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diff --git a/source/know/concept/hydrogen-atom/index.md b/source/know/concept/hydrogen-atom/index.md new file mode 100644 index 0000000..a8443dd --- /dev/null +++ b/source/know/concept/hydrogen-atom/index.md @@ -0,0 +1,579 @@ +--- +title: "Hydrogen atom" +sort_title: "Hydrogen atom" +date: 2023-10-15 +categories: +- Physics +- Quantum mechanics +layout: "concept" +--- + +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](/know/concept/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](/know/concept/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](/know/concept/legendre-polynomials/), +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](/know/concept/laguerre-polynomials/), +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](/know/concept/electromagnetic-wave-equation/): + +$$\begin{aligned} + \hbar \omega + = E_i - E_f + = E_1 \bigg( \frac{1}{n_i^2} - \frac{1}{n_f^2} \bigg) +\end{aligned}$$ + +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: + +$$\begin{aligned} + \boxed{ + \frac{1}{\lambda_0} + = \frac{\omega}{2 \pi c} + = \mathcal{R}\Big( \frac{1}{n_i^2} - \frac{1}{n_f^2} \Big) + } +\end{aligned}$$ + +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: + +$$\begin{aligned} + \mathrm{Ry} + = 2 \pi \hbar c R_\mathrm{H} + = |E_1| + = \frac{\mu}{2 \hbar^2} \bigg( \frac{q^2}{4 \pi \varepsilon_0} \bigg)^2 + \approx 13.61 \:\mathrm{eV} +\end{aligned}$$ + +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. |