Zeeman Shift on Positronium

[QuantumIsHard]

Homework Statement

The magnetic moment for an electron is [itex]\mu_e = -e/m S_e[/itex].
The magnetic moment for a positron is [itex]\mu_e = +e/m S_e[/itex].
In the ground state, the quantum numbers are n=1 and l=0.

a) What is the physical mechanism for the Zeeman shift?
b) Write the Hamiltonian and identify [itex]H₀[/itex] and H'.
c) In the weak-field limit, the two spins couple together to make a total spin F. What are the possible values for F in the ground state of positronium?
d) Continuing in the weak field limit, the Hamiltonian needs to be written in terms of the total spin F. Project each of the spins S onto the total spin F and find the value of the Lande g-factor for each value of F.
e) Sketch the energy level shift as a function of applied B for each value of F in the weak field limit.

Homework Equations

g-factor:
[itex]g_J = 1 + {j*(j+1) – l*(l+1) + 3/4}/{2*j*(j+1)}[/itex]

Energy shift:
[itex]E_Z¹ = \mu_B g_J B_ext m_J[/itex]

The Attempt at a Solution

a) The motion of the electron and positron will produce a magnetic field experienced by the other. The Zeeman shift will factor in this field.

b) $H0 = -\hbar^2/{2m} * ( {\delta^2}/{\delta^2 r1} ) - \hbar^2/{2m} * ( {\delta^2}/{\delta^2 r2} )$ and $H' = {k*e^2}/{|r1-r2|^2}$, with H just being the sum of the two.

c) F=1 when the spins align and F=0 when the spins are opposite.

d) If I knew j and m_j, I believe I could do this with a Clebsh-Gordan table.

e) I think I could just use the above equation once I know [itex]\mu_B, g_J, and, m_J[/itex]

I'm nowhere near confident with (a)-(c) and am stuck entirely on (d) and (e). Any help would be greatly appreciated.

This is my first post on this forum, so my apologies for any formatting issues.

 

[andrien]

see if you can find something useful here
https://www.physicsforums.com/showthread.php?t=656655&highlight=hyperfine

 

[PhysicsGente]

You are explaining hyperfine splitting not the Zeeman effect. The Zeeman effect is the effect on the atom's magnetic moment due to an external magnetic field.

For part b, you can write the hamiltonian as $H=\mu\cdot B_{ext}.$ (think about why).
I don't know what you are referring to by H_0 and H'.

 

[andrien]

I have given a reference in post#3 there,which deals with both.

 

[paranormal]

a) The physical mechanism for the Zeeman shift on positronium is the interaction between the magnetic dipole moments of the electron and positron. As the positronium is placed in an external magnetic field, the electron and positron will experience a torque due to their magnetic dipole moments, causing a shift in the energy levels.

b) The Hamiltonian for positronium can be written as H = H0 + H', where H0 is the unperturbed Hamiltonian and H' is the perturbation term. In this case, H0 represents the kinetic energy of the electron and positron, while H' represents the potential energy due to their mutual interaction.

c) In the weak-field limit, the total spin F can take on values of 0 or 1. This is because the electron and positron can either have their spins aligned (F=1) or opposite (F=0).

d) In order to find the g-factor for each value of F, we can use the equation gF = 1 + \frac{F(F+1)-J(J+1)+S(S+1)}{2F(F+1)}, where J is the total angular momentum and S is the total spin. For F=0, we have gF=1, and for F=1, we have gF=2. This means that the energy levels for F=0 will not be shifted in the presence of an external magnetic field, while the energy levels for F=1 will have a Zeeman shift proportional to the external magnetic field.

e) In the weak-field limit, the energy levels for F=1 will be split into three levels, with the middle level being degenerate. This is because the Zeeman shift is proportional to the magnetic field and the energy levels are spaced by the energy difference between the two spin states. For F=0, the energy levels will remain unchanged. The energy level shift can be sketched as a function of the applied magnetic field, with the energy levels for F=1 being shifted upwards and downwards, while the energy levels for F=0 remain unchanged.