Analyze the Zeeman effect for the n=3states of hydrogen, in the weak, strong, and intermediate field regimes. Construct a table of energies (analogous to Table 6.2), plot them as functions of the external field (as in Figure 6.12), and check that the intermediate-field results reduce properly in the two limiting cases.

Short Answer

Expert verified

Energies in intermediate field are as follows,

ϵ1=E39γ+β

ϵ2=E33γ+2β

ϵ3=E3γ+3β

ϵ4=E36γ+β2+9γ2+βγ+β24

ϵ5=E36γ+β29γ2+βγ+β24

ϵ6=E32γ+32β+γ2+35+β24

ϵ7=E32γ+32βγ2+35+β24

ϵ8=E32γ+β2+γ2+15βγ+β24

.ϵ9=E32γ+β2γ2+15βγ+β24

Other nine energies can be obtained by changing the sign ofβ .Forβ<<γ weak field limit is obtained and forβ>γ strong field limit is obtained.

Step by step solution

01

Given. Step 2: Determination of the Weak field

For weak field, from Equation 6.67.

E3j=13.6eV9[1+α29(3j+1/234)]E3j=13.6eV9[1+α23(1j+1/214)]

gj=1+j(j+1)l(l+1)+342j(j+1)

EZ1=μBgjBextmj

It is known thatl=0, j=12, mj=±12, and gj=2. Substitute these values.

E=E3,1/2+EZ1=13.6eV9[1+α23(114)]±μBBext=13.6eV9[1+α24]±μBBext

It is known that l=1, j=12or32. Substitute these values.

Forj=12,mj=±12,

g1/2=1+12322+3421232=23

E=13.6eV9[1+α24]±13μBBext

Forj=32,mj=±12,±32,

g3/2=1+32522+3423252=43

E=13.6eV9[1+α23(1214)]±μBBext{4312,mj=±124332,mj=±32=13.6eV9[1+α212]±μBBext{23,mj=±122,mj=±32

It is known thatl=2,j=32or52. Substitute these values.

For j=32, mj=±12,±32

g3/2=1+32526+3423252=45

E=13.6eV9[1+α212]±μBBext{4512,mj=±124532,mj=±32=13.6eV9[1+α212]±μBBext{25,mj=±1265,mj=±32

For j=52, mj=±12,±32,±52

g5/2=1+52726+3425272=65

E=13.6eV9[1+α23(1314)]±μBBext{6512,mj=±126532,mj=±326552,mj=±52=13.6eV9[1+α236]±μBBext{35,mj=±1295,mj=±323,mj=±52

02

Step 3: Determination of energies in weak field

Energies in weak field from lowest state to highest are determined as follows,

For,l=0,j=1/2,mj=1/2

E=13.6eV9[1+α24]μBBext

For,l=0,j=1/2,mj=+1/2

E=13.6eV9[1+α24]+μBBext

For,l=1,j=1/2,mj=1/2

E=13.6eV9[1+α24]13μBBext

For,l=1,j=1/2,mj=+1/2

E=13.6eV9[1+α24]+13μBBext

For,l=1,j=3/2,mj=3/2

E=13.6eV9[1+α212]2μBBext

For l=1,j=3/2,mj=1/2,

E=13.6eV9[1+α212]23μBBext

For,l=1,j=3/2,mj=+1/2

E=13.6eV9[1+α212]+23μBBext

For,l=1,j=3/2,mj=+3/2

E=13.6eV9[1+α212]+2μBBext

For ,l=2,j=3/2,mj=3/2

E=13.6eV9[1+α212]65μBBext

For,l=2,j=3/2,mj=1/2

E=13.6eV9[1+α212]25μBBext

For,l=2,j=3/2,mj=+1/2

E=13.6eV9[1+α212]+25μBBext

For,l=2,j=3/2,mj=+3/2

E=13.6eV9[1+α212]+65μBBext

For,l=2,j=5/2,mj=5/2

E=13.6eV9[1+α236]3μBBext

For,l=2,j=5/2,mj=3/2

E=13.6eV9[1+α236]95μBBext

For,l=2,j=5/2,mj=1/2

E=13.6eV9[1+α236]35μBBext

For,l=2,j=5/2,mj=+1/2

E=13.6eV9[1+α236]+35μBBext

For,l=2,j=5/2,mj=+3/2

E=13.6eV9[1+α236]+95μBBext

For,l=2,j=5/2,mj=+5/2

E=13.6eV9[1+α236]+3μBBext

03

Determination of the Strong field 

For Strong field,the total energy is for n=3.

E=13.6eVn2+μBBext(ml+2ms)+13.6eVn3α2[34nl(l+1)mlmsl(l+1)(l+12)]=13.6eV9{1+α23[l(l+1)mlmsl(l+1)(l+12)14]}+μBBext(ml+2ms)

Forl=0,ml=0,ms=1/2,

E=13.6eV9{1+α24}μBBext

Forl=0,ml=0,ms=+1/2,

E=13.6eV9{1+α24}+μBBext

Forl=1,ml=1,ms=1/2,

E=13.6eV9{1+α23[1214]}2μBBext=13.6eV9{1+α212}2μBBext

Forl=1,ml=1,ms=+1/2,

E=13.6eV9{1+α23[5614]}=13.6eV9{1+7α236}

Forl=1,ml=0,ms=1/2,

E=13.6eV9{1+α23[2314]}μBBext=13.6eV9{1+5α236}μBBext

Forl=1,ml=0,ms=+1/2,

E=13.6eV9{1+α23[2314]}+μBBext=13.6eV9{1+5α236}+μBBext

Forl=1,ml=+1,ms=1/2,

E=13.6eV9{1+α23[5614]}=13.6eV9{1+7α236}

Forl=1,ml=+1,ms=+1/2,

E=13.6eV9{1+α23[1214]}+2μBBext=13.6eV9{1+α212}+2μBBext

Forl=2,ml=2,ms=1/2,

E=13.6eV9{1+α23[1314]}3μBBext=13.6eV9{1+α212}3μBBext

Forl=2,ml=2,ms=+1/2,

E=13.6eV9{1+α23[71514]}μBBext=13.6eV9{1+13α2180}μBBext

Forl=2,ml=1,ms=1/2,

E=13.6eV9{1+α23[113014]}2μBBext=13.6eV9{1+7α2180}2μBBext

Forl=2,ml=1,ms=+1/2,

E=13.6eV9{1+α23[133014]}=13.6eV9{1+11α2180}

Forl=2,ml=0,ms=1/2,

E=13.6eV9{1+α23[2514]}μBBext=13.6eV9{1+α220}μBBext

Forl=2,ml=0,ms=+1/2,

E=13.6eV9{1+α23[2514]}+μBBext=13.6eV9{1+α220}+μBBext

Forl=2,ml=+1,ms=1/2,

E=13.6eV9{1+α23[133014]}=13.6eV9{1+11α2180}

Forl=2,ml=+1,ms=+1/2,

E=13.6eV9{1+α23[113014]}+2μBBext=13.6eV9{1+7α2180}2μBBext

Forl=2,ml=+2,ms=1/2,

E=13.6eV9{1+α23[71514]}+μBBext=13.6eV9{1+13α2180}+μBBext

Forl=2,ml=+2,ms=+1/2,

E=13.6eV9{1+α23[1314]}+3μBBext=13.6eV9{1+α212}+3μBBext

04

Step 5:Determination of the Intermediate field

For Intermediate fieldthe fine structure energy is given as follows,

Efs1=(En)22mc2(34nj+1/2)

For En=(E19),

Efs1=(E19)22mc2(312j+1/2)=3E12162mc2(14j+1/2)=E1254mc2(14j+1/2)=E1α2108(14j+1/2)=3γ(14j+1/2)

It is known that γ=13.6eV324α2.

Forj=1/2,Efs1=9γ, forj=3/2,and Efs1=3γforj=5/2,Efs1=γ. These are diagonal elements inW matrixZeeman Write the expression for the Hamiltonian.

HZ'=1β(LZ+2SZ)

Here, β=μBBext.

05

Step 6:Determination of non-zero blocks of matrix

Non-zero blocks ofWmatrix are as follows,

9γβ,9γ+β,3γ2β,3γ+2β,(3γ23β23β23β9γ13β),(3γ+23β23β23β9γ+13β)

06

Step 7:Use ofClebsch-Gordan coefficients

For,l=2

Use Clebsch-Gordan coefficients (in this casej=5/2orj=3/2).

|5252=|22|1212

5/2,5/2HZ'5/2,5/2=5252|β(LZ+2SZ)|5252A=β22|1212|(LZ+2SZ)|22|1212=β22|1212|(21)|22|1212=3β

For ,j=3/2

|5232=45|21|1212+15|22|1212

5/2,3/2HZ'5/2,3/2=5232|β(LZ+2SZ)|5232=β5232|(LZ+2SZ){45|21|1212+15|22|1212}=β5232|{(11)45|21|1212+(2+1)15|22|1212}=β{4521|1212|+1522|1212|}{(2)45|21|121215|22|1212}=β(8515)=95β

Forj=1/2,

.

|5212=35|20|1212+25|21|1212

role="math" localid="1659005505710" 5/2,1/2HZ'5/2,1/2=35β|5212=35|21|121225|20|1212

Forj=1/2,

5/2,1/2HZ'5/2,1/2=35β|5232=15|22|121215|21|12125/2,3/2HZ'5/2,3/2=95β

|5252=|22|12125/2,5/2HZ'5/2,5/2=3β

07

Step 8:Determination of diagonal elements of matrix -W

The Diagonal elements of matrixWforj=5/2are as follows,

γ+3β,γ3β,γ95β,γ+95β,γ35β,γ+35β

For,j=32,

|3232=15|21|121245|22|12123/2,3/2HZ'3/2,3/2=65β

|3212=25|20|121235|21|12123/2,1/2HZ'3/2,1/2=25β

|3212=35|21|121225|20|12123/2,1/2HZ'3/2,1/2=25β

|3232=45|21|121215|22|12123/2,3/2HZ'3/2,3/2=65β

The Diagonal elements of matrixWforj=3/2are as follows,

3γ+65β,3γ65β,3γ+25β,3γ25β

08

Step 10:Determination of non-zero off-diagonal elements

The non-zero off-diagonal elements are as follows,

5/2,3/2HZ'3/2,3/2=3/2,3/2HZ'5/2,3/2=β{4521|1212|+1522|1212|}(LZ+2SZ){15|21|121245|22|1212}=β{4521|1212|+1522|1212|}{(11)15|21|1212(2+1)45|22|1212}=β{4521|1212|+1522|1212|}{(2)15|21|1212+45|22|1212}=β(45+25)=25β

For j=12,

5/2,1/2HZ'3/2,1/2=3/2,1/2HZ'5/2,1/2=65β

5/2,1/2HZ'3/2,1/2=3/2,1/2HZ'5/2,1/2=65β

5/2,3/2HZ'3/2,3/2=3/2,3/2HZ'5/2,3/2=25β

Non-zero off-diagonal elements of matrixW are as follows,

25β,65β

09

Step 11:Determination of Matrix  −W

MatrixW(which has18×18elements) has six1×1 blocksand six2×2 blocks:

9γβ,9γ+β,3γ2β,3γ+2β,γ+3β,γ3β

(3γ23β23β23β9γ13β),(3γ+23β23β23β9γ+13β)

(γ+95β25β25β3γ+65β),(γ95β25β25β3γ65β)(γ35β65β65β3γ25β),(γ+35β65β65β3γ+25β)

10

Step 12:Determination of Eigen values

Eigen values of2×2block is need to be determined. It is required to calculate for3of them as eigen values for other three will be obtained by changing the sign of.

|3γ23βλ23β23β9γ13βλ|=0λ2+λ(β12γ)+γ(27γ7β)=0λ=6γβ2±9γ2+βγ+β24

|γ95βλ25β25β3γ65βλ|=0λ2+λ(3β4γ)+3γ2335βγ+2β2=0λ=2γ32β±γ2+35+β24

|γ35βλ65β65β3γ25βλ|=0λ2+λ(β4γ)+γ(3γ115β)=0λ=2γβ2±γ2+15βγ+β24

11

Step 13:Determination of energies in intermediate field

Finally, energies in intermediate field are as follows,

ϵ1=E39γ+β

ϵ2=E33γ+2β

ϵ3=E3γ+3β

ϵ4=E36γ+β2+9γ2+βγ+β24

ϵ5=E36γ+β29γ2+βγ+β24

ϵ6=E32γ+32β+γ2+35+β24

ϵ7=E32γ+32βγ2+35+β24

ϵ8=E32γ+β2+γ2+15βγ+β24

ϵ9=E32γ+β2γ2+15βγ+β24

Other nine energies can be obtained by changing the sign ofβ<<γ .For weak field limit is obtained and for strong β>γfield limit is obtained.

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Most popular questions from this chapter

When an atom is placed in a uniform external electric field ,the energy levels are shifted-a phenomenon known as the Stark effect (it is the electrical analog to the Zeeman effect). In this problem we analyse the Stark effect for the n=1 and n=2 states of hydrogen. Let the field point in the z direction, so the potential energy of the electron is

H's=eEextz=eEextrcosθ

Treat this as a perturbation on the Bohr Hamiltonian (Equation 6.42). (Spin is irrelevant to this problem, so ignore it, and neglect the fine structure.)

(a) Show that the ground state energy is not affected by this perturbation, in first order.

(b) The first excited state is 4-fold degenerate: Y200,Y211,Y210,Y200,Y21-1Using degenerate perturbation theory, determine the first order corrections to the energy. Into how many levels does E2 split?

(c) What are the "good" wave functions for part (b)? Find the expectation value of the electric dipole moment (pe=-er) in each of these "good" states.Notice that the results are independent of the applied field-evidently hydrogen in its first excited state can carry a permanent electric dipole moment.

Question: Consider a quantum system with just three linearly independent states. Suppose the Hamiltonian, in matrix form, is

H=V0(1-o˙0000o˙0o˙2)

WhereV0is a constant, ando˙is some small number(1).

(a) Write down the eigenvectors and eigenvalues of the unperturbed Hamiltonian(o˙=0).

(b) Solve for the exact eigen values of H. Expand each of them as a power series ino˙, up to second order.

(c) Use first- and second-order non degenerate perturbation theory to find the approximate eigen value for the state that grows out of the non-degenerate eigenvector ofH0. Compare the exact result, from (a).

(d) Use degenerate perturbation theory to find the first-order correction to the two initially degenerate eigen values. Compare the exact results.

Work out the matrix elements of HZ'andHfs'construct the W matrix given in the text, for n = 2.

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.

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?

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