Stationary perturbation theory is concerned with finding the changes in the discrete energy levels and the changes in the corresponding energy eigenfunctions of a system, when the Hamiltonian of a system is changed by a small amount. Let H=H0+H’=H0+λW. H0 is the unperturbed Hamiltonian whose eigenvalues E0p and eigenstates |φip> are known. Let {|φip>}denote an orthonormal eigenbasis of H0, H0|φip>= E0p|φip>. Here i denotes the degeneracy. Assume that the matrix elements of H’ in the eigenbasis of H0 are small compared to the matrix elements of H0.
<φip|H'|φip> << <φip|H0|φip>=E0p.
We write H’=λW
with λ<<1
and <φip|W|φip>≈E0p.
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Example:![]() H=H0+H’
is the Hamiltonian of a perturbed one-dimensional harmonic oscillator.
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We are looking for the eigenvalues E(λ) and the eigenstates |y(λ)> of H(λ)=H0+λW.
H(λ)|yp>=Ep(λ)|yp> or, keeping the dependence on λ in mind without specifically writing it down, H|yp>=Ep|yp>.
Since λW is small, we assume that E and |y> can be expanded as a power series in λ.
Ep=E0p+λE1p+λ2E2p+...
,
|yp>=|yp0>+λ|yp1>+λ2|yp2>+...
.
We may then write
(H0+λW)(|yp0>+λ|yp1>+λ2|yp2>+...)
= (E0p+λE1p+λ2E2p+...)(|yp0>+λ|yp1>+λ2|yp2>+...).
This equation is must be
valid over a continuous range of λ.
Therefore we equate coefficients of equal powers of λ
on both sides to obtain a series of equations that represent successively
higher orders of the perturbation.
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(H0-E0p)|yp0>=0 |
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(H0-E0p)|yp1>=(E1p-W)|yp0> |
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(H0-E0p)|yp2>=(E1p-W)|yp1>+E2p|yp0> |
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(H0-E0p)|yp3>=(E1p-W)|yp2>+E2p|yp1>+E3p|yp0> |
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... |
(H0-E0p)|yp0>=0 implies that |yp0> is a linear combination of unperturbed eigenfuctionns |φip> with the corresponding eigenvalue E0. We choose <yp0|yp0>=1. |yps> is not uniquely defined. We can add an arbitrary multiple of |yp0> to each |yps> without affecting the left hand side of the above equations. Most often this multiple is chosen so that <yp0|yps>=0. The perturbed ket is then not normalized.
We then have
0
=
To calculate the energy
to sth order, we only need to know the state vector to order s-1.
The first-order energy correction therefore is λE1n=<φn|H’|φn>.
We have En=E0n+<φn|H’|φn>+O(l2).
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Example:![]() ![]()
En=E0n
+O(l2).
To first order, the eigenvalues of H are equal to the eigenvalues
of the unperturbed Hamiltonian H0.
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First-order eigenvector corrections:
(H0-E0p)|yp1> =( E1p-W)|yp0> = (E1p-W)|φp>. |yp0> is an eigenstate of the unperturbed Hamiltonian. We may expand
in terms of the basis vectors |φip'>. In the expansion bp=0 because <yp0|ypi>=0.
Multiply from the left by <φip''|.
Therefore
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Example:![]() ![]()
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Since we have found the expression for the state vector to first order, we can now find the expression for the energy to second order.
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Example:![]() ![]()
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Let H=H0+H’=H0+λW. In practice, after having derived the perturbation expansion, we often set λ=1 and let H’=W be small.
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Solution:
The ground state is not degenerate. The potential energy of an electron outside a uniformly charged sphere of radius r0 and total charge qe is ![]() Inside the sphere the potential
energy is From Gauss’ law we know
that
Let
since r0 << a0,
=4´10-9eV.
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(a) Calculate the first-order energy shift of the nth eigenvalue due to the perturbation.
(b) Write out the first three non vanishing terms for the first-order perturbation expansion of the ground state in terms of the unperturbed eigenfunctions of the infinite well.
(c) Calculate the second-order energy shift for the ground state.
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Solution:
H =H0+H’.
The eigenvalues of H0 are not degenerate.
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