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Suppose we perturb the infinite cubical well (Equation 6.30) by putting a delta function 鈥渂ump鈥 at the point(a/4,a/2,3a/4):H'=a3V0(x-a/4)(y-a/2)(z-3a/4).

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

Short Answer

Expert verified

The first order correction to the energy of the ground state areWab=8V0sin2(4)sin(2)sin()sin(32)sin(34)=0.Wac=8V0sin(4)sin(2)sin2(2)sin(32)sin(34)=8V0(12)(1)(1)(-1)(12)=-4V0.Wbc=8V0sin(4)sin(2)sin()sin(2)sin2(34)=0.

Step by step solution

01

Definition of first order correction energy

The anticipated value of the perturbation in the unperturbed state is the first order adjustment to the energy.

02

Finding the first order corrections to the energy of the ground state and the first excited states

Ground state is non degenerate; Eqs. 6.9鈬

En'=n0H'n0 鈥(6.9).

localid="1658148464503" E1=2a3a3V0a0sin2axsin2aysin2azx-a4y-a2z-3a4dxdydz.=8V0sin24sin22sin234=8V012(1)12=2V0

First excited states:

Waa=8V0sin2axsin2aysin22azx-a4y-a2z-3a4dxdydz.=8V012(1)(1)=4V0.

Wbb=8V0sin2axsin22aysin2azx-a4y-a2z-3a4dxdydz.=8V012(0)12=0.

Wcc=8V0sin22axsin2aysin2azx-a4y-a2z-3a4dxdydz.=8V0(1)(1)12=4V0.

Wab=8V0sin24sin2sin()sin32sin34=0.

Wbc=8V0sin4sin2sin()sin2sin234=0.

W=4V010-1000-101=4V0D;det(D-)=1-0-10-0-101-=-(1-)2+=0

=0,or(1-)2=11-=1=0.

So the first-order corrections to the energies are 0,8V0.

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Most popular questions from this chapter

The Feynman-Hellmann theorem (Problem 6.32) can be used to determine the expectation values of1/rand1/r2for hydrogen.23The effective Hamiltonian for the radial wave functions is (Equation4.53)

22md2dr2+22ml(l+1)r2-e2401r

And the eigenvalues (expressed in terms ofl)24are (Equation 4.70)

En=-me432202h2jmax+l+12

(a) Use =ein the Feynman-Hellmann theorem to obtain 1/r. Check your result against Equation 6.55.

(b) Use =lto obtain 1/r2. Check your answer with Equation6.56.

(a) Plugs=0,s=2, and s=3into Kramers' relation (Equation 6.104) to obtain formulas for (r-1),(r),(r-2),and(r3). Note that you could continue indefinitely, to find any positive power.

(b) In the other direction, however, you hit a snag. Put in s=-1, and show that all you get is a relation between role="math" localid="1658216018740" (r-2)and(r-3).

(c) But if you can get (r-2)by some other means, you can apply the Kramers' relation to obtain the rest of the negative powers. Use Equation 6.56(which is derived in Problem 6.33) to determine (r-3) , and check your answer against Equation 6.64.

Suppose we put a delta-function bump in the center of the infinite square well:

H'=伪未(x-a/2)

whereais a constant.

(a) Find the first-order correction to the allowed energies. Explain why the energies are not perturbed for evenn.

(b) Find the first three nonzero terms in the expansion (Equation 6.13) of the correction to the ground state,11.

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.

Find the (lowest order) relativistic correction to the energy levels of the one-dimensional harmonic oscillator. Hint: Use the technique in Example 2.5 .

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