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# $$\S 16.4$$ Hydrogen atom Energy Levels

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### Hydrogen Atom

The classical Hamiltonian for an electron and a nucleus of charge $Ze$ is

$$H = \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} - \frac{Ze^2}{|\vec{r}_1-\vec{r}_2|} \label{EQ01}$$

where $m_1, m_2$ are the masses of the electron and the nucleus and $\vec{r}_1, \vec{r}_2$ denote their respective positions.The case $Z=1$ corresponds to H atom, $Z=2$ singly ionized He atom and $Z=3$ doubly ionized Li atom and so on.

The Schr\"{o}dinger equation for the electron nucleus system takes the form

$$-\frac{\hbar^2}{2m_1}\Big(\PP{x_1}+ \PP{y_1} + \PP{z_1} \Big)\Psi -\frac{\hbar^2}{2m_1}\Big(\PP{x_2}+ \PP{y_2} + \PP{z_2} \Big)\Psi -\frac{Ze^2}{|\vec{r}_1-\vec{r}_2|}\Psi = \Eca \Psi. \label{EQ02}$$

Since the potential depends on relative position  only, the two body problem can be reduced to an equivalent one body problem with reduced mass by changing the frame of reference to the centre of mass frame. Introducing the centre of mass and relative coordinates defined by

$$\vec{R}=\frac{m_1\vec{r}_1 + m_2 \vec{r}_2}{m_1+m_2}, \qquad \vec{r}=\vec{r_1}-\vec{r}_2 . \label{EQ03}$$

The centre of mass will move like a free particle, and the relative motion reduces to that of a particle of reduced  mass $\mu =\frac{m_1m_2}{m_1+m_2}$ Therefore it is not surprising that the separation of variables in the Schr\"{o}dinger equation can be achieved by changing to these new variables $\vec{r}$ and  $\vec{R}$. In terms of these variables the Schr\"{o}dinger equation takes the form

$$-\frac{\hbar^2}{2M}\Big(\PP{X}+ \PP{Y} + \PP{Z} \Big)\Psi(\vec{R},\vec{r}) -\frac{\hbar^2}{2\mu}\Big(\PP{x}+ \PP{y} + \PP{z} \Big)\Psi(\vec{R},\vec{r}) -\frac{Ze^2}{r} \Psi(\vec{R},\vec{r})= \Eca \Psi(\vec{R},\vec{r}). \label{EQ04}$$

Here $M=m_1+m_2$ is the total mass, $\mu$ is the reduced mass. If we now write the full wave function $\Psi(\vec{R},\vec{r})$ as

$$\Psi(\vec{r},\vec{r}) = U(\vec{R})u(\vec{r}) \label{EQ05}$$

and substitute it in \EqRef{EQ04}, the variables $\vec{R}$ and $\vec{r}$ get separated and we would get the following differential equations for $U(\vec{R})$ and $u(\vec{r})$

\begin{eqnarray} -\frac{\hbar^2}{2M}\Big(\PP{X}+ \PP{Y} + \PP{Z} \Big) U(\vec{R}) = E_{\text{cm}} U(\vec{R}) \label{EQ06}\\ -\frac{\hbar^2}{2\mu}\Big(\PP{x}+ \PP{y} + \PP{z} \Big)u(\vec{r}) -\frac{Ze^2}{r} u(\vec{r})= E u(\vec{r}). \label{EQ07} \end{eqnarray}

$E_\text{cm}, E$ are constants appearing from the process of separation of variables so that $E+ E_\text{cm}=\Eca$. The equation \REF{EQ06} is a free particle equation for the centre of mass and \EqRef{EQ07} describes the relative motion of the electron and the nucleus.

The Schr\"{o}dinger equation \REF{EQ07} can now be solved by separation of variables in spherical polar coordinates  $r,\theta,\phi$. The angular part of the wave function is given by the spherical harmonics $Y_{\ell 2m}(\theta,\phi)$ and therefore we write

$$u(\vec{r}) = R(r) Y_{\ell m}(\theta,\phi) \label{EQ08}.$$

The radial equation for $R(r)$ takes the form

\begin{eqnarray} \frac{1}{r^2}\dd{r}\Big(r^2\dd[R]{r} \Big) +\frac{2\mu E}{\hbar^2}\Big( E+ \frac{Ze^2}{r}- \frac{\ell(\ell+1)}{\hbar^2}\Big)R(r)&=&0\label{EQ09},\\ \DD[R(r)]{r} + \frac{2}{r} \dd[R(r)]{r} + \frac{2\mu E}{\hbar^2}\Big( E+ \frac{Ze^2}{r}- \frac{\ell(\ell+1)}{\hbar^2}\Big)R(r)&=&0 \label{EQ10}. \end{eqnarray}

The radial equation involves effective potential

$$V_\text{eff}(r)=-\frac{Ze^2}{r} + \frac{\ell(\ell+1)\hbar^2}{2\mu r^2}.\label{EQ11}$$

Remembering that $\ell(\ell+1)\hbar^2$ is the eigenvalue of the square of orbital angular momentum, $L^2$,the second term is seen to be the centrifugal barrier term that appears in classical mechanics. The effective potentail goes to zero fr large $r$. Hence for $E>0$ the energy eigenvalues will be continuous and the bound states exist only for negative $E$, so we write $E=-|E|$. It is convenient to work with dimensionless variables $\rho$ and $\lambda$ defined by

%FIXME work with fine structure constant and rest mass mc^2 etc.

\begin{eqnarray} \rho = \alpha r, &~& \alpha^2 = \frac{8\mu |E|}{\hbar^2}\label{EQ12} \\ \lambda = \frac{2\mu Ze^2}{\alpha \hbar^2} &=& \frac{Ze^2}{\hbar} \sqrt{\frac{\mu}{2|E|}}  \label{EQ13}. \end{eqnarray}

The equation for radial wave function written in terms of $\rho$ takes the form

$$\DD[R]{\rho} + \frac{2}{\rho}\dd[R]{\rho} + \Big(\frac{\lambda}{\rho} -\frac{1}{4} -\frac{\ell(\ell+1)}{\rho^2}\Big)R = 0. \label{EQ14}$$

The above equation \eqRef{EQ14} can be transformed into a form similar to one dimensional Schr\"{o}dinger equation  by introducing $\chi(\rho)= \rho R(\rho)$ which gives the following equation for $\chi(\rho)$

$$\DD[\chi]{\rho} + \\Big(frac{\lambda}{\rho} -\frac{1}{4} -\frac{\ell(\ell+1)}{\rho^2} \Big)\chi(\rho)=0. \label{EQ15}$$

\subsection*{Large $\rho$ behaviour}

The behaviour of the radial wave function for large $\rho$ can be easily found by taking large $\rho$ limit of \EqRef{EQ15}. Neglecting the terms $\frac{\lambda}{\rho}$ and $\frac{\ell(\ell+1)}{\rho^2}$ compared to $1/4$ we get

$$\DD[\chi(\rho)]{\rho} - \frac{1}{4} \chi{\rho} =0.\label{EQ16}$$

showing that the wave function behaves like $\exp(\pm\rho/2)$ for large $\rho$. The wave function must be bounded everywhere including at infinity, so we must have $\chi(\rho)\approx e^{-\rho/2}$. This suggests that we write $R=e^{-\rho/2} F(\rho)$, and solve for $F(\rho)$. The equation for $F(\rho)$ turns out to be

$$\DD[F(\rho)]{\rho} + \left(\frac{2}{\rho}-1\right) + \left[ \frac{\lambda-1}{\rho}-\frac{\ell(\ell+1}{\rho^2}\right] \label{EQ17}$$

\subsection*{Solution by Frobenius method}

We now find solution of the differential equation  for $F(\rho)$ by the method of series solution. Assuming the form

$$F(\rho) = \sum_{m=0}a_m \rho^{c+m},\label{EQ18},$$

substituting in \EqRef{EQ17}, and equating coefficients of lowest power of $\rho$ to zero we get

$$c(c+1) - \ell(\ell+1) = 0 \Longrightarrow c=-\ell-1, \ell\label{EQ19}$$

Since $\ell> 0$, the value $c=-\ell(\ell+1)$ give solution diverging at $\rho=0$.

Therefore we choose $c=\ell=$ and the recurrence relation for the coefficients $a_m$  turns out to be

$$a_{m+1}= \frac{(m+\ell+1-\lambda}{(m+1)(m+2\ell+2)} a_m . \label{EQ20}$$

The ratio of coefficients for large $m$

$$\frac{a_{m+1}}{a_m} \sim \frac{1}{m} \label{EQ21}$$

coincides with the corresponding value for the series $\rho^k \exp(\rho)$. Hence if the series does not terminate, the solution $F(\rho)$ gives the radial wave function diverging like $\rho^k \exp(\rho/2)$ for large $\rho$. This is unacceptable and hence the series must terminate. This happens if all terms vanish after some $n^\prime$ {\it i.e.} $a_m=0$  for all $m>n^\prime$. For this to happen we must have $a_{n^{\prime+1}}=0$. Hence from \EqRef{EQ20} we get

$$\lambda = n^\prime +\ell+1 \label{EQ22}.$$

The energy is then given by

$$\boxed { E_n=-|E_n|= - \frac{Z^2e^4\mu}{2\hbar^2n^2} = \frac{Z\alpha^2 }{2n^2}(\mu c^2) } \label{EQ23}.$$

where $c$ is velocity of light and $\alpha=\frac{e^2}{\hbar c}\approx\frac{1}{137}$ is the fine structure constant.

\subsection*{Properties of H atom wave functions}

The final expressions for wave functions for hydrogen like problems is given by

\begin{eqnarray} u_{n\ell m}(r,\theta,\phi) &=& R_{n\ell}(r) Y_{\ell m}(\theta,\phi) \label{EQ24}\\ R_{n\ell}(r) &=& N_{n\ell}\, \rho^\ell \, L_{n+\ell}^{2\ell+1}(\rho)e^{-\rho/2} \label{EQ25}\\ N_{n\ell} &=& \sqrt{ \left(\frac{2Z}{na_0}\right)^3 \frac{(n-\ell-1)!}{2n\big((n+\ell)!\big)} }\label{EQ26} \end{eqnarray}

with

$$\rho=\Big(\frac{2Z}{na_0}\Big)r, \qquad a_0=\frac{\hbar^2}{\mu e^2} \label{EQ27}.$$

and $n$ is the principle quantum number.
%FIXME Check $n$ radial or principal QNO??

Here $L_q^p(\rho)$ are associated Laguerre  polynomials and $a_0$ is the radius of first Bohr orbit of the the electron in hydrogen atom. The energy levels are given by

$$E_n=-\frac{Z^2e^4\mu}{2\hbar^2n^2} \label{EQ28}.$$

The first few radial wave functions are

\begin{eqnarray} R_{10} &=& (Z/2a_0)^{\frac{3}{2}} 2 \exp(-Zr/2a_0))\label{EQ29}\\ R_{20}(r) &=&(Z/2a_0)^{\frac{3}{2}} (2-Zr/a_0) \exp(-Zr/2a_0))\label{EQ30}\\ R_{20}(r) &=&(Z/2a_0)^{\frac{3}{2}} (Zr/\sqrt{3}\,a_0) \exp(-Zr/2a_0))\label{EQ31} \end{eqnarray}

\paragraph*{A comment on hydrogen atom energy levels}

Finally we wish to remind you that the non-relativistic result $-R/n^2$ for the energy levels of H-atom is not the end of story for $H$-atom levels. Precision experiments show that each level is not a single level. To understand the experimental facts we must take into account of relativistic effects using Dirac theory of electron

$${ \begin{array}{llll} \mbox{Dirac Theory} & \nearrow \raisebox{0.5em}{\mbox{\hspace{5mm} Spin \ orbit \ coupling \hspace{5mm} }}\searrow & \mbox{Fine Structure} \\ &\searrow \raisebox{-0.25em}{\mbox{Relativistic variation of mass}} \nearrow \end{array}}$$

Also a hyperfine structure, seen in the energy levels,  requires a treatment of the spin-spin interaction of electron with the nucleus and an explanation of a tiny Lamb shift' requires use of  quantum field theory.

$$\begin{array}{ll} \mbox{Hyperfine structure} & \rightarrow \mbox{Effect of Nuclear Spin}\\ \mbox{Lamb shift} & \rightarrow \mbox{Quantum field Theory, Vacuum Polarization Effect}\\[2mm] \end{array}$$

### Accidental Degeneracy

General Properties of Bound State Spectra}

A potential is spherically symmetric if in polar variables it depends only on $r$ and not on $\theta$ and $\phi$ coordinates . We shall now discuss general properties of solution of 3-dimensional Schr\"{o}dinger equation $H\psi = E\psi$ where

$$H = {\vec{p}\,^2\over 2m} + V(r)$$

and the potential  $V(r)$ is spherically symmetric.

Conserved quantities
We note that all the three components of $\vec{L}$ commute with
Hamiltonian
$${} [\vec{L} , H] = 0$$
hence
$${} [\vec{L}^2 , H] = 0\,\,\,.$$

The parity operator $P$
$$P\psi(\vec{r}) = \psi(-\vec{r})$$
also commutes with $L^2L_z$ and $H$, operators. Therefore, the eigen functions of $H$ will also be eigen functions of $L^2, L_z$ and parity and each level can be assigned a definite value of $l, m$  and parity. For a state with definite value of $l$, the value of  parity is $\equiv(-1)^l$. In this case $L^2, L_z$ and $H$ form a complete commuting set.

$(2\ell+1)$ degeneracy
We use the notation $\ket{El,m}$  to denote the simultaneous eigenvector of $H, L^2$ and $L_z$
\begin{eqnarray}  H\ket{E,l m} & = & E\ket{E,l m}\\  L^2\ket{E,l m} &=& l(l+1)\hbar^2\ket{E,l m}\\ \mbox{and}~~ L_z\ket{E,l m} &=& m\hbar\ket{E,l m}\\ P\ket{E,l m} &=& (-1)^l\ket{E,l m} \end{eqnarray}
Applying $L_-$ on $\ket{El,m}$ several times leads successively to
\begin{eqnarray} \ket{E l, m-1>}~,~\ket{E,l,m-2} &\cdots& |E l, -l>\ \end{eqnarray}
and the action of $L_+$ on $\ket{E,l m}$ leads to the states
\begin{eqnarray}  \ket{E l, m+1}~,~\ket{E,l,m+2} &\cdots& |E l, l> \end{eqnarray}

All these states will have the same value of energy. This statement can be proved by making use of the fact that $H$ commutes with $L_\pm$ and that action of $L_+$ (or $L_-$) on $|Elm>$ leads to states $|El,m+1>$ (or $|El,m-1>$). Thus we see that the bound state energy eigenvalues of a spherically symmetric potential problem with have $(2l+1)$ fold degeneracy. ( What about the continuous energy eigenvalues? )

Radial wave function The Schr\"odinger equation for a spherically symmetric potential can be solved by separation of variables in polar coordinates. The angular part of the wave functions turns out to be a spherical harmonic $Y_{lm}(\theta,\phi)$ and the  wave function has the form $$\psi(r,\theta,\phi) = R(r)Y_{lm}(\theta, \phi)$$ where $R(r)$ is radial wave function satisfying the Schr\"odinger equation $$-{\hbar^2\over2m} {1\over r^2} {d\over dr} r^2{d\over dr} R(r) + \left(V(r) + {l(l+1)\hbar^2\over2mr^2}\right) R(r) = ER(r)\,\,\,.$$ If we define $\chi(r)=rR(r)$, the function $\chi$ satisfies the equation $$-{\hbar^2\over2m} {d^2\chi\over dr^2} + \left(V(r) + {l(l+1)\hbar^2\over2mr^2}-E\right)\chi = 0\,\,\,.$$ with boundary condition $\chi(r)|_{r=o}=0$, otherwise $R(r)$, the radial wave function will tend to $\infty$ as $r\to\infty$.

Bound state spectrum
The equation for $\chi$ has the form of one dimensional Schr\"odinger equation. Let $n^\prime$ denote the number of zeros of the radial wave function, excluding $r=0$ and at $r=\infty$. Then for a fixed value of $l$, the energy will increase with $n^\prime$, $n^\prime=0,1,2,\cdots$ will correspond to, for a fixed $l$, the ground' state, first excited state, the second excited state etc. Because of the $l$ dependence  of the term ${l(l+1)\hbar^2\over2mr^2}$ in the potential appearing in equation for $\chi(r)$, we expect that as $l$ is changed, keeping the number of nodes to be the same, $E$ would also change. Increasing $l$ would lead to increase in $E$, when $n^\prime$ is kept fixed.

Thus the spectrum would appear as follows
\begin{tabular}{cccc}
&&&{\bf ---------------}\\
&  & {\bf ---------------}\\
& {\bf ---------------}\\
&&& \underline{$n^\prime=3$~~~~~~~}\\
&& \underline{$n^\prime=3$~~~~~~~}\\
$n^\prime=3$ & {\bf ---------------}\\
&&& \underline{$n^\prime=2$~~~~~~~}\\
&& \underline{$n^\prime=2$~~~~~~~}\\
$n^\prime=2$ & {\bf---------------}\\
&&& \underline{$n^\prime=1$~~~~~~~}\\
&& \underline{$n^\prime=1$~~~~~~~}\\
$n^\prime=1$ & {\bf---------------}\\
&&& \underline{$n^\prime=0$~~~~~~~}\\
&& \underline{$n^\prime=0$~~~~~~~}\\
$n^\prime=0$ & {\bf---------------}\\
& $l=0$        & $l=1$ & $l=2$\\
&nondegenerate & $m=-1,0,1$        & $m=-2,-1,0,1,2$\\
&              & 3 fold degenerate & 5 fold degenerate\\
\end{tabular}

Coulomb problem spectrum
For hydrogen atom the energy levels are given by $$E = -{Z^2 e^4m\over2\hbar^2(n^\prime+l+1)^2}$$ The energy does not depend on $n^\prime$ and $l$ separately but only on the combination $n=(n^\prime+l+1)$. For a fixed $n$, $l$ can have values $0,1,\cdots,n-1$ ( because $n^\prime\ge0$ ) and all these solutions correspond to the same energy eigenvalue. The energy level diagram of H-atom, therefore, appears as shown below.

\vskip0.1cm
{\small
\begin{tabular}{|c|ccccc|}
\hline\hline
& $l=0$ & $l=1$ & $l=2$ & $l=3$ & $l=4$ \\
\hline
$n=7$ & \rule{1.75cm}{0.02in} & \rule{1.75cm}{0.02in} &
\rule{1.75cm}{0.02in} & \rule{1.75cm}{0.02in} & \rule{1.75cm}{0.02in}\\
$n=6$ & \rule{1.75cm}{0.01in} & \rule{1.75cm}{0.01in} &
\rule{1.75cm}{0.01in} & \rule{1.75cm}{0.01in} & \rule{1.75cm}{0.01in}\\
$n=5$ & \rule{1.75cm}{0.01in} & \rule{1.75cm}{0.01in} &
\rule{1.75cm}{0.01in} & \rule{1.75cm}{0.01in} & \rule{1.75cm}{0.01in}\\
$n=4$ & \underline{~~~$n=4$~~~} & \underline{~~~$n=4$~~~} &
\underline{~~~$n=4$~~~} & \underline{~~~$n=4$~~~} & \underline{~~~$n=4$~~~}\\
&                         &               && 7 fold      & 9 fold \\
&                         &               && degenerate  & degenerate\\
$n=3$ & \underline{~~~$n=3$~~~} & \underline{~~~$n=3$~~~} &
\underline{~~~$n=3$~~~} & &\\
&&&5 fold&&\\
&&& degenerate&&\\
$n=2$ & \underline{~~~$n=2$~~~} & \underline{~~~$n=2$~~~} &&&\\
&                         & 3 fold &&&\\
&                         & degenerate&&&\\
$n=1$  & \underline{~~~$n=1$~~~} &&&&\\
& non-degenerate &&&&\\
\hline\hline
\end{tabular}}

Putting all the levels which have the same energy together we get the following schematic representation of energy levels of $H$ atom. This table also shows that the allowed values of $l$ for each $n$, and number of $m$ values for each level. The number of total $m$ values, with the same energy, is $n^2$ and the degeneracy, after taking spin into account, becomes $2n^2$.

\begin{tabular}{ccccc}
%       \cline{3-5}
&& $l$ values  &number of $m$ values & degeneracy \\
\cline{3-5} &&&&\\
&& $0,1,.. n-1$ & $n^2$     & $2n^2$ \\[2mm]
\cline{3-5}

& &&& \\
$n=4$   & \underline{~~~~~32~~~~~} & $l=0,1,2,3$ &
$\sum(2l+1)=1+3+5+7=16$ & $2\times 16=32$\\
$n=3$   & \underline{~~~~~18~~~~~} & $l=0,1,2$ &
$\sum2(l+1)=1+3+5=9$ & $2\times 9=18$\\
$n=2$   & \underline{~~~~~~8~~~~~} & $l=0,1$ &
$\sum(2l+1)=1+3=4$ & $2\times 4 =8$\\
&      &&&\\
&      &&&\\
&      &&&\\
$n=1$   & \underline{~~degen=2~~} & $l=0$ &
$\sum(2l+1)=1$ & $2\times 1 =2$\\
\cline{3-5}
\end{tabular}\\

Accidental degeneracy
Comparing the hydrogen atom levels with those of a general spherically symmetric potential, we find that energies for states with several different  values of $l~ (=0,1,2 \dots n-1)$  are the same. For a general spherically symmetric potential different combinations of $n,l$ values correspond to different bound state energies, and are $(2l+1)$ fold degenerate. Thus there is an extra degeneracy is present for H atom beyond the expected $(2l+1)$ fold degeneracy  this phenomenon present in the case of hydrogen atom is known as {\it accidental degeneracy.} Another well known case of accidental degeneracy is that of isotropic harmonic oscillator ( $V(r)= {1\over 2} kr^2$) in three dimensions.

Remarks
It must be emphasized that the accidental degeneracy is  due to the special symmetry of the Coulomb problem.

Any slight deviation of the potential from ${1\over r}$   will result in splitting of energy levels with different  values of $l$.

It is known that the accidental degeneracy is present whenever the Schr\"odinger equation $H\psi=E\psi$ can be separated into ordinary differential equations in more than one set of coordinate system. For  $H$ atom --- Separation of variables for the Coulomb problem is possible in
(i) spherical polar coordinates $r\theta,\phi$
(ii)  parabolic coordinates $\xi, \eta, \phi$
\begin{eqnarray}  \xi = r-z &=& r(1-\cos\theta)\\         \eta = r+z &=& r(1+\cos\theta)\qquad\qquad \mbox{(Schiff)}\\   \phi &=& \phi         \end{eqnarray}
For isotropic harmonic oscillator, also $V(r) ={1\over 2} k r^2$, the Schrodinger equation can be separated in two set of variables:
(i) Cartesian Coordinates (ii) Spherical polar coordinates.
The isotropic oscillator also exhibits accidental degeneracy.

In these known cases, the accidental degeneracy is hardly accidental. It can be understood in terms of extra symmetries present in the problem.

\

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