# 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 $-L/2$?

• " 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

The propability $P$ that a system in state $|\Psi\rangle$ will eventually transit into state $|\Theta\rangle$ is given by: $$P = | \langle\Theta | \Psi \rangle |^2$$

So for example if you have a System in Spin-X-State "up" denoted by $|\Psi\rangle \hat{=} \frac{1}{\sqrt{2}} (1, 1)^T$, the propability of getting Spin-Z-State "up" (e.g. by measuring along the z-axis) denoted by $|\Theta\rangle \hat{=} (1, 0)^T$ is: $$P = | \frac{1}{\sqrt{2}} (1, 0) \cdot (1, 1)^T |^2 = \frac{1}{2}$$

• Note that this answer is about pure states. For arbitrary mixed states, you'd need the trace (as in the other answer). – Martin Mar 16 '15 at 13:05

Ok, I've got it!

Using the bra-ket notation we have 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
• Comment to the answer: it is not worthwhile to show "manipulations" that are not manpulations, but formal exercizes. You could have said the answer without any of the intermediate steps, as there are no intermediate steps. The definition of the inner product of two states is the amplitude for one state to be the same as the other, and the absolute square of this is by definition the probability.. – Ron Maimon Jun 26 '12 at 21:46