21.3 The hydrogen atom

21.3 The hydrogen atom

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The Coulomb potential problem provides a good description of the quantum physics of the hydrogen atom, but it is missing an important feature of that system, the fact that electrons are spin $\frac{1}{2}$ systems. To describe this, one really needs to take as space of states two-component wavefunctions

$$|\psi ⟩=(\genfrac{}{}{0ex}{}{{\psi }_1(\mathbf{q})}{{\psi }_2(\mathbf{q})})$$

(or, equivalently, replace our state space $H$ of wavefunctions by the tensor product $\mathcal{H}\otimes {\mathbf{C}}^2)$ in a way that we will examine in detail in chapter 34.

The Hamiltonian operator for the hydrogen atom acts trivially on the ${\mathbf{C}}^2$ factor, so the only efect of the additional wavefunction component is to double the number of energy eigenstates at each energy. Electrons are fermions, so antisymmetry of multi-particle wavefunctions implies the Pauli principle that states can only be occupied by a single particle. As a result, one finds that when adding electrons to an atom described by the Coulomb potential problem, the first two fill up the lowest Coulomb energy eigenstate (the ${\psi }_{100}$ or 1S state at $n$ = 1), the next eight fill up the $n=2$ states (two each for ${\psi }_{200}$, ${\psi }_{211},{\psi }_{210},{\psi }_{21-1})$ etc. This goes a long ways towards explaining the structure of the periodic table of elements.

When one puts a hydrogen atom in a constant magnetic field $B$, for reasons that will be described in section 45.3, the Hamiltonian acquires a term that acts only on the ${\mathbf{C}}^2$ factor, of the form

$$\frac{2e}{mc}\mathbf{B}\cdot \sigma$$

This is exactly the sort of Hamiltonian we began our study of quantum mechanics with for a simple two-state system. It causes a shift in energy eigenvalues proportional $\mathrm{to}\pm |\mathbf{B}|$ for the two diferent components of the wavefunction, and the observation of this energy splitting makes clear the necessity of treating the electron using the two-component formalism.

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