# QM - Measuring Angular Momentum

Hi,

I have a quick homework problem because I am confused. So a wave function with l = 1 in the state:

$$|\psi> = \frac{1}{\sqrt{14}} $\left( \begin{array}{ccc} 1 \\ 2 \\ 3i \end{array} \right)$$$

and i have to find the probability to be in state $$\hbar, -\hbar, 0$$ in $$L_z$$, so i applied $$L_z$$ so $$|\psi>$$ and got

$$L_z|\psi> = \frac{\hbar}{\sqrt{14}} $\left( \begin{array}{ccc} 1 \\ 0 \\ -3i \end{array} \right)$$$

How do i find the probabilities from that?

Thanks,

Biest

Always, the probability of finding a state $$|\psi\rangle$$ in a state $$|\phi\rangle$$ is $$\left| \left<\phi \middle| \psi \right>\right|^2$$.

But how do i take the $$\hbar, -\hbar 0$$ condition into account?

assuming your $$|\psi\rangle$$ is in the basis of normalised angular momentum eigenstates, then your probabilities just correspond to $$\frac{1}{14}$$ for $$\hbar$$, $$\frac{2}{7}$$ for 0 and $$\frac{9}{14}$$ for $$-\hbar$$.

so what happens is that for $$|\phi\rangle$$ to return you an eigenvalue of say $$\hbar$$, we need

$$|\phi\rangle = $\left( \begin{array}{ccc} 1 \\ 0 \\ 0 \end{array} \right)$$$

and taking $$\left| \left<\phi \middle| \psi \right>\right|^2$$, we get $$\frac{1}{14}$$.

hope it helps. QM was never my strong point. haha