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Although the Schr枚dinger equation for helium itself cannot be solved exactly, there exist 鈥渉elium-like鈥 systems that do admit exact solutions. A simple example is 鈥渞ubber-band helium,鈥 in which the Coulomb forces are replaced by Hooke鈥檚 law forces:

H=-22m(12+22)+12m2|r1-r1|2(8.78).

(a) Show that the change of variables from

r1,r2,tor1,r2,tou12(r1+r2),v12(r1+r2) (8.79).

turns the Hamiltonian into two independent three-dimensional harmonic oscillators:

H=[-2mu2+12m2u2]+[-2mu2+121-m2u2](8.80)

(b) What is the exact ground state energy for this system?

(c) If we didn鈥檛 know the exact solution, we might be inclined to apply the method of Section 7.2 to the Hamiltonian in its original form (Equation 7.78). Do so (but don鈥檛 bother with shielding). How does your result compare with the exact answer? Answer:(H)=3(1-/4)a.

Short Answer

Expert verified

(a) Required equation is proved.

(b)Egs=32魔蝇(1+1-)

(c)Egs32(1+1-12)=3(1-4)

Step by step solution

01

(a)Showing the changes of variables

r1=12(u+v);r2=12(u-v);r12+r22=12(u2+2uv+v2+u2-2uv+v2)=u2+v2(12+22)f(r1,r2)=2fx12+2fy12+2fz122fx22+2fy22+2fz22fx1=fuxuxx1+fvxvxx1=12fux+fvx;fx2=fuxuxx2+fvxfx1=12x1fu1+fvx=122fux2uxx1+2fuxvxvxx1+2fuxux

=122fux2+22fuxvx+2fvx2.2x22=12x2fux-fvx=122fux2uxx2+2fuxvxvxx2-2fv2uxuxx2-2fv2xvxx2.=122fux2-22fuxvx+2fvx2.So2fx12+2fx22=2fux2+2fvx2,andlikewiseforyandz:12+22=u2+v2.H=22mu2+v2+12m蝇2u2+v2-4m蝇22v2H=-22mu2+12m蝇2u2+22mv2+12m蝇2u2-12位m蝇2v2

02

(b) The exact ground state energy 

The energy is

32fortheupartand321-forthevpart:Egs=321+1-.

03

(c) The ground state for one dimension oscillator 

The ground state for a one-dimensional oscillator is

0(x)=m1/4e-max2/2(Eq.2.60)So,fora3-Doscillator,thegroundstateis颈蝉蠄0(r)=m蝇3/4e-m蝇r2/2andfortwoparticlesr1,r2=m蝇3/4e-m蝇2r21,r22(ThisistheanalogtoEq.7.18.)r1,r2100r1100r2=8蟺补3e-2r1,r2/aH=32+32+Vee=3+Vee(TheanalogtoEq.7.20).H=8E1+Vee(8.20).

localid="1655995511764" Vee=-4m2m蟺魔3e-mr12+r22r1-r22r12-2r1-r2+r22d3r1d3r2Vee=e2400,8蟺补32e-4r1+r2/ar1-r2d3r1d3r2Ther1,r2termintegratestozero,bysymmetry,andther22termisthesameasther12term,soVee=-4m2m蟺魔3e-mr12+r22r12d3r1d3r2.=-4m2m蟺魔342000e-mr22/r22dr2000e-mr12/r14dr1.=-8m45蟺魔314m蟺魔m38m2蟺魔m=-34魔蝇.H=3魔蝇-34位魔蝇=3魔蝇1-4.Thevariationprinciplesaysthismustexceedtheexactground-stateenergy(b);letscheckit:

31-4>321+1-2-2>1+1-1-2>1-1--24>1-.Itchecks.Infact,expandingtheexactanswerinpowersof,Egs32魔蝇1+1-12=3魔蝇1-4,werecoverthevariationresult.

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Most popular questions from this chapter

InProblem 7.7we found that the trial wave function with shielding (Equation 7.27), which worked well for helium, is inadequate to confirm the existence of a bound state for the negative hydrogen ion.

1(r1,r2)z3a3e-z(r1+r2)/a (7.27)

Chandrasekhar used a trial wave function of the form

(r1,r2)A[1(r1)2(r2)+2(r1)1(r2)] (7.62).

Where

1(r)z13蟺补3e-z1/a,1(r)z23a3e-z2/a,(7.63)

In effect, he allowed two different shielding factors, suggesting that one electron is relatively close to the nucleus, and the other is farther out. (Because electrons are identical particles, the spatial wave function must be symmetries with respect to interchange. The spin state鈥攚hich is irrelevant to the calculation鈥攊s evidently anti symmetric.) Show that by astute choice of the adjustable parameters Z1and Z2you can get<H>less than -13.6.

Answer: :<H>=E1x6+y6(-x8+2x7+12x6y2-12x5y2-18x3y4+118xy6-12y8).

Wherexz1+z2.y2z1z2Chandrasekhar usedZ1=1.039

(Since this is larger than 1, the motivating interpretation as an effective nuclear charge cannot be sustained, but never mindit鈥檚 still an acceptable trial wave function) andZ2=0.283.

(a) Use a trial wave function of the form

(x)={Acos(x/a),if(a/2<x<a/2)0otherwise

to obtain a bound on the ground state energy of the one-dimensional harmonic oscillator. What is the "best" value ofa? CompareHminwith the exact energy. Note: This trial function has a "kink" in it (a discontinuous derivative) ata/2; do you need to take account of this, as I did in Example 7.3?

(b) Use(x)=Bsin(x/a)on the interval(a,a)to obtain a bound on the first excited state. Compare the exact answer.

Find the best bound on Egsfor the delta-function potentialV(x)=-(x), using a triangular trial function (Equation 7.10, only centered at the origin). This time a is an adjustable parameter.

(a) GeneralizeProblem 7.2, using the trial wave function

(x)=A(x2+b2)n,

For arbitrary n. Partial answer: The best value of b is given by

localid="1658300238725" b2=hm[n(4n-1)(4n-3)2(2n+1)]1/2

(b) Find the least upper bound on the first excited state of the harmonic oscillator using a trial function of the form

(x)=Bx(x2+b2)n.

Partial answer: The best value of b is given by

localid="1658300555415" b2=hm[n(4n-5)(4n-3)2(2n+1)]1/2.

(c) Notice that the bounds approach the exact energies as n 鈫掆垶. Why is that? Hint: Plot the trial wave functions for n = 2 , n = 3 , and n = 4, and compare them with the true wave functions (Equations 2.59 and 2.62). To do it analytically, start with the identity

ez=limn(1+zn)n0(x)=(mh)1/4e-m2hx2 (2.59).

1(x)=A1a^+0=A12hm(-hddx+mx)(mh)1/4e-m2hx21(x)=A1(mh)1/42mhxe-m2hx2(2.62).

As an explicit example of the method developed inProblem 7.15, consider an electron at rest in a uniform magnetic field B=B2kfor which the Hamiltonian is (Equation 4.158):

H=-纬叠 (4.158).

H0=eBzmSz (7.57).

The eigenspinors, xaarelocalid="1655969802629" xb,andthecorrespondingenergies,EaandEb,aregiven in Equation 4.161. Now we turn on a perturbation, in the form of a uniform field in the x direction:

{x+,withenergyE+=-B0/2x-,withenergyE-=+-B0/2 (4.161).

H'=eBxmSx (7.58).

(a) Find the matrix elements of H鈥, and confirm that they have the structure of Equation 7.55. What is h?

(b) Using your result inProblem 7.15(b), find the new ground state energy, in second-order perturbation theory.

(c) Using your result inProblem 7.15(c), find the variation principle bound on the ground state energy.

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