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Categories: Physics, Quantum mechanics.

Hydrogen atom

The quantum-mechanical calculation of the hydrogen atom is, in my opinion, the single most important model in all of physics: miraculously, it is possible to find closed-form solutions for the wave function of an electron in a proton’s potential well. The results are highly educational, and also qualitatively tell us a lot about all other chemical elements.

We start from the time-independent Schrödinger equation, where $\mu$ is the reduced mass of the electron-proton system, and $V$ is the proton’s Coulomb potential:

$$\begin{aligned} E \psi = - \frac{\hbar^2}{2 \mu} \nabla^2 \psi + V \psi \end{aligned}$$

In spherical coordinates $(r, \theta, \varphi)$ it becomes as follows, where $V$ only depends on $r$:

$$\begin{aligned} E \psi = - \frac{\hbar^2}{2 \mu} \bigg( \pdvn{2}{\psi}{r} + \frac{2}{r} \pdv{\psi}{r} + \frac{1}{r^2} \pdvn{2}{\psi}{\theta} + \frac{1}{r^2 \tan{\theta}} \pdv{\psi}{\theta} + \frac{1}{r^2 \sin^2{\theta}} \pdvn{2}{\psi}{\varphi} \bigg) + V \psi \end{aligned}$$

We will use the method of separation of variables by making the following ansatz, such that the Schrödinger equation takes the form below:

$$\begin{aligned} \psi(r, \theta, \varphi) = R(r) \: Y(\theta, \varphi) \end{aligned}$$ $$\begin{aligned} E R Y &= - \frac{\hbar^2}{2 \mu} \bigg( R'' Y + \frac{2 R' Y}{r} + + \frac{R Y_{\theta\theta}}{r^2} + \frac{R Y_\theta}{r^2 \tan{\theta}} + \frac{R Y_{\varphi\varphi}}{r^2 \sin^2{\theta}} \bigg) + V R Y \end{aligned}$$

After multiplying by $- 2 \mu r^2 / (\hbar^2 R Y)$, each term depends on $r$ or $(\theta, \varphi)$, but not both:

$$\begin{aligned} 0 &= \bigg( r^2 \frac{R''}{R} + 2 r \frac{R'}{R} - \frac{2 \mu}{\hbar^2} r^2 V + \frac{2 \mu}{\hbar^2} r^2 E \bigg) + \bigg( \frac{Y_{\theta\theta}}{Y} + \frac{1}{\tan{\theta}} \frac{Y_\theta}{Y} + \frac{1}{\sin^2{\theta}} \frac{Y_{\varphi\varphi}}{Y} \bigg) \end{aligned}$$

Since these two groups are independent, this equation can only hold if there exists a separation constant $C$ such that:

$$\begin{aligned} C &= r^2 \frac{R''}{R} + 2 r \frac{R'}{R} - \frac{2 \mu}{\hbar^2} r^2 ( V - E ) \\ &= - \frac{Y_{\theta\theta}}{Y} - \frac{1}{\tan{\theta}} \frac{Y_\theta}{Y} - \frac{1}{\sin^2{\theta}} \frac{Y_{\varphi\varphi}}{Y} \end{aligned}$$

Now we have two simpler equations than the one we started with. We multiply them by $R$ and $Y$ respectively, and define $C \equiv \ell (\ell + 1)$ to help us later ($\ell$ is unknown for now). The results are the radial equation and the angular equation:

$$\begin{aligned} \boxed{ \begin{aligned} \ell (\ell + 1) R &= r^2 R'' + 2 r R' - \frac{2 \mu}{\hbar^2} r^2 (V - E) R \\ - \ell (\ell + 1) Y &= Y_{\theta\theta} + \frac{1}{\tan{\theta}} Y_\theta + \frac{1}{\sin^2\theta} Y_{\varphi\varphi} \end{aligned} } \end{aligned}$$

Note that this calculation has not really been specific to hydrogen so far: it is applicable to all spherically symmetric quantum systems.

Angular equation

Let us keep this generality, by keeping $V$ unspecified for now, In that case, the radial equation cannot be solved yet, but the angular one can. We separate the variables again:

$$\begin{aligned} Y(\theta, \varphi) = \Theta(\theta) \: \Phi(\varphi) \end{aligned}$$

Insert this into the equation and multiply by $\sin^2\theta / (\Theta \Phi)$ to get a clean separation:

$$\begin{aligned} \sin^2{\theta} \frac{\Theta''}{\Theta} + \sin{\theta} \cos{\theta} \frac{\Theta'}{\Theta} + \ell (\ell + 1) \sin^2{\theta} = - \frac{\Phi''}{\Phi} \end{aligned}$$

Each term depends on $\theta$ or $\varphi$ but not both, so there exists a constant $m^2$ such that:

$$\begin{aligned} m^2 &= \sin^2{\theta} \frac{\Theta''}{\Theta} + \sin{\theta} \cos{\theta} \frac{\Theta'}{\Theta} + \ell (\ell + 1) \sin^2{\theta} \\ &= - \frac{\Phi''}{\Phi} \end{aligned}$$

These are two distinct equations; multiplying by $\Theta$ and $\Phi$ respectively yields:

$$\begin{aligned} \boxed{ \begin{aligned} m^2 \Theta &= \sin^2{\theta} \:\Theta'' + \sin{\theta} \cos{\theta} \:\Theta' + \ell (\ell + 1) \sin^2{\theta} \:\Theta \\ - m^2 \Phi &= \Phi'' \end{aligned} } \end{aligned}$$

Clearly the latter is the simplest, so we start there. It is an eigenvalue problem for $m^2$, but it looks like a harmonic oscillator equation, so the solutions are easily found to be:

$$\begin{aligned} \boxed{ \Phi(\varphi) = e^{i m \varphi} } \end{aligned}$$

Because the coordinate $\varphi$ is only defined in the interval $[0, 2\pi]$, we demand periodic boundary conditions $\Phi(0) = \Phi(2 m \pi)$, which tells us that $m$ is an integer.

The other equation, for $\Theta$, needs a bit more work. We write it out like so:

$$\begin{aligned} 0 &= \dvn{2}{\Theta}{\theta} + \frac{\cos{\theta}}{\sin{\theta}} \dv{\Theta}{\theta} + \Big( \ell (\ell + 1) - \frac{m^2}{\sin^2{\theta}} \Big) \Theta \end{aligned}$$

And then perform a change of variables $\theta \to \xi$ where $\xi \equiv \cos{\theta}$, leading to:

$$\begin{aligned} 0 &= - \dv{}{\theta} \bigg( \sin{\theta} \dv{\Theta}{\xi} \bigg) - \cos{\theta} \dv{\Theta}{\xi} + \bigg( \ell (\ell + 1) - \frac{m^2}{\sin^2{\theta}} \bigg) \Theta \\ &= \sin^2{\theta} \dvn{2}{\Theta}{\xi} - 2 \cos{\theta} \dv{\Theta}{\xi} + \bigg( \ell (\ell + 1) - \frac{m^2}{\sin^2{\theta}} \bigg) \Theta \\ &= (1 - \xi^2) \dvn{2}{\Theta}{\xi} - 2 \xi \dv{\Theta}{\xi} + \bigg( \ell (\ell + 1) - \frac{m^2}{1 - \xi^2} \bigg) \Theta \end{aligned}$$

This result can be recognized as Legendre’s generalized equation, a known eigenvalue problem for $\ell (\ell + 1)$, which has solutions when $\ell$ is a non-negative integer. Those solutions are called the associated Legendre polynomials $P_\ell^m(x)$ of degree $\ell$ and order $m$. For a given $\ell$, there exist $2 \ell + 1$ such “polynomials” (they actually contain square roots too) indexed by the integer $m$ in the range $[-\ell, \ell]$, so e.g. for $\ell = 2$ there is $m = -2, -1, 0, 1, 2$. We now have:

$$\begin{aligned} Y_\ell^m(\theta, \varphi) \propto P_\ell^m(\cos{\theta}) \: e^{i m \varphi} \end{aligned}$$

We are still missing a constant factor, found by imposing the normalization condition:

$$\begin{aligned} \int_0^{2 \pi} \int_0^\pi |Y_\ell^m|^2 \sin{\theta} \dd{\theta} \dd{\varphi} = 1 \end{aligned}$$

Calculating the normalization constant (not shown here) leads to the following definition of the so-called spherical harmonics $Y_\ell^m$ of degree $\ell$ and order $m$:

$$\begin{aligned} \boxed{ Y_\ell^m(\theta, \varphi) = (-1)^m \sqrt{\frac{(2 \ell + 1) (\ell - m)!}{4 \pi (\ell + m)!}} \: P_\ell^m(\cos{\theta}) \: e^{i m \varphi} } \end{aligned}$$

These are important functions: the wave function of any spherically symmetric quantum system is a superposition of $Y_\ell^m$ with $r$-dependent coefficients. And, as befits a (component of a) wave function, they form an orthonormal basis, specifically:

$$\begin{aligned} \int_0^{2 \pi} \int_0^\pi Y_\ell^m \:Y_{\ell'}^{m'} \:\sin\theta \:d\theta \:d\varphi = \delta_{\ell\ell'} \delta_{mm'} \end{aligned}$$

Radial equation

With the angular part solved, we now turn to the radial part. Introducing $u(r) = r R(r)$, such that the derivatives of $R(r)$ become:

$$\begin{aligned} R' = \frac{r u' - u}{r^2} \qquad\qquad R'' = \frac{r^2 u'' - 2 r u' + 2 u}{r^3} \end{aligned}$$

Inserting this into the radial equation and cancelling some of the terms:

$$\begin{aligned} \ell (\ell + 1) \frac{u}{r} &= \frac{r^2 u'' - 2 r u' + 2 u}{r} + \frac{2 r u' - 2 u}{r} - \frac{2 \mu}{\hbar^2} (V - E) r u \\ &= r u'' - \frac{2 \mu}{\hbar^2} (V - E) r u \end{aligned}$$

After multiplying by $\hbar^2 / (2 \mu r)$ and rearranging, this turns into:

$$\begin{aligned} E u = - \frac{\hbar^2}{2 \mu} u'' + \bigg( V + \frac{\hbar^2}{2 \mu} \frac{\ell (\ell + 1)}{r^2} \bigg) u \end{aligned}$$

Here it is useful to define an effective potential $V_{\mathrm{eff}}(r)$ as below. Keep in mind that $\ell$ is known after solving the angular equation:

$$\begin{aligned} V_{\mathrm{eff}} \equiv V + \frac{\hbar^2}{2 \mu} \frac{\ell (\ell + 1)}{r^2} \end{aligned}$$

This yields a relation of the same form as the time-independent Schrödinger equation, just with $V$ replaced by $V_{\mathrm{eff}}$. This is the “true” radial equation, an eigenvalue problem for $E$:

$$\begin{aligned} \boxed{ E u = - \frac{\hbar^2}{2 \mu} u'' + V_{\mathrm{eff}} u } \end{aligned}$$

Now, finally, we specialize for the hydrogen atom. Coulomb’s law tells us the attractive force $F(r)$ between the electron and the proton, which we integrate to find the potential energy $V(r)$:

$$\begin{aligned} F(r) = \frac{q^2}{4 \pi \varepsilon_0 r^2} \qquad \implies \qquad V(r) = - \frac{q^2}{4 \pi \varepsilon_0 r} \end{aligned}$$

Where $q < 0$ is the electron’s charge, and $\varepsilon_0$ is the permittivity of free space. Note that $V < 0$, so there is a natural distinction between bound states $E < 0$ (where the electron is trapped in the proton’s well), and scattering states $E > 0$ (where the electron is free). The true radial equation, after dividing by $E$, is now given by:

$$\begin{aligned} u = - \frac{\hbar^2}{2 \mu E} u'' + \bigg( \frac{\hbar^2}{2 \mu E} \frac{\ell (\ell + 1)}{r^2} - \frac{\hbar^2 \mu}{\hbar^2 \mu} \frac{q^2}{4 \pi \varepsilon_0 r E} \bigg) u \end{aligned}$$

For brevity, let us introduce new constants $\kappa$ and $\rho_0$, defined as follows:

$$\begin{aligned} \kappa \equiv \frac{\sqrt{-2 \mu E}}{\hbar} \qquad\qquad \rho_0 \equiv \frac{\mu q^2}{2 \pi \varepsilon_0 \hbar^2 \kappa} \end{aligned}$$

Where $E < 0$, as we are interested in bound states. Now the radial equation has become:

$$\begin{aligned} 0 = \frac{1}{\kappa^2} u'' + \Big( \frac{\rho_0}{\kappa r} - \frac{\ell (\ell + 1)}{\kappa^2 r^2} - 1 \Big) u \end{aligned}$$

To clean this up further, we switch to the dimensionless variable $\rho \equiv \kappa r$, yielding:

$$\begin{aligned} 0 = u'' + \Big( \frac{\rho_0}{\rho} - \frac{\ell (\ell + 1)}{\rho^2} - 1 \Big) u \end{aligned}$$

We then choose the following ansatz for $u(\rho)$, where $v(2 \rho)$ is unknown:

$$\begin{aligned} u(\rho) &= w(2 \rho) \: \rho^{\ell + 1} \: e^{- \rho} \end{aligned}$$

For reference, we also calculate its first and second derivatives:

$$\begin{aligned} u'(\rho) &= \Big( 2 \rho w' + (\ell + 1 - \rho) w \Big) \rho^\ell \: e^{- \rho} \\ u''(\rho) &= \bigg( 4 \rho^2 w'' + 4 (\ell + 1 - \rho) \rho w' + \big( \rho^2 - 2 \rho (\ell + 1) + \ell (\ell + 1) \big) w \bigg) \rho^{\ell-1} \: e^{- \rho} \end{aligned}$$

Inserting this into the radial equation and dividing out all common factors gives:

$$\begin{aligned} 0 &= 4 \rho w'' + 4 (\ell + 1 - \rho) w' + \big( \rho_0 - 2 (\ell + 1) \big) w \end{aligned}$$

Let us rearrange this to put it in a more suggestive form. Keep in mind that $w = w(2 \rho)$:

$$\begin{aligned} 0 &= (2 \rho) w'' + \Big( (2 \ell + 1) + 1 - (2 \rho) \Big) w' + \Big( \frac{\rho_0}{2} - \ell - 1 \Big) w \end{aligned}$$

This can be recognized as Laguerre’s generalized equation, a well-known eigenvalue problem for $\lambda \equiv (\rho_0 / 2 \!-\! \ell \!-\! 1)$. It has solutions when $\lambda$ is a non-negative integer, in other words for $\rho_0 = 2n$ with $n = 1, 2, 3,...$, which also tells us that $\ell$ cannot be larger than $n - 1$. Then the solutions are the so-called associated Laguerre polynomials $L_{n - \ell - 1}^{2 \ell + 1}(2 \rho)$, therefore:

$$\begin{aligned} u(\rho) \propto \rho^{\ell + 1} \: e^{-\rho} \: L_{n - \ell - 1}^{2 \ell + 1}(2 \rho) \end{aligned}$$

We are still missing a constant factor, found by imposing the normalization condition:

$$\begin{aligned} \int_0^\infty R^2 \: r^2 \dd{r} = 1 \end{aligned}$$

Calculating the normalization constant (not shown here) leads to this radial solution $R_{n\ell}(r)$:

$$\begin{aligned} R_{n \ell}(r) = \sqrt{\frac{(n - \ell - 1)!}{2 n (n + \ell)!}} \: (2 \kappa)^{3/2} \: (2 \kappa r)^\ell \: e^{-\kappa r} \: L_{n - \ell - 1}^{2 \ell + 1}(2 \kappa r) \end{aligned}$$

Meanwhile, by isolating the definitions of $\kappa$ and $\rho_0$ for $E$, we find the eigenenergies to be:

$$\begin{aligned} E = - \frac{\hbar^2}{2 \mu} \kappa^2 = - \frac{\hbar^2}{2 \mu} \bigg( \frac{\mu q^2}{2 \pi \varepsilon_0 \hbar^2 \rho_0} \bigg)^2 \end{aligned}$$

Since $\rho_0 = 2 n$, these allowed Bohr energies $E_n$ of the electron are as follows:

$$\begin{aligned} \boxed{ E_n = - \frac{1}{n^2} \frac{\mu}{2 \hbar^2} \bigg( \frac{q^2}{4 \pi \varepsilon_0} \bigg)^2 } \end{aligned}$$

At this point, it is customary to also define the reduced Bohr radius $a_0^*$, given by:

$$\begin{aligned} \boxed{ a_0^* \equiv \frac{1}{n \kappa} = \frac{4 \pi \varepsilon_0 \hbar^2}{\mu q^2} } \approx 5.295 \times 10^{-11} \:\mathrm{m} = 0.5295 \:\mathrm{\AA} \end{aligned}$$

The non-reduced Bohr radius $a_0$ simply uses the electron’s raw mass $m_e$ instead of $\mu$. Roughly speaking, $a_0^*$ is the most probable electron-proton distance after a measurement of the electron’s position while it is in its ground state. This is often to used to write $R_{n \ell}(r)$ as:

$$\begin{aligned} \boxed{ R_{n \ell}(r) = \sqrt{\frac{(n - \ell - 1)!}{2 n (n + \ell)!}} \bigg( \frac{2}{n a_0^*} \bigg)^{3/2} \bigg( \frac{2 r}{n a_0^*} \bigg)^\ell e^{- r / n a_0^*} \: L_{n - \ell - 1}^{2 \ell + 1}\Big(\frac{2 r}{n a_0^*}\Big) } \end{aligned}$$

Quantum numbers

Multiplying the angular and radial parts together, we thus arrive at the following expression for the full wave function $\psi_{n \ell m}$:

$$\begin{aligned} \boxed{ \psi_{n \ell m}(r, \theta, \varphi) = R_{n \ell}(r) \: Y_\ell^m(\theta, \varphi) } \end{aligned}$$

The indices $n$, $\ell$, and $m$ are the quantum numbers, which describe the state of the electron. There is also a fourth not shown here, the spin quantum number, which is $+1/2$ or $-1/2$ for spin-up or spin-down electrons respectively.

The principal quantum number $n$, often called the shell number, gives the energy level (shell) of the electron, because the other numbers do not appear in $E_n$’s formula. Since $E_n = E_1 / n^2$, the energy differences decrease with increasing $n$, so electrons in higher shells can be excited more easily (i.e. they need less energy to get excited).

The azimuthal quantum number $\ell$ gives the subshell of shell $n$ in which the electron is located. It takes integer values from $0$ to $n - 1$ inclusive, with $0$, $1$, $2$, and $3$ respectively also called the $s$, $p$, $d$, and $f$ subshells. The electron’s total angular momentum is given by $\hbar \sqrt{\ell (\ell + 1)}$.

The magnetic quantum number $m$ splits the electrons in each subshell into orbitals, and takes integer values from $-\ell$ to $\ell$. The $z$-component of the electron’s angular momentum is $\hbar m$.

The total degeneracy of each energy level $n$ can be calculated as the sum of an arithmetic series, and is found to be $n^2$ excluding spin (or $2 n^2$ with spin).

Unsurprisingly, all these wave functions form an orthonormal basis (although not a complete one unless the scattering states with $E > 0$ are included):

$$\begin{aligned} \int_0^{2 \pi} \int_0^\pi \int_0^\infty \psi_{n \ell m}^* \: \psi_{n' \ell' m'} \: r^2 \sin{\theta} \dd{r} \dd{\theta} \dd{\varphi} = \delta_{nn'} \delta_{\ell\ell'} \delta_{mm'} \end{aligned}$$

When an excited electron drops from a state with energy $E_i$ to a lower level $E_f$, it emits a photon with energy $\hbar \omega$, where $\omega$ is the angular frequency of the resulting electromagnetic wave:

$$\begin{aligned} \hbar \omega = E_i - E_f = E_1 \bigg( \frac{1}{n_i^2} - \frac{1}{n_f^2} \bigg) \end{aligned}$$

The corresponding vacuum wavelength is $\lambda_0 = 2 \pi c / \omega$, leading to the Rydberg formula, which was discovered empirically before the hydrogen atom had been solved:

$$\begin{aligned} \boxed{ \frac{1}{\lambda_0} = \frac{\omega}{2 \pi c} = R_\mathrm{H} \bigg( \frac{1}{n_i^2} - \frac{1}{n_f^2} \bigg) } \end{aligned}$$

Quantum mechanics then successfully gave a theoretical value to the experimentally determined Rydberg constant $R_\mathrm{H}$ (or $R_\infty$ if the raw electron mass $m_e$ is used):

$$\begin{aligned} \boxed{ R_\mathrm{H} = \frac{|E_1|}{2 \pi \hbar c} = \frac{\mu}{4 \pi \hbar^3 c} \bigg( \frac{q^2}{4 \pi \varepsilon_0} \bigg)^2 } \approx 1.097 \times 10^7 \:\mathrm{m}^{-1} \end{aligned}$$

The transitions from excited states to the ground state $n_f = 1$ correspond to ultraviolet spectral lines known as the Lyman series. Similarly, transitions to $n_f = 2$ give visible lines known as the Balmer series, and transitions to $n_f = 3$ explain the Paschen series of infrared lines.

The Rydberg constant is not to be confused with the Rydberg energy $\mathrm{Ry}$, which is the ionization energy of ground-state hydrogen, and is sometimes used as a unit in calculations:

$$\begin{aligned} \mathrm{Ry} \equiv 2 \pi \hbar c R_\mathrm{H} = |E_1| = \frac{\mu}{2 \hbar^2} \bigg( \frac{q^2}{4 \pi \varepsilon_0} \bigg)^2 \approx 13.61 \:\mathrm{eV} \end{aligned}$$

The point is that the hydrogen atom’s solution gave clear explanations for known experimental data, and settled the mystery of what an atom actually looks like. While other elements’ atoms generally do not have such closed-form solutions (because they have more than one electron), their orbitals are qualitatively very similar. In short, this model is the foundation of our modern understanding of atoms.

References

1. D.J. Griffiths, D.F. Schroeter, Introduction to quantum mechanics, 3rd edition, Cambridge.
2. R. Shankar, Principles of quantum mechanics, 2nd edition, Springer.

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