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Unformatted text preview: VII. Time Independent
Perturbation Theory
Griﬃths Chapter 6 1 A. Nondegenerate Perturbation Theory
1. Formulation
Suppose we have solved the time independent
Schrödinger equation for some potential.
0
H 0 n0 = En n0 Therefore we know the eigenvalues En0 and the
orthonormal eigenfunctions 〈n0m0〉 = δnm.
Now we perturb the potential slightly.
What are the new eigenvalues and eigenfunctions?
2 A. Nondegenerate Perturbation Theory In general, we won’t be able to solve the new Schrödinger
equation, but we can use the old eigenvalues and eigenfunctions
to get an approximate solution.
H n = En n
new Hamiltonian Let’s write the new Hamiltonian, corresponding to the perturbed
potential, in terms of the old Hamiltonian. H = H0 + H1
Next, we assume the new eigenvalues and eigenfunctions can
be expanded about the old ones. The s uperscripts g ives the power of
m atrix element of H1 t hat we expect
t he term to come out proportional to. 0
1
2
E n = En + En + En + · · · n = n0 + n1 + n2 + · · ·
3 A. Nondegenerate Perturbation Theory 0
1
2
E n = En + En + En + · · · Substitute everything in... H = H0 + H1 H n = En n n = n0 + n1 + n2 + · · · 2. Zeroth Order Theory
0
H 0 n0 = En n0 We recover the
u nperturbed system 3. First Order Theory
1st order in the
m atrix element of
H1, as expected. See board VII.A.3 1
En = n0 H 1 n0 The first order correction to the
e nergy is the expectation value o f
t he perturbation in the unperturbed
s tate. m0 H 1 n0
n1 =
m0
0
0
(En − Em )
m=n
4 A. Nondegenerate Perturbation Theory 4. An Example
Griﬃths Problem 6.1: Suppose we put a delta function
bump in the centre of an inﬁnite square well:
H 1 = αδ (x − a/2) where α is a constant.
(a) Find the ﬁrst order correction to the allowed
energies. Explain why the energies are not
perturbed for even n.
(b) Find the ﬁrst three nonzero terms in the
expansion of the correction to the ground state
wave function. Solution: See board VII.A.4
5 A. Nondegenerate Perturbation Theory 5. Second Order Energies
The second order energies are found, using the
same method, to be
m0 H 1 n0 2
2
En =
0
0
En − Em See board VII.A.5 m=n Most books stop here (Shankar, Griﬃths, for
example) ignoring the 2nd order wave functions.
We’ll stop here too. For those interested, Liboﬀ
gives the 2nd order wave functions. 6 A. Nondegenerate Perturbation Theory 6. Another Example
Griﬃths Problem 6.4: Suppose we put a delta
function bump in the centre of an inﬁnite square
well:
1
H = αδ (x − a/2) where α is a constant. Find the second order correction to the energies. Solution: See board VII.A.6 7 B. Degenerate Perturbation Theory
Following Griffiths, we’ll focus on energy corrections for this case. 1. The Problem
We found the ﬁrst order perturbation correction for the
wave function was
m0 H 1 n0
1
n =
m0
0
0
(En − Em )
m=n If two diﬀerent states, n0〉 and m0〉 have the same energy,
En0 = Em0, then this expression blows up.
If the ﬁrst order wave function expression is not reliable in this
case, there’s no reason to trust the ﬁrst order energy either.
8 B. Degenerate Perturbation Theory 2. The Solution
Instead of using the basis set {n0〉}, let’s transform to a
diﬀerent basis set where the energy denominator
doesn’t blow up.
In other words, we’ll ﬁnd a basis set that is nondegenerate.
Then, we can use nondegenerate perturbation
theory without a problem.
This is possible because perturbations lift the
degeneracy of unperturbed states.
9 B. Degenerate Perturbation Theory This is possible because perturbations lift the
degeneracy of unperturbed states. 10 B. Degenerate Perturbation Theory Operationay:
1. Find the matrix elements of H1 with respect to the
degenerate elements of {n0〉}.
• H1nm = 〈n0H1m0〉.
2. Diagonalize the matrix H1nm.
• That is, ﬁnd the eigenvalues and eigenvectors.
3. The eigenvalues are the ﬁrst order energy
corrections, En1.
The eigenvectors are the “good” zeroth order wave
functions. Justiﬁcation: See board VII.B.2
11 B. Degenerate Perturbation Theory Mathematically, this corresponds to choosing a new
basis set, built from the unperturbed set {n0〉}, that
diagonalizes the matrix H1mn.
n =
¯
i ain i0 1
nH 1 m = Hnm δnm
¯
¯
¯¯ ¯¯ mn = δmn
¯¯
¯¯ As shown on board VII.B.2 This gives for the ﬁrst order energy correction:
1
En = nH 1 n
¯
¯ This is the same form of energy correction we had for nondegenerate perturbation theory.
If we used the unperturbed basis set {n0〉} however instead of
{n〉}, we would get the wrong answer.
12 B. Degenerate Perturbation Theory 3. An Example
Griﬃths Example 6.2: Consider the three
dimensional inﬁnite cubic well.
Let’s introduce the perturbation
H=
1 V0
0 ,
, if 0 < x < a/2 and 0 < y < a/2;
otherwise. If we calculate the ﬁrst excited state energy corrections using
nondegenerate perturbation theory, we get V0/4 for all three
states.
Using degenerate perturbation theory, we ﬁnd, correctly, that
the three states all have diﬀerent energies. Solution: See board VII.B.3
13 B. Degenerate Perturbation Theory 4. An Easier Way
Choose a Hermitian operator A that commutes
with both H0 and H1 (if you can ﬁnd one). Use
simultaneous eigenfunctions of A and H0 as the
basis set n〉. This set will automatically diagonalize
H1.
The energies are then easily found by using
1
En = nH 1 n
¯
¯ Proof: See board VII.B.4 Example: The Fine Structure of Hydrogen. 14 C. The Fine Structure of Hydrogen
The Bohr Formula gave us the spectrum of hydrogen.
2 2
2
E1
mc
e
1
En = 2 = −
,
n = 1, 2, 3, · · ·
2c2 4π
2
n
2
n
0
on the order of α2mc2 e2
1
α≡
≈
4π0 c
137 The Bohr energy is on the order of α2mc2. 15 C. The Fine Structure of Hydrogen The Bohr energy is on the order of α2mc2.
Relativistic corrections to the electron in the hydrogen atom
are on the order of α4mc2.
Spinorbit coupling corrections are also on the order of
α4mc2.
Taken together, these two corrections give the ﬁne
structure of the hydrogen atom.
Higher order corrections, that we will not study, include the
Lamb shift on the order of α5mc2 and the hyperﬁne
structure, on the order of (m/mp)α4mc2. 16 C. The Fine Structure of Hydrogen 1. The Relativistic Correction
The Hamiltonian for the electron in the hydrogen atom is
written, as always, as the kinetic energy plus the potential
energy: H = T+V. P2
nonrelativistic kinetic energy
T=
2m
Correcting this kinetic energy for the electron using special
relativity gives, to lowest order,
P2
P4
T=
−
+ ···
3 c2
2m 8m with special relativity Therefore our ﬁrst order perturbation can be taken as
P4
1
denotes correction due to relativity
Hr = − 3 2
8m c
See board VII.C.1.a
17 C. The Fine Structure of Hydrogen
We can put this perturbation into our perturbation theory formula:
1
En = nH 1 n
¯
¯ 1
Hr P4
=− 3 2
8m c What basis set
should we use?
Instead of diagonalizing the matrix, we can note that Lz and L2 are both operators
that commute with Hr1 and with the unperturbed hydrogen Hamiltonian H0 that
we previously derived.
Therefore the eigenfunctions of L2 and Lz will automatically diagonalize Hr1 and
these form our “good” basis set.
We found these eigenfunctions when we solved the unperturbed hydrogen atom.
They are just nlm〉 or ψnlm(r).
Thus the unperturbed wave functions of the hydrogen atom form a “good” basis
set, which is to say, n, l, m are good quantum numbers for this problem.
See board VII.C.1.b C. The Fine Structure of Hydrogen The ﬁrst order, relativistic, correction to the Bohr energies is
1
Er = − (En )
2mc2 2 4n
l + 1 /2
−3 See board VII.C.1.c 19 C. The Fine Structure of Hydrogen 2. SpinOrbit Coupling
Imagine an electron “orbiting” a proton.
From the electron frame, the proton is orbiting the electron.
The moving positive charge sets up a magnetic ﬁeld that the
“spinning” electron feels as a torque.
This tends to align the magnetic moment μ of the electron
in the direction of the magnetic ﬁeld B.
The Hamiltonian of the electron due to this eﬀect will be
H = −µ · B
20 C. The Fine Structure of Hydrogen H = −µ · B The magnetic ﬁeld created by the proton is found using the
BiotSavard law. It is:
1
e
L
B=
See board VII.C.2.a
2 r3
4π0 mc
orbital angular momentum of the electron The magnetic dipole moment of the electron is e
µ=− S
m See board VII.C.2.b spin angular momentum of the electron Combining these gives us the ﬁrst order perturbation correction
for the hydrogen atom electron due to spinorbit coupling.
2
denotes correction due
e
1
1
Hso =
S · L t o spinorbit coupling
2 c2 r 3
8π0 m
21 C. The Fine Structure of Hydrogen
We can put this perturbation into our perturbation theory formula:
1
Hso = 2 e
8π0 1
S·L
2 c2 r 3
m 1
En = nH 1 n
¯
¯ What basis set
should we use?
L2 also commutes with this spinorbit H1so, (so l is a good quantum number), but
Lz does not commute, so m (or ml) is not a good quantum number.
S2 commutes, as does J2 and Jz. Taken together, J2, Jz, L2, S2 form a complete set of
commuting observables, that is, they share eigenfunctions. They also all commute
with H0.
Therefore the common eigenfunction of J2, Jz, L2, S2, jmjls〉 are our basis set.
Happily, we know these from solving the L2 eigenvalue equation.
See board VII.C.2.c
22 C. The Fine Structure of Hydrogen The ﬁrst order, spinorbit, correction to the Bohr energies is
1
Eso = (En )
mc2 2 n[j (j + 1) − l(l + 1) − 3/4]
l(l + 1/2)(l + 1) See board VII.C.2.d Combining this with the relativistic ﬁrst order correction
gives
denotes total fine
2
(En )
4n
1
s tructure correction
Ef s =
3−
2mc2
j + 1/2 See Grifﬁths problem 6.17 Combining this with the Bohr formula gives the corrected
energy levels for the hydrogen atom.
2
13.6eV
α
n
3
Enj = −
1+ 2
−
See board
2
n
n
j + 1/2 4
23 VII.C.2.e C. The Fine Structure of Hydrogen Enj
2
13.6eV
α
n
3
=−
1+ 2
−
2
n
n
j + 1/2 4 The ﬁne structure
breaks the degeneracy
in l so that for a given n,
diﬀerent l no longer all
have the same energy.
It preserves degeneracy
in j.
The good quantum
numbers are n, l, s, j,
and mj.
24 D. The Zeeman Eﬀect
The energy levels of an atom placed in an external magnetic
ﬁeld Bext will be shifted.
This is called the Zeeman Eect.
This is represented by the Hamiltonian
HZ = −(µl + µs ) · Bext
magnetic dipole
e
µl = −
L moment due to the
2m e lectron orbital motion magnetic dipole
moment due to the
e lectron spin The Hamiltonian therefore becomes
e
HZ =
(L + 2S) · Bext
2m
25 µs = − e
S
m D. The Zeeman Eﬀect
e
HZ =
(L + 2S) · Bext
2m The nature of the Zeeman eﬀect for a hydrogen
atom in a magnetic ﬁeld depends on the relative
strengths of the external ﬁeld compared with the
internal magnetic ﬁeld.
1
e
B= 4π0 mc2 r3 L For Bext << Bint, ﬁne structure dominates and the Zeeman eﬀect
can be treated as a perturbation on the ﬁne structure results.
For Bext >> Bint , then the Zeeman eﬀect dominates and ﬁne
structure should be treated as a perturbation on the Zeeman
results.
26 D. The Zeeman Eﬀect 1. WeakField Zeeman Eect
In the weak ﬁeld, the Zeeman Hamiltonian can be treated as a
perturbation on the solved ﬁne structure hydrogen atom.
The set of operators, J2, Jz, L2, S2 from the the ﬁne structure
hydrogen atom commute with the Zeeman Hamiltonian, so the
appropriate wave functions for our perturbation theory are jmjls〉.
See Grifﬁths problem 6.16 The formula for the ﬁrst order corrections to the energy of the
ﬁne structure hydrogen atom due to a weak external magnetic
ﬁeld would be
added n to match Griffiths 1
1
EZ = nljmj HZ nljmj left off s as a spectator variable 27 1
HZ D. The Zeeman Eﬀect e
=
(L + 2S) · Bext
2m 1
1
EZ = nljmj HZ nljmj e
=
Bext · nljmj L + 2Snljmj
2m
Solving this, we ﬁnd an energy correction of
choosing zaxis along
1
EZ 1
EZ where
and = µB gJ Bext mj t he direction of the
m agnetic field e
µB =
= 5.788 × 10−5 eV/T
2m j (j + 1) − l(l + 1) + 3/4
gJ = 1 +
2j (j + 1) See board VII.D.1.a
28 Bohr magneton Landé gfactor D. The Zeeman Eﬀect
Adding this energy correction to the total, ﬁne structure
corrected, Bohr energy formula, gives
2
−13.6eV
α
n
3
Enjmj =
1+ 2
−
+ µB gJ Bext mj
2
n
n
j + 1/2 4
The nondegenerate ground state of the hydrogen atom splits into
two when placed in such a magnetic ﬁeld.
m = +1/2
j E1,1/2,±1/2 = −13.6eV 1 + 2 α
4 ± µB Bext
mj = 1/2 See board VII.D.1.b
29 D. The Zeeman Eﬀect 2. StrongField Zeeman Eect
Let’s assume that Bext >> Bint. In fact, let’s begin by taking Bint = 0.
The Zeeman Hamiltonian
e
HZ =
(L + 2S) · Bext
2m
becomes just
eBext
HZ =
(Lz + 2Sz )
2m
if we let Bext lie along the zaxis.
The energy levels of the hydrogen atom due to this PaschenBack Eect, ignoring ﬁne structure, are
Perturbation theory
Enml ms 13.6eV
=−
+ µB Bext (ml + 2ms )
2
n
30 n ot used here! See board VII.D.2.a D. The Zeeman Eﬀect
Griﬃths then adds ﬁne structure as a perturbation.
He doesn’t fully justify his use of basis set – he proceeds as if Lz and
Sz commute with H1so.
He does this in an average sense. He is using “almost good” quantum
numbers. Note his typo however in equation 6.81.
See board VII.D.2.b Overall, Griﬃths gets
1
Efs = 13.6eV 2
α
3
n 3
l(l + 1) − ml ms
−
4n
l(l + 1/2)(l + 1) See Grifﬁths problem 6.22
31 ...
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This note was uploaded on 09/28/2011 for the course PHYS 334 taught by Professor Resch during the Spring '08 term at Waterloo.
 Spring '08
 RESCH

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