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Consider the (eight) n=2states,|2lmlms⟩.Find the energy of each state, under strong-field Zeeman splitting. Express each answer as the sum of three terms: the Bohr energy, the fine-structure (proportional toa2), and the Zeeman contribution (proportional toμBBext.). If you ignore fine structure altogether, how many distinct levels are there, and what are their degeneracies?

Short Answer

Expert verified

Only five of the eight distinct energies remain when fine structure is ignored: There are three degenerate energies.

Step by step solution

01

Definition of Zeeman splitting.

When radiation (such as light) originates in a magnetic field, Zeeman splitting occurs, which is the splitting of a single spectral line into two or more lines of different frequencies.

02

Step2: Derivation.

For n=2, there are eight different states |2lmlms⟩:

i)l=0   ml=0   ms=-12,ii)l=0   ml=0   ms=12,iii)l=1   ml=-1   ms=-12,iv)l=1   ml=-1   ms=12,v)l=1   ml=0   ms=-12,vi)l=1   ml=0   ms=12,vii)l=1   ml=1   ms=-12,viii)l=1   ml=1   ms=1/2i)l=0   ml=0   ms=-12,ii)l=0   ml=0   ms=12,iii)l=1   ml=-1   ms=-12,iv)l=1   ml=-1   ms=12,v)l=1   ml=0   ms=-12,vi)l=1   ml=0   ms=12,vii)l=1   ml=1   ms=-12,viii)l=1   ml=1   ms=12

Total energy is equal to:

E=-13.6eVn2+μBBext(ml+2ms)+13.6eVn3α234n-l(l+1)-mlmsll+12(l+1)

role="math" localid="1656064856692" i)E=-13.6eVn2-μBBext+13.6eVn3α234n-1.ii)E=-13.6eVn2+μBBext+13.6eVn3α234n-1.iii)E=-13.6eVn2-2μBBext+13.6eVn3α234n-32.iv)E=-13.6eVn2+13.6eVn3α234n-56.v)E=-13.6eVn2-μBBext+13.6eVn3α234n-23.vi)E=-13.6eVn2+μBBext+13.6eVn3α234n-23.vii)E=-13.6eVn2+2μBBext+13.6eVn3α234n-56.viii)E=-13.6eVn2+13.6eVn3α234n-12.

If fine structure is ignored,a=0.

Therefore, the following energies:

E=-13.6eVn2-μBBextDegree of degeneracy: 2

E=-13.6eVn2+μBBextDegree of degeneracy: 2

E=-13.6eVn2Degree of degeneracy: 2

E=-13.6eVn2-2μBBextDegree of degeneracy: 1

E=-13.6eVn2+2μBBextDegree of degeneracy: 1

Now only there are five distinct energies instead of eight.

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Most popular questions from this chapter

Question: In a crystal, the electric field of neighbouring ions perturbs the energy levels of an atom. As a crude model, imagine that a hydrogen atom is surrounded by three pairs of point charges, as shown in Figure 6.15. (Spin is irrelevant to this problem, so ignore it.)

(a) Assuming that r≪d1,r≪d2,r≪d3show that

H'=V0+3(β1x2+β2y2+β3z2)-(β1+β2+β3)r2

where

βi≡-e4πε0qidi3,andV0=2(β1d12+β2d22+β3d32)

(b) Find the lowest-order correction to the ground state energy.

(c) Calculate the first-order corrections to the energy of the first excited states Into how many levels does this four-fold degenerate system split,

(i) in the case of cubic symmetryβ1=β2=β3;, (ii) in the case of tetragonal symmetryβ1=β2≠β3;, (iii) in the general case of orthorhombic symmetry (all three different)?

(a) Find the second-order correction to the energies(En2)for the potential in Problem 6.1. Comment: You can sum the series explicitly, obtaining -for odd n.

(b) Calculate the second-order correction to the ground state energy(E02)for the potential in Problem 6.2. Check that your result is consistent with the exact solution.

Problem 6.6 Let the two "good" unperturbed states be

ψ±0=α±ψa0+β±ψb0

whereα±andβ±are determined (up to normalization) by Equation 6.22(orEquation6.24). Show explicitly that

(a)are orthogonal;role="math" localid="1655966589608" (⟨ψ+0∣ψ-0⟩=0);

(b) ⟨ψ+0|H'|ψ-0⟩=0;

(c)⟨ψ±0|H'|ψ±0⟩=E±1,withE±1given by Equation 6.27.

Two identical spin-zero bosons are placed in an infinite square well (Equation 2.19). They interact weakly with one another, via the potential

V(x1,x2)=-aV0δ(x1-x2). (2.19).

(where V0is a constant with the dimensions of energy, and a is the width of the well).

(a)First, ignoring the interaction between the particles, find the ground state and the first excited state—both the wave functions and the associated energies.

(b) Use first-order perturbation theory to estimate the effect of the particle– particle interaction on the energies of the ground state and the first excited state.

Starting with Equation 6.80, and using Equations 6.57, 6.61, 6.64, and 6.81, derive Equation 6.82.

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