The Hamiltonian for the
positronium atom in the ^{1}*S* state in a magnetic field * B*
is, to a good approximation,

,

where 1 labels the electron and 2 the positron. Here *S _{1}* and

(a) Choose the *z*-axis along* B*. Which of the following
are constants of motion?

1. *S _{1}^{2}* and

2. S

3. S

4.

5. S

(b) Find the energy eigenvalues of* H* in terms of the constants *A*
and .

- Solution:
(a) The positronium atom consists of two spin ½ particles. We neglect the spatial motion of the particles and consider only spin interactions. The first term in the Hamiltonian is the spin-spin interaction, and the second term is the interaction of the spins with an external magnetic field.

.

In the {

*|S,S*} basis the matrix of_{z}>**S**_{1}**×***S*_{2}is diagonal, while the matrix of is diagonal in the {*|++>,|+->,|-+>,|-->*} basis. Let us work in the {*|++>,|+->,|-+>,|-->*} basis.Then the matrix of

**S**_{1}**×***S*_{2}is,

and the matrix of S

_{1z}- S_{2z}is.

The matrix of

*H*therefore is,

where .

To find out if an observable is a constant of motion, we have to find out if it commutes with the Hamiltonian

*H*. If it commutes, it is a constant of motion.1. .

2. The matrix of

*S*is ._{1z}The matrix of

*S*is ._{2z}*S*, and_{1z}, S_{2z}*H*do not have a set of common eigenvectors. Therefore .3. The matrix of

*S*^{2}is . .4. The matrix of

*S*_{1x}*+S*is ._{2x}The matrix of

*S*_{1y}*+S*is ._{2y}The matrix of

*S*_{1z}*+S*is ._{2z}.

is not a constant of motion.**S**_{1}+**S**_{2 }5.

(b) The eigenvalue of*S*is a constant of motion._{z}=S_{1z}+S_{2z}*H*are found from*E=C*or from.

(-C+D-E)(-C-D-E)-4C^{2}=0

The value*E=C*is twofold degenerate.*E*.^{2}+2CE+C^{2}-D^{2}-4C^{2}=0.

An electron (charge* -e*, and magnetic moment m) moves through a magnetic field.

(a) Write the Hamiltonian for this particle (assume non-relativistic kinematics). Give m for the electron in Bohr magnetons.

(b) Now ignore the charge and spatial motion of the particle. Assume a constant
magnetic field * B_{0}* in the

(c) Suppose that a neutron in this spin state passes between the two poles of a magnet.
If * B* is inhomogeneous and in the

- Solution:
(a) The intrinsic magnetic moment of a particle is .

For an electron , where is the Bohr magneton in Gaussian units.

In SI units the Hamiltonian of the particle is

,

where

, , .

(b) .The matrix of

*H*is .The energy eigenvalues are found from .

Let .

Let .

.

.

.

(c) A neutron is in the state.

It is passed through the analyzer, which measures the S

_{x}component. The probability that the neutron will be found in the spin down state is .

An atom with *L=0* and *S=½* is initially in its ground
state with* S _{z} =-½* . A field is turned on at

,

where *H _{0}* is the atomic Hamiltonian with

(a) What is the probability *P*, to lowest order in *H _{1}*, that the
atom remains in the state

Here , *e* is the
magnitude of the electron charge, *m _{e}* is the electron mass, and

(b) Under what conditions on* t* will *P=0*?

- Solution:
(a) Assume that the energy of the twofold degenerate ground state is much lower than the energy of any excited state, and approximate the system by a two-level system.

Denote the two levels by*|+>*and*|->*.*H*, the system is degenerate._{0}|+>=E_{0}|+>, H_{0}|->=E_{0}|->

{*|+>,|->*} are the eigenstates of*S*. ._{z}At

*t=0*the system is in the state*|->*, but the Hamiltonian is*H*, with_{0}+W=H.

In the {

*|+>,|->*} the matrix of*W*is.

The matrix of

*H*is.

The eigenvalues of

*H*are , and the corresponding eigenfunctions are.

In this problem .

.

.

.

If then .

(b) If with n an odd integer, then .

The Hamiltonian of a system of two spin ½ particles may be written as

,

where **S**_{1} and **S**_{2} are the spin
matrices for particle 1 and 2 respectively.

(a) Show that *H* commutes with the square of the total spin
operator and the *z*-component of the total spin.

(b) Find the eigenspinors and eigenenergies of this Hamiltonian.

- Solution
(a) The first term (A) is the part of the Hamiltonian that does not depend on spin, the second term is the spin-spin interaction, and the third term is the interaction of the spins with an external magnetic field. In the {|S,S

_{z}>} basis the matrices of are diagonal.,

.

, .

The matrix of

*H*is therefore diagonal and the matrix of*S*is diagonal.^{2}, .

Since all these matrices are diagonal, they all commute with each other.

(b) The eigenvalues of*H*are the diagonal matrix elements. The eigenvectors are the corresponding basis vectors, and .