# Spin |1/2,0>

1. Apr 27, 2010

### LostConjugate

I forget. Assuming you are working with S and S_z what prevents S_z from taking on a value of 0 say when the angular momentum happens to be along the x and/or y axis only.

2. Apr 27, 2010

### SpectraCat

What do you mean? Are you talking about the raising and lowering operators as defined below?

$$S_{\pm}=S_x \pm iS_y$$

Recall that for a given eigenstate |s,ms>, the raising (lowering) operator changes it into the next higher (lower) eigenstate, where ms is increased (decreased) by one, provided that it is not at its maximum (minimum) value of +s (-s).

The relevant parts of that are that the raising and lowering operators can only change ms by integer steps, and that the max and min values are equal to s. Thus, if s is a half-integer value, then ms must also be a half-integer value in the range -s to +s. Therefore it cannot be zero.

3. Apr 28, 2010

### tom.stoer

Spin is just the physical expression of the mathematical structure SO(3) (3-dim. rotation group) or its universal cover SU(2) (2-dim. complex rotation group). One can derive the SO(3) representations = the angular momentum J=0, 1, 2, ... with Jz = -J, -J, ..., +J; the ladder operators act between Jz. If one uses the SU(2) one finds in addition the half integer representations, that means J=0, 1/2, 1, 3/2, 2, ... again with Jz = -J, -J, ..., +J.

That means for half-integer spin Jz=0 is forbidden by group representation theory.

4. Apr 28, 2010

### LostConjugate

Now I am more confused. I don't understand why the intrinsic angular momentum must be in the Z direction, I thought it was an arbitrary direction. For example in a standard angular momentum problem the total angular momentum can be 1 while the Z component can be 0. This would be a mass moving around in the zx or zy plane.

Last edited: Apr 28, 2010
5. Apr 28, 2010

### SpectraCat

You are right that any arbitrary spatial axis can be chosen to define the projections of the spin ... the choice of the z-axis as the "principal axis" is by convention only. However, we know from the HUP and the Stern-Gerlach experiments that we can only ever know one component of the spin at a time. So, once we choose to define Sz as the operator that commutes with the Hamiltonian, that restricts what can be sone in terms of Sx and Sy .. you cannot define wavefunctions where the projection of the spin on the z-axis and the x-axis are simultaneously well-defined.

In terms of your second point, I am not sure what you are saying ... S=1, MS=0 would define a spin-vector of length sqrt(2) having no projection on the z-axis, and therefore with a random orientation in the xy-plane.

Also, remember that with spin, we are talking about intrinsic angular momentum .. a purely QM phenomenon that has no relation to moving bodies.

6. Apr 28, 2010

### LostConjugate

Understood it is different, I would say it has relations though. It has a magnetic field just as an electric charge with angular momentum would. I just don't understand why the Z field has to be + or - 1/2 since the entire angular momentum is always 1/2. There is no room to have angular momentum in any other basis vector. HUP would imply that not all the angular momentum could be found along a single vector.

7. Apr 28, 2010

### SpectraCat

This is fundamental to the quantum mechanics of angular momentum ... you clearly have issues with your understanding of that subject, and I don't have time to post a review of the relevant derivations. I suggest that you review that subject in a textbook, and then come back and ask specific questions if you still have confusion.

8. Apr 28, 2010

### LostConjugate

So intrinsic angular momentum favors our definition of a Z axis? Makes no sense.

Last edited: Apr 28, 2010
9. Apr 28, 2010

### SpectraCat

No .. not at all ... read my response in post #5. Z-axis is just a convention .. you might as well call it the "Highlander axis", since there can be only one. That is the important part, as I said above.

10. Apr 28, 2010

### LostConjugate

So we could choose S_x and S_y and get the same result? However some of the spin can't be in S_x and some in S_z and so on?

11. Apr 28, 2010

### Zarqon

If you make a measurement (Sz) on a single spin 1/2 particle, it is clear that you will get either mz = +1/2 OR mz = -1/2.

However, if the particle that you are looking at is free, i.e. not near any extrernal field that can provide a real reference axis (in terms of making one spin direction lower energy than the other), then if you perform the above measurement on many identically prepared particles, the average of those measurements will yield mz = 0. Just as it would yield zero for any direction, as a free particle without a sense of direction should yield spherically symmetric results.

So, on average you can certainly measure mz = 0. Is this what you were thinking of perhaps?

12. Apr 28, 2010

### LostConjugate

This might be my confusion. When we measure spin we always measure it in the presence of an external magnetic field? And if so, we normally take the direction of the mag field to be along the Z axis? The spin then lines up along the magnetic field?

13. Apr 28, 2010

### Zarqon

Yes, by convention, z is the axis of the external field. What happens when an external magnetic field is applied is that there is now an energy difference between spin up and spin down. A spin projection on this axis (z), pointing anti-parallel to to the magnetic field has a lower energy than the parallel pointing spin projection. If the particle is in a thermally cold enough state (such that kT is less than the energy difference) the particle is typically found in the lower energy state, but in the presence of heating effects both spin states could still be populated. In either case though the magnetic field will split the degeneracy that was there before (this is the essence of the Zeeman effect for example).

14. Apr 28, 2010

### LostConjugate

Thanks Zarqon and Spectra

Edit: Oh and Tom :)