# Natural Eigenbasis composite 2 X Spin 1/2 system

1. Jun 8, 2008

### cathalcummins

The way I am being taught in my course is that instead of

!s_1= 1/2, m_1 =1/2 ; s_2= 1/2, m_2 =1/2 > = !+ + >
!s_1= 1/2, m_1 =1/2 ; s_2= 1/2, m_2 =-1/2 > = !+ - >
!s_1= 1/2, m_1 =-1/2 ; s_2= 1/2, m_2 =1/2 > = !- + >
!s_1= 1/2, m_1 =-1/2 ; s_2= 1/2, m_2 =-1/2 > = !- - >

We have

¦s=1,m_z=1>= ¦++>
¦s=1,m_z=0>= 1/sqrt(2) (¦+->+¦-+>)
¦s=1,m_z=-1>= ¦++>

¦s=0,m_z=0>= 1/sqrt(2) (¦+->-¦-+>)

I understand the spin-1 triplet. My question is "how do you compute ¦+->-¦-+> to get composite s=0,m=0". Is it essentially

1/sqrt(2) (¦1 0> - ¦1 0> )= ¦0 0>

?

Thanks.

2. Jun 8, 2008

### lbrits

The idea is that you derive this new basis by requiring that it be eigenstates of total angular momentum. However, if you are just given the states, you should still be able to show that it is an eigenstate.

For example, show that the state is an eigenstate of $$S_1^z + S_2^z$$ and also $$S_1^2 + S_2^2$$, and what are the eigenvalues?

3. Jun 8, 2008

### pam

The spin zero state is chosen to be orthogonal to the spin one state.

4. Jun 8, 2008

### lbrits

Looks like I misunderstood the question. Orthogonality is usually the way to go but you can do the Clebsch-Gordan decomposition more systematically I believe.

5. Jun 8, 2008

### cathalcummins

Of course! orthogonality, thank you Pam. When reading David J Griffiths Introduction to quantum mechanics - (Prentice Hall, 1995) p166, the way it is presented almost seems to insinuate a "nice consequence" as opposed to a necessary condition.

$$\left( S_{1z} \otimes \mathbb{I} + \mathbb{I} \otimes S_{2z} \right)\frac{1}{\sqrt{2}}\{ \ket{\uparrow\downarrow}-\ket{\downarrow\uparrow}\}=\frac{\hbar}{2\sqrt{2}} \left( \ket{\uparrow\downarrow}+\ket{\downarrow\uparrow}-\ket{\uparrow\downarrow}-\ket{\downarrow\uparrow} \right)$$

So that

$$\left( S_{1z} \otimes \mathbb{I} + \mathbb{I} \otimes S_{2z} \right)\frac{1}{\sqrt{2}}\{ \ket{\uparrow\downarrow}-\ket{\downarrow\uparrow}\}=\left(0 \right)\frac{1}{\sqrt{2}} \left( \ket{\uparrow\downarrow}-\ket{\downarrow\uparrow}\right)$$

Meaning that

$$\{ \ket{\uparrow\downarrow}-\ket{\downarrow\uparrow}\}$$

is an eigenvector of the $$S_z=\left{ S_{1z} \otimes \mathbb{I} + \mathbb{I} \otimes S_{2z}\right}$$ operator with corresponding eigenvalue $$0$$. Similarly, but with more algebra, we should arrive with the corresponding eigenvalue $$s(s+1)\hbar^2=0(0+1)\hbar^2=0$$ for the $$S^2$$ operator.

Technically, the zero spin state is orthogonal to every other eigenvector yes?

Last edited: Jun 8, 2008
6. Jun 9, 2008

### pam

All non-degenerate evectors are orthogonal.

7. Jun 9, 2008

### cathalcummins

To motivate the Clebsch-Gordon, and prove $$m=m_1+m_2$$ my lecturer done the obvious exploitation of the fact:

$$J_z=J_{1z}+J_{2z}$$

so that the expectation value of $$J_z-J_{1z}-J_{2z}$$ is zero. That makes sense to a certain extent. Though, and here is the problem: the operators above do not posess simultaneous eigenkets. But, regardless, he goes on to calculate it wrt two seperate eigenbases; $$| \phi \rangle$$ and $$| \psi \rangle$$ where

$$| \phi \rangle = | j_1 j_2 m_1 m_2\rangle$$

and

$$| \psi \rangle = | j_1 j_2 j m \rangle$$

This formalism is that of Modern Quantum Mechanics - J.J Sakurai (Addison-Wesley Rev .ed, 1994). The Expectation value is found by (according to my lecturers notes)

$$\langle \phi | J_z-J_{1z}-J_{2z} | \psi \rangle$$

which must return zero. Okay I'm gonna stop there because I simply don't agree that this is the expectation value.

You can still prove $$m=m_1+m_2$$ by the elementary steps outlined in Modern Quantum Mechanics - J.J Sakurai (Addison-Wesley Rev .ed, 1994) p 208 . This way makes sense to me and it mentions nothing of expectation value. Sorry to post such a trivial question but I'm just trying to weed out any mistakes in my lecturers notes.

Last edited: Jun 9, 2008