# What do the 4 spin states of two particles represent

1. Jun 8, 2010

### LostConjugate

I can only imagine 3 states, the angular momentum adding constructively either in the positive or negative, or acting destructively.

specifically what is the difference between the state with no total angular momentum and destructive angular momentum (singlet state) and the one with a positive angular momentum and destructive angular momentum (the second triplet state).

2. Jun 8, 2010

### phyzguy

They have different quantum numbers. The singlet state has L=0 and M=0, while the triplet state has L=1 and M=0. Depending on the system, they can have different energies and the outcome of measurements can be different.

3. Jun 8, 2010

### LostConjugate

But the quantum numbers are proportional to the actual angular momentum energy eigenvalue. l(l + 1) in adjusted units.

So what I am asking is what is the difference between the states physically. Is one state trying to say that the angular momentum is canceled in the z direction but not in the x direction? While the other is saying that it is canceled in all directions?

4. Jun 9, 2010

### Zarqon

This is correct. The tripplet states all have an angular momentum, while the singlet state does not (think length of the vector is really zero). The second quantum number is simply a measure of the projection of that spin onto one chosen axis (e.g. z-axis). In the case of the projection being zero for the second tripplet case, it means that the vector length is still non-zero, but it's pointing in a direction perpendicular to the z-axis (i.e. in the xy-plane).

5. Jun 9, 2010

### LostConjugate

So then why is there not more states?

For example:

No Angular Momentum in the z-axis, but in the x-axis, and not in the y-axis.

What about in both the x-axis and the y-axis but not the z-axis.

In the Z and X axis, but not the y axis.

The spin can be constructive or destructive on any axis it is measured on.

6. Jun 9, 2010

### Staff: Mentor

Do you know about the theorem which says that in order for a system to have simultaneously definite values for two quantities, the operators corresponding to those quantities must commute?

The operators for Lx, Ly and Lz do not commute with each other, in any pair. Therefore only one of them can have a definite value.

However, each of those three operators commutes with the operator for the magnitude L. (Actually we usually talk about L2 because it's easier to deal with, mathematically.) So L and Lx can be simultaneously definite, as can L and Ly, or L and Lz, with the other two components being indefinite in each case. By custom, we call the "definite" component Lz, although it could actually be any one of the three.

7. Jun 9, 2010

### LostConjugate

Most of that I understand, except that I thought no direction was definite, usually described as probability cones.

So what your saying is that if it is not in the direction we are measuring we don't know what it is in the other directions at all? Not even to some uncertainty? So all we care about is 1 direction.

8. Jun 9, 2010

### Klockan3

Nonono, you know the spin in one direction exactly, but the other directions you don't. That doesn't mean that you know the direction, for example you know that the spins z component is 1/2 then there is a bit left over for the y and z components that is uncertain which creates this cone, but the z component is exactly known and the total is exactly known so what is left is just a circle.

Also the state is in all of that circle at once, it is not a set of states but a single state. So for L=1, z=0 you got a circle in the y-x plane, while for L=0 Z=0 you got just a point.

9. Jun 9, 2010

### LostConjugate

A bit left over? I thought the spin along any direction is always 1/2. So it is 1/2 in all directions all the time. plus or minus. Now when you have two electrons they are either 1 or 0 in any direction.

Which to me means there is the possibility of being 1 in x, y, and/or z and everything else is just a linear combination. So there should be 2x2x2 states, 8.

10. Jun 9, 2010

### Klockan3

No, firstly they are fermions. Fermions can't share states with each other even if they just partly share them. All eigenstates of x and y can be written as linear combinations of z eigenstates thus the z eigenstates takes up all of the possible states.

Then the total angular momentum is not 1/2, it is sqrt(3/4) for a spin 1/2 particle and thus you got a bit left over for the x and y components to share which creates the uncertainty circle.

11. Jun 10, 2010

### LostConjugate

How can you write eigenstates of x and y as linear combinations of z if they are orthogonal?

12. Jun 10, 2010

### Zarqon

With "constructive or destructive spin" I assume you mean the addition of two angular momenta, either in the same direction, in which case you get the three tripplet states, or in opposite direction, in which case you get the singlet state. It can thus be constructive "on different axes" but destructive addition can happen only in one way, i.e. with spin S = 0.

Constrive addition leads to S = 1 (assuming spin 1/2 particles) and this can be pictured as a vector with length 1. This vector can point in different directions, but it is found that if you project the possible states it can have onto a chosen axis (can be any axis of your choice, but is by convention chosen to be z) then this projeciton is also quantized with values $$m_{Sz}$$ = +1, 0, and -1. Here, 0 means the vector was found to point in the xz-plane of the Bloch sphere. It still has length one, it just points in a direction that gives no reading when you're measuring along the z-axis.

Note also that the states corresponding to $$m_{Sz}$$ = +1, 0, and -1 are degenerate in the absence of an external field, because there would be nothing to break the spherical symmetry.

13. Jun 10, 2010

### LostConjugate

So even though you would normally get 1/2 when projecting a vector with length 1 onto the z axis from the xz axis because of the quantization you instead get 0. That means if the spin vector is slightly off of your measurement vector you will always measure 0? Seems like we would always measure 0, but we never do, or I should say we never do for the single electron.

14. Jun 10, 2010

### peteratcam

Like this:
Eigenstates of Pauli matrix $$\sigma_z$$:
$$\begin{pmatrix}1\\0\end{pmatrix},\begin{pmatrix}0\\1\end{pmatrix}$$
Eigenstates of Pauli matrix $$\sigma_x$$:
$$\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix},\frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}$$ <--- note, linear combinations of eigenstates of $$\sigma_z$$.

Just because the z-direction is perpendicular to the x-direction does not mean the respective eigenstates are orthogonal in the quantum mechanical sense.

15. Jun 10, 2010

### LostConjugate

$$C\begin{pmatrix}1\\0\end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}$$ What is C?

16. Jun 10, 2010

### peteratcam

C would have to be a rotation matrix by 45 degrees perhaps, but this question seems totally irrelevant, I don't know why you are asking it.

17. Jun 10, 2010

### LostConjugate

I am trying to figure out how they are linearly dependent.

18. Jun 10, 2010

### peteratcam

$$\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 1\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 0\end{pmatrix}+\frac{1}{\sqrt{2}}\begin{pmatrix}0 \\ 1\end{pmatrix}$$
$$\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ -1\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 0\end{pmatrix}-\frac{1}{\sqrt{2}}\begin{pmatrix}0 \\ 1\end{pmatrix}$$
This is basic linear algebra - it sounds like you need to sort out your understanding of a single spin-1/2 degree of freedom with its 2-dimensional hilbert space, before you consider combining two of them to get the singlet and triplet basis of the 4-dimensional hilbert space.

19. Jun 11, 2010

### LostConjugate

Yea I need to go back and look at it all from the beginning.