The Quantum Hamiltonian Including a B-field

We will quantize the Hamiltonian

$$\bgroup\color{black}H={1\over 2m}\left(\vec{p}+{e\over c}\vec{A}\right)^2-e\phi\egroup$$

in the usual way, by replacing the momentum by the momentum operator, for the case of a constant magnetic field. (Later we will quantize the field in section 10.) Note that the momentum operator will now include momentum in the field, not just the particle's momentum. The computation yields

$$\bgroup\color{black}{-\hbar^2\over 2m}\nabla^2\psi+{e\over 2m... ...r^2B^2-(\vec{r}\cdot\vec{B})^2\right)\psi=(E+e\phi)\psi.\egroup$$

The usual kinetic energy term, the first term on the left side, has been recovered. The standard potential energy of an electron in an Electric field is visible on the right side. We see two additional terms due to the magnetic field. An estimate of the size of the two B field terms for atoms shows that, for realizable magnetic fields, the first term is fairly small (down by a factor of ${B\over 2.4\times 10^9}\ \mathrm{gauss}$ compared to hydrogen binding energy), and the second can be neglected. The second term may be important in very high magnetic fields like those produced near neutron stars or if distance scales are larger than in atoms (like in a plasma).

So, for atoms, the dominant additional term is the one we anticipated classically in section 1.5,

$$\bgroup\color{black}H_B={e\over 2mc}\vec{B}\cdot\vec{L}=-\vec{\mu}\cdot\vec{B},\egroup$$

where $\vec{\mu}=-{e\over 2mc}\vec{L}$. This is, effectively, the magnetic moment due to the electron's orbital angular momentum. In atoms, this term gives rise to the Zeeman effect: otherwise degenerate atomic states split in energy when a magnetic field is applied. Note that the electron spin which is not included here also contributes to the splitting and will be studied later.

The Zeeman effect, neglecting electron spin, is particularly simple to calculate because the the hydrogen energy eigenstates are also eigenstates of the additional term in the Hamiltonian. Hence, the correction can be calculated exactly, (and easily).

$$\bgroup\color{black}\Delta E=\mu_BBm_\ell\egroup$$

where $m_\ell$ is the usual quantum number for the z component of orbital angular momentum. The Zeeman splitting of Hydrogen states, with spin included, was a powerful tool in understanding Quantum Physics and we will discuss it in detail in 5.