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# Posts Tagged ‘L_z operator’

## A problem on spherical harmonics.

Posted by peeterjoot on January 10, 2011

# Motivation.

One of the PHY356 exam questions from the final I recall screwing up on, and figuring it out after the fact on the drive home. The question actually clarified a difficulty I’d had, but unfortunately I hadn’t had the good luck to perform such a question, to help figure this out before the exam.

From what I recall the question provided an initial state, with some degeneracy in $m$, perhaps of the following form

\begin{aligned}{\lvert {\phi(0)} \rangle} = \sqrt{\frac{1}{7}} {\lvert { 12 } \rangle}+\sqrt{\frac{2}{7}} {\lvert { 10 } \rangle}+\sqrt{\frac{4}{7}} {\lvert { 20 } \rangle},\end{aligned} \hspace{\stretch{1}}(1.1)

and a Hamiltonian of the form

\begin{aligned}H = \alpha L_z\end{aligned} \hspace{\stretch{1}}(1.2)

From what I recall of the problem, I am going to reattempt it here now.

## Evolved state.

One part of the question was to calculate the evolved state. Application of the time evolution operator gives us

\begin{aligned}{\lvert {\phi(t)} \rangle} = e^{-i \alpha L_z t/\hbar} \left(\sqrt{\frac{1}{7}} {\lvert { 12 } \rangle}+\sqrt{\frac{2}{7}} {\lvert { 10 } \rangle}+\sqrt{\frac{4}{7}} {\lvert { 20 } \rangle} \right).\end{aligned} \hspace{\stretch{1}}(1.3)

Now we note that $L_z {\lvert {12} \rangle} = 2 \hbar {\lvert {12} \rangle}$, and $L_z {\lvert { l 0} \rangle} = 0 {\lvert {l 0} \rangle}$, so the exponentials reduce this nicely to just

\begin{aligned}{\lvert {\phi(t)} \rangle} = \sqrt{\frac{1}{7}} e^{ -2 i \alpha t } {\lvert { 12 } \rangle}+\sqrt{\frac{2}{7}} {\lvert { 10 } \rangle}+\sqrt{\frac{4}{7}} {\lvert { 20 } \rangle}.\end{aligned} \hspace{\stretch{1}}(1.4)

## Probabilities for $L_z$ measurement outcomes.

I believe we were also asked what the probabilities for the outcomes of a measurement of $L_z$ at this time would be. Here is one place that I think that I messed up, and it is really a translation error, attempting to get from the english description of the problem to the math description of the same. I’d had trouble with this process a few times in the problems, and managed to blunder through use of language like “measure”, and “outcome”, but don’t think I really understood how these were used properly.

What are the outcomes that we measure? We measure operators, but the result of a measurement is the eigenvalue associated with the operator. What are the eigenvalues of the $L_z$ operator? These are the $m \hbar$ values, from the operation $L_z {\lvert {l m} \rangle} = m \hbar {\lvert {l m} \rangle}$. So, given this initial state, there are really two outcomes that are possible, since we have two distinct eigenvalues. These are $2 \hbar$ and $0$ for $m = 2$, and $m= 0$ respectively.

A measurement of the “outcome” $2 \hbar$, will be the probability associated with the amplitude $\left\langle{{ 1 2 }} \vert {{\phi(t)}}\right\rangle$ (ie: the absolute square of this value). That is

\begin{aligned}{\left\lvert{ \left\langle{{ 1 2 }} \vert {{\phi(t) }}\right\rangle }\right\rvert}^2 = \frac{1}{7}.\end{aligned} \hspace{\stretch{1}}(1.5)

Now, the only other outcome for a measurement of $L_z$ for this state is a measurement of $0 \hbar$, and the probability of this is then just $1 - \frac{1}{7} = \frac{6}{7}$. On the exam, I think I listed probabilities for three outcomes, with values $\frac{1}{7}, \frac{2}{7}, \frac{4}{7}$ respectively, but in retrospect that seems blatently wrong.

## Probabilities for $\mathbf{L}^2$ measurement outcomes.

What are the probabilities for the outcomes for a measurement of $\mathbf{L}^2$ after this? The first question is really what are the outcomes. That’s really a question of what are the possible eigenvalues of $\mathbf{L}^2$ that can be measured at this point. Recall that we have

\begin{aligned}\mathbf{L}^2 {\lvert {l m} \rangle} = \hbar^2 l (l + 1) {\lvert {l m} \rangle}\end{aligned} \hspace{\stretch{1}}(1.6)

So for a state that has only $l=1,2$ contributions before the measurement, the eigenvalues that can be observed for the $\mathbf{L}^2$ operator are respectively $2 \hbar^2$ and $6 \hbar^2$ respectively.

For the $l=2$ case, our probability is $4/7$, leaving $3/7$ as the probability for measurement of the $l=1$ ($2 \hbar^2$) eigenvalue. We can compute this two ways, and it seems worthwhile to consider both. This first method makes use of the fact that the $L_z$ operator leaves the state vector intact, but it also seems like a bit of a cheat. Consider instead two possible results of measurement after the $L_z$ observation. When an $L_z$ measurement of $0 \hbar$ is performed our state will be left with only the $m=0$ kets. That is

\begin{aligned}{\lvert {\psi_a} \rangle} = \frac{1}{{\sqrt{3}}} \left( {\lvert {10} \rangle} + \sqrt{2} {\lvert {20} \rangle} \right),\end{aligned} \hspace{\stretch{1}}(1.7)

whereas, when a $2 \hbar$ measurement of $L_z$ is performed our state would then only have the $m=2$ contribution, and would be

\begin{aligned}{\lvert {\psi_b} \rangle} = e^{-2 i \alpha t} {\lvert {12 } \rangle}.\end{aligned} \hspace{\stretch{1}}(1.8)

We have two possible ways of measuring the $2 \hbar^2$ eigenvalue for $\mathbf{L}^2$. One is when our state was ${\lvert {\psi_a} \rangle}$ (, and the resulting state has a ${\lvert {10} \rangle}$ component, and the other is after the $m=2$ measurement, where our state is left with a ${\lvert {12} \rangle}$ component.

The resulting probability is then a conditional probability result

\begin{aligned}\frac{6}{7} {\left\lvert{ \left\langle{{10}} \vert {{\psi_a}}\right\rangle }\right\rvert}^2 + \frac{1}{7} {\left\lvert{ \left\langle{{12 }} \vert {{\psi_b}}\right\rangle}\right\rvert}^2 = \frac{3}{7}\end{aligned} \hspace{\stretch{1}}(1.9)

The result is the same, as expected, but this is likely a more convicing argument.