Hydrogen Atom in Uniform Electric Field

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 Schrödinger 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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Frank0 said:
(a)Is there any intuitive explanation for <1s|H|1s>=<1s|H0|1s> and same for 2pz?
Symmetry. If you put the (undisturbed, symmetric) ground state in a uniform electric field, the increased potential at one side is canceled by the lowered potential at the other side.

(b)Is there any intuitive explanation about why H=H0+Frcosθ is Hermitian?
It is just another potential shape plus the usual kinetic term.

(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...
Does the book give an analytic expression? Maybe it is just a rounding error.
 
mfb said:
Symmetry. If you put the (undisturbed, symmetric) ground state in a uniform electric field, the increased potential at one side is canceled by the lowered potential at the other side.


It is just another potential shape plus the usual kinetic term.


Does the book give an analytic expression? Maybe it is just a rounding error.

Thanks and the book does not give analytic expression, I put exactly what the book says between the dash line.
 

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