# State of a system after $L_z^2$ been measured

A system is initially in a state given by $$|{\psi_i}\rangle = \begin{pmatrix}\frac12\\1\over2\\1\over{\sqrt2}\end{pmatrix}$$ corresponding to the angular momentum $l=1$ in the $L_z$ basis of states with $m=+1,0,-1$. If $L_z^2$ is measured in this state yielding a result 1, what is the state after the measurement?

First, I don’t really understand the phrase “angular momentum $l=1$ in the $L_z$ basis of states with $m=+1,0,-1$”. Does $l=1$ not necessarily imply $m=+1,0,-1$?

Second, I think since the measurement leaves the system in an eigenstate of $L_z$, the state after the measurement will be $$|{\psi_f}\rangle = \begin{pmatrix}0\\0\\1\end{pmatrix}$$

Is it correct? Or do I need to use the ladder operators?

"In the basis $m$ = +1, 0, -1" means that each row of your vector corresponds to one of these $m$. Column vectors like that make no sense unless you specify a basis; it's like if I gave you a vector in Euclidean space $(1,2,3)$ without telling you whether the basis is Cartesian $(\mathbf{i},\mathbf{j},\mathbf{k})$ or Spherical $(\mathbf{r},\mathbf{\phi},\mathbf{\theta})$.

In the same way as a vector $(1,2,3)$ in the Cartesian basis can be written as $1\mathbf{i} + 1\mathbf{j} + 1\mathbf{k}$,

then your $|{\psi_i}\rangle = \begin{pmatrix}\frac12\\1\over2\\1\over{\sqrt2}\end{pmatrix}$ can be written as $$\frac{1}{2}|m=1\rangle + \frac{1}{2}|m=0\rangle + \frac{1}{\sqrt{2}}|m=-1\rangle.$$

$l=1$ means that the total angular momentum is $l(l+1) = 2$ ($\hbar$). Unless you make another measurement to break the $m$ degeneracy then you have to assume a superposition of the three possible states (in your case the m=-1 is twice more likely than any of the other two). Given $l$, $m$ ranges from $-l$ to $+l$. Hence the three numbers.

If $L_z$ gives you one, it meant that $m=1$ and the state after the measurement is $$|{\psi_f}\rangle = \begin{pmatrix}1\\0\\0\end{pmatrix},$$ according to my convention above.

However you are saying that a measurement of $L_z^2$ gives you one, which means that you could have either $m=-1$ or $m=1$. So in that case you throw away the $m=0$ solution and renormalise your state: $$|{\psi_i}\rangle = N \begin{pmatrix}\frac12\\0\\1\over{\sqrt2}\end{pmatrix},$$ which gives $$|{\psi_i}\rangle = \sqrt{\frac{4}{3}} \begin{pmatrix}\frac12\\0\\1\over{\sqrt2}\end{pmatrix}.$$

• "m=-1 is twice more likely than any of the other two" can you clarify? "Unless you make another measurement to break the m degeneracy" do you mean in two successive measurements I will have a complete knowledge of $L_z$? – Sayontön Vöttacharjo Feb 19 '17 at 16:54
• I mean that the probability of finding $m=1$ is $(\frac{1}{2})^2 = 0.25$, same for $m=0$ but for $m=-1$ it's $(\frac{1}{\sqrt{2}})^2 = 0.5$. You can just make one measurement of $L^2$ first to know what's $l$, then make a measuremnt of $L_z$ to know what's $m$. One measuremnt per quantum number usually. But if you measure $L_z^2$ then you are not uniquely determining $m$, but only $m^2$ which could still be $-1$ or $1$. – SuperCiocia Feb 19 '17 at 16:58