Phy5646/Rayleigh Ritz Variational Principle Ex

Submitted by Team 1 (Joon-Il Kim, Jorge Barreda, Muhandis)

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Questions:

Show that if an unperturbed Hamiltonian

${\mathcal {H}}_{0}=-{\frac {\hbar ^{2}}{2m}}\left({\boldsymbol {\nabla }}_{1}^{2}+{\boldsymbol {\nabla }}_{2}^{2}\right)-{\frac {Z^{'}e^{2}}{|{\vec {r}}_{1}|}}-{\frac {Z^{'}e^{2}}{|{\vec {r}}_{2}|}},$

where $Z^{'}=Z-\sigma =2-{\frac {5}{16}},$ is chosen, the first-order perturbation of the ground state of helium vanishes.

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Solution:

If we choose ${\mathcal {H}}_{0}\!$ as above, from the eq. #4.5.3.1 the perturbed term can be written by

${\mathcal {H}}'={\frac {\sigma e^{2}}{|{\vec {r}}_{1}|}}-{\frac {\sigma e^{2}}{|{\vec {r}}_{2}|}}+{\frac {e^{2}}{|{\vec {r}}_{1}-{\vec {r}}_{2}|}},$ where $\sigma ={\frac {5}{16}}.\!$

Then, the first-order perturbation energy of the ground state of helium is

$\left\langle {\mathcal {H}}'\right\rangle =\int {d^{3}r_{1}}\int {d^{3}r_{2}}\psi _{100}^{*}({\vec {r}}_{1})\psi _{100}^{*}({\vec {r}}_{2})\left[-{\frac {\sigma e^{2}}{|{\vec {r}}_{1}|}}-{\frac {\sigma e^{2}}{|{\vec {r}}_{2}|}}+{\frac {e^{2}}{|{\vec {r}}_{1}-{\vec {r}}_{2}|}}\right]\psi _{100}({\vec {r}}_{1})\psi _{100}({\vec {r}}_{2}).$

The wave function is

$\psi _{100}\left({\vec {r}}_{1,2}\right)=\left({\frac {Z^{3}}{\pi {a_{0}}^{3}}}\right)^{1/2}\exp \left[-{\frac {Z|{\vec {r}}_{1,2}|}{a_{0}}}\right],$

where $a_{0}={\frac {\hbar ^{2}}{me^{2}}}.\!$

The first and second terms in the integrand will give

$E_{1,2}^{(1)}(\sigma )=-{\sigma e^{2}}\left\langle {\frac {1}{r_{1}}}\right\rangle =-{\sigma e^{2}}{\frac {\left(Z-\sigma \right)}{a_{0}}}=-{\frac {me^{4}}{\hbar ^{2}}}\sigma \left(Z-\sigma \right).$

The third term will give

$E_{3}^{(1)}(\sigma )={\frac {5\left(Z-\sigma \right)me^{4}}{8\hbar ^{2}}}.$

Therefore, the first-order perturbation is

$E_{0}^{(1)}(\sigma )=-{\frac {me^{4}}{\hbar ^{2}}}\left(2\sigma \left(Z-\sigma \right)-{\frac {5\left(Z-\sigma \right)}{8}}\right).$

Therefore, if we put $Z=2\!$ and $\sigma ={\frac {5}{16}}\!$ into the above equation,

$E_{0}^{(1)}\left({\frac {5}{16}}\right)=-{\frac {me^{4}}{\hbar ^{2}}}\left\{2{\frac {5}{16}}\left(2-{\frac {5}{16}}\right)-{\frac {5}{8}}\left(2-{\frac {5}{16}}\right)\right\}=0.$

Thus, the first-order perturbation energy of the ground state of helium vanishes. Q.E.D

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(Note: This problem is excerpted from Quantum Mechanics, 3nd edition, p479, which is written by Eugen Merzbacher.)