Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Question: Derive the fine structure formula (Equation 6.66) from the relativistic correction (Equation 6.57) and the spin-orbit coupling (Equation 6.65). Hint: Note tha j=l±12t; treat the plus sign and the minus sign separately, and you'll find that you get the same final answer either way.

Short Answer

Expert verified

The fine structure formula is Er'=(En)22mc2[3-4n(j+12)].

Step by step solution

01

Formula Used

The relativistic correction in the energy levels is:

Er'=-(En)2(2mc2)4nl+12-3

And, the spin-orbit coupling: E50'=(En)22mc2[n[j(j+1)-l(l+1)-34]l(l+12)(l+1)]

Where,

j=l±12l=j±12

The equations 6.65, 6.66, 6.67 are

role="math" localid="1658296542883" Es01=(En)2mc2[n[j(j+1)-l(l+1)-34]l(l+12)(l+1)]......(6.65)Efs1=(En)22mc2[3-4nj+12]....(6.66)E=13.6eVn2[a+α2n2(4nj+12-34)]....(6.67)

02

The Spin-orbit coupling form

Takel=j-12 and substitute into the relativistic correction equation:

Er'=-(En)22mc2[4nl-12+12-3]

And the spin-orbit coupling has the form:

Eso'=(En)2mc2[n[j(j+1)-j-12(j-12+1)]-34(j-12)(j-12+12)(j-12+1)]=(En)2mc2[n[j(j+1)-(j-12)(j-12)]-34j(j-12)(j+12)]=(En)2mc2[n[j2+j-(j2-14)]-34j(j2+14)]=(En)2mc2[n[j-12]j(j-12)(j+12)]=(En)2mc2[nj(j-12)]

03

The fine structure formula

Now, calculate the fine structure formula

Efs'=Er'+Eso'=(En)2mc24nj-3+(En)2mc22njj+12=(En)2mc22njj+12-4nj+3=(En)2mc22nj-4nj+12j2j+12+3

Solve further the equation

Efs'=(En)2mc22nj-4n2-2njj2j+12+3=(En)2mc23-4nj+12

04

Again, calculate spin-orbit coupling

Now take l=j+12 and substitute into the relativistic correction equation:

Er'=-(En)22mc24nj+12+12-3=-(En)22mc24nj+1-3

And the spin-orbit coupling has the form,

Eso'=(En)2mc2njj+1-j+12j+12+1-34j+12j+12+12j+12+1=(En)2mc2nj2+j-j2-32j-12j-34-34j+12j+1j+32=(En)2mc2n-j-32j+12j+1j+32=(En)2mc2-nj+32j+12j+1j+32=(En)2mc2-nj+12j+1

05

The fine structure formula

Now, calculate the fine structure formula:

Efs'=Er'+Eso'=-(En)22mc24nj+1-3-(En)22mc2-2nj+1j+12=(En)22mc23-4nj+1+2nj+1j+12=(En)22mc23-4nj+1j+12+2nj+1j+1j+1j+12

Solve the equation further

=(En)22mc23-4n+2n+2nj+1j+12=(En)22mc23-4nj+1j+1j+12=(En)22mc23-4nj+12

This is the same answer for l=j-12.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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+0ψ-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.

Suppose we perturb the infinite cubical well (Equation 6.30) by putting a delta function “bump” at the point(a/4,a/2,3a/4):H'=a3V0δ(x-a/4)δ(y-a/2)δ(z-3a/4).

Find the first-order corrections to the energy of the ground state and the (triply degenerate) first excited states.

(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.

Question: Consider the Stark effect (Problem 6.36) for the states of hydrogen. There are initially nine degenerate states, ψ3/m (neglecting spin, as before), and we turn on an electric field in the direction.

(a) Construct the matrix representing the perturbing Hamiltonian. Partial answer: <300|z|310>=-36a,<310|z|320>=-33a,<31±1|z|32±1>=-(9/2)a,,

(b) Find the eigenvalues, and their degeneracies.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free