Recall that energy for a particle on a sphere looks like \(E_\ell=\ell(\ell+1)\frac{\hbar^2}{2I}\). Because of the degenerate nature of energy for a sphere, each term in our superposition with \(\ell=17\) needs to be considered. This means that we will need to include all possible \(m\) values for our particular \(\ell\). Our probability will then be the sum over \(m\) of the square of the coefficients for terms with \(\ell=17\). The coefficients can be found by taking inner products as shown below. \begin{equation} \mathcal{P}(E_{17})=\sum_{m'=-17}^{m'=17}|c_{17m'}|^2=\sum_{m'=-17}^{m'=17}\left|\left\langle {17,m'}\middle|{\Psi}\right\rangle \right|^2=\sum_{m'=-17}^{m'=17}\left|\left\langle {17,m'}\right|\sum_{\ell=0}^\infty\sum_{m=-\ell}^\ell c_{\ell m}\left|{\ell, m}\right\rangle \right|^2 \end{equation} Our probability could contain contributions from up to \(35\) different superposition states due to the degeneracy of \(\ell=17\).
For a particle on a sphere, the \(z\)-component of angular momentum takes the form \(m\hbar\). For our particular case, that means we need to take into consideration all terms in our superposition that have \(m=5\). Picking out the coefficients, \(c_{\ell5}\), associated with our \(m\) value is a little more complicated than in the previous part. While before knowing \(\ell\) gave us restrictions for what \(m\) could be, knowing \(m\) does not as tightly restrict \(\ell\) values that need to be included. We know that because \(m\) cannot be greater than \(\ell\) that we won't have any terms where \(\ell = 1,2,3,4\). \begin{equation} \mathcal{P}(L_z=5\hbar)=\sum_{\ell'=5}^{\infty}|c_{\ell5}|^2=\sum_{\ell'=5}^{\infty}\left|\left\langle {\ell',5}\middle|{\Psi}\right\rangle \right|^2=\sum_{\ell'=5}^{\infty}\left|\left\langle {\ell',5}\right|\sum_{\ell=0}^\infty\sum_{m=-\ell}^\ell c_{\ell m}\left|{\ell, m}\right\rangle \right|^2 \end{equation} This probability could could contain contributions from any term \(\ell > 5\) so we have hypothetically an infinite sum.
\begin{align} \left|{\psi}\right\rangle &\doteq\psi(\theta,\phi)\\ &=\frac{1}{\sqrt{2}}Y_1^{-1} + \frac{1}{\sqrt{3}}Y_1^0 + \frac{i}{\sqrt{6}}Y_0^0 \end{align}
The state \(\left\vert \psi\right\rangle\) is written in terms of the eigenstates of two operators (i.e. they have two labels). The first label tells us the eigenvalue of \(\ell\) and the second of \(m\). We are asked to find the probability that \(m=2\), \(-1\), or \(0\). \begin{align} \mathcal{P}(m=2)&\doteq\sum_{\ell=0}^{\infty} \left\vert\left\langle \ell, 2|\psi\right\rangle \right\vert^2\\ &=\sum_{\ell=0}^{\infty} \left\vert\left\langle \ell,2| \left(\frac{1}{\sqrt{2}}\left\vert 1, -1\right\rangle + \frac{1}{\sqrt{3}}\left\vert 1, 0\right\rangle + \frac{i}{\sqrt{6}}\left\vert 0,0\right\rangle\right) \right\rangle\right\vert^2 \\ &=\sum_{\ell=0}^{\infty} \left\vert\left( \frac{1}{\sqrt{2}}\left\langle \ell,2 | 1,-1\right\rangle +\frac{1}{\sqrt{3}}\left\langle \ell,2 | 1, 0\right\rangle + \frac{i}{\sqrt{6}}\left\langle \ell,2 | 0,0\right\rangle \right)\right\vert^2\\ &=0 \end{align} \begin{align} \mathcal{P}(m=-1)&\doteq\sum_{\ell=0}^{\infty} \left\vert\left\langle \ell, -1|\psi\right\rangle \right\vert^2\\ &=\sum_{\ell=0}^{\infty} \left\vert\langle \ell,-1| \left(\frac{1}{\sqrt{2}}\left\vert 1, -1\right\rangle + \frac{1}{\sqrt{3}}\left\vert 1, 0\right\rangle + \frac{i}{\sqrt{6}}\left\vert 0,0\right\rangle\right) \right\vert^2 \\ &=\sum_{\ell=0}^{\infty} \left\vert\left( \frac{1}{\sqrt{2}}\left\langle \ell,-1 | 1,-1\right\rangle +\frac{1}{\sqrt{3}}\left\langle \ell,-1 | 1, 0\right\rangle + \frac{i}{\sqrt{6}}\left\langle \ell,-1 | 0,0\right\rangle \right)\right\vert^2\\ &=\sum_{\ell=0}^{\infty} \left\vert\frac{1}{\sqrt{2}}\, \delta_{\ell,1}\right\vert^2\\ &=\frac{1}{2} \end{align}
\begin{align} \mathcal{P}(m=0)&\doteq\sum_{\ell=0}^{\infty} \left\vert\left\langle \ell,0|\psi\right\rangle \right\vert^2\\ &=\sum_{\ell=0}^{\infty} \left\vert\langle \ell,0| \left(\frac{1}{\sqrt{2}}\left\vert 1, -1\right\rangle + \frac{1}{\sqrt{3}}\left\vert 1, 0\right\rangle + \frac{i}{\sqrt{6}}\left\vert 0,0\right\rangle\right) \right\vert^2 \\ &=\sum_{\ell=0}^{\infty} \left\vert\left( \frac{1}{\sqrt{2}}\left\langle \ell,0 | 1,-1\right\rangle +\frac{1}{\sqrt{3}}\left\langle \ell,0 | 1, 0\right\rangle + \frac{i}{\sqrt{6}}\left\langle \ell,0 | 0,0\right\rangle \right)\right\vert^2\\ &=\sum_{\ell=0}^{\infty} \left\vert\frac{1}{\sqrt{3}}\, \delta_{\ell,1} +\frac{i}{\sqrt{6}}\, \delta_{\ell,0}\right\vert^2\\ &=\left\vert\frac{1}{\sqrt{3}}\right\vert^2 +\left\vert\frac{i}{\sqrt{6}}\right\vert^2\\ &=\frac{1}{2} \end{align} Be careful in the last last line of the last case. You need to take the square of the norm of each of the individual coefficients and then add, rather than adding the coefficients and then squaring. Why?
In order to find the state after taking a measurement, we have to use Postulate 5 (the projection postulate): \begin{equation} \left|{\psi'}\right\rangle =\frac{P_n\left|{\psi}\right\rangle }{\sqrt{\left\langle {\psi}\right|P_n\left|{\psi}\right\rangle }} \end{equation} When \(L_z=-\hbar\), the projection operator corresponds to states where \(m=-1\), so for this situation the projection operator is \begin{equation} P_{-1}=\sum_{\ell=0}^{\infty}\left|{\ell,-1}\right\rangle \left\langle {\ell,-1}\right|=\left|{1,-1}\right\rangle \left\langle {1,-1}\right|. \end{equation} The new state is \begin{align} \left|{\psi'}\right\rangle &= \frac{\left|{1,-1}\right\rangle \left\langle {1,-1}\right| \left(\frac{1}{\sqrt{2}}\left|{1,-1}\right\rangle +\frac{1}{\sqrt{3}}\left|{1,0}\right\rangle +\frac{i}{\sqrt{6}}\left|{0,0}\right\rangle \right)}{\sqrt{\left\langle {\psi}\right|\left(\left|{1,-1}\right\rangle \left\langle {1,-1}\right|\right)\left(\frac{1}{\sqrt{2}}\left|{1,-1}\right\rangle +\frac{1}{\sqrt{3}}\left|{1,0}\right\rangle +\frac{i}{\sqrt{6}}\left|{0,0}\right\rangle \right)}}\\ &=\frac{\frac{1}{\sqrt{2}}\left|{1,-1}\right\rangle } {\sqrt{\left\langle {\psi}\right|\frac{1}{\sqrt{2}}\left|{1,-1}\right\rangle }} =\frac{\frac{1}{\sqrt{2}}\left|{1,-1}\right\rangle }{\sqrt{\left(\frac{1}{\sqrt{2}}\right)^2}}\\ &=\left|{1,-1}\right\rangle , \end{align} which is what we would expect, since there is only one state that has \(m=-1\).
When we make a measurement of \(L_z=0\hbar\), \(m=0\) and there are two states that are possible, \(\left|{1,0}\right\rangle \) and \(\left|{0,0}\right\rangle \), so the projection operator is \begin{equation} P_{0}=\sum_{\ell=0}^{\infty}\left|{\ell,0}\right\rangle \left\langle {\ell,0}\right|=\left|{1,0}\right\rangle \left\langle {1,0}\right|+\left|{0,0}\right\rangle \left\langle {0,0}\right| \end{equation} and the new state is \begin{align} \left|{\psi'}\right\rangle &= \frac{\left(\left|{1,0}\right\rangle \left\langle {1,0}\right|+\left|{0,0}\right\rangle \left\langle {0,0}\right|\right) \left(\frac{1}{\sqrt{2}}\left|{1,-1}\right\rangle +\frac{1}{\sqrt{3}}\left|{1,0}\right\rangle +\frac{i}{\sqrt{6}}\left|{0,0}\right\rangle \right)} {\sqrt{\left\langle {\psi}\right|\left(\left|{1,0}\right\rangle \left\langle {1,0}\right|+\left|{0,0}\right\rangle \left\langle {0,0}\right|\right) \left(\frac{1}{\sqrt{2}}\left|{1,-1}\right\rangle +\frac{1}{\sqrt{3}}\left|{1,0}\right\rangle +\frac{i}{\sqrt{6}}\left|{0,0}\right\rangle \right)}}\\ &=\frac{\frac{1}{\sqrt{3}}\left|{1,0}\right\rangle +\frac{i}{\sqrt{6}}\left|{0,0}\right\rangle } {\sqrt{\left\langle {\psi}\right|\left(\frac{1}{\sqrt{3}}\left|{1,0}\right\rangle +\frac{i}{\sqrt{6}}\left|{0,0}\right\rangle \right)}} =\frac{\frac{1}{\sqrt{3}}\left|{1,0}\right\rangle +\frac{i}{\sqrt{2}}\left|{0,0}\right\rangle }{\sqrt{\frac{1}{3}+\frac{1}{6}}}\\ &= \sqrt\frac{2}{3}\left|{1,0}\right\rangle +i\frac{1}{\sqrt{3}}\left|{0,0}\right\rangle , \end{align} which is just a re-normalized superposition of the the two \(m=0\) states.
Since we have already calculated the probabilities and eigenvalues, it is quickest to use the weighted average notation. \begin{align} \left\langle\psi\vert L_z\vert \psi \right\rangle &=\sum_{m=-\infty}^\infty m\hbar \mathcal{P}_{L_z=m\hbar} =0\hbar \left(\frac{1}{2}\right)+\left(-1\hbar\right) \left(\frac{1}{2}\right) =-\frac{1}{2} \hbar\\ \end{align} Alternatively, the expectation value of \(L_z\) is given by: \begin{align} \left\langle\psi\vert L_z\vert \psi \right\rangle &=\left\langle\psi \right\vert L_z\vert \left(\frac{1}{\sqrt{2}}\left\vert 1, -1\right\rangle + \frac{1}{\sqrt{3}}\left\vert 1, 0\right\rangle + \frac{i}{\sqrt{6}}\left\vert 0,0\right\rangle\right)\\ &=\left\langle\psi\right\vert \left(\frac{1}{\sqrt{2}}(-1\hbar)\left\vert 1, -1\right\rangle +\frac{1}{\sqrt{3}}(0\hbar)\left\vert 1, 0\right\rangle +\frac{i}{\sqrt{6}}(0\hbar)\left\vert 0,0\right\rangle \right)\\ &=(-1\hbar)\left( \frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\left\langle 1,-1\vert 1,-1\right\rangle +\frac{1}{\sqrt{3}}\frac{1}{\sqrt{2}}\left\langle 1,0\vert 1,-1\right\rangle +\frac{-i}{\sqrt{6}}\frac{1}{\sqrt{2}}\left\langle 0,0\vert 1,-1\right\rangle\right)\\ &+(0\hbar)\left( \frac{1}{\sqrt{2}}\frac{1}{\sqrt{3}}\left\langle 1,-1\vert 1,0\right\rangle +\frac{1}{\sqrt{3}}\frac{1}{\sqrt{3}}\left\langle 1,0\vert 1,0\right\rangle +\frac{-i}{\sqrt{6}}\frac{1}{\sqrt{3}}\left\langle 0,0\vert 1,0\right\rangle\right)\\ &+(0\hbar)\left( \frac{1}{\sqrt{2}}\frac{i}{\sqrt{6}}\left\langle 1,-1\vert 0,0\right\rangle +\frac{1}{\sqrt{3}}\frac{i}{\sqrt{6}}\left\langle 1,0\vert 0,0\right\rangle +\frac{-i}{\sqrt{6}}\frac{i}{\sqrt{6}}\left\langle 0,0\vert 0,0\right\rangle\right)\\ &=-\frac{\hbar}{2} \end{align}
Notice that I've put a lot more steps in than you need so that you can see where all the terms go. Many of the terms are zero!!
The expectation value of \(L^2\) is given by: \begin{align} \left\langle\psi\vert L^2\vert \psi \right\rangle &=\left\langle\psi \right\vert L^2\vert \left(\frac{1}{\sqrt{2}}\left\vert 1, -1\right\rangle + \frac{1}{\sqrt{3}}\left\vert 1, 0\right\rangle + \frac{i}{\sqrt{6}}\left\vert 0,0\right\rangle\right)\\ &=\left\langle\psi\right\vert \left(\frac{1}{\sqrt{2}}(2\hbar^2)\left\vert 1, -1\right\rangle +\frac{1}{\sqrt{3}}(2\hbar^2)\left\vert 1, 0\right\rangle +\frac{i}{\sqrt{6}}(0\hbar^2)\left\vert 0,0\right\rangle \right)\\ &=(2\hbar^2)\left( \frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\left\langle 1,-1\vert 1,-1\right\rangle +\frac{1}{\sqrt{3}}\frac{1}{\sqrt{2}}\left\langle 1,0\vert 1,-1\right\rangle +\frac{-i}{\sqrt{6}}\frac{1}{\sqrt{2}}\left\langle 0,0\vert 1,-1\right\rangle\right)\\ &+(2\hbar^2)\left( \frac{1}{\sqrt{2}}\frac{1}{\sqrt{3}}\left\langle 1,-1\vert 1,0\right\rangle +\frac{1}{\sqrt{3}}\frac{1}{\sqrt{3}}\left\langle 1,0\vert 1,0\right\rangle +\frac{-i}{\sqrt{6}}\frac{1}{\sqrt{3}}\left\langle 0,0\vert 1,0\right\rangle\right)\\ &+(0\hbar^2)\left( \frac{1}{\sqrt{2}}\frac{i}{\sqrt{6}}\left\langle 1,-1\vert 0,0\right\rangle +\frac{1}{\sqrt{3}}\frac{i}{\sqrt{6}}\left\langle 1,0\vert 0,0\right\rangle +\frac{-i}{\sqrt{6}}\frac{i}{\sqrt{6}}\left\langle 0,0\vert 0,0\right\rangle\right)\\ &=\left(\frac{1}{2}+\frac{1}{3}\right)\, 2\hbar^2\\ &=\frac{5}{3}\, \hbar^2 \end{align} Notice that I've put a lot more steps in than you need so that you can see where all the terms go. Many of the terms are zero, but fewer than in the previous case.
For the rigid rotor, the Hamiltonian \(\hat H\) is proportional to \(L^2\), by construction. (This will not be the case for the hydrogen atom.) We can see this by looking at the differential operators: \begin{align} \hat H &= -\frac{\hbar^2}{2\mu r_0^2} \left[ \frac{1}{\sin\theta} \frac{\partial}{\partial\theta} \left(\sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\right]\\ &=\frac{1}{2\mu r_0^2}\, L^2 \end{align}
Therefore, the expectation value of \(\hat H\) is proportional to the expectation value of \(L^2\) (see the previous part) with this same proportionality constant. \begin{equation} \left\langle\psi\vert \hat H\vert\psi\right\rangle = \frac{1}{2\mu r_0^2}\left\langle\psi\vert L^2\vert\psi\right\rangle =\frac{5}{3}\frac{\hbar^2}{2\mu r_0^2} \end{equation}
Find the coefficients of the \(\left|\ell,m\right\rangle=\left|0,0\right\rangle\), \(\left|1,-1\right\rangle\), \(\left|1,0\right\rangle\), and \(\left|1,1\right\rangle\) terms in the spherical harmonic expansion of \(f(\theta,\phi)\). It is helpful to remember that Mathematica has a built-in function for spherical harmonics: SphericalHarmonicY[l, m, \[Theta], \[Phi]].
We want to write this expression in terms of spherical harmonics. Unlike particle on a ring, where all of the terms in an expansion look like \(e^{im\phi}\), the spherical harmonics are not as easily recognized by inspection. Because it is challenging to look at this function and notice algebraic steps one could take to massage it into a superposition of spherical harmonics, we instead will identify terms that could be in our superposition and find their expansion coefficients using inner products.
Any of our coefficients can be found using the following inner product: \begin{equation} c_{\ell m}=\int^{2\pi}_{0}\int^{\pi}_{0}Y_\ell^{m}(\theta,\phi)^*f(\theta,\phi) \sin(\theta)d\theta d\phi \end{equation} These integrals are most easily found using technology. For our given function \(f(\theta, \phi)\) (remembering that since our function is piecewise) the inner products will be: \begin{align} c_{0,0} &= 0.623806\\ c_{1,-1} &= 0\\ c_{1,0} &= 0.706438 \\ c_{1,1} &= 0\\ \end{align} To check reasonableness we note that each of these is less than one. And we can further look at the square norm of these coefficents to ensure that their sum is also less than one. \begin{align} |c_{0,0}|^2 &= 0.389 \approx 40\%\\ |c_{1,-1}|^2 &= 0\\ |c_{1,0}|^2 &= 0.499 \approx 50\%\\ |c_{1,1}|^2 &= 0 \end{align} We can see that we have about \(89\%\) of our function probability accounted for with these terms of its expansion. This means that there are other non-zero terms in this expansion that we have not yet found. Making these checks allows us to write our expansion as: \begin{equation} f(\theta, \phi) = 0.623806~Y_0^0(\theta,\phi) + 0.706438~Y_1^0(\theta,\phi) + \dots \end{equation}
First, let us recall what the \(L^2\) operator tells us about possible measurement values of the square of total angular momentum. \begin{equation} L^2 \left|{\ell,m}\right\rangle = \ell(\ell+1)\hbar^2\left|{\ell,m}\right\rangle \end{equation} This means that \(2\hbar^2\) would be associated with \(\ell =1\). To find the probability of this measurement, we take the sum of all probabilities associated with \(Y_1^m\). \begin{equation} {\cal P}(L^2=2\hbar^2)= |c_{1,-1}|^2 + |c_{1,0}|^2+|c_{1,1}|^2= 0.499 \end{equation} Similarly, we can look at \(4\hbar^2\) and see that it is associated with a non integer value of \(\ell\). This means that it is not a possible measurement value.
If a quantum particle, confined to the surface of a sphere, is in the state above, what is the probability that the particle can be found in the region \(0<\theta<\frac{\pi}{6}\) and \(0<\phi<\frac{\pi}{6}\)?
To find the probability that our particle is in the desired region, we need to take the square of the norm of our function and integrate that over the region. \begin{equation} \int_0^{\frac{\pi}{6}}\int_0^{\frac{\pi}{6}} f^*(\theta,\phi)f(\theta,\phi) \sin\theta ~d\theta~ d\phi = 0.0269 \end{equation}