Master LOM David Clement & Antoine BrowaeysAtom-Light Interaction

Breit-Rabi formula in the ground-state of alkali atoms - solution

The goal of this problem is to calculate the energy shift of the atomic states in theground-state of alkali atoms in the presence of a static magnetic field B = Bez. Weconsider alkali atoms with a nuclear spin I = 3/2 and a single electron in the outershell (S = 1/2). This configuration is that of Lithium-7 , Sodium-23, Potassium-39 andRubidium-87 atoms.

1. Prerequisite. Consider the operators I and J and the ladder operators I± andJ± defined as

I± = Ix ± iIy J± = Jx ± iJy .

Show that

I.J = IzJz +1

2

(I+J− + I−J+

). (1)

2. What is the contribution of fine structure interaction Hamiltonian to the energysplitting in the ground-state induced by B ?

3. Express the hyperfine splitting in the ground-state as a function of AHF wherethe hyperfine Hamiltonian writes

HHF = AHF I.J. (2)

4. To obtain the energy shifts of the atomic state in the presence of the magneticfield, one needs to find the eigenstates and eigenvalues of the Hamiltonian in thepresence of B. We consider the Hamiltonian

H = AHF I.J− µs.B = AHF I.J− gsµBBJz. (3)

In the (mI ,mJ) basis there are two atomic states that are decoupled from allother states. Which are they ? Write their energy as a function of B ?

5. Show that the Hamiltonian of eq. 3 couples the other atomic states two by two.

6. Take one sets of coupled states and calculate the corresponding energies as func-tion of B.

7. Plot the Breit-Rabi diagram of these alkali atoms in their ground-state.

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Correction Breit-Rabi alkali atoms in the ground-state.

There is no fine structure in the ground-state (L = 0) and the hyperfine splittingbetween the F = 1 and F = 2 atomic state writes

∆VHF = EHF(F = 2)− EHF(F = 1) =3

4AHF −

(−5

4AHF

)= 2AHF. (4)

The coupling Hamiltonian preserves the total projection of angular momentummI + mJ . The states | mI = 3

2 ,mJ = 12〉 and | −3

2 ,−12〉 are thus not coupled and

there exists three pairs of coupled states for the values mI +mJ = 0, 1, 2.

The non-coupled states have a linear dependency with B :

E±( 32, 12) =

3

4AHF ±

gsµB2

B (5)

Let us consider the states coupled with mI + mJ = 1 : | 32 ,−

12〉 and | 1

2 ,12〉. Recall

that

J+ | J,mJ〉 = ~√J(J + 1)−mJ(mJ + 1) | J,mJ + 1〉 (6)

J− | J,mJ〉 = ~√J(J + 1)−mJ(mJ − 1) | J,mJ − 1〉. (7)

In the 2× 2 basis the Hamiltonian writes(−3

4AHF − gsµB2 B

√32 AHF√

32 AHF

14AHF + gsµB

2 B

)(8)

leading to the eigen-energies

−AHF

4±

√3

4A2

HF +

(AHF + gsµBB

2

)2

. (9)

Proceeding accordingly with the other coupled states one obtains the following Breit-Rabi diagram depicted in Fig. 1(a).

Figure 1 – (a) Breit-Rabi diagram for the ground-state of Sodium atoms 32S1/2 consi-dered here. (b) Breit-Rabi diagram for an excited of Sodium atoms 32P3/2.

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