I just did exercise 1.22 in Modern Quantum Chemistry by Szabo and Ostlund. This is a practice problem about linear variational method.(adsbygoogle = window.adsbygoogle || []).push({});

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Question: The Schrodinger equation(in atomic units) for a hydrogen atom in a uniform electric field F in the z direction is

(-1/2∇^{2}-1/r+Frcosθ)|ψ>=(H_{0}+Frcosθ)|ψ>=ε(F)|ψ>

Use the trial function |ψ>=c_{1}|1s>+c_{2}|2p_{z}>

where |1s> and |2p_{z}> are normalized eigenfunctions of H_{0}, i.e.,

|1s>=exp(-r)/sqrt(pi)

|2p_{z}>=r*exp(-r/2)*cosθ/sqrt(32pi)

to find an upper bound to ε(F). In constructing the matrix representation of H, you can avoid a lot of work by noting that

(H_{0})|1s>=-1/2|1s>

(H_{0})|2p_{z}>=-1/8|2p_{z}>

Using sqrt(1+x)=1+x/2, expand your answer in a Taylor series in F, i.e.,

E(F)=E(0)-αF^{2}/2+...

Show that the coefficient α, which is approximate dipole polarizability of the system, is equal to 2.96. The exact result is 4.5.

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I've done the integral and got

<1s|H|1s>=-1/2

<1s|H|2p_{z}>=<2p_{z}|H|1s>=F*128sqrt(2)/243

<2p_{z}|H|2p_{z}>=-1/8

so matrix representation of H is

(-1/2 F*128sqrt(2)/243)

(F*128sqrt(2)/243 -1/8)

Solve the eigenvalue problem for H we get

E(F)=-5/16-3/16*(1+8388608/534681 F^{2})^{1/2}(lower eigenvalue)

and α=524288/178227

My question is

(a)Is there any intuitive explaination for <1s|H|1s>=<1s|H_{0}|1s> and same for 2p_{z}?

(b)Is there any intuitive explaination about why H=H_{0}+Frcosθ is Hermitian?

(c)If anyone is willing to repeat the calculation could you please check the result for me? The book says α=2.96 but I get α=524288/178227=2.94...

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# Hydrogen Atom in Uniform Electric Field

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