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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.

Short Answer

Expert verified
  1. The Hamiltonian=e value of1r=1an2
  2. The Hamiltonian=l value of1r2=1a2n3l+12

Step by step solution

01

Define Feynman-Hellmann theorem

The Hellmann鈥揊eynman theorem connects the derivative of total energy with respect to a parameter with the expectation value of the Hamiltonian's derivative with respect to the same parameter. All the forces in the system can be estimated using classical electrostatics once the spatial distribution of the electrons has been known by solving the Schr枚dinger equation, according to the theorem.

02

Use λ=e in the Feynman-Hellmann theorem to obtain ⟨1/r⟩

Hamiltonian and associated energies are accessible:

H=-2md2dr2+2ml(l+1)r2-e2401rEn=-me4322022jmax+l+12

a) =e

Ene=He-me382022jmax+l+12=-e201r

-me382022jmax+l+12=-e201r1r=me2402jmax+l+12a=42me21r=1an2

The Hamiltonian=evalue of1r=1an2

03

Use λ=l to obtain 1/r2

b) =l

Enl=Hlme4162022jmax+l+13=22mr2(2l+1)me4162022jmax+l+13=2(2l+1)2m1r2

1r2=2m2(2l+1)me4162022n3a=42me2=2a2n3(2l+1)1r2=1a2n3l+12

The Hamiltonian=lvalue of1r2=1a2n3l+12

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Most popular questions from this chapter

Question: The most prominent feature of the hydrogen spectrum in the visible region is the red Balmer line, coming from the transition n = 3to n = 2. First of all, determine the wavelength and frequency of this line according to the Bohr Theory. Fine structure splits this line into several closely spaced lines; the question is: How many, and what is their spacing? Hint: First determine how many sublevels the n = 2level splits into, and find Efs1for each of these, in eV. Then do the same for n = 3. Draw an energy level diagram showing all possible transitions from n = 3to n = 2. The energy released (in the form of a photon) is role="math" localid="1658311193797" (E3-E2)+E, the first part being common to all of them, and the E(due to fine structure) varying from one transition to the next. Find E(in eV) for each transition. Finally, convert to photon frequency, and determine the spacing between adjacent spectral lines (in Hz- -not the frequency interval between each line and the unperturbed line (which is, of course, unobservable), but the frequency interval between each line and the next one. Your final answer should take the form: "The red Balmer line splits into (???)lines. In order of increasing frequency, they come from the transitionsto (1) j =(???),toj =(???) ,(2) j =(???) to j =(???)鈥︹. The frequency spacing between line (1)and line (2)is (???) Hz, the spacing between line (2)and (3) line (???) Hzis鈥︹..鈥

For the harmonic oscillator[Vx=1/2kx2], the allowed energies areEN=(n+1/2)魔蝇,(n=0.1.2,..),whererole="math" localid="1656044150836" =k/mis the classical frequency. Now suppose the spring constant increases slightly:k(1+')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 in,, up to second order.

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

Consider the (eight) n=2states, |2ljmj. Find the energy of each state, under weak-field Zeeman splitting, and construct a diagram like Figure 6.11 to show how the energies evolve asBext increases. Label each line clearly, and indicate its slope.

If I=0, then j=s,mj=ms, and the "good" states are the same (nms)for weak and strong fields. DetermineEz1(from Equation) and the fine structure energies (Equation 6.67), and write down the general result for the I=O Zeeman Effect - regardless of the strength of the field. Show that the strong field formula (Equation 6.82) reproduces this result, provided that we interpret the indeterminate term in square brackets as.

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