5.2 Solving the Hydrogen Atom

To find the eigenfunctions of the hydrogen Hamiltonian, we will have to solve all three equations 5.10, 5.13, and 5.14, find the possible values of the separation constants, and calculate the normalisation coefficients. Let’s do the simplest one, the azimuthal equation, first.

5.2.1 The Azimuthal Equation

Equation 5.14 is the simple harmonic motion equation, and its solutions are

\Phi(\phi)=Ce^{\pm im\phi}.

(5.15)

For this wavefunction to be normalised, we must have

\int_{0}^{2\pi}\lvert\Phi(\phi)\rvert^{2}\operatorname{\mathrm{d}}\!\phi=\int_% {0}^{2\pi}\lvert C\rvert^{2}=1,

(5.16)

which implies

C=\frac{1}{\sqrt{2\pi}}.

(5.17)

What are the possible values of $m$ ? Note that $\Phi(\phi)$ must be single-valued, and $\phi$ is a periodic variable, which enforces the following constraint for any value of $\phi$ :

	$\displaystyle\Phi(\phi)$	$\displaystyle=\Phi(\phi+2\pi)$		(5.18)
	$\displaystyle\frac{1}{\sqrt{2\pi}}e^{im\phi}$	$\displaystyle=\frac{1}{\sqrt{2\pi}}e^{im(\phi+2\pi)}=\frac{1}{\sqrt{2\pi}}e^{% im\phi}e^{2\pi im}.$		(5.19)

This implies that $e^{2\pi im}$ must be $1$ , and therefore $m$ must be an integer. The condition that $\Phi(\phi)$ must be periodic is our boundary condition that we are using to constrain $m$ .

$m$ is called the magnetic quantum number. We will come back to why it is called this later.

Finally, note that since $\lvert e^{im\phi}\rvert^{2}=1$ for all values of $\phi$ , the final probability density $\lvert\psi\rvert^{2}$ will be independent of $\phi$ .

5.2.2 The Polar Equation

To solve equation 5.13, we will first make the substitution $u=\cos\theta$ along with

	$\displaystyle\frac{\operatorname{\mathrm{d}}\!u}{\operatorname{\mathrm{d}}\!\theta}$	$\displaystyle=-\sin\theta$		(5.20)
	$\displaystyle\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!\theta}$	$\displaystyle=\frac{\operatorname{\mathrm{d}}\!u}{\operatorname{\mathrm{d}}\!% \theta}\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!u}=-\sin% \theta\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!u}=-\sqrt{% 1-u^{2}}\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!u},$		(5.21)

where in the last equality we have used the identity $\sin\theta=\sqrt{1-\cos^{2}\theta}$ over the range $0\leq\theta\leq\pi$ . With these substitutions,

\sin\theta\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!\theta% }=\sqrt{1-u^{2}}\left(-\sqrt{1-u^{2}}\frac{\operatorname{\mathrm{d}}\!}{% \operatorname{\mathrm{d}}\!u}\right)=(u^{2}-1)\frac{\operatorname{\mathrm{d}}% \!}{\operatorname{\mathrm{d}}\!u},

(5.22)

so equation 5.13 becomes

	$\displaystyle(u^{2}-1)\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm% {d}}\!u}\left((u^{2}-1)\frac{\operatorname{\mathrm{d}}\!\Theta}{\operatorname{% \mathrm{d}}\!u}\right)=(m^{2}-\lambda^{2}(1-u^{2}))\Theta$		(5.23)
	$\displaystyle\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!u}% \left((1-u^{2})\frac{\operatorname{\mathrm{d}}\!\Theta}{\operatorname{\mathrm{% d}}\!u}\right)=\left(\frac{m^{2}}{1-u^{2}}-\lambda^{2}\right)\Theta,$		(5.24)

where in the last line we have divided by $-(u^{2}-1)$ .

This is a well-known differential equation called the general Legendre equation. Its solutions are the associated Legendre polynomials $P_{\ell}^{m}(u)$ , where $\ell$ is a non-negative integer related to the separation constant $\lambda^{2}$ by

\lambda^{2}=\ell(\ell+1).

(5.25)

This can be derived by a power series solution.

The form of the associated Legendre polynomials can be given in terms of the (regular) Legendre polynomials $P_{\ell}(u)$ , which are solutions of the (regular) Legendre equation, which is the general Legendre equation with $m=0$ . The Legendre polynomials can be defined by Rodrigues’ formula:

P_{\ell}(u)=\frac{1}{2^{\ell}\ell!}\frac{\operatorname{\mathrm{d}}\!^{\ell}}{% \operatorname{\mathrm{d}}\!u^{\ell}}[(u^{2}-1)^{\ell}].

(5.26)

From this, the associated Legendre polynomials are given by

	$\displaystyle P_{\ell}^{m}(u)$	$\displaystyle=(-1)^{m}(1-u^{2})^{\frac{m}{2}}\frac{\operatorname{\mathrm{d}}\!% ^{m}P_{\ell}(u)}{\operatorname{\mathrm{d}}\!u^{m}}$		(5.27)
		$\displaystyle=\frac{(-1)^{m}}{2^{\ell}\ell!}(1-u^{2})^{\frac{m}{2}}\frac{% \operatorname{\mathrm{d}}\!^{\ell+m}}{\operatorname{\mathrm{d}}\!u^{\ell+m}}[(% u^{2}-1)^{\ell}].$		(5.28)

This formula implies that we have the constraint

0\leq\lvert m\rvert\leq\ell,

(5.29)

which is also something that falls out from the series solution.

$\ell$ is called the orbital angular momentum quantum number, and again we will discuss its name later.

5.2.3 Spherical Harmonics

Taking the solutions to the azimuthal and polar equations together, the solutions to the whole angular part of the TISE for the hydrogen atom (equation 5.11) are

Y_{\ell}^{m}(\theta,\phi)=Ne^{im\phi}P_{\ell}^{m}(\cos\theta),

(5.30)

where $N$ is a normalisation constant such that

\int_{0}^{2\pi}\int_{0}^{\pi}\lvert Y_{\ell}^{m}(\theta,\phi)\rvert^{2}\sin% \theta\operatorname{\mathrm{d}}\!\theta\operatorname{\mathrm{d}}\!\phi=1.

(5.31)

$N$ turns out to be given by

N=\sqrt{\frac{(2\ell+1)(\ell-m)!}{4\pi(\ell+m)!}}.

(5.32)

The allowed values of the orbital angular momentum and magnetic quantum numbers are

	$\displaystyle\ell$	$\displaystyle=0,1,2,3,\dots$		(5.33)
	$\displaystyle m$	$\displaystyle=-\ell,-\ell+1,\dots,-1,0,1,\dots,\ell-1,\ell.$		(5.34)

Since the angular equation 5.11 did not depend on the potential at all, the angular part of the TISE would be the same for any problem where the potential energy only depended on radial distance. For this reason, the functions $Y_{\ell}^{m}(\theta,\phi)$ are quite important and they appear in any problem with spherical symmetry. They are known as the spherical harmonics and are to the surface of a sphere what sine and cosine are to a circle, i.e. they form a complete and orthonormal basis for functions defined on the surface of the unit sphere.

The first few (normalised) spherical harmonics are given by

$\displaystyle Y_{0}^{0}$	$\displaystyle=\frac{1}{\sqrt{4\pi}}$	(5.35)
$\displaystyle Y_{1}^{0}$	$\displaystyle=\sqrt{\frac{3}{4\pi}}\cos\theta,\quad Y_{1}^{1}=-\sqrt{\frac{3}{% 8\pi}}\sin\theta e^{i\phi}$	(5.36)
$\displaystyle Y_{2}^{0}$	$\displaystyle=\sqrt{\frac{5}{16\pi}}(3\cos^{2}\theta-1),\quad Y_{2}^{1}=-\sqrt% {\frac{15}{8\pi}}\sin\theta\cos\theta e^{i\phi},\quad Y_{2}^{2}=\sqrt{\frac{15% }{32\pi}}\sin^{2}\theta e^{2i\phi},$	(5.37)

the spherical harmonics with negative $m$ are given by

Y_{\ell}^{-m}(\theta,\phi)=(-1)^{m}(Y_{\ell}^{m}(\theta,\phi))^{\ast}.

(5.38)

The magnitude squared of the spherical harmonics gives the probability distribution of finding the electron in a certain direction from the proton. If we plot the magnitude squared, looking at the distance of the surface from the origin in a given direction gives us an idea of the likelihood of finding an electron in that direction (the larger the magnitude squared, the higher the probability). As we mentioned before, since the azimuthal part of the spherical harmonics is always just a complex exponential, we have that $\lvert Y_{\ell}^{m}\rvert^{2}$ does not depend on $\phi$ , which means that all the spherical harmonics are rotationally symmetric about the $z$ -axis. Furthermore, we have that

\lvert Y_{\ell}^{m}\rvert^{2}=\lvert Y_{\ell}^{-m}\rvert^{2},

(5.39)

we will come back to the implications of this later.

The state with $\ell=m=0$ is the only one that is spherically symmetric, the electron is equally likely to be found in any direction. For $\ell>0$ we see that when $m=0$ the electron is most likely to be found on the $z$ -axis, whereas for $\lvert m\rvert=\ell$ it is most likely to be found in the $x$ - $y$ plane.

5.2.4 The Radial Equation

Finally we turn to solving the radial equation 5.10. Substituting the separation constant that we found earlier $\lambda^{2}=\ell(\ell+1)$ , we have

\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!r}\left(r^{2}% \frac{\operatorname{\mathrm{d}}\!R(r)}{\operatorname{\mathrm{d}}\!r}\right)+% \frac{2\mu r^{2}}{\hbar^{2}}\left(\frac{e^{2}}{4\pi\varepsilon_{0}r}+E\right)R% (r)=\ell(\ell+1)R(r),

(5.40)

The solutions can then be derived explicitly using a series solution to be

R_{n\ell}(r)=\sum_{p=0}^{n-\ell-1}c_{p}\left(\frac{r}{a_{0}}\right)^{p+\ell}e^% {-\frac{r}{na_{0}}},

(5.41)

where the $c_{p}$ ’s are coefficients, $n$ is a positive integer, and $a_{0}$ is the Bohr radius, given by

a_{0}=\frac{4\pi\varepsilon_{0}\hbar^{2}}{\mu e^{2}}.

(5.42)

Technically this is the reduced Bohr radius as the official definition has the electron mass instead of the reduced mass, but the difference is incredibly small since the electron is so much less massive than the proton.

The series solution imposes another constraint that

\ell\leq n-1.

(5.43)

We find that the energy eigenvalues only depend on $n$ , and are given by

E_{n}=-\frac{\hbar^{2}}{2\mu a_{0}^{2}n^{2}}=-\frac{R_{H}}{n^{2}},

(5.44)

where we have defined the Rydberg constant $R_{H}=\frac{\hbar^{2}}{2\mu a_{0}^{2}}$ . Because $E_{n}$ depends only on the quantum number $n$ , it is called the principal quantum number.

These radial parts of the wavefunction should be normalised using the condition

\int_{0}^{\infty}\lvert R_{n\ell}\rvert^{2}r^{2}\operatorname{\mathrm{d}}\!r=1,

(5.45)

and the first few eigenfunctions are:

	$\displaystyle R_{10}(r)$	$\displaystyle=N_{10}e^{-\frac{r}{a_{0}}}$		(5.46)
	$\displaystyle R_{20}(r)$	$\displaystyle=N_{20}\left(1-\frac{r}{2a_{0}}\right)e^{-\frac{r}{2a_{0}}},\quad R% _{21}(r)=N_{21}\frac{r}{a_{0}}e^{-\frac{r}{2a_{0}}},$		(5.47)

where $N_{n\ell}$ are the normalisation constants.

Multiplying the radial part with the spherical harmonics found earlier, we finally get the whole spatial part of the energy eigenstates as

u_{n\ell m}(r,\theta,\phi)=R_{n\ell}(r)Y_{\ell}^{m}(\theta,\phi).

(5.48)

The spherical harmonics have no units, so since the magnitude squared of the whole spatial part should have units of ${\mathrm{m}}^{-3}$ (or “probability per unit volume”) the magnitude squared of the radial part $\lvert R_{n\ell}\rvert^{2}$ also has units of ${\mathrm{m}}^{-3}$ . This means that the radial part $R_{n\ell}$ itself has units of ${\mathrm{m}}^{-\frac{3}{2}}$ .

5.2.5 Radial Probability Density

The radial part of the wavefunction gives information about the likelihood of finding the electron a certain distance from the proton. Let’s calculate the probability of finding the electron in a spherical shell between $r_{1}$ and $r_{2}$ .

$\displaystyle P(r_{1}<r<r_{2})$	$\displaystyle=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{r_{1}}^{r_{2}}\lvert u(r,% \theta,\phi)\rvert^{2}r^{2}\sin\theta\operatorname{\mathrm{d}}\!r\operatorname% {\mathrm{d}}\!\theta\operatorname{\mathrm{d}}\!\phi$	(5.49)
	$\displaystyle=\int_{0}^{2\pi}\int_{0}^{\pi}\lvert Y_{\ell}^{m}(\theta,\phi)% \rvert^{2}\sin\theta\operatorname{\mathrm{d}}\!\theta\operatorname{\mathrm{d}}% \!\phi\int_{r_{1}}^{r_{2}}\lvert R_{n\ell}(r)\rvert^{2}r^{2}\operatorname{% \mathrm{d}}\!r$	(5.50)
	$\displaystyle=\int_{r_{1}}^{r_{2}}\lvert R_{n\ell}(r)\rvert^{2}r^{2}% \operatorname{\mathrm{d}}\!r,$	(5.51)

where in the last line we have used the fact that the spherical harmonics are normalised. The integrand $\lvert R_{n\ell}\rvert^{2}r^{2}$ has units of ${\mathrm{m}}^{-1}$ or “probability per radial metre”, and so we must interpret this quantity as the radial probability density, not the radial part of the wavefunction. This means that the most likely radial distance to find the electron from the proton is given by the maximum of $\lvert R_{n\ell}\rvert^{2}r^{2}$ , not $\lvert R_{n\ell}\rvert^{2}$ .

Example 5.1.

What is the most likely distance to find the electron in the ground state $R_{10}$ ?

The radial part of the wavefunction is given by $R_{10}(r)=N_{10}e^{-\frac{r}{a_{0}}}$ , so the most likely distance is given by the maximum of

\lvert R_{10}(r)\rvert^{2}r^{2}=\lvert N_{10}\rvert^{2}e^{-\frac{2r}{a_{0}}}r^% {2}.

(5.52)

We can find this by differentiating and setting the derivative to zero:

	$\displaystyle\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!r}% \left(\lvert N_{10}\rvert^{2}e^{-\frac{2r}{a_{0}}}r^{2}\right)=2\lvert N_{10}% \rvert^{2}re^{-\frac{2r}{a_{0}}}-\frac{2\lvert N_{10}\rvert^{2}}{a_{0}}r^{2}e^% {-\frac{2r}{a_{0}}}$	$\displaystyle=0$		(5.53)
	$\displaystyle\implies 2r\left(1-\frac{r}{a_{0}}\right)$	$\displaystyle=0.$		(5.54)

One solution to this equation is $r=0$ which is a minimum, and the other solution is $r=a_{0}$ , which must be the maximum.

This justifies the common statement that the “size” of a hydrogen atom is $a_{0}$ .

In general, the radial probability density has $n-\ell$ maxima. For a fixed $n$ , the radius of the first maximum increases with increasing $\ell$ . The maximum values also increase with increasing radius, which makes sense because there is more volume, so we are more likely to find the electron there.