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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)=-aV0(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鈥攂oth the wave functions and the associated energies.

(b) Use first-order perturbation theory to estimate the effect of the particle鈥 particle interaction on the energies of the ground state and the first excited state.

Short Answer

Expert verified

(a) Ground state:

10(x1,x2)=1(x1)1(x2)=2asin(x1a)sin(x2a);E10=2E1=22ma2

First excited state:

20(x1,x2)=12[1x12x2+2x11x2].=E20=E1+E2=5222ma2

(b)-32V0(38-516)=-2V0

Step by step solution

01

(a) Finding the ground state and the first excited state

In terms of the one-particle states (Eq. 2.28) and energies (Eq. 2.27):

nx=2asin苍蟺ax (2.28).

En=2kn22m=n2222ma2 (2.27).

Ground state:10x1x2=1x11x2=2asin蟺虫1asin蟺虫2a;E10=2E1=22ma2

First excited state:20x1x2=121x12x2+2x11x2

02

 Step2: (b) estimating the effect of the particle– particle interaction on the energies of the ground state and the first excited state

E11=10H'10=-aV02a20a0asin2x1asin2x2ax1-x2dx1dx2=-4V0a0asin4xadx=-4V0aa0sin4ydy=-4V0,38=-32V0E21=20H'20

=-aV02a20asinx1asin2x2a+sin2x1asinx2a2x1-x2dx1dx2.=-2V0a0asinxasin2xa+sin2xa+sin2xasinxa2dx=-8V0a0asinxasin2xadx=-8V0a.a0sin2ysin22ydy=-8V0a.40sin2ysin2ycos2ydy=-32V00(sin4y-sin6y)dy-32V038-516=-2V0.

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Most popular questions from this chapter

Question: In Problem 4.43you calculated the expectation value ofrsin the state321. 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.

Problem 6.6 Let the two "good" unperturbed states be

0=a0+b0

whereandare determined (up to normalization) by Equation 6.22(orEquation6.24). Show explicitly that

(a)are orthogonal;role="math" localid="1655966589608" (+0-0=0);

(b) +0|H'|-0=0;

(c)0|H'|0=E1,withE1given by Equation 6.27.

Question: Evaluate the following commutators :

a)[LS,L]

b)[LS,S]

c)role="math" localid="1658226147021" [LS,J]

d)[LS,L2]

e)[LS,S2]

f)[LS,J2]

Hint: L and S satisfy the fundamental commutation relations for angular momentum (Equations 4.99 and 4.134 ), but they commute with each other.

[LX,LY]=ihLz;[Ly,Lz]=ihLx;[Lz,Lx]=ihLy.......4.99[SX,SY]=ihSz;[Sy,Sz]=ihSx;[Sz,Sx]=ihSy........4.134

(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: Sometimes it is possible to solve Equation 6.10 directly, without having to expand in terms of the unperturbed wave functions (Equation 6.11). Here are two particularly nice examples.

(a) Stark effect in the ground state of hydrogen.

(i) Find the first-order correction to the ground state of hydrogen in the presence of a uniform external electric fieldEext (the Stark effect-see Problem 6.36). Hint: Try a solution of the form

(A+Br+Cr2)e-r/acos

your problem is to find the constants , and C that solve Equation 6.10.

(ii) Use Equation to determine the second-order correction to the ground state energy (the first-order correction is zero, as you found in Problem 6.36(a)). Answer:-m(3a2eEext/2)2 .

(b) If the proton had an electric dipole moment p the potential energy of the electron in hydrogen would be perturbed in the amount

H'=-epcos4o0r2

(i) Solve Equation 6.10 for the first-order correction to the ground state wave function.

(ii) Show that the total electric dipole moment of the atom is (surprisingly) zero, to this order.

(iii) Use Equation 6.14 to determine the second-order correction to the ground state energy. What is the first-order correction?

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