Hydrogen Atom - The final take?

The study of atoms is a mysterious one, especially when one begins to study this in high school, hearing the abstract concepts of wavefunctions and probability. In a first course on Quantum Mechanics one finally tackles this atom rigorously - as the most simple spherically symmetric system and also as a rare example to solve exactly. In this article we uncover some more secrets about this atom. High resolution spectrometers observed a fine splitting over the degeneracy one is familiar with from high school. What follows is a story about this more richer and broken energy structure.

*Hydrogen Atom under Magnification: Direct Observation of the Nodal Structure of Stark States.*

Let’s first recall some basic quantum mechanical ideas and methods behind understanding the hydrogen atom.

In the case of a spherically symmetric system like Hydrogen, with a single electron under the influence of proton field at the center viz. $V \equiv V(r)$, the schrodinger equation takes the special form, splitting the wavefunction into Radial and angular parts:

\[\psi\left( r,\theta,\phi \right) = R(r)Y\left( \theta, \phi \right).\]

We solve for these using the time-independent Schrodinger equation:

\[-\frac{\hbar^2}{2m}\nabla ^2\psi + V\psi = E\psi,\]

$\psi$ being the spatial wavefunction and $E$ the energy. The radial part $u(r) \equiv rR(r)$ obeys,

\[- \frac{\hbar^2}{2m} \frac{d^2u}{dr^2} + \left[ V + \frac{\hbar^2}{2m}\frac{l\left( l+1 \right)}{r^2}\right]u = Eu.\]

And the angular part is solved completely independent of the potential $V(r)$ by the spherical harmonics,

\[Y_l^m\left(\theta,\phi\right) = \sqrt{\frac{\left( 2l+1 \right) }{4\pi} \frac{\left( l-m \right)! }{\left( l+m \right)!}}e^{im\phi}P_l^m\left( \cos\theta \right).\]

$P_l^m$ is the associated Legendre function, $l = 0, 1, 2, \ldots$ appears as a separation constant, and $m = 0, \pm 1, \pm 2, \ldots$ appears due to the $\phi \to \phi + 2\pi$ symmetry.

Now diving into the hydrogen atom potential, $V = -\frac{e^2}{4\pi\epsilon_0}\frac{1}{r}$ and solving for the radial part, we find the allowed energy levels to be

\[E_n = - \left[\frac{m_e}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2\right]\cdot \frac{1}{n^2}, ~~~ n = 1,2,3,\ldots\]

with a ground state of $-13.6 eV.$

We find that for an arbitrary $n$ above, the possible values of $l$ include $0,1,\ldots,n-1$ and for each $l$ we have $(2l + 1)$ possible values for $m: -l,\ldots,0,\ldots,l$. This gives a degeneracy in each energy level $n$ to be,

\[deg.(n) = \sum_{l=0}^{n-1} (2l + 1) = n^2.\]

We can summarise the entirety of this Hydrogen atom via the following form of the spatial wavefunction, and the different quatum numbers.

\[\psi_{nlm} = A \left(\frac{r}{a_0}\right)^l \left[1 + \beta\left(\frac{r}{a_0}\right) + \ldots + (\sim)\left(\frac{r}{a_0}\right)^N\right] \cdot \exp\left({-\frac{r}{na_0}}\right)\cdot Y_{lm}(\theta,\phi),\] \[n = N + l + 1, ~~~~ N = 0,1,2,\ldots; l = 0,1,\ldots,n-1; m = -l,\ldots,-,\ldots,l.\]

Energy in terms of the fine structure.

Compared to what we are going to soon d scuss, this is is retarded. As we will see, the hydrogen atom has much richer energy structure and the above serves just as a good base.

The algebraic methods in quantum mechanics are far reaching and it’s important to recap some of it in the context of Hydrogen atom. Not only that the algebraic structure is powerful on its own, it often also gives us a way to formalise and gain insights on the more abstruse elements of solving a Schrodinger equation. In particular, the symmetries can be captured more accessibly.

The algebra of Hydrogen atom

By algebra, I mostly mean the algebra of angular momentum that’s going to play a crucial role in reconciling the fine structure.

The $L^2$, $L_z$ commutation, simultaneous egienvalues and the angular momentum bases. Griff
Abstruse part of schrodinger equation into symmetric. Griff
Spin, orbital, total angular momentum - basis. Zwe
Coupled and uncoupled basis for H-atom. Zwe

(In progress…)

Baby Dirac equation

How do we reconcile the interaction of electron in a hydrogen atom with the magnetic field? Here's how people have tried formulating.

For a loop carrying a current I, we write the magnetic moment $\bar{\mu} = \frac{I \bar{A}}{C}$. So, an electron with orbital angular momentum $\bar{L}$ has $\bar{\mu} = \frac{q}{2mC}\bar{L}$.

Being reasonably intuitive, people have tried writing $\bar{\mu} = g \frac{e}{2m_eC} \bar{S}$ for the magnetic moment due to the quantum spin of electron. Clearly we were ambitious about the nature of magnetic moment in quantum regime in writing it almost same as classical one. But we have also parametrized all our ignorance in the factor $g$. Experiments show that $g \approx 2$, a simple 2 for the electron! So quantum spin is not that strange from classical after all! But nevertheless this is the first glance at why the quantum spin is not about electron spinning around it's axis, although telling us its probably okay to think on those lines (with a whispering $g$ accompanying every time).

So, Pauli in 1927 went on to formulate this whole business. He noticed something funny, that \begin{align} (\bar{\sigma} .\bar{a}) (\bar{\sigma} .\bar{b}) = (\bar{a}.\bar{b} +i\bar{\sigma}. (\bar{a}\times\bar{b})) \tag{1} \end{align} gives $(\bar{\sigma} .\bar{p})^2 = \bar{p}^2$. With prior learning from Stern-Gerlach eperiment regarding the two spin states of electron, he started out with writing the Hamiltonian eigenstate as a column vector, a 2-spinor; $\psi = \begin{bmatrix} \chi \\ \eta \end{bmatrix}$.

Rewriting the hamiltonian this way replacing $\bar{p}^2$ with $(\bar{\sigma}.\bar{p})^2$ and in present of the E.M fields, $\bar{p} \rightarrow \bar{\pi} = \bar{p} - \frac{q}{C}\bar{A}$, we have \begin{align} H_{Pauli} &= {\frac {1}{2m}}\left[{\bar {\sigma }}\cdot (\bar {p} -q\bar {A} )\right]^{2}+q\phi \tag{2} \\ &= \frac{1}{2m}\left(\bar {p} -q\bar {A} \right)^{2}-2\times\frac{q\hbar {\bar {\sigma }}\cdot \bar {B}}{2m}+q\phi \\ &= {\frac {1}{2m}}\left[|\bar {p} |^{2}-q(\bar {L} +2\bar {S} )\cdot \bar {B} \right]+q\phi \tag{3} \end{align} Where we have assumed magnetic field is weak in writing the third line. The g-factor is naturally seen to be 2 in the Pauli equation as the coefficient of $\bar{S}$.

The Dirac equation

The Electromagnetic interaction is reconciled due to Pauli, and even the Lande’s g-factor for the electron is explained! In 1928 Dirac figured a way to blend the relativistic artifacts correctly with the Pauli’s equation. This is a nice point to discuss the issues in relativistic formulation that physicists faced which ultimately lead to QFT. I divert all such subtleties to a blog titled The beginnings of QFT, but for now I will assume Dirac equation is a nice way to reconcile the relativistic mechanics in the quantum world. To escape the infinite orders of terms when one expands the relativistic energy, Dirac wrote $p^2C^2 + m^2C^4$ as a square of some linear function of $p$. Clearly there are no mixing terms, so Dirac went crazy. He wrote, $p^2C^2 + m^2C^4 = (C\bar{\alpha}.\bar{p} + \beta mC^2)$ Doing the obvious sanity checks leaves us with $(\alpha_i)^2 = \beta^2 = 1; \alpha_i\beta + \beta\alpha_i = 0 = \alpha_i\alpha_j + \alpha_j\alpha_i$. With nice little discussion on when will such properties be satisfied, one ends up with the following representation, \begin{align} \bar{\alpha} = \begin{bmatrix} 0 & \bar{\sigma}
\bar{\sigma} & 0 \end{bmatrix}; \beta = \begin{bmatrix} \mathbb{1} & 0 \ 0 & -\mathbb{1}\end{bmatrix} \end{align}

Dirac Equation before and after coupling with electromagnetic fields.

Foldy–Wouthuysen transformation

I understand only the following as of now:

Writing $\psi = \Omega^{-1}\chi$ with $\Omega = 1 + \frac{(\bar{\sigma}.\bar{\pi})^2}{8m^2C^2}$ gives the right Fine structure Hamiltonian.

The Fine structure of Hydrogen atom

Cut to post Barton Zwiebach lecture binge. Here’s the Hydrogen Atom Fine structure summarized.

The Fine Structure of Spectrum of the Hydrogen Atom.

My complete handwritten notes of the lectures by Barton Zwebach, MIT OCW, 8.06 on the Fine structure of Hydrogen atom can be found here. A not-so-complete notes on Foldy–Wouthuysen transformation oof lectures from P40X, NISER can be found here.

Zeeman Structure

In explaining the Fine strcture of Hydrogen atom, we used the Dirac equation and added an electromagnetic interaction term, eventually assuming only the internal coloumbic electric field to be present i.e. no other external electric or magnetic field. In this section we will introduce an external magnetic field.

(In progress…)