Introduction to Quantum Mechanics · Problem 7.50

Time-Independent Perturbation Theory — David J. Griffiths

Physics — Quantum Mechanics · diamagnetism_helium

Problem

For an atom in a uniform magnetic field

\mathbf{B}_0 = B_0\,\hat{z}

, Equation 4.230 gives

H = H_{\text{atom}} - B_0(\gamma_o L_z + \gamma S_z) + \frac{e^2}{8m}B_0^2 \sum_{i=1}^{Z}\left(x_i^2 + y_i^2\right),

where

L_z

and

S_z

refer to the total orbital and spin angular momentum of all the electrons. (a) Treating the terms involving

B_0

as a perturbation, compute the shift of the ground state energy of a helium atom to second order in

B_0

. Assume that the helium ground state is given by

\psi_0 = \psi_{100}(r_1)\,\psi_{100}(r_2)\,\frac{|\!\uparrow\downarrow\rangle - |\!\downarrow\uparrow\rangle}{\sqrt{2}}

where

\psi_{100}

refers to the hydrogenic ground state (with

Z = 2

). (b) Use the results of Problem 7.49 to calculate the magnetic susceptibility of helium. Given a density of

0.166\,\text{kg/m}^3

, obtain a numerical value for the susceptibility. *Note:* The experimental result is

-1.0 \times 10^{-9}

(the negative sign means that helium is a diamagnet). The results can be brought closer by taking account of screening, which increases the orbital radius (see Section 8.2).

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