Show thatΘ=AIn[tan(θ2)]satisfies the θequation (Equation 4.25), for l = m = 0. This is the unacceptable "second solution" -- whats wrong with it?

Short Answer

Expert verified

Θθ=AIn[tanθ2]does satisfy the equation but this is the unacceptable second solution since Θblows up at θ=0and atθ=π

Step by step solution

01

Define the Schrödinger equation

A differential equation describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer is related to a particle's probability density in space and time.

02

Calculation

We need to show that,

Θθ=AIn[tanθ2]

Satisfies the equation

sinθddθ(sinθdΘdθ+[II+1sin2θ-m2]Θ=0

So now take the derivative then we get,

dΘdθ=Atan(θ2)12sec2(θ2)dΘdθ=A21sin(θ2)cos(θ2)=Asinθ

Therefore,

ddθ(sinθdΘdθ)=ddθA=0

Then, we get,

role="math" localid="1656064432888" Θ0=AIn0=A-Θπ=AIn(tanπ2)=AIn-=A

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

(a)Derive Equation 4.131 from Equation 4.130. Hint: Use a test function; otherwise you're likely to drop some terms.

(b)Derive Equation 4.132 from Equations 4.129 and 4.131 .Hint : Use Equation 4.112.

A hydrogenic atom consists of a single electron orbiting a nucleus with Z protons. (Z=1 would be hydrogen itself,Z=2is ionized helium ,Z=3is doubly ionized lithium, and so on.) Determine the Bohr energies En(Z), the binding energyE1(Z), the Bohr radiusa(Z), and the Rydberg constant R(Z)for a hydrogenic atom. (Express your answers as appropriate multiples of the hydrogen values.) Where in the electromagnetic spectrum would the Lyman series fall, for Z=2and Z=3? Hint: There’s nothing much to calculate here— in the potential (Equation 4.52) Ze2, so all you have to do is make the same substitution in all the final results.

V(r)=-e24πo0˙1r (4.52).

An electron is at rest in an oscillating magnetic field

B=B0cos(ωt)k^

whereB0 andω are constants.

(a) Construct the Hamiltonian matrix for this system.

(b) The electron starts out (at t=0 ) in the spin-up state with respect to the x-axis (that is:χ(0)=χ+(x)). Determine X(t)at any subsequent time. Beware: This is a time-dependent Hamiltonian, so you cannot get in the usual way from stationary states. Fortunately, in this case you can solve the timedependent Schrödinger equation (Equation 4.162) directly.

(c) Find the probability of getting-h/2 , if you measure Sx. Answer:

sin2(γB02ωsin(ωt))

(d) What is the minimum field(B0) required to force a complete flip inSx ?

Work out the radial wave functions R30,R31,andR32using the recursion formula. Don’t bother to normalize them.

Consider the earth–sun system as a gravitational analog to the hydrogen atom.

(a) What is the potential energy function (replacing Equation 4.52)? (Let be the mass of the earth, and M the mass of the sun.)

V(r)=-e24π00,1r

(b) What is the “Bohr radius,”ag,for this system? Work out the actual number.

(c) Write down the gravitational “Bohr formula,” and, by equating Ento the classical energy of a planet in a circular orbit of radius r0, show that n=r0/ag.From this, estimate the quantum number n of the earth.

(d) Suppose the earth made a transition to the next lower level(n-1) . How much energy (in Joules) would be released? What would the wavelength of the emitted photon (or, more likely, gravitation) be? (Express your answer in light years-is the remarkable answer a coincidence?).

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