Question: In Problem 4.43you calculated the expectation value ofrsin the stateψ321. Check your answer for the special cases s = 0(trivial), s = -1(Equation 6.55), s = -2(Equation 6.56), and s = -3(Equation 6.64). Comment on the case s = -7.

Short Answer

Expert verified

for:s=0:1s=-1:19as=-2:1135a2s=-3:1405a3

s = -7 Or smaller that the result will be undefined due to

negative numbers! = Undefined quantity

Step by step solution

01

Given information:

Equation 6.55, 6.56, 6.64 are

1r=1n2a1r2=1(l+1/2)n3a21r3=1l(l+1/2)l+1n3a3

Problem 4.43-(a) Construct the spatial wave function ψfor hydrogen in the state n = 3 , l=2 . m = 1 Express your answer as a function of r,θ,ϕ, and a (the Bohr radius) only - no other variables, (p,z etc.) or ,functions, ( y , v etc.), or constants, (A,c0etc.), or derivatives, allowed ( πis okay, and e, and 2 , etc.).

(b) Check that this wave function is properly normalized, by carrying out the appropriate integrals over r,θ,and ϕ.

(c) Find the expectation value of rsin this state. For what range of s (positive and negative) is the result finite?

02

Step 2:Check for trivial or non-trivial solution

For n = 3,l = 2,m = 1

Then:

- For s = 0:

1=6!6!1……….[Trivial]

- For s = -1

1r=5!6!3a2-1=1623a=133a=1n2a

- For s = -2:

1r2=4!6!3a22=13023a2=2135a2=12+1233a2=11+12n3a2

03

solve further

-For s = -3

1r3=3!6!3a2-3=1120827a3=122+122+133a3=1ll+12l+1n3a3

- For s = -7

1r7=-1!6!3a2-7-1!

undentified

Then for s = -7 or smaller that the result will be undefined due to

negative numbers! = Undefined quantity

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Most popular questions from this chapter

(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.

Let aand bbe two constant vectors. Show that

(a.r)(b.r)sinθdθdϕ=4π3(a.b)

(the integration is over the usual range:0<θ<π,0<ϕ<2π). Use this result to demonstrate that

(3Sp.rSe.r-Sp.Ser3)=0

For states with I=0. Hint:r=sinθcosϕi+sinθsinϕΦ+cosθk.

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.

For the harmonic oscillator[Vx=1/2kx2], the allowed energies areEN=(n+1/2)ħω,(n=0.1.2,..),whererole="math" localid="1656044150836" ω=k/mis the classical frequency. Now suppose the spring constant increases slightly:k(1+ο')k(Perhaps we cool the spring, so it becomes less flexible.)

(a) Find the exact new energies (trivial, in this case). Expand your formula as a power series inο,, up to second order.

(b) Now calculate the first-order perturbation in the energy, using Equation 6.9. What ishere? Compare your result with part (a).

Hint: It is not necessary - in fact, it is not permitted - to calculate a single integral in doing this problem.

Consider a particle of mass m that is free to move in a one-dimensional region of length L that closes on itself (for instance, a bead that slides frictionlessly on a circular wire of circumference L, as inProblem 2.46).

(a) Show that the stationary states can be written in the formψn(x)=1Le2πinx/L,(-L/2<x<L/2),

wheren=0,±1,±2,....and the allowed energies areEn=2mnπL2.Notice that with the exception of the ground state (n = 0 ) – are all doubly degenerate.

(b) Now suppose we introduce the perturbation,H'=-V0e-x2/a2where aLa. (This puts a little “dimple” in the potential at x = 0, as though we bent the wire slightly to make a “trap”.) Find the first-order correction to En, using Equation 6.27. Hint: To evaluate the integrals, exploit the fact that aLato extend the limits from ±L/2to±after all, H′ is essentially zero outside -a<x<a.

E±1=12Waa+Wbb±Waa-Wbb2+4Wab2(6.27).

(c) What are the “good” linear combinations ofψnandψ-n, for this problem? Show that with these states you get the first-order correction using Equation 6.9.

En'=ψn0H'ψn0(6.9).

(d) Find a hermitian operator A that fits the requirements of the theorem, and show that the simultaneous Eigenstates ofH0and A are precisely the ones you used in (c).

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