(Sec. 1) I NONDEGENERATE PERTURBATION THEORY ' 165
where H0 is the operator, derived for the unperturbed system with known
eigenfunctions (/13 and eigenvalues W2. That is,
Hs/1?; = WM. [72
The term H is the perturbation term, derived by the usual operator
(Chap. 7) PROBLEMS - 181
correction to the nth level,
W: 2* H $3. (17211;, [7I2
We repeat the above operation, except using #1 (m # n), obtaining a set of
"1
equations which gives, by [7IO], the correction to the wave function of the
nth level,
M." H' 8"?
(Sec. 2) A SAMPLE CALCULATION ' 179
x1, y1, 21, and the other on x2, y2, 22. (#0 is the product of two hydrogen-like
wave functions, each dependent upon one set of coordinates. The zero-order
energy W0 is the sum of the individual energies of the two elec
(Sec. 1) NONDEGENERATE PERTURBATION THEORY - 167
shall only consider the case where, even when A = l, (i.e., the perturbation is
set at the actual magnitude required in the problem), the A2 terms in Figure 7 .2
are of small magnitude compared to the A-dep
(Sec. 1) NONDEGENERATE PERTURBATION THEORY - 171
eigenfunctions. We assume, therefore, that the correction 911;, to the zero-order
wave function 902, is given by the series
in. = z: a? [7404
That is, the correction terms added to the nth zero-order eigenf
5.1 Fine structure of hydrogenic atoms 239
corrections involving the spin operator such as those arising from H g we start
from the unperturbed equation
H0 Wnlmlmx = En Wnlmlm, (5.8)
where E" are the Schrodinger eigenvalues (3.29) (with ,u = m) and the ze
5.1 Fine structure of hydrogenic atoms 243
We may now combine the effects of H1, H3 and H; to obtain the total energy
shift AE = AE1 + AE2 + AE3 due to relativistic corrections. From (5.15), (5.26) and
(5.27) we have for all 1 (Problem 5.3)
n,- mc
2
- .5 One-electron atoms: fine structure and
5.1
hyperfine structure
Our discussion of the energy levels and wave functions of one-electron atoms in
Chapter 3 was based on the simple, non-relativistic Hamiltonian
H p2 26.2
=_ 5.1
2,11 (41teo)r ( )
wher
(Sec. 2) A SAMPLE CALCULATION - 177
With the aid of Figure 7.4d, one can see at once that H41 2 0, and there-
fore a4=0.
As higher ajs are calculated, one should use exact integration in the
calculation of the intensity of the oddnumbered components, beca
(Chap. 7) NONDEGENERATE PERTURBATION THEORY - 163
simple standing-wave pattern that will occur in the upper diagram depends
upon the fact that the plane waves propagating to the right are superimposed
upon the reected plane waves propagating to the left.
5.1 Fine structure of hydrogenic atoms 241
In our case V(r) = Ze2/(4rt0r), so that
1 Ze2 1
60) = 2m2c2 41:80 r3 (5.18)
Since the operator L2 does not act on the radial variable r nor on the spin
variable, and commutes with the components of L, we see
(Sec. 1) NONDEGENERATE PERTURBATION THEORY - 169
tion will differ only slightly from the zero-order wave function, and the true
eigenvalue will differ only slightly from the zero-order eigenvalue.
Figure 7.3b shows the correct shape for the true eigenfunc
5.2 The Lamb shift 251
ribbon), from which the atoms in the metastable state 251,2 can eject electrons
by giving up their excitation energy. Atoms in the ground state are not detected,
the measured electronic current being proportional to the number of me
5.2 The Lamb shift 249
populated. In this case the ratios of line intensities are the same as those of the
corresponding transition probabilities. The relative intensities of the ne structure
components of the H0, line are shown in Fig. 5.5.
Comparison wi
254
One-electron atoms: fine structure and hyperfine structure
5.3
j= 3/2 293/2 - 29.6 keV
. 251.2 34.1 keV
1=1/2"' 2131/2 34.2 keV
75 eV
Ly-ot
M1
458 eV
151,2 132 keV
1': 1/2 -
Dirac QED T
Figure 5.9 The n = 1 and n = 2 energy levels of U"1+ according to
246
One-electron atoms: fine structure and hyperfine structure
ndm
npm "dz/2
"Pi/2
"Pm
, , 1
n 51/2 n Puz
(a) (b)
Figure 5.3 Allowed transitions in the multiplets (a) np - ns and (b) nd - np.
electric dipole operator D = er does not depend on the spin,
in both summations includes a term that would be zero, and we dont like that. We need another
approach for a degenerate system.
An example application is an infinite square well with a
brick of height V0 as pictured at the right. If we use x = 0
and x = L
2. First, use second-order nondegenerate perturbation theory to compute the second-order corrections to the energies of the bound states of the infinitely deep square well with a delta function perturbation in the center of the well:
(a) Write down the in
A common notation is to place a prime on the summation to denote m 6= n, and to leave the
limit of the summation understood as , so the result is written
|n(1)> =
(0)
(0)
X0 < m
|H1 |n >
(0)
En
m
(0)
Em
(0)
|m
>.
To find the second order correction for th
2.(a) Using the potential, the wave functions, and the perturbation of Problem 1, we find that
the position space integrals are given by
< m|H1 |n > =
Z
0
a
r
mx
2
a
sin
x
a
a
2
2
< m|H1 |n > =
a
Z
a
sin
0
r
nx
2
sin
dx
a
a
mx
nx
a
x
sin
dx.
a
Chapter 15
Time-Independent Perturbation Theory (TIPT)
Time-independent perturbation theory (TIPT) is an approximation method used for systems
which have small variations from systems we can solve or have already solved. It often gives good
eigenvalues bu
The function is in a location where the odd n wave functions are non-zero. If the function
barrier were infinite, the wave function could not alter itself by going over so would have to adjust by having a zero value at the location of the function, i.e.,
Having established background and meaning of degeneracy, and the fact that a linear combination of degenerate states is an eigenstate, we return to equation (8),
H0 |n(1)> +H1 |n(0)> = En(0) |n(1)> +En(1) |n(0)> .
(8)
(0)
Forming an inner product with < a
1.(a)
For
a
H1 = x
2
n0 (x)
and
=
r
the integral of interest in position space is given by
nx
2
sin
,
a
a
En1 = < n|H|n > = < n0 (x)|H|n0 (x) >
r
Z r
nx
nx
2
a 2
=
sin
x
sin
dx
a
a
2
a
a
En1
2
=
a
Z
a
sin2
0
nx
a
x
dx
a
2
for the well that goe
5. The Zeeman splitting of the energy levels due to an applied magnetic field is extremely important! Static magnetic fields are one of the best tools we have for controlling the level splitting
of a great variety of systems. In this problem, you will cal