# Probability of getting a particular spin

I'm a beginner in quantum mechanics, and I'm a bit confused about states and the probability to measure certain values. I would like to understand at least the following simplified situation:

Suppose the operator $S_z$ is measuring the spin in the $z$ direction of a free particle. Let $$\{e_s\},\quad s = -L, -L+1, \dots, L-1, L$$ be a basis of eigenstates for $S_z$, with $S_z(e_s) = s e_s$.

If at time $t<0$ the system is constantly in the pure state $P_{\psi}$, where $\psi$ is one basis vector, how can I compute the probability that the measuring the spin in the $z$ direction has result $-L/2$?

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" S_z is measuring the spin in the x direction" - did you mean the z direction? Is the particle completely free or is there some sort of magnetic field its in? –  DJBunk Apr 27 '12 at 14:27
Yes, it was a mistake. It is z. By the way the particle is free –  Abramodj Apr 27 '12 at 14:31
Why $L/2$ specifically? –  David Z Apr 28 '12 at 0:51
Therefore the required probability is given by the square norm of the inner product between $\psi$ and $e_{-L/2}$.
Comment to the answer(v1): There seems to be something wrong with the spin normalization. If say, $L=1/2$, then $e_{s=-L/2=-1/4}$. But $s$ is supposed to be half-integral, cf. the question (v3). –  Qmechanic Apr 27 '12 at 19:51