# Average value of square spin

1. Jun 23, 2015

### Gvido_Anselmi

Hello everyone.
Let us consider two fermions whose spin state wave functions are certain (normalized) $\psi_{1}$ and $\psi_{2}$.
So the overall normalized spin state wave function of this system is of the form:

$\psi = (1/\sqrt{2})(\psi_{1} \psi_{2} - \psi_{2} \psi_{1})$

Is there any difference between calculating the total square spin operator $\hat S^{2}$ eigenvalue and the average value of $S^{2}$?

Last edited: Jun 23, 2015
2. Jun 23, 2015

### Gvido_Anselmi

The problem was to calculate average value of $S^2$ for given spin state wave functions of two electrons. Honestly I don't understand the meaning of this problem. We know that total square spin operator's eigenvalues are $S^{2}=S(S+1)$ and take values $0$ or $2$. Eigenvalues of linear Hermitian operator are observables or average values of physical quantities by definition. So what does it mean to take average value of average value or there is something I understand wrong?

Last edited: Jun 23, 2015
3. Jun 24, 2015

### JorisL

You are right that the (vector) addition of the individual spins gives us either $S^2=0 \Rightarrow S=0$ or $S^2=2 \Rightarrow S = 1$.
Now these states correspond to the singlet and triplet states of this composite system.
The singlet state has (as the name suggests) 1 state ($S_z$).
But for the triplet state we can have spin along the z-axis $-1,\, 0,\, +1$.

Now if you say that each of these states is equally probable i.e. the density matrix is proportional to the unit matrix.
In that case $0\leq \langle S^2 \rangle \leq 2$, check that.

4. Jun 24, 2015

### Gvido_Anselmi

Obvious! I have forgotten that spin-state wave function is not always antisymmetric. In general case we should expand it in a linear combination of antisymmetric (singlet) and symmetric (triplet) parts. I can guess that $S^2=2P$ where $P$ is the norm of triplet state and the proof is not too difficult.
Thank you very much!