# Quantum Mechanics: Three Spin Observables

1. Feb 16, 2015

### Robben

When trying to solve $\mathbb{S}^2 =\hbar^2s(s+1)\mathbb{I},$
I got that $\mathbb{S}^2 = \mathbb{S}^2 _x+\mathbb{S}^2_y+\mathbb{S}^2_z = \frac{3\hbar^2}{4} \left[\begin{array}{ c c }1 & 0\\0 & 1\end{array} \right] = \frac{3\hbar^2}{4}\mathbb{I},$ but how does $\frac{3\hbar^2}{4} = \hbar^2s(s+1)?$

2. Feb 16, 2015

### dextercioby

I don't understand, It's a matrix equality => 4 equalities.

3. Feb 16, 2015

### Robben

4. Feb 16, 2015

### dextercioby

Well, S^2 is equal to a 2x2 matrix in 2 different ways. One is through the eigenvalue equation, and the other is through the sum of the squares of the 3 Pauli matrices. So you can equal these 2 matrices and obtain 4 equalities *one for each matrix element*.

5. Feb 16, 2015

### Robben

Hm, so how does $\frac{3\hbar^2}{4} = \hbar^2s(s+1)?$

6. Feb 16, 2015

### Strilanc

Can you explain what $\mathbb{S}$ and $s$ are supposed to be? Is $\mathbb{S}$ required to be unitary? Hermitian? A linear combination of Pauli matrices? Is $s$ a scaling factor? Real? Complex? Known ahead of time?

If $\mathbb{S}$ is a linear combination of Pauli matrices like $\hat{v} \cdot \vec{\sigma}$, where $\hat{v}$ is a unit vector, then it is unitary and Hermitian and thus its own inverse. So $\mathbb{S}^2$ will be just $\mathbb{I}$, meaning $\hbar^2 s(s+1)$ must equal 1, and all that's left is a quadratic equation.

7. Feb 16, 2015

### Robben

$s$ is the eigenvalues and $\mathbb{S}$ is Hermitian.

8. Feb 17, 2015

### Strilanc

In that case I think it reduces to $\left( s \hat{v} \cdot \vec{\sigma} \right)^2 = \hbar^2 s(s+1) \mathbb{I}$, where $\hat{v}$ is an arbitrary unit vector and $s$ is a root of $s^2 (1 - \hbar^2) - s \hbar^2 = 0$.

9. Feb 17, 2015

### samalkhaiat

Because $s = \frac{1}{2}$:
1) Don't you have $\frac{3}{4} \hbar^{2} = \frac{1}{2} ( \frac{1}{2} + 1 ) \hbar^{2} , \ \ \Rightarrow s = \frac{1}{2}$ ?
2) For what $s > 0$, do you have $s^{2} + s - \frac{3}{4} = ( s - \frac{1}{2} ) ( s + \frac{3}{2} ) = 0$ ?

10. Feb 17, 2015

### Robben

I see, thank you very much guys!