6. Time-Independent (Stationary State) Perturbation Theory
So far, we have been looking at systems that we can solve completely analytically. The potential
energies and Hamiltonians have been of a form that makes the Schroedinger equation solvable without
any approximation techniques. What if we don’t have such a “nice” Hamiltonian. In general, one must
rely on numerical methods to solve the Schroedinger equation (a differential equation). However, if the
Hamiltonian is not too different from one of the nice, solvable Hamiltonians, then we can use the results
of perturbation theory to find approximate energies and eigenfunctions of the system. This theory is
what we will develop here.
Perturbation theory works best when the system’s total energy is “perturbed” by a small additional
potential energy. The total Hamiltonian can be expressed as a sum of an unperturbed Hamiltonian( H o )
and a perturbation ( H ' ):
H Ho H ' (6.1)
The unperturbed Hamiltonian is one of the nice, solvable Hamiltonians while the perturbation is a
relatively small adjustment to the total energy.
The plan is to expand the system Hamiltonian, and hence the system energy, into a sum of terms
involving higher and higher powers of H ' , similar to a Taylor series expansion of a function around
some value. Since the perturbation is small, we only have to keep the first few terms (typically two or
three terms) to obtain a close approximation to the actual energy.
Because of the analogy to a Taylor series, it is useful to review a Taylor expansion for some function
f(x). Let’s take the function
f ( x ) (1 x ) 1/ 2
The first three terms of the Taylor series expansion about x=0 are
f ( x) 1 x x 2 ....
We view x as the perturbation to the unperturbed value of f(x) which is one. If x is very small compared
to one, then the two terms added to one in the expansion are adequate to obtain the value of f(x). The
first term (involving x) added to one is the first order correction and the second term (involving x2) is
the second order correction. As an example, let x = 0.2. The value of f(x = 0.2) = 1.0954451. Using the
first and second order correction, we obtain f(0.2) = 1 + 0.1 - 0.005 = 1.095. This approximate value is
only 0.04% less than the actual value.
We will attempt to do a similar expansion to find the corrections to the unperturbed energy to obtain
an approximate but very accurate value for the energy of the perturbed system.
B. Nondegenerate State Perturbation
The first procedure that we obtain will only work if a system is in a perturbed state that is very close to
a nondegenerate, unperturbed eigenstate of the unperturbed Hamiltonian. This probably won’t be
obvious at first, but we will see later why this is so. We will discuss what to do if the unperturbed state
is degenerate in a later section.
Let the unperturbed eigenstate be denoted by (n0) . The superscript of (0) means that this state is the
unperturbed eigenstate of H o . Let’s let the energy eigenvalue of this state be E n( 0) so that
Ho ( 0) E n 0) ( 0)
The subscript n refers to this one particular eigenstate of H o . Now for some time-saving notation. Let’s
just refer to the eigenstate by its label n; we drop the so that the eigenvalue equation (6.2) looks like
Ho n ( 0 ) E n 0 ) n ( 0 )
We assume that the actual eigenstate of the total Hamiltonian is very close to the unperturbed
state n ( 0) and that it can be written as a series expansion of the form
n ( 0) a b ... (6.4)
where a is the first order correction and b the second order correction to n ( 0) . Likewise, we
assume that the actual energy E is very close to the unperturbed energy E n( 0) and can be written as a
series expansion of the form
E E n 0) 1 2 ...
where 1 is the first order correction and 2 the second order correction to E n( 0) . The question is how
to find these corrections, particularly the energy corrections since finding the energy is usually of
To assist us in this endeavor, recall that what gives rise to these series expansions is the perturbation
H ' of the Hamiltonian. If the perturbation is “turned off”, then the correction terms vanish; if the
perturbation is “turned on”, then the correction terms appear. Let’s introduce an integer variable q such
that we write
H H o qH ' (6.6)
This q can have two values: if q = 0 then the perturbation is turned off, and if q = 1 then the perturbation
is turned on. Using this q, we can rewrite (6.4) and (6.6) as
n ( 0) q a q 2 b ... (6.7)
and E E n 0) q 1 q 2 2 ...
Notice, again, that if q = 0 the perturbation is off and we get the unperturbed eigenstate and energy, but
if q = 1 the perturbation is on and we get the corrections to the eigenstate and energy. Notice also that
the second order corrections involve the square of q. This is logical since we expect a second order term
involving the square of H ' to give rise to the second order corrections to the eigenstate and energy.
Now the Schroedinger equation for the system is
H E (6.9)
Substituting (6.6-6.8) into this equation gives
qH ' n ( 0) q a q 2 b ... E n 0) q 1 q 2 2 ... n ( 0) q a q 2 b ...
Now we want any first order corrections to only depend on terms involving H ' , and second order
corrections to only depend on terms involving H ' 2 , etc. This is accomplished if we simply require that
the sum of terms involving q to a certain power on the left side of (6.10) equal the sum of terms
involving q to that same power on the right side of the equation. This is sometimes referred to as
“equating the powers” of q. Let’s do this for the first three powers of q.
Equating powers of q0 gives:
Ho n ( 0 ) E n 0 ) n ( 0 )
This is simply the Schroedinger equation for the unperturbed Hamiltonian that we wrote in (6.2)
Equating powers of q1 gives:
Ho a H ' n ( 0) En0) a 1 n ( 0)
Equating powers of q2 gives:
Ho b H ' a En0) b 1 a 2 n ( 0)
We will stop here as a second order correction is sufficient for most systems of interest.
First Order Correction to Energy
We take the scalar product of the four vectors in (6.12) with the unperturbed eigenstate. Using Dirac
notation, this looks like
n ( 0) Ho a n ( 0) H ' n ( 0) n ( 0) En 0) a n ( 0) 1 n ( 0)
Let’s look at the first scalar product in this equation in more detail. Recall that the adjoint of the
Hamiltonian is equal to the Hamiltonian since it is an operator corresponding to a dynamic variable
(total energy). So we can replace H with H to get
n ( 0) Ho a n ( 0) Ho a (6.15)
By definition of the adjoint (refer to (2.21)), the right side of this equation can be written as
n ( 0) Ho a a Ho n ( 0) *
E n 0) * a n ( 0) *
n ( 0) Ho a En0) * n ( 0) a
Now the energy E n( 0) is real since we can measure it (or more formally, the eigenvalues of operators
corresponding to dynamic variables are always real), so the complex conjugate of the energy is simply
the energy. Thus, we have shown that
n ( 0 ) Ho a E n 0 ) n ( 0 ) a
(Notice that we can think of H o as acting on the bra vector n ( 0) to give E n( 0) . This is fine since the
Hamiltonian is self-adjoint. Since any operator of a dynamic variable is self-adjoint, you can do the
same for any operator inside a bracket.)
The energies inside the brackets of the two terms on the right of (6.14) can be simply taken outside the
brackets to yield
En0) n ( 0) a n ( 0) H ' n ( 0) En0) n ( 0) a 1 n ( 0) n ( 0)
Assuming that n ( 0) is normalized, we can solve this equation for the first order correction to the
1 n ( 0) H ' n ( 0) (6.19)
Second Order Correction to Energy
Since we found a successful method to obtain 1 let’s try the same method to find 2 . We take the
scalar product of the five vectors in the second order equation (6.13) with the unperturbed eigenstate:
n ( 0) Ho b n ( 0) H ' a n ( 0) En0) b n ( 0) 1 a n ( 0) 2 n ( 0)
Evaluating these five scalar products gives
En0) n ( 0) b n ( 0) H ' a En 0) n ( 0) b 1 n ( 0) a 2 n ( 0) n ( 0)
and solving for 2 gives
2 n ( 0) H ' a 1 n ( 0) a (6.22)
Let’s simplify things by requiring that the corrections to the unperturbed eigenstate be orthogonal to the
eigenstate. Thus n ( 0) a 0 and we have
2 n ( 0) H ' a (6.23)
Hmm...notice that we need the first order correction to the eigenstate to find the second order correction
to the energy. How can we find a ?
Look again at the first order equation (6.12). We can rearrange it to yield
Ho a E n 0) a H ' 1 n ( 0)
This is a differential equation for the function a. We could simply try to solve it for each particular
system. A more general approach would be to use as series solution to solve the equation. You can
always use a series solution for a differential equation. So let’s try this series approach. We assume that
the function a can be expanded as a sum of coefficients times the eigenfunctions of the unperturbed
Hamiltonian. This is valid since these eigenfunctions form a complete set of basis vectors. Any vector
in the same vector space should be able to formed by the appropriate superposition of these basis
vectors. Thus, we write function a as
a cm m( 0) (6.25)
Some notes about this expression:
1) These eigenfunctions m( 0) are the eigenfunctions of the unperturbed Hamiltonian H o .
Remember that the eigenfunction of the total Hamiltonian, , is very close in form to one of these
unperturbed eigenfunctions which we called n ( 0) .
2) We do not include this particular eigenfunction in the expansion ( m n ) since we have required
that a be orthogonal to n ( 0) . Thus, it does not contribute to the formation of a .
3) The cm are the coefficients of the expansion. Recall in Chapter 2 that we required these
coefficients to be the scalar products between the vector we are forming and the basis vectors, so
cm m( 0) a (6.26)
Thus, we can write the expansion for the function a as
a m( 0) m( 0) a (6.27)
Now let’s take the scalar product of the vectors in (6.24) with one of the unperturbed eigenstates:
m( 0) Ho a m( 0) E n 0) a m( 0) H ' 1 n ( 0)
Em0) m( 0) a En0) m( 0) a m( 0) H ' n ( 0) 1 m( 0) n ( 0)
The last scalar product on the right side of this equation is zero since the unperturbed eigenstates are
orthogonal. We can then solve this equation for the coefficient cm:
m ( 0) H ' n ( 0)
m ( 0)
E m0) E n 0)
Okay, we’re now ready to combine our results. Let’s substitute (6.30) into (6.27) to get
m ( 0) H ' n ( 0)
a m ( 0)
m 0 E m0) E n 0)
and let’s substitute this into (6.23) to get an expression for 2 :
m ( 0) H ' n ( 0)
2 n ( 0)
H ' m
m 0 E m0) E n 0)
n ( 0) H ' m ( 0) m ( 0) H ' n ( 0)
m 0 E m0) E n 0)
n ( 0) H ' m ( 0)
m 0 E m0) E n 0)
Here is the second order correction to the energy. Notice that it does involve the square of the
perturbation H ' as we assumed! The energy correction is a sum of potentially an infinite number of
terms. How many terms do we need? In practice, you take only a few terms to get an approximate value
of 2 . This is ironic since 2 itself is one of the few terms you use to get an approximate value of the
We need to make a very important point about this expression. Notice that E m0) cannot equal E n( 0) or
2 will blow up. Now we already require that m n which avoids one potential blow up. But if one of
the other eigenstates is degenerate with the eigenstate n ( 0) , then E m0) E n 0) and a blow up will occur.
Thus, this procedure for finding the second order energy correction is only valid for a perturbed state that
closely resembles a nondegenerate, unperturbed eigenstate. That is, n ( 0) cannot be a degenerate state.
C. Almost Harmonic Oscillator
We know that the potential energy of a simple harmonic oscillator is a parabola. Suppose that the
potential energy of a system is almost parabolic but with a slight perturbation. Suppose that the
potential energy is of the form
V ( x) Cx x 4 (6.33)
where the fourth order term is very small compared to the second order term. We can use the procedure
outlined above to find the first order correction to the energy for such an oscillator. Recall that all of the
states of the unperturbed oscillator are nondegenerate so that we could use the procedure for any of the
states. But let’s focus on the ground state.
The Hamiltonian is H H o H ' where the perturbation is
H ' x 4 (6.34)
We assume that the actual state closely resembles the unperturbed ground state, i.e.
n ( 0) 0 (6.35)
and the unperturbed ground state energy is
E n 0)
The first order correction to the energy can be found using (6.19) (Problem [6.1]). The result is
4m 2 2
Notice that the first order correction is proportional to but inversely proportional to the square of the
mass. One can go on to show that the second order correction to the energy is zero for this potential
energy perturbation. You can also show that for any state, the first order correction to the energy is
1 (6n 2 6n 3) (6.38)
In Problem [6.2], we take a look at the size of this correction for a model of a diatomic molecule.
D. Stark Effect for Ground State Hydrogen Atom
When a Hydrogen atom is situated in an electric field, there is a shift in its energy. This phenomenon
is called the Stark effect. (Recall that a shift in energy also occurs if the atom is placed in a magnetic
field. This is called the Zeeman effect.) The potential energy of the electron due to the electric field is
typically much smaller than the potential energy due to the Coulomb attraction of the proton. Thus, this
potential energy due to the field is the perturbation. If we consider the effect of this perturbation on an
atom in the ground state, then we can use our nondegenerate perturbation theory since the ground state
of a Hydrogen atom is a nondegenerate state.
The unperturbed Hamiltonian is simply the Hamiltonian of an ordinary Hydrogen atom; it is the sum
of the kinetic energy and the Coulomb potential energy:
p 2 ke 2
where k is Coulomb’s constant. The unperturbed eigenstate that closely resembles the perturbed
eigenstate is the ground state:
n ( 0)
100 e r / ao (6.40)
We have written out the function for the ground state since we will have to actually perform integrations
in evaluating scalar products. (Sorry..there is no way to avoid this!) Table 7.2 on page 243 in Eisberg &
Resnick lists the first few eigenfunctions. (That table is also duplicated in the problem section of this
chapter.) Recall that ao is the Bohr radius and equal to 0.53 Å. The unperturbed energy is
E ( 0)
n E100 13.6 eV (6.41)
What is the form of the perturbation H ' ? This will be the potential energy due to the interaction of the
electron with the electric field. Let’s call the electric field . Recall from Physics II that the work done
by the electric field in moving a charged particle along some path is given by
W F ds e ds (6.42)
Work is also the difference between the potential energy at the initial position and the potential energy at
the final position:
W Vi V f (6.43)
Over the small displacement of ds r f ri , the change in potential energy is then
Vi V f e ds e rf ri e ri rf (6.44)
So we conclude that the potential energy at some position r is given by
V e r (6.45)
Let’s assume that the electric field is uniform with a magnitude of and a direction along the +z axis.
Then the perturbation to the Hamiltonian is
H ' V ez er cos (6.46)
We are now ready to apply our recipes for finding the corrections to the unperturbed energy of the
atom as summarized in (6.19) and (6.32). You will do this in Problems [6.3] and [6.4]. Let’s discuss the
results of these problems here.
Discussion of Stark Effect in Ground State Hydrogen
First, let’s look at the size of this perturbation to the potential energy. Taking the radius of the ground
state atom to be the Bohr radius, the Coulomb potential energy is
ke 2 (9 10 9 Nm 2 / C 2 )(16 10 19 C) 2
4.35 10 18 J 27.2 eV
r (0.53 10 m)
(Do you remember that you showed that the potential energy is twice the total energy for a 1/r2 force?
This is the “virial theorem”.) Let’s pick a realistic electric field size of 10000 V/m. This could be
produced easily in the lab by applying a voltage difference of 100 V across a set of parallel plates
separated by 1 cm. With this value for the field, the maximum potential energy of the electron would be
er (16 10 19 C)(10000 V / m)(0.53 10 10 m) 8.48 10 26 J 5 10 7 eV
This is approximately 50 million times smaller than the Coulomb potential energy! This is indeed a
small perturbation, perhaps so small that the perturbation is negligible? We will see. A quick
calculation shows that if the potential energy due to the field is to be comparable to the Coulomb
potential energy, then the field has to be on the order of 500 billion Volts per meter (5 1011 V/m)!
Thus, we can say with some confidence that any electric field we can apply to the atom can be treated as
You will show in Problem [6.3] that the first order correction to the ground state energy is zero. This
tells us that the perturbation to the energy is going to be small. You will show in Problem [6.4] that the
second order correction is
Note that the electric field lowers the energy of the electron since 2 is negative and the unperturbed
energy is also negative (-13.6 eV). Also note that the second order correction to the energy depends on
the square of the field as we expect (since the Hamiltonian perturbation depends linearly on the field
Taking our field value of 10000 V/m from above, the energy is lowered by
(0.53 10 10 m) 3 (10000 V / m) 2
. 2.8 10 33 J 17 10 14 eV
9 10 Nm / C
9 2 2
This shift is undetectable! What kind of shift is detectable? Recall that the spin-orbit interaction
produced an energy shift on the order of 10-4 eV. This is detectable using optical spectroscopy and
measuring the transition wavelengths between shifted energy levels. The wavelength of a photon
emitted in a transition is given by
where E is the transition energy. Differentiating both sides of this equation will tell us how a
perturbation to the energy affects the wavelength:
For example, if the atom makes a transition from the n=2 state (E2 = -3.4 eV) to the ground state (E1 = -
13.6 eV) , the transition energy in the absence of any perturbations is 10.2 eV. If the transition energy is
lowered by 10-4 eV, then the transition wavelength increases by
( 104 eV) 0.01 A
This is about the limit of resolution in spectroscopy so that energy shifts smaller than 10-4 eV are just too
small to detect.
Suppose we did just want to see the perturbation due to the Stark effect. Solving (6.47) for the electric
field necessary to give a second order correction of 2 = -10-4 eV = -1.6 10-23 J gives
1/ 2 1/ 2
k 2 (9 10 9 Nm 2 / C 2 )( 16 10 23 J )
10 8 108 V / m
. 168(0.53 10 m) 3
This is an extremely large field! (Air experiences dielectric breakdown at about 3 106 V/m.) Wouldn’t
this extremely large field exert a tremendous force on the electron and actually ionize the atom (i.e. strip
the electron from the proton)? No! This electric field would exert a force of
F e (16 10 19 C)(8 108 N / C) 13 10 10 N
on the electron. The Coulomb force holding the electron to the proton is
ke 2 (9 109 Nm2 / C 2 )(16 10 19 C 2 )
FC 2 10
8.2 10 8 N
ao (0.53 10 m) 2
The Coulomb force is approximately 600 times stronger than the electric force, even with this extremely
large field. The atom will not ionize!
What can we say in summary about the Stark effect in ground state Hydrogen? The effect is extremely
small. For it to be easily detected requires extremely large fields on the order of hundreds of millions of
Volts per meter.
A final note of interest should be made between magnetic field effects and electric field effects on the
Hydrogen atom. Recall that magnetic field effects are readily observable. The spin-orbit interaction
produces a detectable energy splitting with an internal magnetic field produced by the orbiting electron
that is approximately 1 Tesla. An external field that is comparable to this or greater can produce larger
Why is the atom so sensitive to magnetic fields but not so sensitive to electric fields when it comes to
energy perturbations? Is it because the magnetic forces are typically much larger than the electric
forces? Consider the relative sizes of the electric and magnetic forces exerted by the fields. The electric
force depends only on the product of charge and the electric field. The magnetic force depends on the
product of charge, magnetic field, and the speed of the electron. In the ground state, the average speed
of the electron is ~ 2.2 106 m/s. For a magnetic field of B = 1 T, the magnetic force is on the order of
FM evB (16 1019 C)(2.2 106 m / s)(1T ) 35 1013 N
This is roughly 370 times smaller than the electric force required to produce an energy shift comparable
to the shift produced by the magnetic field. (Compare this force to the 1.3 10-10 N we calculated
above.) So the answer isn’t the size of the forces involved. Then what does cause this difference ?
The answer is electron spin! The magnetic field interacts with the electron spin to produce these energy
shifts. Recall that we found the energy shifts by modeling the electron spinning on its axis as producing
a magnetic dipole moment which interacts with the magnetic field. The electric field has no interaction
with the spin of the electron. Thus, the electric field’s effect on energy is much less noticeable.
E. Degenerate State Perturbation
F Stark Effect for Excited Hydrogen Atom
Show that the first order correction to the ground state energy for the almost harmonic oscillator is
1 . Use the Hamiltonian perturbation in (6.34).
4m 2 2
Note: You have two options in evaluating the scalar product to find 1 . You could use the functional
form for the unperturbed ground state which is 0 ( x) A0 e mx / 2 and perform the actual integration.
You do have look up or find the normalization constant A0. The second option, which I suggest you use,
does not require the functional form of the ground state nor involves an actual integration. Here, we
replace x in H ' by x
a a . Now you do have to raise x to the fourth power which gives 16
terms involving a and a ! But the only terms that will not give zero in the resulting scalar products are
the ones that give back 0 . These are the terms that involve the same number of a and a operators.
It is then just a matter of keeping track of these terms and their operations. Remember that we know
how a and a operate on the eigenfunctions as summarized in (3.32) and (3.38). It is this technique
which allows one to find that the first order energy correction for the nth state is the expression in (6.38)
A typical constant for C for a diatomic molecule is 1000 J/m2. Assume the mass is 4.1x10-26 kg.
Suppose that the potential energy is perturbed as in (6.34) with = 1022 J/m4.
(a) Show that the unperturbed ground state energy of the molecule is 51.4 meV.
(b) Suppose that the maximum displacement of one of the atoms from its equilibrium position is 0.5 Å.
Show that the maximum potential energy due to the x2 term is 7.81 eV while the x4 perturbation term has
a value of 0.39 eV. This verifies that the perturbation is small and that our results can be trusted.
(c) Evaluate the first order correction to the energy and show that the energy increases by approximately
0.024% (0.013 meV).
Show that the first order correction to the energy for the Stark effect for the ground state of the Hydrogen
atom is zero. Do this by evaluating the integral expressed in (6.19) and showing that it is zero. Note
that this is an integration in spherical polar coordinates over a volume.
Hint: Recall that the radial integral never gives zero. Don’t worry about evaluating it. Examine the
angular integrals and see if one of them gives zero. You can save even more time by noting if the
integrands are odd or even functions over the ranges formed by the limits of integration.
Show that the second order correction to the energy is the expression in (6.47). To do this, you must
evaluate the first few terms in the sum in (6.32) for 2 :
n ( 0) H ' m ( 0)
m 0 E m0) E n 0)
Remember that n ( 0) is the ground state eigenfunction of the unperturbed Hydrogen atom and E n( 0) is
the unperturbed ground state energy ( E n( 0) ke 2 2a 0 ). The other m( 0) states are the unperturbed
excited states of Hydrogen which have energies of E m0)
ke 2 2a o where n is the principal
quantum number of the state. For each excited state, you can evaluate the integral n ( 0) H ' m( 0) , square
the result, and divide by the difference in energies between the excited and ground states. The question
is how many terms should you use? Look at the first fourteen eigenfunctions which are tabulated at the
end of this section. Let’s evaluate the first thirteen terms in the sum for 2 which involve the ground
state and the first thirteen excited states.
At this point you may be dreading doing thirteen complicated integrals but, as you might expect, some of
those integrals yield zero and it’s not too much work to show this. An integral will yield zero if the
integral over from 0 to gives zero and/or the integral over from 0 to 2 gives zero.
Now the ground state does not involve . The excited states either don’t involve or have an eiml
dependence. If an excited state doesn’t involve , then we know that the integral over from 0 to
2will give 2If the excited state does involve , then we know that the integral will give zero since
d 0 . Thus, we can discard the eight integrals that involve excited states with ml 1,2 .
The ground state does not involve . You do pick up a cos from H ' and a sin from dV. Now
cossin is an odd function over the interval from 0 to and will yield zero upon integration. If an
excited state has no dependence on then the integral n ( 0) H ' m( 0) will still yield zero. This allows
us to discard the two integrals involving 200 and 300. If an excited state involves an even function of
over the interval from 0 to 2, then we can also discard it since the integrand in the integral will still
be odd. This observation allows us to discard 320. We have reduced our thirteen integrations to just
Now you must perform the two integrals involving the remaining excited states. First take m( 0) 210
and evaluate the first term in the sum for 2 to find 148 a o 2 k . Then take m( 0) 310 and
evaluate the second term to find 0.20 a o 2 k . The sum of these two terms is the value of 2 using the
first thirteen terms in its expansion.
Some useful integrals:
cos3 x n!
cos x sin xdx x e
dx for a 0, n 0
a n 1
From Table 7.2 Eisberg & Resnick (pg. 243)