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Frank0
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I just did exercise 1.22 in Modern Quantum Chemistry by Szabo and Ostlund. This is a practice problem about linear variational method.
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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>=(H0+Frcosθ)|ψ>=ε(F)|ψ>
Use the trial function |ψ>=c1|1s>+c2|2pz>
where |1s> and |2pz> are normalized eigenfunctions of H0, i.e.,
|1s>=exp(-r)/sqrt(pi)
|2pz>=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
(H0)|1s>=-1/2|1s>
(H0)|2pz>=-1/8|2pz>
Using sqrt(1+x)=1+x/2, expand your answer in a Taylor series in F, i.e.,
E(F)=E(0)-αF2/2+...
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|2pz>=<2pz|H|1s>=F*128sqrt(2)/243
<2pz|H|2pz>=-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 F2)1/2 (lower eigenvalue)
and α=524288/178227
My question is
(a)Is there any intuitive explanation for <1s|H|1s>=<1s|H0|1s> and same for 2pz?
(b)Is there any intuitive explanation about why H=H0+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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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>=(H0+Frcosθ)|ψ>=ε(F)|ψ>
Use the trial function |ψ>=c1|1s>+c2|2pz>
where |1s> and |2pz> are normalized eigenfunctions of H0, i.e.,
|1s>=exp(-r)/sqrt(pi)
|2pz>=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
(H0)|1s>=-1/2|1s>
(H0)|2pz>=-1/8|2pz>
Using sqrt(1+x)=1+x/2, expand your answer in a Taylor series in F, i.e.,
E(F)=E(0)-αF2/2+...
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|2pz>=<2pz|H|1s>=F*128sqrt(2)/243
<2pz|H|2pz>=-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 F2)1/2 (lower eigenvalue)
and α=524288/178227
My question is
(a)Is there any intuitive explanation for <1s|H|1s>=<1s|H0|1s> and same for 2pz?
(b)Is there any intuitive explanation about why H=H0+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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