Central Forces: Spring-2025
HW 09 (SOLUTION): Due W5 D2
- Sphere Table
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Attached, you will find a table showing different representations of physical
quantities associated with a quantum particle confined to a sphere. Fill in all of the missing entries. Hint: You may look ahead. We filled out a number of the entries throughout the table to give you hints about what the forms of the other entries might be.
pdf link for the Table
or doc link for the Table
pdf link for the completed Table
- QM Sphere with Time Dependence
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Consider a quantum particle on a sphere. At t = 0, the particle is in state:\[\left|{\psi(t=0)}\right\rangle =\frac{1}{\sqrt{2}}\left(\left|{2,0}\right\rangle +\left|{1,0}\right\rangle \right)\]
Calculate the following quantities for some later time, \(t>0\), and identify whether each quantity is time-dependent.
- \(\left|{\psi(t)}\right\rangle \).
Each ket \(|\ell,m\rangle\) gets its own exponential energy term. The first ket has energy \(E_2=\frac{\hbar^2}{2I}\, 2(2+1)=6\frac{\hbar^2}{2I}\) and the second ket has energy \(E_1=\frac{\hbar^2}{2I}\, 1(1+1)=2\frac{\hbar^2}{2I}\). Therefore \[\left|{\psi(t)}\right\rangle =\frac{1}{\sqrt{2}}\left(\left|{2,0}\right\rangle e^{-iE_2t/\hbar} +\left|{1,0}\right\rangle e^{-iE_1t/\hbar}\right)\] This expression depends on time.
- \(\langle L_z\rangle\)
Both kets, even as functions of time, are eigenstates of \(L_z\) with eigenvalue \(0\hbar\), therefore the expectation value is \(0\), independent of time.
We can check that the calculation methods give this same answer:
Using the sum over probabilities method, we see that \begin{align} \langle L_z\rangle&=0\hbar\, \mathcal{P}(\left|{2,0}\right\rangle )+0\hbar\, \mathcal{P}(\left|{1,0}\right\rangle )\\ &=0\hbar\,\left\vert\frac{1}{\sqrt{2}}\left|{2,0}\right\rangle e^{-iE_2t/\hbar}\right\vert^2 +0\hbar\,\left\vert\frac{1}{\sqrt{2}}\left|{1,0}\right\rangle e^{-iE_1t/\hbar}\right\vert^2\\ &=0\hbar\left(\frac{1}{2}+\frac{1}{2}\right)\\ &=0 \end{align}
Using the bra/ket method, we see that \begin{align} \langle L_z\rangle &=\left\langle {\psi(t)}\right|\, L_z\, \left|{\psi(t)}\right\rangle \\ &=\frac{1}{\sqrt{2}}\left(\left\langle {2,0}\right|e^{iE_2t/\hbar} +\left\langle {1,0}\right|e^{iE_1t/\hbar}\right)\, L_z\,\frac{1}{\sqrt{2}}\left(\left|{2,0}\right\rangle e^{-iE_2t/\hbar} +\left|{1,0}\right\rangle e^{-iE_1t/\hbar}\right)\\ &=\frac{1}{\sqrt{2}}\left(\left\langle {2,0}\right|e^{iE_2t/\hbar} +\left\langle {1,0}\right|e^{iE_1t/\hbar}\right) \frac{1}{\sqrt{2}}\left(0\hbar\left|{2,0}\right\rangle e^{-iE_2t/\hbar} +0\hbar\left|{1,0}\right\rangle e^{-iE_1t/\hbar}\right)\\ &=0 \end{align}
- \(\mathcal{P}(L^2=6\hbar^2)\)
The eigenvalue \(L^2=6\hbar^2\) corresponds to \(\ell=2\) which happens only for the first eigenket. \begin{align} \mathcal{P}(L^2=6\hbar^2)&=\left\vert\left\langle {2,0}\middle|{\psi(t)}\right\rangle \right\vert^2\\ &=\left\vert \frac{1}{\sqrt{2}}e^{-iE_2t/\hbar}\right\vert^2\\ &=\frac{1}{2} \end{align} The time dependence cancels. Any probability like this will cancel as long as the operator for which the physical property is being measured (in this case, \(L^2\)) commutes with the Hamiltonian. In that case, every eigenket has only one exponential time dependence in it and the time dependence will cancel for each separate probability.
- The probability that the particle can be found in the “southern” hemisphere.
To calculate a probability in space, we need to use the wave function representation. We can look up the spherical harmonics in a table: \begin{align} \psi(\theta, \phi, t)&=\left\langle {\theta, \phi}\middle|{\psi(t)}\right\rangle \\ &=\frac{1}{\sqrt{2}}\left(Y_2^0(\theta,\phi)\, e^{-iE_2t/\hbar} +Y_1^0(\theta,\phi)\, e^{-iE_1t/\hbar}\right)\\ &=\frac{1}{\sqrt{2}}\left(\sqrt{\frac{5}{16\pi}}\left(3\cos^2\theta-1\right)\, e^{-iE_2t/\hbar} +\sqrt{\frac{3}{4\pi}}\left(\cos\theta\right)\, e^{-iE_1t/\hbar}\right)\\ \end{align} Now we need to integrate the probability density \(\vert\psi(t)\vert^2\) over the relevant part of the sphere. Don't forget to use the area element for the sphere (i.e. include a factor of \(\sin\theta\)). \begin{align} \mathcal{P}&=\int_0^{2\pi}\int_{\pi/2}^{\pi} \left[\frac{1}{\sqrt{2}}\left(\sqrt{\frac{5}{16\pi}}\left(3\cos^2\theta-1\right)\, e^{-iE_2t/\hbar} +\sqrt{\frac{3}{4\pi}}\left(\cos\theta\right)\, e^{-iE_1t/\hbar}\right)\right]^*\\ &\qquad\left[\frac{1}{\sqrt{2}}\left(\sqrt{\frac{5}{16\pi}}\left(3\cos^2\theta-1\right)\, e^{-iE_2t/\hbar} +\sqrt{\frac{3}{4\pi}}\left(\cos\theta\right)\, e^{-iE_1t/\hbar}\right)\right] \sin\theta\, d\theta\, d\phi\\ \end{align} Aaaackkk!! Don't panic. Take a deep breath, foil-like-mad, and recognize when you need to use the exponential definition of cosine: \begin{align} \mathcal{P}&=\frac{1}{2}\int_0^{2\pi}\int_{\pi/2}^{\pi} \left[\frac{5}{16\pi}\left(3\cos^2\theta-1\right)^2 +\frac{3}{4\pi}\left(\cos\theta\right)\right.\\ &\qquad\left.\sqrt{\frac{15}{64\pi^2}}\,\cos\theta\left(3\cos^2\theta-1\right)\, \left(e^{+iE_2t/\hbar}e^{-iE_1t/\hbar}+e^{+iE_1t/\hbar}e^{-iE_2t/\hbar}\right) \right] \sin\theta\, d\theta\, d\phi\\ &=\frac{1}{2}\int_0^{2\pi}\int_{\pi/2}^{\pi} \left[\frac{5}{16\pi}\left(3\cos^2\theta-1\right)^2 +\frac{3}{4\pi}\left(\cos\theta\right)\right.\\ &\qquad\left.\sqrt{\frac{15}{64\pi^2}}\,\cos\theta\left(3\cos^2\theta-1\right)\, 2\cos\left(\frac{(E_2-E_1)t}{\hbar}\right) \right] \sin\theta\, d\theta\, d\phi\\ \end{align} While this expression looks messy, it is the sum of a bunch of powers of \(\cos\theta\) times \(\sin\theta\), so a simple \(u\)-substitution will work. You can finish the problem.
Because the cross-terms are non-zero, this expression will clearly be time dependent. The probability of where the mass is depends on time! Hurray! Finally the universe has become interesting.
- The probability that the particle can still (at the time \(t\)) be found in the state
\[\left|{\psi}\right\rangle =\frac{1}{\sqrt{2}}\left(\left|{2,0}\right\rangle +\left|{1,0}\right\rangle \right)\]
\begin{align} \mathcal{P}&=\left\vert\left\langle {\psi(0)}\middle|{\psi(t)}\right\rangle \right\vert^2\\ &=\left\vert\frac{1}{\sqrt{2}}\left(\left\langle {2,0}\right|+\left\langle {1,0}\right|\right) \frac{1}{\sqrt{2}}\left(\left|{2,0}\right\rangle e^{-iE_2t/\hbar} +\left|{1,0}\right\rangle e^{-iE_1t/\hbar}\right)\right\vert^2\\ &=\left\vert\frac{1}{2}\left(e^{-iE_2t/\hbar}+e^{-iE_1t/\hbar}\right)\right\vert^2\\ &=\frac{1}{4}\left(e^{iE_2t/\hbar}+e^{iE_1t/\hbar}\right) \left(e^{-iE_2t/\hbar}+e^{-iE_1t/\hbar}\right)\\ &=\frac{1}{4}\left(1+1 +\left(e^{+iE_2t/\hbar}e^{-iE_1t/\hbar}+e^{+iE_1t/\hbar}e^{-iE_2t/\hbar}\right)\right)\\ &=\frac{1}{2}\left(1+\cos\left(\frac{(E_2-E_1)t}{\hbar}\right)\right) \end{align} The state oscillates with time and returns to the initial state periodically.
- \(\left|{\psi(t)}\right\rangle \).
- Confidence Rating
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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?