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Next: Quantum numbers Up: Quantum Mechanics of Atoms Previous: A full Quantum Mechanical

Solving the Schrödinger equation for hydrogen-like atoms

Consider the Schrödinger (time independent) Wave Equation
\begin{displaymath}
-\frac{\hbar ^2}{2m}\nabla ^2 \psi + U\psi = E\psi
\end{displaymath} (1)

which is expanded as
\begin{displaymath}
\left( \frac{\partial ^2}{\partial x^2} +
\frac{\partial ^2}...
...^2}{\partial z^2}
\right)\psi + \frac{2m}{\hbar^2}(E-U)\psi =0
\end{displaymath} (2)

in Cartesian co-ordinates.

When applied to the hydrogen atom, the wave function should describe the behaviour of both the nucleus and the electron, $\psi \Rightarrow \psi ({\bf r}_n,{\bf r}_e)$. This means we have a two body problem, which is very difficult to solve. We can fortunately convert this two-body problem to an effective one-body problem by transforming from the Laboratory System to the Centre of Mass System.

\begin{displaymath}
({\bf r}_n,{\bf r}_e) \Rightarrow ({\bf R},{\bf r}).
\end{displaymath} (3)

In this system, the wave function is separable, $\psi \Rightarrow \psi ({\bf R})\psi ({\bf r})$. (Can you think why ?) In addition, we note that the CofM motion is constant in the absence of external forces. Thus we need only solve the wave equation for the behaviour of $\psi({\bf r})$. So for studying hydrogen-like atoms themselves, we need only consider the relative motion of the electron with respect to the nucleus. Note that in this case the appropriate mass to use in the wave equation will be the reduced mass of the electron, $m_e \Rightarrow m_r = m_n m_e/(m_n+m_e)$. Here ${\bf R},{\bf r}$ are the position vectors to the CofM and the electron w.r.t. the nucleus respectively and the meaning of the other symbols are obvious.

The appropriate potential is of course the simple radial electrostatic potential for a point charge nucleus of charge $Ze$. (Note that if we are not dealing with hydrogen, then we are dealing with hydrogen-like atoms which are fully stripped of all but one electron.)

\begin{displaymath}
U(x,y,x) = U({\bf r}) = U(r) =-\frac{Ze^2}{4\pi \epsilon_0 r}
\end{displaymath} (4)

has spherical symmetry. One could write $r=\sqrt{x^2+y^2+z^2}$ and solve the wave equation in Cartesian co-ordinates. This would work but it would be very tedious, as the mathematics does not display the symmetry of the physics. Accordingly we rather exploit the spherical symmetry of the potential, and perform a co-ordinate transformation from Cartesian Co-ordinates (efficient for rectangle shapes) to Spherical Polar Co-ordinates (efficient for spherical shapes)
\begin{displaymath}
(x,y,z) \longrightarrow (r,\theta,\phi).
\end{displaymath} (5)

These new co-ordinates are defined in figure 1. Under this co-ordinate transformation, the wave equation takes the form,
\begin{displaymath}
\frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2\frac{\par...
...rtial^2\psi}{\partial \phi^2}
+ \frac{2m}{\hbar^2}(E-U)\psi =0
\end{displaymath} (6)

Exercise 1 Writing the double differential operator in spherical polar co-ordinates is not trivial. Check Electromagnetic Fields and Waves by Lorrain and Carson, Chapter 1, if you require insight into the last step. You are not expected to be able to do this transformation. However, make sure, using a sketch, that you can show how an infinitesimal volume element behaves under this transformation.

\begin{displaymath}
dv = dx\,dy\,dz = r^2 \sin\theta\,dr\,d\theta\,d\phi = d^3r
\end{displaymath} (7)

Figure 1: Cartesian and Spherical Polar co-ordinates
\includegraphics[width=0.5\textwidth]{Sph_Polar.eps}

Exercise 2
Write each of the variables $(r,\theta,\phi)$ in terms of the variables $(x,y,z)$, also perform the inverse mapping. You may use figure 1. Note that the potential is radial, $U({\bf r})=U(r,\theta,\phi)=U(r)$, which means it depends only on $r$, and not on $\theta$ or $\phi$.

The wave function necessarily is separable into radial, polar and azimuthal factors under a radial potential as follows : $\psi (r,\theta ,\phi) = R(r)\Theta(\theta)\Phi(\phi)$. (Once again, can you think why ?) Substituting the above expression and the potential into the spherical polar representation of the wave equation, we find, after some manipulation,

\begin{displaymath}
\frac{\sin^2 \theta}{R}\frac{\partial}{\partial r}\left( {r^...
...E\right)
=-\frac{1}{\Phi}\frac{\partial^2\Phi}{\partial\phi^2}
\end{displaymath} (8)

Exercise 3
Ensure you can achieve the last result with your own pencil and paper. Doing these exercises and tutorials properly helps to make you familiar with quantum mechanics, via your fingertips, into the marrow of your bones. You may use your textbooks the first time you do this.

This equation has on the left functions of $(r,\theta)$ only, and on the right a function of $(\phi)$ only. Accordingly, it can only be satisfied for all values of the independent variables by requiring that both sides are equal to the same constant value. Hence it follows the the left hand side, which is called the azimuthal equation, is equal to a constant, which is called the azimuthal constant. We give the azimuthal constant the symbol $m_l^2$, for reasons which will become clear with hindsight.

\begin{displaymath}
-\frac{1}{\Phi}\frac{d^2\Phi}{d\phi^2} = \mbox{const(azimuthal)} = m_l^2
\end{displaymath} (9)

The solution for the azimuthal equation (by the ansatz method) is

\begin{displaymath}
\Phi_{m_l}(\phi)=Ae^{im_l\phi}
\end{displaymath} (10)

Exercise 4
Check this with your own pencil and paper....

Imposing the boundary conditions, we must have $\Phi(\phi)$ single valued, and $\Phi(\phi) = \Phi(\phi+2\pi)$. Therefore, we require the constant $m_l$ to be quantised as follows.

\begin{displaymath}
m_l = 0, \pm 1, \pm 2, \pm 3, \cdots
\end{displaymath} (11)

We call $m_l$ the magnetic quantum number. You may imagine that the quantity $\frac{d\Phi}{d\phi}$ measures the change of the azimuthal part of the wave function with change of the azimuthal co-ordinate. Thus this quantity can sense differences under a rotation about the $z$-axis. Intuitively, if the wave function of the electron changed under such a rotation, one would be able to discern it, and classically, a rotating charge has a magnetic moment. This is a heuristic justification of the appellation of ``magnetic quantum number'' for $m_l$.

Clearly, not only is the separation constant quantised, but also the azimuthal wave function becomes part of a family of wave functions labelled by the quantum number $m_l$.

$\displaystyle m$ $\textstyle \longrightarrow$ $\displaystyle m_l$ (12)
$\displaystyle \Phi(\phi)$ $\textstyle \longrightarrow$ $\displaystyle \Phi_{m_l}(\phi)$  

Congratulations !

You have just seen that the quantisation of the magnetic quantum number arises naturally from the condition that the wave function must be single valued and satisfy its boundary conditions. The imposition of boundary conditions also lead to quantisation of the wave function in the previous examples we have seen (particle in a box). This is just a mathematical way of saying we have successfully trapped a wave function in an attractive potential. The value of the prefactor $A$ will be set later via a normalisation condition. Thus fortified, we proceed to the polar wave function. Substituting the constant, (known as a separation constant), $m_l^2$ back into the wave equation above and re-arranging terms
\begin{displaymath}
\frac{1}{R}\frac{\partial}{\partial r}\left( {r^2}\frac{\par...
...}
\left(\sin\theta\frac{\partial\Theta}{\partial\theta}\right)
\end{displaymath} (13)

Again, we have an equation which has on the left functions of $r$ only, and on the right functions of $\theta$ only. Accordingly, it can only be satisfied for all values of the independent variables by requiring that both sides are equal to the same constant value. Hence it follows,
$\displaystyle \frac{1}{R}\frac{d}{dr}\left( {r^2}\frac{dR}{dr}\right) +
\frac{2mr^2}{\hbar^2} \left(\frac{Ze^2}{4\pi \epsilon_0 r} + E\right)$ $\textstyle =$ $\displaystyle \mbox{const(polar)}$ (14)
$\displaystyle \frac{m_l^2}{\sin^2 \theta} - \frac{1}{\Theta \sin \theta} \frac{d}{d\theta}
\left(\sin\theta\frac{d\Theta}{d\theta}\right)$ $\textstyle =$ $\displaystyle \mbox{const(polar)}$  

We have achieved so far separated equations for the last two wave functions, viz. the radial wave equation for $R(r)$ and the polar wave equation for $\Theta(\theta)$. The solution of these two equations is beyond the scope of this course. Rest assured, it proceeds as in the case for the azimuthal wave function. That is, imposing the boundary conditions causes the separation constant to become quantised and also the radial wave function and the polar wave function to become part of a family labelled by the appropriate quantum number.

$\displaystyle R(r)$ $\textstyle \longrightarrow$ $\displaystyle R_{n,l}(r)$ (15)
$\displaystyle \Theta(\theta)$ $\textstyle \longrightarrow$ $\displaystyle \Theta_{l,m_l}(\theta)$  

(The angle dependent wave functions are known as the Spherical Harmonic Functions, and the radial wave functions as Laguerre polynomials. These solutions are tabulated below in figure 2.)

Figure 2: Normalised wave functions of the hydrogen atom for $n=1,2$ and 3.

Hence we find, on solving the polar wave equation for $\Theta(\theta)$ that

\begin{displaymath}
\mbox{const(polar)} = l(l+1) \mbox{\ \ where \ \ } l \ge \vert m_l\vert \mbox{ is an integer}
\end{displaymath} (16)

so that
\begin{displaymath}
m_l = 0, \pm 1, \pm 2, \pm 3, \cdots ,\pm l
\end{displaymath} (17)

We call $l$ the orbital quantum number. In a later section, we will evidence the relationship of this quantum number to orbital angular momentum. This is then the heuristic justification of its appellation.

Exercise 5
What do you think the boundary conditions are for the polar wave equation ?

In the introductory course on Quantum Mechanics, we saw that $E$ was also a separation constant. It arose when we separated the time and space parts of the Time dependent wave equation to arrive at the Time independent wave equation, which we have presented at the top of this section. So it will come as no surprise now to find that $E$ becomes quantised on requiring that the solutions of the radial wave equation above also obey boundary conditions. In the case of the radial wave equation, we obviously require that $R(r=0)$ is finite and $R(r=\infty) = 0$. We get ,

\begin{displaymath}
E_n = -\frac{mZ^2e^4}{32\pi^2\epsilon^2_0\hbar^2}
\left( \frac{1}{n^2} \right) = \frac{E_1}{n^2}, \ \ \ n=1,2,3,\cdots
\end{displaymath} (18)

with the ancillary condition that
\begin{displaymath}
n \ge l+1 \mbox{ or } l = 0, 1, 2, \cdots ,(n-1)
\end{displaymath} (19)

We call $n$ the principal quantum number.

Exercise 6
This is the same expression as in the Bohr model. Muse on what this correspondence with the Bohr model implies.

Exercise 7
How would these results change if we were dealing with positronium and not hydrogen ?

We may summarise our results so far :
The Schrödinger Wave Equation for hydrogen like atoms in three dimensions is best treated in spherical polar co-ordinates

\begin{displaymath}
(x,y,z) \longrightarrow (r,\theta,\phi)
\end{displaymath} (20)

because the Coulomb potential for this case
\begin{displaymath}
U(x,y,x) = U({\bf r}) = U(r) =-\frac{Ze^2}{4\pi \epsilon_0 r}
\end{displaymath} (21)

is spherically symmetric. Consequently, the wave function will be separable
\begin{displaymath}
\psi ({\bf r}) = R(r)\Theta(\theta)\Phi(\phi)
\end{displaymath} (22)

and the Schrödinger Wave Equation reduces to the three equations
\begin{displaymath}
\frac{d^2\Phi}{d\phi^2} - m_l^2 \Phi = 0 ,
\end{displaymath} (23)


\begin{displaymath}
\frac{\hbar^2}{2m}
\frac{1}{r^2}\frac{d}{dr}\left( {r^2}\fra...
..._0 } \frac{1}{r} -\frac{l(l+1)\hbar^2}{2mr^2} -E \right]R = 0
\end{displaymath} (24)

and
\begin{displaymath}
\frac{1}{\sin \theta} \frac{d}{d\theta}
\left(\sin\theta\fra...
...eft [
l(l+1) - \frac{m_l^2}{\sin^2 \theta} \right]\Theta = 0.
\end{displaymath} (25)

Solving these equations subject to the appropriate boundary conditions leads to the three sets of quantum numbers :

\begin{eqnarray*}
\mbox{ Principal Quantum Number \ \ } & n = 1,2,3, \cdots \\
...
...antum Number \ \ } & m_l = 0, \pm 1, \pm 2, \pm 3, \cdots ,\pm l
\end{eqnarray*}



In this process, we also find a family of wave functions, labelled by the quantum numbers :
\begin{displaymath}
\psi_{nlm_l} ({\bf r}) = R_{nl}(r)\Theta_{lm_l}(\theta)\Phi_{m_l}(\phi).
\end{displaymath} (26)

We can think of the set of quantum numbers $(n,l,m_l)$ as identifying a wave function for a particular state $\psi_{nlm_l} ({\bf r})$. It is typical that quantum numbers appear naturally when quantum particles are trapped in a particular region of space by an attractive potential. A selection of the lowest energy wave functions have been collected in the table above. These wave functions are normalised so that the probability density for finding an electron in a particular state $(n,l,m_l)$ represented by $\vert\psi_{n,l,m_l} (r,\theta ,\phi)\vert^2$ is unity when integrated over all space.
\begin{displaymath}
\int^{\infty}_0 \int^{\pi}_0 \int^{2\pi}_0 \vert\psi_{n,l,m_l} (r,\theta ,\phi)\vert^2 \,
dr.rd\theta .r\sin\theta d\phi = 1
\end{displaymath} (27)

Observation
Schrödinger sure deserved the Nobel Prize !

Example
Suppose we want to verify the energy of the ground state wave function of the hydrogen atom, $Z=1$ and $(n=1,l=0,m_l=0)$. We note that it is only the radial wave equation which contains $E$, the energy of the state. The appropriate radial wave function is

\begin{displaymath}
R(r) = \frac{2}{a_0^{3/2}}e^{-r/a_0}.
\end{displaymath} (28)

We can substitute this wave function into the radial wave equation
\begin{displaymath}
\frac{\hbar^2}{2m}
\frac{1}{r^2}\frac{d}{dr}\left( {r^2}\fra...
...2} +E \right]
\left(\frac{2}{a_0^{3/2}}e^{-r/a_0} \right) = 0.
\end{displaymath} (29)

Using $l=0$ and $E = E_1$, we simplify to find
\begin{displaymath}
\left [ \left( \frac{2}{a_0^{7/2}} + \frac{4mE_1}{\hbar^2a_0...
...\frac{4}{a_0^{5/2}} \right)\frac{1}{r}
\right]e^{-r/a_0} =0 .
\end{displaymath} (30)

Each parenthesis must equal nought for the entire equation to equal nought. We therefore find an expression for $a_0$, the Bohr radius,
\begin{displaymath}
a_0 = \frac{4\pi\epsilon_0\hbar}{me^2} = r_1
\end{displaymath} (31)

and the ground state energy
\begin{displaymath}
E_1 = -\frac{\hbar^2}{2ma_0^2} = -\frac{me^2}{32\pi^2\epsilon_0^2\hbar^2}
\end{displaymath} (32)


next up previous
Next: Quantum numbers Up: Quantum Mechanics of Atoms Previous: A full Quantum Mechanical
Simon Connell 2004-10-04