How to Find Probabilities in Quantum Mechanics?

[Biest]

Hi,

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

|\psi&gt; = \frac{1}{\sqrt{14}} \[ \left( \begin{array}{ccc}<br /> 1 \\<br /> 2 \\<br /> 3i \end{array} \right)\]

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

L_z|\psi&gt; = \frac{\hbar}{\sqrt{14}} \[ \left( \begin{array}{ccc}<br /> 1 \\<br /> 0 \\<br /> -3i \end{array} \right)\]

How do i find the probabilities from that?Thanks,

Biest

 

[genneth]

Always, the probability of finding a state |\psi\rangle in a state |\phi\rangle is \left| \left&lt;\phi \middle| \psi \right&gt;\right|^2.

 

[Biest]

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

 

[jae05]

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&lt;\phi \middle| \psi \right&gt;\right|^2, we get \frac{1}{14}.

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