Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
3rd Edition
ISBN: 9781107189638
Author: Griffiths, David J., Schroeter, Darrell F.
Publisher: Cambridge University Press
Question
Book Icon
Chapter 4.1, Problem 4.7P
To determine

The spherical harmonics Yll(θ,ϕ) and Y32(θ,ϕ) and whether they satisfy the angular equation for appropriate value of m and l.

Expert Solution & Answer
Check Mark

Answer to Problem 4.7P

The spherical harmonics Yll(θ,ϕ) is 1l!(2l+1)!4π(12eiϕsinθ)l_ and Y32(θ,ϕ) is 141052πe2iϕsin2θcosθ_ and it satisfies with the angular equation for appropriate value of m and l.

Explanation of Solution

Write the general expression for the spherical harmonics Yll(θ,ϕ).

    Yll(θ,ϕ)=(2l+1)4π1(2l)!eilϕPll(cosθ)        (I)

Here, Pll(cosθ) is the associated Legendre function.

Write the expression for the associated Legendre function.

    Pll(x)=(1)l(1x2)l/2(ddx)lPl(x)        (II)

Here, Pl(x) is the Legendre function.

Write the expression for the Legendre function.

    Pl(x)=12ll!(ddx)l(x21)l        (III)

Use equation (III) in (II) to solve for Pll(x).

    Pll(x)=(1)l2ll!(1x2)l/2(ddx)2l(x21)l        (IV)

The (x21)l=x2l+, where all other terms involve powers of x less than 2l, and hence give zero when differentiated 2l times.

Then the equation (IV) becomes,

    Pll(x)=(1)l2ll!(1x2)l/2(ddx)2lx2l        (V)

The expansion of (ddx)nxn can be written as

    (ddx)nxn=n!        (VI)

Use equation (VI) in (V) and it becomes,

    Pll(x)=(1)l(2l)!2ll!(1x2)l/2        (VII)

Use equation (VII) in (I) to solve for the spherical harmonics Yll(θ,ϕ).

    Yll(θ,ϕ)=(1)l(2l+1)!4πeilϕ(2l)!2ll!(sinθ)l=1l!(2l+1)!4π(12eiϕsinθ)l        (VIII)

Use equation (VIII) to find Y32(θ,ϕ).

    Y32(θ,ϕ)=74π15!e2iϕP32(cosθ)=74π15!15e2iϕcosθsin2θ=141052πe2iϕsin2θcosθ        (IX)

Consider 1l!(2l+1)!4π(12)l=A and the spherical harmonics Yll(θ,ϕ) can be written as

    Yll(θ,ϕ)=A(eiϕsinθ)l        (X)

Differentiate equation (X) with respect to θ and it becomes,

    Yllθ=Aeilϕl(sinθ)l1cosθsinθYllθ=AeilϕlcosθYllsinθθ(sinθYllθ)=lcos(sinθYllθ)lsin2θYll=(l2cos2θlsin2θ)Yll        (XI)

Take the second derivative of Yll(θ,ϕ) with respect to ϕ and it becomes,

    2Yllϕ2=l2Yll        (XII)

The sum of the equations (XII) and (XI) can be written as

    sinθθ(sinθYllθ)+2Yllϕ2=(l2cos2θlsin2θ)Ylll2Yll=l(l+1)sin2θYll        (XIII)

The equation (XIII) satisfies with the equation given in (4.18).

Consider B=141052π and the spherical harmonics Y32(θ,ϕ) can be written as

    Y32=Be2iϕsin2θcosθ        (XIV)

Take the partial derivative of equation (XIV) with respect to θ and it becomes,

    Y32θ=Be2iϕ(2sinθcos2θsin2θ)        (XV)

Use equation (XV) in (XIII) and it can be written as

    sinθθ(sinθY32θ)=Be2iϕsinθθ(2sin2θcos2θsin4θ)=Be2iϕsinθ(4sinθcos2θ4sin3θ4sin3θcosθ)=4Be2iϕsin2θcosθ(cos2θ2sin2θ)=4(cos22sin2θ)Y32        (XVI)

Take the second partial derivative of Y32 with respect to ϕ.

    2Y32ϕ2=4Y32        (XVII)

Use equation (XVII) and (XVI) in (XIII) and compare,

    4(cos2θ2sin2θ1)Y32=4(3sin2θ)Y32=l(l+1)sin2θY32        (XVIII)

In equation (XVIII), where l=3 fits the RHS of equation 4.18 and hence it is proved.

Conclusion:

Therefore, The spherical harmonics Yll(θ,ϕ) is 1l!(2l+1)!4π(12eiϕsinθ)l_ and Y32(θ,ϕ) is 141052πe2iϕsin2θcosθ_ and it satisfies with the angular equation for appropriate value of m and l.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Answers with -1.828, -1.31 or 939.3 are not correct.
Three slits, each separated from its neighbor by d = 0.06 mm, are illuminated by a coherent light source of wavelength 550 nm. The slits are extremely narrow. A screen is located L = 2.5 m from the slits. The intensity on the centerline is 0.05 W. Consider a location on the screen x = 1.72 cm from the centerline. a) Draw the phasors, according to the phasor model for the addition of harmonic waves, appropriate for this location. b) From the phasor diagram, calculate the intensity of light at this location.
A Jamin interferometer is a device for measuring or for comparing the indices of refraction of gases. A beam of monochromatic light is split into two parts, each of which is directed along the axis of a separate cylindrical tube before being recombined into a single beam that is viewed through a telescope. Suppose we are given the following, • Length of each tube is L = 0.4 m. • λ= 598 nm. Both tubes are initially evacuated, and constructive interference is observed in the center of the field of view. As air is slowly let into one of the tubes, the central field of view changes dark and back to bright a total of 198 times. (a) What is the index of refraction for air? (b) If the fringes can be counted to ±0.25 fringe, where one fringe is equivalent to one complete cycle of intensity variation at the center of the field of view, to what accuracy can the index of refraction of air be determined by this experiment?
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Classical Dynamics of Particles and Systems
Physics
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Cengage Learning
Text book image
University Physics Volume 3
Physics
ISBN:9781938168185
Author:William Moebs, Jeff Sanny
Publisher:OpenStax
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Modern Physics
Physics
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Cengage Learning
Text book image
University Physics Volume 2
Physics
ISBN:9781938168161
Author:OpenStax
Publisher:OpenStax