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).
\[ H\begin{vmatrix}\left|{m}\right\rangle &~&~&\left|{\ell,m}\right\rangle &~&~&\left|{n,\ell,m}\right\rangle \\ \underline{\text{particle on a ring}}&~&~& \underline{\text{particle on a sphere} }&~&~& \underline{\text{Hydrogen atom}}\\\\ \\ \\ \end{vmatrix} \]
\[ H\begin{vmatrix}\left|{m}\right\rangle &~&~&\left|{\ell,m}\right\rangle &~&~&\left|{n,\ell,m}\right\rangle \\ \underline{\text{particle on a ring}}&~&~& \underline{\text{particle on a sphere} }&~&~& \underline{\text{Hydrogen atom}}\\\\ -\frac{\hbar^2}{2I}\frac{\partial^2}{\partial\phi^2}&~& ~&-\frac{\hbar^2}{2I}\left(\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right) - \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\right) &~&~&-\frac{\hbar^2}{2\mu}\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)-\frac{\hbar^2}{2\mu r^2}(-\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right) -\frac{1}{\sin^2\theta}(\frac{\partial^2}{\partial\phi^2}))-\frac{e^2}{4\pi \epsilon_o r}\\ \end{vmatrix} \]
A hydrogen atom is initially in the state \(\left|{\Psi(t=0)}\right\rangle =\frac{1}{\sqrt{2}}\left(\vert 1,0,0\rangle +\vert 2,1,0\rangle\right)\).
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}
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}
\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}
Consider a quantum particle confined to the surface of a cylinder (not including the endcaps). Let the height of the cylinder be equal to half its circumference.
The particle is confined to the surface of a cylinder, so the radius is fixed: \begin{equation} r=r_0. \end{equation} The particle can move in the angular direction \(\phi\) and along the height of the cylinder in the \(z\) direction. The surface element for the cylinder is \begin{equation} dA=r_0,d\phi,dz. \end{equation} The relevant coordinates are \begin{equation} 0\leq \phi \leq2\pi,\qquad 0\leq z \leq L. \end{equation} Since the height of the cylinder is equal to half its circumference, \begin{equation} L=\frac{1}{2}(2\pi r_0)=\pi r_0. \end{equation}
Since \(r\) is fixed, there is no radial kinetic energy term. The Hamiltonian is \begin{equation} \hat H=-\frac{\hbar^2}{2\mu}\left(\frac{1}{r_0^2}\frac{\partial^2}{\partial \phi^2}+\frac{\partial^2}{\partial z^2}\right). \end{equation} The energy eigenvalue equation is \begin{equation} -\frac{\hbar^2}{2\mu}\left(\frac{1}{r_0^2}\frac{\partial^2 \psi}{\partial \phi^2}+\frac{\partial^2 \psi}{\partial z^2}\right)=E\psi. \end{equation}
The boundary condition in the angular direction is periodic: \begin{equation} \psi(\phi+2\pi,z)=\psi(\phi,z). \end{equation} The boundary conditions in the \(z\) direction are the same as for an infinite square well: \begin{equation} \psi(\phi,0)=0,\qquad \psi(\phi,L)=0. \end{equation}
This system looks like a combination of a quantum ring in the \(\phi\) direction and an infinite square well in the \(z\) direction. Therefore we guess a product solution: \begin{equation} \psi(\phi,z)=\Phi(\phi)Z(z). \end{equation}
The normalized angular part is the same as the quantum ring: \begin{equation} \Phi_m(\phi)=\frac{1}{\sqrt{2\pi r_0}}e^{im\phi},\qquad m=0,\pm1,\pm2,\ldots \end{equation}
The normalized \(z\)-dependent part is the same as the infinite square well: \begin{equation} Z_{n_z}(z)=\sqrt{\frac{2}{L}}\sin\left(\frac{n_z\pi z}{L}\right),\qquad n_z=1,2,3,\ldots \end{equation}
Therefore the normalized energy eigenfunctions are \begin{equation} \psi_{m,n_z}(\phi,z)=\Phi_m(\phi)Z_{n_z}(z). \end{equation} Substituting the two normalized pieces gives \begin{equation} \psi_{m,n_z}(\phi,z)=\frac{1}{\sqrt{2\pi r_0}}\sqrt{\frac{2}{L}}e^{im\phi}\sin\left(\frac{n_z\pi z}{L}\right). \end{equation} So \begin{equation} \psi_{m,n_z}(\phi,z)=\frac{1}{\sqrt{\pi r_0 L}}e^{im\phi}\sin\left(\frac{n_z\pi z}{L}\right). \end{equation} Using \(L=\pi r_0\), this can also be written as \begin{equation} \psi_{m,n_z}(\phi,z)=\frac{1}{\pi r_0}e^{im\phi}\sin\left(\frac{n_z z}{r_0}\right). \end{equation}
The energy eigenvalues are \begin{equation} E_{m,n_z}=\frac{\hbar^2}{2\mu}\left(\frac{m^2}{r_0^2}+\frac{n_z^2\pi^2}{L^2}\right). \end{equation} Since \(L=\pi r_0\), \begin{equation} \frac{\pi^2}{L^2}=\frac{1}{r_0^2}. \end{equation} Therefore \begin{equation} E_{m,n_z}=\frac{\hbar^2}{2\mu r_0^2}\left(m^2+n_z^2\right). \end{equation} Let \begin{equation} \epsilon=\frac{\hbar^2}{2\mu r_0^2}. \end{equation} Then \begin{equation} E_{m,n_z}=\epsilon(m^2+n_z^2). \end{equation}
Now we list the lowest energy states by finding the smallest allowed values of \(m^2+n_z^2\), where \begin{equation} m=0,\pm1,\pm2,\ldots,\qquad n_z=1,2,3,\ldots \end{equation}
The ground state is \begin{equation} m=0,\qquad n_z=1. \end{equation} Therefore \begin{equation} E_{0,1}=\epsilon. \end{equation} The state in ket notation is \begin{equation} \left|{0,1}\right\rangle . \end{equation} The full time-dependent wave function is \begin{equation} \Psi_{0,1}(\phi,z,t)=\frac{1}{\pi r_0}\sin\left(\frac{z}{r_0}\right)e^{-i\epsilon t/\hbar}. \end{equation} This state is not degenerate.
The first excited energy level has \begin{equation} m=\pm1,\qquad n_z=1. \end{equation} Therefore \begin{equation} E_{\pm1,1}=2\epsilon. \end{equation} The states in ket notation are \begin{equation} \left|{1,1}\right\rangle ,\qquad \left|{-1,1}\right\rangle . \end{equation} The full time-dependent wave functions are \begin{equation} \Psi_{1,1}(\phi,z,t)=\frac{1}{\pi r_0}e^{i\phi}\sin\left(\frac{z}{r_0}\right)e^{-i(2\epsilon)t/\hbar}. \end{equation} \begin{equation} \Psi_{-1,1}(\phi,z,t)=\frac{1}{\pi r_0}e^{-i\phi}\sin\left(\frac{z}{r_0}\right)e^{-i(2\epsilon)t/\hbar}. \end{equation} These two states are degenerate because clockwise and counterclockwise motion around the cylinder have the same kinetic energy.