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Question: In the text I asserted that the first-order corrections to an n-fold degenerate energy are the eigen values of the Wmatrix, and I justified this claim as the "natural" generalization of the case n = 2.

Prove it, by reproducing the steps in Section 6.2.1, starting with

ψ0=∑j=1nαjψj0

(generalizing Equation 6.17), and ending by showing that the analog to Equation6.22 can be interpreted as the eigen value equation for the matrix W.

Short Answer

Expert verified

The analog function can be interpreted as the eigenvalue equation for the matrix W.

Step by step solution

01

Significance of Schrodinger’s equation

The Schrodinger wave equation is a linear partial differential equation that governs the wave function of the Quantum Mechanical System.

The following can be interpreted from the equation.

H^ψ=Eψand H^0ψ0=E0ψ0andH^=H^0+λH^

02

Determination of the eigenvalues

Write equation 6.17.

ψ0=αψa0+βψb0

Write equation 6.22.

αWaa+βWab=αE1

It is known that En=En6+λEn1+λ2En2+...and ψ=ψ0+λψ1+λ2ψ2+...and ψ0=j=1nαfψj0

Apply Schrodinger's equation.

H^0+λH^=ψ0+λψ1+λ2ψ2+...=(En0+λEn1+λ2En2+...)ψ0+λψ1+λ2ψ2+...=H^0ψ0+λH^0ψ1+H^'ψ0+λ2H^0ψ2+H^'ψ1+...=En0ψ0+λEn0ψ0+En1ψ0+λ2En0ψ2+En1ψ1+En2ψ0+...

Obtain the first-order correction for the energy eigen values by equating the coefficients of the same order.

=H0ψ1+H^'ψ0=En0ψ1+E1ψ0

Multiply both sides by ψ_j0*.

ψj0H^0ψ1+ψj0H^'ψ0=En0ψj0ψ1+En1ψj0ψ0

It is known thatψ0=∑j=1nαfψj0 and(ψ0H^0=En0*(ψ0,H^'=H^0* , . Then the

role="math" localid="1658211098621" En0=En0(ψ0H^0=En0*(ψ0

Write the generalized form of the equation.

En0ψj0ψ1+ψj0H^'ψ0=En0ψj0ψ1+En1ψj0ψ0∑j=1nαjψj0H^'ψ0=En1∑j=1nαjψj0ψj0∑j=1nαjWij=En1∑j=1nαjWij

Thus, the analog function can be interpreted as the eigenvalue equation for the matrix W.

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Most popular questions from this chapter

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.

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

(b) Now calculate the first-order perturbation in the energy, using Equation 6.9. What ishere? Compare your result with part (a).

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

(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'=αδ(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,Ψ11.

Sometimes it is possible to solve Equation 6.10 directly, without having to expand ψ1nin 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 field (the Stark effect-see Problem 6.36). Hint: Try a solution of the form

(A+Br+Cr2)e-r/acosθ

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/2h)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'=-±ð±è³¦´Ç²õθ4Ï€00r2~

(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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