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Time Independent Perturbation Theory
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Most of the time in quantum mechanics, exact solutions of the Schrodinger
equation are not possible. This is where perturbation theory comes into
play. The basic technique is:
- Divide Hamiltonian into 2 parts: H = H^{0} + εH^{1}
- H^{0} is unperturbed Hamiltonian. The Schrodinger equation
can be solved exactly.
- H' is small additive term that corrects H^{0}
- ε turns perturbation on and off.
The basic idea of perturbation theory is to write out both sides of
the perturbed equation as power series in ε and set coefficients
of the same powers of ε to be equal.
Consider the Schrodinger Equation
H^{0}|ψ_{n}^{0}> = E_{n}^{0}|ψ_{n}^{0}>
Now perturb it
{H^{0} + εH^{1}}|ψ_{n}> = E_{n}|ψ_{n}>
Take the LHS and RHS and expand them as Taylor series
ψ_{n} = ψ_{n}^{0} + εψ_{n}' + ε^{2}ψ_{n}''/2! + ... where ψ_{n}' = dψ_{n}/dε
and
E_{n} = E_{n}^{0} + εE_{n}' + ε^{2}E_{n}''/2! + ... where E_{n}' = dE_{n}/dε
If we substitute these back into the perturbed SE and set the
terms containing ε's of the same power to be equal we get
(the Dirac notation has been dropped for simplicity):
ε^{0}: H^{0}|ψ_{n}^{0}> = E_{n}|ψ_{n}^{0}> ... 0 order
ε^{1}: H^{0}|ψ_{n}'> + H'|ψ_{n}^{0}> = E_{n}^{0}|ψ_{n}'> + E_{n}'|ψ_{n}^{0}> ... 1st order
ε^{2}: H^{0}|ψ_{n}''> + H'|ψ_{n}'> = E_{n}^{0}|ψ_{n}''> + E_{n}'|ψ_{n}'> + E_{n}''|ψ_{n}^{0}> ... 2nd order
If we assume that ψ_{0}' does not differ greatly from the
wavefunction for ψ_{n}^{0} we can expand ψ_{0}' as:
ψ_{n}' = Σ_{n}a_{n}ψ_{n}^{0}
We can then substitute this into the ε^{1} equation to get
H^{0}|Σ_{n}a_{n}ψ_{n}^{0}> + H'|ψ_{n}^{0}> = E_{n}^{0}|Σ_{n}a_{n}ψ_{n}^{0}> + E_{n}'|ψ_{n}^{0}>
Now we can write H^{0}|Σ_{n}a_{n}ψ_{n}^{0}> as Σ_{n}a_{n}H^{0}|ψ_{n}^{0}> which equals Σ_{n}a_{n}E_{n}^{0}|ψ_{n}^{0}>
Substituting we get
H'|ψ_{n}^{0}> + Σ_{n}a_{n}E_{n}^{0}|ψ_{n}^{0}> = E_{n}^{0}Σ_{n}a_{n}|ψ_{n}^{0}> + E_{n}'|ψ_{n}^{0}> ... 1.
Now if we multiply everything by ψ_{n}^{0}^{*} and integrate to get:
<ψ_{n}^{0}^{*}|H'|ψ_{n}^{0}> + Σ_{n}a_{n}E_{n}^{0}<ψ_{n}^{0}^{*}|ψ_{n}^{0}> = E_{n}^{0}Σ_{n}a_{n}<ψ_{n}^{0}^{*}|ψ_{n}^{0}> + E_{n}'<ψ_{n}^{0}^{*}|ψ_{n}^{0}> ... 1.
∴ E_{n}' = <ψ_{n}^{0}|H'|ψ_{n}^{0}>
To obtain the first correction to the nth eigenstate, multiply
both sides of 1. by ψ_{m}^{0}^{*}, where m is an unperturbed orthogonal
state to n. (Recall that states are orthogonal if they correspond
to different eigenvalues of an Hermitian operator).
<ψ_{m}^{0}|H'|ψ_{n}^{0}> + Σ_{n}a_{n}E_{n}^{0}<ψ_{m}^{0}|ψ_{n}^{0}> = E_{n}^{0}Σ_{n}a_{n}<ψ_{m}^{0}|ψ_{n}^{0}> + E_{n}'<ψ_{m}^{0}|ψ_{n}^{0}>
<ψ_{m}^{0}|H'|ψ_{n}^{0}> + a_{m}E_{m}^{0} = E_{n}^{0}a_{m}
∴ a_{m} = <ψ_{m}^{0}|H'|ψ_{n}^{0}>/(E_{n}^{0} - E_{m}^{0})
This can be interpreted as "how much of ψ_{m}^{0} is in the perturbed
state ψ_{n}'.
Example: The He atom
The He atom is an example of a '3 body problem' that cannot be solved
analytically.
H^{0} = [-(h^{2}/2m)∇_{1}^{2} - 2e^{2}/4πε_{0}r_{1}] + [(-h^{2}/2m)∇_{2}^{2} - 2e^{2}/4πε_{0}r_{2}]
H' = e^{2}/4πε_{0}r_{12} electron-electron perturbation
From the H atom we get for each electron (1s):
ψ_{1}^{0} = (1/√π)(2/a_{0})^{3/2}exp(-2r_{1}/a_{0}) (a_{0} = Bohr radius)
and
ψ_{2}^{0} = (1/√π)(2/a_{0})^{3/2}exp(-2r_{2}/a_{0})
Where E_{1}^{0} = E_{2}^{0} = -Z^{2}R_{H}/n^{2} = -4R_{H} = -4(2.18 x 10^{-18} J) (R_{H} = Rydberg constant)
∴ E = E_{1}^{0} + E_{2}^{0} = -1.744 x 10^{-17} J
In the ground state, the inner shell is full with 2 electrons that have opposite
spins. Since the spins are antisymmetric, the spatial part of the wavefunction
must be symmetric. Thus, using the Hartree product we can write:
ψ_{n}^{0} = ψ_{1}^{0}ψ_{2}^{0}
Now let's calculate the correction to the energy when we factor in the
electron-electron interaction term.
E_{n}' = <ψ_{n}^{0}|H'|ψ_{n}^{0}>
This results in the following double integral:
E' = ∫∫(1/√π)(2/a_{0})^{3/2}(1/√π)(2/a_{0})^{3/2}exp(-2r_{1}/a_{0})exp(-2r_{2}/a_{0})(e^{2}/4πε_{0}r_{12})
(1/√π)(2/a_{0})^{3/2}(1/√π)(2/a_{0})^{3/2}exp(-2r_{1}/a_{0})exp(-2r_{2}/a_{0})dτ_{1}dτ_{2}
= ∫∫(1/π)^{2}(2/a_{0})^{3}exp(-2r_{1}/a_{0})exp(-2r_{2}/a_{0})(e^{2}/4πε_{0}r_{12})(2/a_{0})^{3}exp(-2r_{1}/a_{0})exp(-2r_{2}/a_{0})dτ_{1}dτ_{2}
This has the solution:
E' = (5/4)(e^{2}/4πε_{0}a_{0})
= 5.44 x 10^{-18} J
Therefore, the corrected energy is -1.744 x 10^{-17} + 5.44 x 10^{-18} = -1.265 x 10^{-17} J
This is in fairly close agreement with actual experimental measurements.