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Hydrogen Atom in Uniform Electric Field

  1. May 25, 2012 #1
    I just did exercise 1.22 in Modern Quantum Chemistry by Szabo and Ostlund. This is a practice problem about linear variational method.
    Question: The Schrodinger equation(in atomic units) for a hydrogen atom in a uniform electric field F in the z direction is
    Use the trial function |ψ>=c1|1s>+c2|2pz>
    where |1s> and |2pz> are normalized eigenfunctions of H0, i.e.,
    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
    Using sqrt(1+x)=1+x/2, expand your answer in a Taylor series in F, i.e.,
    Show that the coefficient α, which is approximate dipole polarizability of the system, is equal to 2.96. The exact result is 4.5.
    I've done the integral and got
    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 explaination for <1s|H|1s>=<1s|H0|1s> and same for 2pz?
    (b)Is there any intuitive explaination 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...
    Last edited: May 25, 2012
  2. jcsd
  3. May 25, 2012 #2


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    Staff: Mentor

    Symmetry. If you put the (undisturbed, symmetric) ground state in a uniform electric field, the increased potential at one side is cancelled 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.
  4. May 25, 2012 #3
    Thanks and the book does not give analytic expression, I put exactly what the book says between the dash line.
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