard to the spin, if for example you have a particle in a magnetic field, the field will interact with the spin and can change its value, kind of like the gravitational field in the above example will interact with the particle’s momentum and will change its value. From the mathematical perspective, the Hamiltonian will have a term involving the magnetic field, which will cause it not to commute with the spin. For a free particle there is no such term, and the spin, as well as the momentum, will be conserved.

Answer 2

One idea I had was that if we assume the wavefunctions only evolve due to the Hamiltonian in the TISE, if we assume a free particle Hamiltonian, this would indeed affect the spatial part of the wavefunciton but not the spin part.

Exactly. The Hamiltonian of a free particle commutes with the angular momentum operators, and so angular momentum (in particular spin angular momentum) is conserved. On the other hand, the Hamiltonian does not commute with the position operator, so position is not conserved. The latter statement is essentially a quantum version of Newton's first law -- an object with no external forces will maintain its velocity, so it can't be the case that a free particle starting with some non-zero velocity maintains the same position for all time.

Answer 3

This depends on the forces, or more appropriately, torques, acting on the spinning particle, just as in the case of a Newtonian spinning object. If there is an external torque, then the spin will evolve; otherwise, it will not. Quantum mechanics does not alter this principle. The same goes for linear momentum. Momentum requires an external force to cause it to change, and this applies as much to quantum mechanics as to Newtonian mechanics.

The difference is that with a spinning elementary particle in quantum mechanics, the magnitude of the spin cannot change - only the direction of the spin axis, and with the further caveat the direction is never fully detailed: the maximum amount of information that can be held in it is $\lg(2|s| + 1)$ bits, where $s$ is the spin quantum number, so an electron (spin 1/2) has only one bit, but a photon (spin 1) carries about 1.58 bits in the direction, and a hypothetical graviton (spin 2) holds a bit more than 2.32 bits.