Central Forces: Spring-2025
HW 10 (SOLUTION): Due W5 D5
- Eigenvalues for Different Systems
S1 5254S
Fill in the following table with the appropriate eigenvalues for each operator for each system.
\[ \begin{matrix} ~\\L_z\\~\\L^2\\~\\H \end{matrix} \begin{vmatrix}\left|{m}\right\rangle &\left|{\ell,m}\right\rangle &\left|{n,\ell,m}\right\rangle \\ \underline{\text{particle on a ring}}& \underline{\text{particle on a sphere} }& \underline{\text{Hydrogen atom}}\\~\\ \\~\\ \\~\\~\\ \end{vmatrix}\]
\[ \begin{matrix} ~\\L_z\\~\\L^2\\~\\H \end{matrix} \begin{vmatrix} \underline{\text{particle on a ring}}& \underline{\text{particle on a sphere} }& \underline{\text{Hydrogen atom}}\\m\hbar&m\hbar&m\hbar\\ \\m^2\hbar^2&\ell(\ell+1)\hbar^2&\ell(\ell+1)\hbar^2\\ \\\frac{m^2\hbar^2}{2I}&\frac{\ell(\ell+1)\hbar^2}{2I}&-\frac{m_e e^4}{2(4\pi\epsilon_o)^2\hbar^2 }\frac{1}{n^2}=-\frac{13.6 eV}{n^2}\\\\ \end{vmatrix}\]
Write the Hamiltonian for each of the following systems explicitly in the position representation (i.e., differential operators).
\[ H\begin{vmatrix}\left|{m}\right\rangle &~&~&\left|{\ell,m}\right\rangle &~&~&\left|{n,\ell,m}\right\rangle \\ \underline{\text{particle on a ring}}&~&~& \underline{\text{particle on a sphere} }&~&~& \underline{\text{Hydrogen atom}}\\\\ \\ \\ \end{vmatrix} \]
\[ H\begin{vmatrix}\left|{m}\right\rangle &~&~&\left|{\ell,m}\right\rangle &~&~&\left|{n,\ell,m}\right\rangle \\ \underline{\text{particle on a ring}}&~&~& \underline{\text{particle on a sphere} }&~&~& \underline{\text{Hydrogen atom}}\\\\ -\frac{\hbar^2}{2I}\frac{\partial^2}{\partial\phi^2}&~& ~&-\frac{\hbar^2}{2I}\left(\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right) - \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\right) &~&~&-\frac{\hbar^2}{2\mu}\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)-\frac{\hbar^2}{2\mu r^2}(-\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right) -\frac{1}{\sin^2\theta}(\frac{\partial^2}{\partial\phi^2}))-\frac{e^2}{4\pi \epsilon_o r}\\ \end{vmatrix} \]
- SP not Hybrid
S1 5254S
A hydrogen atom is initially in the state \(\left|{\Psi(t=0)}\right\rangle =\frac{1}{\sqrt{2}}\left(\vert 1,0,0\rangle +\vert 2,1,0\rangle\right)\).
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If you measure the energy of this state, what possible values could
you obtain?
The only possible results of a quantum measurement are the eigenvalues of the operator corresponding to the quantity that is measured. The results of a measurement of energy will yield the energy eigenvalues which depend on the principle quantum number \(n\). \begin{equation} E_n = \frac{E_1}{n^2} \end{equation} In this case, There are two different values of \(n\) that contribute to \(\Psi\), \(n=1\) and \(n=2\). Therefore the values of the energy that can be measured are: \begin{align} E_1 &=-13.6 eV.\\ E_2 &= \frac{E_1}{4} \end{align}
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What is \(\left|{\Psi(t)}\right\rangle \), where \(t>0\)?
This state is made up of two eigenfunctions of the Hamiltonian for the hydrogen atom, with principle quantum numbers \(n=1\) and \(n=2\). This means that the energy of these two eigenstates are different (remember, the energy only depends on the principle quantum number, not \(\ell\) or \(m\)) so we must introduce different phase factors for each eigenstate. \begin{align} e^{-\frac{iE_1}{\hbar}t}\qquad\qquad\qquad&\textrm{for $n=1$}\\ e^{-\frac{iE_2}{\hbar}t} = e^{-\frac{iE_1}{4\hbar}t}\qquad&\textrm{for $n=2$} \end{align} \begin{equation} |\Psi(t)\rangle = \left(\frac{1}{\sqrt{2}}\right) \left( |1,0,0\rangle e^{-\frac{iE_1}{\hbar}t} + |2,1,0\rangle e^{-\frac{iE_1}{4\hbar}t}\right) \end{equation}
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Calculate the expectation value \(\langle\hat L^2\rangle\) in this
state, as a function of time. Did
you expect this answer? Please explain your reasoning.
\begin{align} \hat L^2 |n,\ell,m\rangle &= \ell(\ell+1)\hbar^2 |n,\ell,m\rangle \\ \Rightarrow& L^2 |n,0,m\rangle = 0\hbar^2 |n,0,m\rangle \\ \Rightarrow& L^2 |n,1,m\rangle = 2\hbar^2 |n,1,m\rangle \\ \end{align} \begin{align} \langle\hat L^2\rangle &= \langle\Psi(t)|\hat L^2|\Psi(t)\rangle\\ &= \frac{1}{\sqrt{2}} \left(\; \langle 1,0,0|e^{\frac{iE_1}{\hbar} t} + \langle2,1,0|e^{\frac{iE_1}{4\hbar} t} \;\right) \;\hat L^2\; \frac{1}{\sqrt{2}}\left(\;|1,0,0\rangle e^{-\frac{iE_1}{\hbar} t} + |2,1,0\rangle e^{-\frac{iE_1}{4\hbar} t}\;\right)\\ &= \frac{1}{2} \left(\; \langle 1,0,0|\hat L^2|1,0,0\rangle + \langle 1,0,0|\hat L^2|2,1,0\rangle e^{\frac{iE_1}{\hbar} t}e^{-\frac{iE_1}{4\hbar} t}\right.\\ & \textrm{} \; \left. + \langle 2,1,0|\hat L^2|1,0,0\rangle e^{\frac{iE_1}{4\hbar} t} e^{-\frac{iE_1}{\hbar} t} + \langle 2,1,0|\hat L^2|2,1,0\rangle \right)\\ &= \frac{1}{2} \left(\; \langle 1,0,0|0\hbar^2|1,0,0\rangle + \langle 1,0,0|2\hbar^2|2,1,0\rangle e^{\frac{iE_1}{\hbar} t} e^{-\frac{iE_1}{4\hbar} t}\right.\\ & \textrm{} \; \left. + \langle 2,1,0|0\hbar^2|1,0,0\rangle e^{\frac{iE_1}{4\hbar} t} e^{-\frac{iE_1}{\hbar} t} + \langle 2,1,0|2\hbar^2|2,1,0\rangle \right)\\ &= \frac{1}{2} \; 2\hbar^2\left(\; \langle 1,0,0|2,1,0\rangle e^{\frac{iE_1}{\hbar} t} e^{-\frac{iE_1}{4\hbar} t} + \langle 2,1,0|2,1,0\rangle \right)\\ \langle\hat L^2\rangle &= \hbar^2 \end{align} Notice that the time-dependence does not all disappear until the second to the last line. The expectation value of \(\hat L^2\) is time-independent, which is expected because this operator commutes with the Hamiltonian. The expectation value is also the average between the two possible values of the square of the angular momentum, which follows from the fact that the probability amplitudes for each term in this case happen to be the same.
Write \(\left|{\Psi(t)}\right\rangle \) in wave function notation.
The solutions of the hydrogen atom can generally be written as follows: \begin{equation} |n,l,m\rangle = R_{nl}(r) \, Y_l^m(\theta,\phi) \end{equation} The specific states we're interested in: \begin{align} |1,0,0\rangle &= R_{10}(r) \, Y_0^0(\theta,\phi)\\ &= 2\,a^{-\frac{3}{2}}\; e^{-\frac{r}{a}}\, \sqrt{\frac{1}{4\pi}}\\ &= \sqrt{\frac{1}{\pi a^3}}\, e^{-\frac{r}{a}}\\ |2,1,0\rangle &= R_{2,1}(r)\, Y_1^0(\theta,\phi)\\ &= \frac{1}{\sqrt{24}}\, a^{-\frac{3}{2}}\, e^{-\frac{r}{2a}} \, \left(\frac{r}{a}\right)\, \sqrt{\frac{3}{4\pi}}\cos{\theta}\\ &= \sqrt{\left(\frac{1}{32\pi a^3}\right)} \left(\frac{r}{a}\right) e^{-\frac{r}{2a}}\,\cos{\theta} \end{align} The eigenstates have different time-dependent phase factors: \(e^{-\frac{iE_1}{\hbar}t}\) and \(e^{-\frac{iE_1}{4\hbar}t}\), respectively. Putting these results together, we see that state in the position representation, i.e. the wave function, is given by: \begin{equation} \Psi(t)=\sqrt{\frac{1}{\pi a^3}} \left[\frac{1}{\sqrt{2}}e^{-\frac{r}{a}}\, e^{-\frac{iE_1}{\hbar}t} +\frac{r}{8a}\cos{\theta} e^{-\frac{r}{2a}}\, e^{-\frac{iE_1}{4\hbar}t}\right] \end{equation}
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If you measure the energy of this state, what possible values could
you obtain?
- Hydrogen, Version 1
S1 5254S
A hydrogen atom is initially in the superposition state \begin{equation} \vert \psi(t=0) \rangle = \frac{1}{\sqrt{14}}\vert 2,1,1\rangle - \frac{2}{\sqrt{14}}\vert 3,2,-1\rangle + \frac{3}{\sqrt{14}}\vert 4,2,2\rangle . \end{equation}
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What are the possible results of a measurement of the energy and with what
probabilities would they occur? Plot a histogram of the measurement results.
Calculate the expectation value of the energy.
The form of our answer should be a list of values for the quantity of interest (\(E\), \(L^2\), \(L_z\)) and the corresponding probabilities, which can also be presented in graphical form.
The possible results of a measurement of energy correspond to the values of \(n\) represented in the superposition. These are \(n = 2,3,4\), where \(E_n = -13.6 eV/n^2\) The probabilities are found by taking appropriate inner products that collapse to the coefficients of the relevant terms, followed by taking the modulus square. \begin{eqnarray*} &{\cal P}_{E_2} = 1/14 \\ &{\cal P}_{E_3} = 4/14 \\ &{\cal P}_{E_4} = 9/14 \end{eqnarray*}
The expectation value is the weighted average of the eigenvalues, which for this system gives:
\(\langle E \rangle = -13.6 eV (1/14(1/4) + 4/14(1/9) + 9/14(1/16) = -13.6 eV (181/144)\)
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What are the possible results of a measurement of the angular momentum
operator \(L^2\) and with what probabilities would they occur? Plot a histogram
of the measurement results. Calculate the expectation value of \(L^2\).
The possible results of a measurement of \(L^2\) correspond to the values of \(\ell\) represented in the superposition. These are \(\ell = 1,2\), where \(L^2 = \hbar^2\ell(\ell+1)\) The probabilities are found by taking appropriate inner products that collapse to the coefficients of the relevant terms, followed by taking the modulus square. Because the superposition contains more than one term with the same value of \(\ell\), it is necessary to sum probabilities corresponding to degenerate states. \begin{eqnarray*} &{\cal P}_{2\hbar^2} = 1/14 \\ &{\cal P}_{6\hbar^2} = 13/14 \end{eqnarray*}
The expectation value is the weighted average of the eigenvalues, which for this system gives:
\(\langle L^2 \rangle = 1/14(2\hbar^2) + 13/14(6\hbar^2) = 40\hbar^2/7\)
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What are the possible results of a measurement of the angular momentum
component operator \(L_z\) and with what probabilities would they occur? Plot a
histogram of the measurement results. Calculate the expectation value of
\(L_z\).
The possible results of a measurement of \(L_z\) correspond to the values of \(m\) represented in the superposition. These are \(m = -1,1,2\), where \(L_z = \hbar m\) The probabilities are found by taking appropriate inner products that collapse to the coefficients of the relevant terms, followed by taking the modulus square. \begin{eqnarray*} &{\cal P}_{-\hbar} = 4/14 \\ &{\cal P}_{\hbar} = 1/14 \\ &{\cal P}_{2\hbar} = 9/14 \end{eqnarray*}
The expectation value is the weighted average of the eigenvalues, which for this system gives:
\(\langle L_z \rangle = 5/14(-\hbar) +1/14\hbar + 9/14(3\hbar) = 23\hbar/14\)
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How do the answers to (a), (b), and (c) depend upon time?
Since the operators \(L_z\), \(L^2\) and \(H\) itself all commute with the Hamiltonian, none of the previous answers will depend on time.
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What are the possible results of a measurement of the energy and with what
probabilities would they occur? Plot a histogram of the measurement results.
Calculate the expectation value of the energy.
- Confidence Rating
S1 5254S
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?