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\title{PHYS 2380 - Assignment 5}
\author{Due April 17, 2017}
\maketitle

\bigbreak

\textbf{1.)} In this problem, we will examine some properties of the wavefunctions of hydrogen-like atoms.
\bigbreak
\textbf{a.)} When we investigated the quantum simple harmonic oscillator we showed that the eigenstates are orthogonal. The hydrogen-like wavefunctions also form an orthogonal set, and must obey the relationship
\begin{equation}
\int_0^{2 \pi} \int_0^\pi \int_0^\infty \psi_{n' l' m'}^* \psi_{nlm} r^2 \sin \theta dr d\theta d\phi = \delta_n^{n'} \delta_l^{l'} \delta_m^{m'}
\end{equation}
where the Kronecker $\delta_a^{a'}$ gives $1$ if $a=a'$ and $0$ if $a \neq a'$. We will not prove the orthogonality relationship in general, but simply demonstrate it for the following pairs: \\
i.) $\psi_{100}$ and $\psi_{200}$, \\
ii.) $\psi_{200}$ and $\psi_{210}$, \\
iii.) $\psi_{210}$ and $\psi_{211}$. \\
When evaluating these expressions, the following integral identity may be useful:
\begin{equation}
\int_0^\infty x^q e^{-\alpha x} dx = \frac{	q!}{a^{q+1}}
\end{equation}  
\\ 

\textbf{b.)} Any eigenstate of a hydrogen-like atom is degenerate with respect to several other states with the same $n$ (except for the ground state, $n=1$, $l=0$, $m=0$). For example, the $n=2$ state has a four-fold degeneracy since this state has orbital quantum number $l=0$ (with magnetic quantum number $m=0$) and $l=1$ (with $m=-1,0,1)$. The most general probability density for atoms in the $n=2$ state is a superposition of these four states with equal amplitudes, 
\begin{equation}
\Psi_2 = \frac{1}{2} \psi_{200} + \frac{1}{2} \psi_{210} + \frac{1}{2} \psi_{211} + \frac{1}{2} \psi_{21-1}.
\end{equation} 
Using the orthogonality condition from part a, find the probability density $P_2(r,\theta,\phi)$ that comes from the state $\Psi_2$ above. Does the probability density have any symmetry (ie, does it depend on all of the coordinates)?\\

\textbf{c.)} Now consider the superposition of the $n=3$ eigenstates. Do an analogous calculation for $P_3(r,\theta,\phi)$ where $\Psi_3$ is written as an evenly-weighted superposition. Recall from the last assignment that the squares of the real amplitude coefficients must sum to $1$ to give a realistic superposition. You do not have to simplify the calculation of $P_3$ fully, just find which coordinates it depends on. You may also find it easier to complete this problem by working on groups of terms arranged by orbital quantum number $l$.\\

\textbf{d.)} Do the solutions to parts b and c have the same type of symmetry? \\



\textbf{2 a.)} Calculate the location at which the radial probability density is a maximum for the $n=2$, $l=1$ state of the hydrogen atom. \\

\textbf{b.)} Calculate the expectation value of the radial coordinate in this state. \\

\textbf{c.)} Explain the physical significance of the difference in the answers to a and b. According to the Bohr model, the radius of an electron orbital is given by 
\begin{equation}
r_{Bohr} = \frac{n^2 a_0}{Z}
\end{equation}
\\

\textbf{3 a.)} Calculate the expectation value $<V>$ of the potential energy in the $n=2$, $l=0$ state of the hydrogen atom. \\

\textbf{b.)} Do the same for the $n=2$, $l=1$ state. \\

\textbf{c.)} Discuss the results of part $a$ and $b$ in connection with the Virial theorem, and explain how they relate to the degeneracy of the hydrogen atom energy levels. 

\bigbreak
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\newpage

The first $10$ wave functions of hydrogen-like atoms:

\begin{equation}
\psi_{100}=\frac{1}{\sqrt{\pi}} \left( \frac{Z}{a_0}\right)^{3/2} e^{-Zr/a_0}
\end{equation}

\begin{equation}
\psi_{200}=\frac{1}{4\sqrt{2 \pi}} \left( \frac{Z}{a_0}\right)^{3/2} \left( 2-\frac{Zr}{a_0} \right) e^{-Zr/2a_0}
\end{equation}

\begin{equation}
\psi_{210}=\frac{1}{4\sqrt{2 \pi}} \left( \frac{Z}{a_0}\right)^{3/2} \frac{Zr}{a_0} e^{-Zr/2a_0} \cos\theta
\end{equation}

\begin{equation}
\psi_{21 \pm 1}=\frac{1}{8\sqrt{\pi}} \left( \frac{Z}{a_0}\right)^{3/2} \frac{Zr}{a_0} e^{-Zr/2a_0} \sin\theta e^{\pm i \phi}
\end{equation}

\begin{equation}
\psi_{300}=\frac{1}{81\sqrt{3 \pi}} \left( \frac{Z}{a_0}\right)^{3/2} \left( 27-18\frac{Zr}{a_0} + 2\frac{Z^2r^2}{a_0^2} \right) e^{-Zr/3a_0}
\end{equation}

\begin{equation}
\psi_{310}=\frac{\sqrt{2}}{81 \sqrt{\pi}} \left( \frac{Z}{a_0}\right)^{3/2}\left( 6- \frac{Zr}{a_0} \right) \frac{Zr}{a_0} e^{-Zr/3a_0} \cos\theta
\end{equation}

\begin{equation}
\psi_{31 \pm 1}=\frac{1}{81\sqrt{\pi}} \left( \frac{Z}{a_0}\right)^{3/2} \left( 6- \frac{Zr}{a_0} \right) \frac{Zr}{a_0} e^{-Zr/3a_0} \sin \theta e^{\pm i \phi}
\end{equation}

\begin{equation}
\psi_{320}=\frac{1}{81\sqrt{6\pi}} \left( \frac{Z}{a_0}\right)^{3/2}\frac{Z^2r^2}{a_0^2}e^{-Zr/3a_0}\left( 3 \cos^2\theta -1 \right)
\end{equation}

\begin{equation}
\psi_{32 \pm 1}=\frac{1}{81 \sqrt{\pi}} \left( \frac{Z}{a_0}\right)^{3/2}\frac{Z^2r^2}{a_0^2}e^{-Zr/3a_0} \sin\theta \cos\theta e^{\pm i \phi}
\end{equation}

\begin{equation}
\psi_{32 \pm 2}=\frac{1}{162 \sqrt{\pi}} \left( \frac{Z}{a_0}\right)^{3/2}\frac{Z^2r^2}{a_0^2}e^{-Zr/3a_0} \sin^2\theta e^{\pm 2i \phi}
\end{equation}
\\

These wavefunctions are given in terms of the Bohr radius, 
\begin{equation}
a_0 = \frac{4 \pi \epsilon_0 \hbar^2}{\mu e^2} = 5.29 \times 10^{-11} m
\end{equation}

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