Central Forces: Spring-2025
HW 08 (SOLUTION): Due W4 D5
- Time Dependence for a Quantum Particle on a Ring
S1 5250S
Consider a quantum particle on a ring. At t = 0, the particle is in state:
\begin{equation}
\left|{\Phi(t=0)}\right\rangle =\sqrt{\frac{2}{3}}\left|{-3}\right\rangle +\frac{1}{\sqrt{6}}\left|{-1}\right\rangle +\frac{i}{\sqrt{6}}\left|{3}\right\rangle
\end{equation}
- You carry out a measurement to determine the \(z\)-component of the angular momentum of the particle at some time, \(t>0\). Calculate the probability that you measure the \(z\)-component of the angular momentum to be \(3\hbar\).
If we want to find probabilities as a function of time, first we have to find the time dependence for the state. The given kets are eigenfunctions of the Hamiltonian (eigenfunctions of energy), so we can immediately multiply each ket by the appropriate time dependence: \begin{align} \left|{\Phi}\right\rangle &=\sqrt{\frac{2}{3}}\left|{-3}\right\rangle +\frac{1}{\sqrt{6}}\left|{-1}\right\rangle +\frac{i}{\sqrt{6}}\left|{3}\right\rangle \\ \Rightarrow \left|{\Phi(t)}\right\rangle &=\sqrt{\frac{2}{3}}\left|{-3}\right\rangle e^{-iE_{-3}t/\hbar} +\frac{1}{\sqrt{6}}\left|{-1}\right\rangle e^{-iE_{-1}t/\hbar} +\frac{i}{\sqrt{6}}\left|{3}\right\rangle e^{-iE_{3}t/\hbar} \end{align}
To find the probability of measuring \(L_z=3\hbar\), we project the state onto eigenstate with that value of angular momentum and take the square of the norm: \begin{align} {\cal P}(L_z=3\hbar)&=\vert\left\langle {3}\middle|{\Phi(t)}\right\rangle \vert^2\\ &=\left\vert\left\langle {3}\right|\left[\sqrt{\frac{2}{3}}\left|{-3}\right\rangle e^{-iE_{-3}t/\hbar} +\frac{1}{\sqrt{6}}\left|{-1}\right\rangle e^{-iE_{-1}t/\hbar} +\frac{i}{\sqrt{6}}\left|{3}\right\rangle e^{-iE_{3}t/\hbar} \right]\right\vert^2\\ &=\left\vert \frac{i}{\sqrt{6}} e^{-iE_{3}t/\hbar}\right\vert^2\\ &=\frac{1}{6} \end{align} Notice that the probability does not depend on time. I used the ket notation in the energy basis because the energy eigenstates have definite angular momentum.
- You carry out a measurement to determine the energy of the particle at some time, \(t>0\). Calculate the probability that you measure the energy to be \(\frac{9\hbar^2}{2I}\).
The time dependent state is the same as we found in the previous part of the problem.
To find the probability of measuring \(E=\frac{9\hbar^2}{2I}\), we project the state separately onto each of the eigenstates with that value of energy and take the square of the norm. Then we ADD the separate probabilities. The energies of the ring are degenerate. Both the kets \(\left|{3}\right\rangle \) and \(\left|{-3}\right\rangle \) have the required energy. \begin{align} {\cal P}(E=\frac{9\hbar^2}{2I})&=\vert\left\langle {3}\middle|{\Phi(t)}\right\rangle \vert^2 +\vert\left\langle {-3}\middle|{\Phi(t)}\right\rangle \vert^2\\ &=\left\vert\left\langle {3}\right|\left[\sqrt{\frac{2}{3}}\left|{-3}\right\rangle e^{-iE_{-3}t/\hbar} +\frac{1}{\sqrt{6}}\left|{-1}\right\rangle e^{-iE_{-1}t/\hbar} +\frac{i}{\sqrt{6}}\left|{3}\right\rangle e^{-iE_{3}t/\hbar} \right]\right\vert^2\\ &\qquad +\left\vert\left\langle {-3}\right|\left[\sqrt{\frac{2}{3}}\left|{-3}\right\rangle e^{-iE_{-3}t/\hbar} +\frac{1}{\sqrt{6}}\left|{-1}\right\rangle e^{-iE_{-1}t/\hbar} +\frac{i}{\sqrt{6}}\left|{3}\right\rangle e^{-iE_{3}t/\hbar} \right]\right\vert^2\\ &=\left\vert \frac{i}{\sqrt{6}} e^{-iE_{3}t/\hbar}\right\vert^2 +\left\vert \sqrt{\frac{2}{3}} e^{-iE_{-3}t/\hbar}\right\vert^2\\ &=\frac{1}{6}+\frac{2}{3}\\ &=\frac{5}{6} \end{align} Notice that the probability does not depend on time. I used the ket notation in the energy basis because the energy eigenstates have definite energy.
- Calculate the probability that the particle can be found in the region \(0<\phi<\frac{\pi}{3}\) at some time, \(t>0\).
To calculate position probabilities, I must use the position representation, so for each ket I will substitute the position representation. It can be helpful in these calculations to plug in the energies \(E_m=\frac{\hbar^2}{2I} m^2\) early. \begin{equation} \left|{m}\right\rangle \rightarrow \frac{1}{\sqrt{2\pi r_0}}e^{im\phi} \end{equation} Although I am using the position representation, I am still using the energy basis, so the time dependence is the same as the previous two problems. \begin{equation} \left|{m}\right\rangle e^{-iE_{m}t/\hbar} \rightarrow \frac{1}{\sqrt{2\pi r_0}}e^{im\phi}e^{-iE_{m}t/\hbar} \end{equation} To find the probability of being in the range \(0\le \phi\le \frac{\pi}{3}\), I integrate the probability density (square of the norm of the wave function) over this range. \begin{align} {\cal P}\left(0\le\phi\le\frac{\pi}{3}\right) &=\int_0^{\pi/3} \Phi^*(\phi)\, \Phi(\phi)\, r_0d\phi\\ &=\frac{1}{2\pi \cancel{r_0}}\int_0^{\pi/3} \left[\sqrt{\frac{2}{3}}e^{3i\phi}e^{+iE_{-3}t/\hbar} +\frac{1}{\sqrt{6}}e^{i\phi}e^{+iE_{-1}t/\hbar} -\frac{i}{\sqrt{6}}e^{-3i\phi}e^{+iE_{3}t/\hbar}\right]\\ &\qquad\left[\sqrt{\frac{2}{3}}e^{-3i\phi}e^{-iE_{-3}t/\hbar} +\frac{1}{\sqrt{6}}e^{-i\phi}e^{-iE_{-1}t/\hbar} +\frac{i}{\sqrt{6}}e^{3i\phi}e^{-iE_{3}t/\hbar}\right]\, \cancel{r_0} d\phi\\ &=\frac{1}{2\pi}\int_0^{\pi/3} \left[1 +\frac{1}{3}\cos(2\phi+(E_{-3}-E_{-1})t/\hbar)\right.\\ &\left.\qquad-\frac{1}{3}\sin(6\phi+(E_{-3}-E_{3})t/\hbar) -\frac{1}{6}\sin(4\phi+(E_{-1}-E_{3})t/\hbar) \right] \, d\phi\\ &=\frac{1}{2\pi}\left[\phi +\frac{1}{6}\sin(2\phi+(E_{-3}-E_{-1})t/\hbar)\right.\\ &\left.\qquad+\frac{1}{18}\cos(6\phi+(E_{-3}-E_{3})t/\hbar) +\left.\frac{1}24\cos(4\phi+(E_{-1}-E_{3})t/\hbar) \right]\right\vert_0^{\pi/3}\\ &=\frac{1}{2\pi}\left[\frac{\pi}{3} +\frac{1}{6}\sin\left(\frac{2\pi}{3}+\frac{(E_{-3}-E_{-1})t}{\hbar}\right) -\frac{1}{6}\sin\left(\frac{(E_{-3}-E_{-1})t}{\hbar}\right)\right.\\ &\qquad+\cancel{\frac{1}{18}\cos\left(\frac{(E_{-3}-E_{3})t}{\hbar}\right)} -\cancel{\frac{1}{18}\cos\left(\frac{(E_{-3}-E_{3})t}{\hbar}\right)}\\ &\qquad+\left.\frac{1}{24}\cos\left(\frac{4\pi}{3}+\frac{(E_{-1}-E_{3})t}{\hbar}\right) -\frac{1}{24}\cos\left(\frac{(E_{-1}-E_{3})t}{\hbar}\right) \right]\\ &=\frac{1}{2\pi}\left[\frac{\pi}{3} +\frac{1}{6}\sin\left(\frac{2\pi}{3}+\frac{8\hbar t}{2I}\right) -\frac{1}{6}\sin\left(\frac{8\hbar t}{2I}\right)\right.\\ &\qquad+\cancel{\frac{1}{18}\cos\left(\frac{(E_{-3}-E_{3})t}{\hbar}\right)} -\cancel{\frac{1}{18}\cos\left(\frac{(E_{-3}-E_{3})t}{\hbar}\right)}\\ &\qquad+\left.\frac{1}{24}\cos\left(\frac{4\pi}{3}-\frac{8\hbar t}{2I}\right) -\frac{1}{24}\cos\left(\frac{8\hbar t}{2I}\right) \right]\\ \end{align} Whew! This is a lot of algebra. There are several different correct versions of the solution depending on whether you factor out the \(\phi\) dependence before integrating.
I'm making you do all this algebra because this is the same algebra that happens every time you have resonance or beats. You will do essentially this same calculation a bizillion times in your life, so it's time for you to start seeing the patterns:
- Don't forget the cross terms! You are not integrating over the entire range of the wave function, so you can not use orthogonality of the eigenstates.
- I arranged the cross terms in complex conjugate pairs. They will always come this way, so it's a good check on the messy algebra.
- You will need to recognize the complex conjugate pairs as forming either sines or cosines.
- Under what circumstances do measurement probabilities change with time?
Probabililties for any physical quantity that shares (energy) eigenstates with the Hamiltonian (the technical description is if the operator commutes with the Hamiltonian) do NOT depend on time. The position eigenstates are different from the energy eigenstates, so position probabilities typically depend on time.
- You carry out a measurement to determine the \(z\)-component of the angular momentum of the particle at some time, \(t>0\). Calculate the probability that you measure the \(z\)-component of the angular momentum to be \(3\hbar\).
- Quantum Numbers on the Sphere
S1 5250S
Consider an arbitrary state for a quantum particle confined to the surface of a sphere written as a superposition of spherical harmonics:
\begin{equation}
\left|{\Psi}\right\rangle =\sum_{\ell=0}^\infty\sum_{m=-\ell}^\ell c_{\ell m}\left|{\ell, m}\right\rangle
\end{equation}
- Suppose you wanted to calculate the probability that an energy measurment of this state would yield \(E_{17}\). What coefficients \( c_{\ell m} \) would you need to find to calculate this? Write an expression for this probability.
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 \(\ell=17\).
- Suppose you wanted to calculate the probability that a measurement of the \(z\)-component of angular momentum would yield \(5\hbar\). What coefficients \( c_{\ell m} \) would you need to find to calculate this? Write an expresson for this probability.
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.
- Suppose you wanted to calculate the probability that an energy measurment of this state would yield \(E_{17}\). What coefficients \( c_{\ell m} \) would you need to find to calculate this? Write an expression for this probability.
- Sphere Questions
S1 5250S
Consider the following normalized state for the rigid rotor given by:
\begin{equation}
\left|\psi\right\rangle=\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
\end{equation}
-
Write the state as a superposition of spherical harmonics \(Y_l^m\)
\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}
-
What is the probability that a measurement of \(L_z\) will yield
\(2\hbar\)? \(-\hbar\)? \(0\hbar\)?
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?
-
If you measured the z-component of angular momentum to be \(-\hbar\), what would
the state of the particle be immediately after the measurement is made? What about if it yields \( 0\hbar \)?
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.
- What is the expectation value of \(L_z\) in the original state \(\left|{\psi}\right\rangle \)?
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!!
-
What is the expectation value of \(L^2\) in the original state \(\left|{\psi}\right\rangle \)?
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.
-
What is the expectation value of the energy in the original state \(\left|{\psi}\right\rangle \)?
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}
-
Write the state as a superposition of spherical harmonics \(Y_l^m\)
- Confidence Rating
S1 5250S
After solving each problem on the assignment, indicate your answers to the following questions for each problem. Answer for the problem as a whole, even if the problem has multiple parts.
- Question Confidence How confident are you that you are interpreting the problem the way the instructor intends?
For the rest of the questions, assume you have interpreted the problem correctly - Problem Confidence How confident are you that you could independentl come up with a correct solution process to a similar problem on a future problem set?
- Answer Confidence How confident are you that your final answer to this question is correct (not solution process)?
- Makes Sense To what degree do you understand how your answer fits (or does not fit) the physical or mathematical situation of the problem?
- Question Confidence How confident are you that you are interpreting the problem the way the instructor intends?