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Find the lowest bound on the ground state of hydrogen you can get using a Gaussian trial wave function

(r)=Ae-br2,

where A is determined by normalization and b is an adjustable parameter. Answer-11.5eV

Short Answer

Expert verified

The lowest bound on the ground state of hydrogen using Gaussian trial wave function isHmin=-11.5eV.

Step by step solution

01

Step 1: Gaussian trail wave function

A Gaussian function is proposed as a trial wave function in a variational calculation on the hydrogen atom. Determine the optimum value of the parameter and the ground state energy of the hydrogen atom. Use atomic units

h=2蟿蟿,me=1,e=1

(r,):=(2)34exp(-r2)

Thegiventrial wave function is of the form:

(r)=Ae-br2

02

Finding the lowest bound on the ground state of hydrogen.

First, we find the normalization constant A:

2(r)r2sindrd=1A20e-2br2r2dr0xsind02xd=1A2182(b)3(2)(2)=1A=2b3/4

V=-e2400A240e-2br21rr2dr=-e24002b3/2414b=-e240022b.

03

Step 3: Finding the value of T

Now we find <T>

T=-h22mA2e-br22e-br2r2sindrddBut2e-br2=1r2ddrr2ddre-br2=1r2ddr-2br3e-br2=-2br2(3r2-2br4)e-br2=-h22m蟺产42b3/24(-2b)0(3r2-2br4)e-2br2dr=h2m蟺产42b3/2318b2b-2b332b22b.=h2m4蟺产2b38b-316b=3h2b2m.

04

Step 4: Finding the value of H

The last two calculations of the results to get <H>

H=3h2b2m-e24o022b;Hb=3h22m-e24o021b=0b=e24o022m3h2.Hmin=3h22me24o0224m29h2-e24o022e24o022m3h2=e24o02mh243-83.=-m2h2e24o0283=83E1=-11.5eV.

Thus the lowest bound on the ground state of hydrogen using Gaussian trial wave function isHmin=-11.5eV.

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Most popular questions from this chapter

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.

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.

Using Egs=-79.0eV for the ground state energy of helium, calculate the ionization energy (the energy required to remove just one electron). Hint: First calculate the ground state energy of the helium ion, He+, with a single electron orbiting the nucleus; then subtract the two energies.

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.

Quantum dots. Consider a particle constrained to move in two dimensions in the cross-shaped region.The 鈥渁rms鈥 of the cross continue out to infinity. The potential is zero within the cross, and infinite in the shaded areas outside. Surprisingly, this configuration admits a positive-energy bound state

(a) Show that the lowest energy that can propagate off to infinity is

Ethreshold=2h28ma2

any solution with energy less than that has to be a bound state. Hint: Go way out one arm (say xa), and solve the Schr枚dinger equation by separation of variables; if the wave function propagates out to infinity, the dependence on x must take the formexp(ikxx)withkx>0

(b) Now use the variation principle to show that the ground state has energy less than Ethreshold. Use the following trial wave function (suggested by Jim Mc Tavish):

(x,y)=A{cos(蟺虫/2a)+cos(蟺测/2a)e-xaandyacos(x/2a)e-y/axaandy>acos(y/2a)e-y/ax.aandya0elsewhere

Normalize it to determine A, and calculate the expectation value of H.
Answer:

<H>=h2ma2[28-1-(/4)1+(8/2)+(1/2)]

Now minimize with respect to 伪, and show that the result is less thanEthreshold. Hint: Take full advantage of the symmetry of the problem鈥 you only need to integrate over 1/8 of the open region, since the other seven integrals will be the same. Note however that whereas the trial wave function is continuous, its derivatives are not鈥攖here are 鈥渞oof-lines鈥 at the joins, and you will need to exploit the technique of Example 8.3.

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