Lecture 11
1 The Time-Dependent and Time-Independent Schrödinger Equations
The time-dependent Schrödinger equation involves the Hamiltonian operator Ĥ and is formulated thus:
ĤΨ(x, t) = ih̄
∂Ψ
∂t
(1)
x stands for all the coordinates. If we define the energy operator Ê by
Ê ≡ ih̄
∂
∂t
(2)
we see that we can write the time-dependent Schrödinger equation as
ĤΨ(x, t) = ÊΨ(x, t)
(3)
Do not confuse this with an eigenvalue equation: the right hand side has an operator Ê, not a scalar value E.
For time-independent problems the Hamiltonian operator does not explicitly depend on the time t, i.e, Ĥ ≡ Ĥ(x, y, z).
We must have the probability density ΨΨ∗ independent of time. This requires that we write Ψ(x, t) as a product of
factors, one involving the time only, and the other involving the other coordinates. If this function is to satisfy the
time-dependent Schrödinger equation, it is easy to show that the time-dependent part must be of the form e −iEt/h̄ . We
then have
Ψ(x, t) = ψ(x) × e−iEt/h̄
∗
∗
Ψ (x, t) = ψ (x) × e
iEt/h̄
(4)
(5)
Note that Ψ(x, t)Ψ∗ (x, t) = ψ(x)ψ ∗ (x) is independent of time. Note also that since Et/h̄ must be dimensionless,
and h̄/t has units of energy, the parameter E must have units of energy. We can substitute eq 4 in eq 1 and show that
Ĥψ(x) = Eψ(x)
(6)
This is the time-independent Schrödinger equation. We see that in this case the wave-function ψ is an eigenfunction
of the Hamiltonian operator with E as its eigenvalue. Eq 6 is called the time-independent Schrödinger equation and ψ
time-independent wave function. This is the equation that we use when the Hamiltonian operator does not explicitly
depend on the time and the system does not change with time (stationary). In cases like the interaction of molecules
with light, the Hamiltonian operator depends explicitly on the time, i.e, Ĥ ≡ Ĥ(x, y, z, t), the wave function Ψ(x, t)
cannot be factored according to eq 4 and we now have to use the time-dependent wave equation. We then calculate the
energy of the system by the recipe
E ≡ hĤi
(7)
= hΨ(x, y, z, t)|Ĥ(x, y, z, t)|Ψ(x, y, z, t)i
(8)
This is in accordance with the recipe in quantum mechanics that any measured observable O for a system described by
a wave function Ψ is to be compared with the quantum mechanical average of the corresponding operator Ô, written
as hÔiΨ , or more briefly, hÔi which stands for (in Dirac’s notation) hΨ|Ô|Ψi.
1
You will recall that in Dirac’s notation
hf |Ô|gi ≡
Z
f ∗ (x)[Ôg(x)]dτ
(9)
dτ is the volume element for the coordinate system considered: dxdydz for Cartesian coordinates, r 2 sin θdrdθdφ
for spherical coordinates. The integral will be a multiple integral, and you have to use the limits of integration. You
will see that the integrand is obtained by first operating on g with the operator Ô and multiplying the result with the
complex conjugate of f .
2 Operators
You will recall that the Hamiltonian operator is obtained from the Hamiltonian of classical mechanics by substituting
for the momenta and coordinates in the classical Hamiltonian by the corresponding operators. You will also recall
that any function of the coordinates and momenta has a corresponding operator, obtained by substituting for any
coordinate qi the corresponding operator qˆi ≡ qi (i.e., operation by qˆi is the same as multiplying by the value of qi )
h̄ ∂
∂
=
where qi is the coordinate conjugate to the momentum pi . It is
and the momentum operator pˆi ≡ −ih̄
∂qi
i ∂qi
convenient to introduce the time operator t̂ ≡ t, i.e., operation with t̂ has the same effect as multiplying by t.
You will also recall that the result of successive operation by two operators Oˆ1 and Oˆ2 depends on the order, i.e.
Oˆ2 Ô1 6= Oˆ1 Ô2 . The difference between these is referred to as the commutator of the two operators, denoted by
[Oˆ1 , Ô2 ]:
[Ô1 , Oˆ2 ] ≡ Ô1 Oˆ2 − Oˆ2 Ô1
(10)
When the commutator of two operators is zero, then the order of operation does not matter and the operators are said
to commute. You will recall that
Lˆ2 Y`m = h̄2 `(` + 1)Y`m
Lˆz Y`m = m` h̄Y`m
(11)
(12)
We see that the spherical harmonic function Y`m is eigenfunction of both the operator corresponding to the square of
the angular momentum, Lˆ2 , and the operator corresponding to the z− component of the angular momentum operator
Lˆz . Using this you should be easily able to verify that [L2 , Lz ] = 0. This is a general feature: when two operators
commute, it is possible to find a set of functions that are simultaneously eigenfunctions of both operators. This has the
consequence that the quantities corresponding to both operators can be measured simultaneously without error.
When two operators do not commute, it is not possible to find functions which are eigenfunctions of both operators,
and the quantities corresponding to the operators cannot be measured simultaneously with accuracy. As examples of
operators that do not commute, we have
[x̂, pˆx ] = ih̄
(13)
[Ê, t̂] = ih̄
[Lˆx , Lˆy ] = ih̄Lˆz
(14)
[Lˆx , Lˆy ] = ih̄Lˆz
[Lˆy , Lˆz ] = ih̄Lˆx
(16)
[Lˆz , Lˆx ] = ih̄Lˆy
(18)
2
(15)
(17)
Particularly important are operators that commute with the Hamiltonian operator. Stationary state wave functions are
eigenfunctions of the Hamiltonian operator, so that only quantities whose operators commute with the Hamiltonian
can be measured accurately. For operators that do not commute with the Hamiltonian, we seek the averages of these
operators, and the standard deviation will give a measure of the “spread”. Thus, for the x̂ operator, which does not
commute with the Hamiltonian,
hxop i ≡ hψ|x|ψi
hx2op i
2
(19)
≡ hψ|x |ψi
(20)
hpx,op i ≡ hψ| − ih̄
(21)
∂
|ψi
∂x
2
∂
hp2x,op i ≡ hψ| − h̄2 2 |ψi
∂x
(22)
(23)
We can calculate ∆x ≡ [hx2op i − hxop i2 ]1/2 and ∆px ≡ [hp2x,op i − hpx,op i2 ]1/2 and verify, for a case for which ψ is
known (for example, the particle in a box), Heisenberg’s uncertainty principle:
∆x∆px ≥ h̄/2
(24)
3 Perturbation Theory
Perturbation Theory is a method for solving a problem in terms of the solutions for a very similar problem. Suppose
that we has solved the time-independent Schrödinger wave equation for a problem with Hamiltonian Ĥ (0) . Let the
(0)
(0)
(0)
(0)
(0)
(0)
(0)
solutions be ψ1 , ψ2 , ψ3 . . . with corresponding energy levels E1 , E2 , E3 , . . .. This means that Ĥ (0) ψ1 =
(0) (0)
(0)
(0) (0)
(0)
(0) (0)
E1 ψ1 , Ĥ (0) ψ2 = E2 ψ2 , . . . Ĥ (0) ψi = E0 ψi . We assume that the solutions are not degenerate; when
there is degeneracy the equations given below will have to be modified slightly.
The superfix (0) represent the unperturbed problem, and subscripts 1, 2, . . . represent the ground state, the 1st excited
(0)
state, etc. Thus Ĥ (0) is the Hamiltonian for the unperturbed problem, ψ1 is the ground state wave function for the
(0)
unperturbed problem, and E3 is the energy of the 2nd excited state for the unperturbed problem.
Let the Hamiltonian for the problem we are interested in (the perturbed problem) be of the form Ĥ = Ĥ (0) + Ĥ (1) ,
where Ĥ (1) is small compared to Ĥ (0) . It is reasonable to suppose that the solutions for the perturbed problem will
be close in some sense to the solutions of the unperturbed problem. We make use of the following fact: the solutions
(0)
ψn for the unperturbed problem form a complete set, i.e., any arbitrary function, in particular the wave functions for
the perturbed problem, can be written as a linear sum of these. Our assumption can be formulated thus:
X
(0)
ψn = ψn(0) +
c i ψi
(25)
i6=n
En =
En(0)
+ En(1) + En(2) + . . .
(26)
In effect we are writing down a correction for both ψn and En in the form of a series. We will see that many of the
higher correction terms will be small so that we are left with a few corrections terms only.
Note that ψn is not normalized.
We summarize the final results.
3
• As a result of the perturbations the wave functions are a “mixture” of the solutions for the unperturbed problem.
(0)
For every n, the perturbed wave function ψn is largely the unperturbed wave function ψn with a little admixture
of the other unperturbed wave functions.
(0)
(0)
• The coefficient ci is a measure of how much ψi makes a contribution to (has got “mixed into”) ψn . Its
contribution to the probability density function will be proportional to c 2i .
• The value of ci may be calculated from
(0)
ci =
(0)
hψn |Ĥ (1) |ψn i
(0)
(0)
En − E i
i 6= n
(27)
(0)
Thus the coefficient ci , the coefficient of ψi in ψn , is equal to the matrix element of the perturbation Ĥ (1)
(0)
(0)
between the unperturbed wave functions ψi and ψn divided by the energy difference between the ith and nth
unperturbed levels. If the matrix element is zero (for symmetry reasons, for example), then c i is zero. Because
of the energy term in the denominator, the levels close to n make a greater contribution than those further away.
(1)
• The first order correction to the energy, En is given by the average of the perturbation Ĥ (1) over the unper(0)
turbed wave function ψn :
En(1) = hĤ (1) iψ(0)
(28)
n
= hψn(0) |Ĥ (1) |ψn(0) i
(29)
(2)
If this is zero for reasons of symmetry, we would be interested in the 2nd order correction E n .
(2)
• The 2nd order correction En is given by
En(2) =
X |hψ (0) |Ĥ (1) |ψn(0) i|2
i
(0)
(0)
En − E i
i6=n
(30)
Since Ĥ (1) is small and it occurs in two factors, the second order correction is smaller than the first order
correction. Here too we see that levels closest to the nth level make the greatest contribution. The levels higher
than n make a positive contribution (push the energy up) while those levels lower than n make a negative
contribution (push the energy down).
3.1 Example
We treat the polarization of the H atom in its ground state by a uniform electric field. In the presence of an electric
field the H atom will develop an induced dipole in the direction of the field which is, to first order, directly proportional
to the electric field:
µind = αE
(31)
Both the dipole moment and the electric field are vector quantities. The constant of proportionality is referred to as the
polarizability. The energy due to the polarization is given by
1
Energy of polarization = − µind .E
2
1
1
= − αE.E = − αE 2
2
2
4
(32)
(33)
where E is the magnitude of the electric field. Since the unit of E is V/m (volt per meter) and that of µ is C m, the
unit of α is C m/(V/m) = C m2 V−1 . In the c.g.s or Gaussian system of units the electric permeability ε 0 is taken as
unitless while in SI units it has a value of 8.854 × 10−12 C2 N−1 m−2 . In the Gaussian system polarizability has units
of m3 . You can convert the SI polarizability to Gaussian polarizability by dividing the SI polarizability by 4πε 0 .
We need to to figure out the perturbing Hamiltonian Ĥ (1) . Recall that the energy of a charge q at a point where the
electric potential is V is qV and that E is related to V by the gradient:
Ez = −
∂V
;
∂z
Ex = −
∂V
;
∂x
Ey = −
∂V
∂y
(34)
Consider the uniform electric field to be in the z direction. Then the potential (with respect to the origin) is
V = −zEz = −r cos θEz = −r.E
(35)
The additional energy due to the field is zero for the nucleus (since V at the origin is zero) and the energy for the
electron is −eV = er cos θEz = ezEz . We therefore have
Ĥ (1) = er cos θEz = ezEz
(36)
We see that Ĥ (1) is directly proportional to z and is therefore odd: Ĥ (1) (z) = −Ĥ (1) (−z)
p
We consider only the 1s, 2s, and 2p unperturbed levels. The 1s and 2s orbitals depend only on r = x2 + y 2 + z 2
and are therefore even. The px , py and pz orbitals are proportional, respectively, to x, y and z and are therefore odd.
e2
1
.
The energy of En is −Rhc0 /n2 = −
4πε0 2a0 n2
Recall that in the unperturbed state the H atom by assumption is in the 1s state. The first order correction to the energy
is, by eq 29
E (1) = h1s|ezEz |1si
(37)
This is zero since the product of the two 1s functions is even and z is odd: the integrand is odd in z and even in y and
x so that the whole integral is zero. We therefore have to calculate the second order corrections. In doing this we have
to bear in mind that the integrals will be of the form
ZZZ
Z
Z
Z
f (x)g(y)h(z)dxdydz = f (x)dx g(y)dy h(z)dz
(38)
and will be zero if the integrand is odd in any one of the variables x, y or z. If we restrict ourselves only to the 2s and
2p levels, the integrals involved in the second order correction are
h2s|ezEz |1si = 0
h2px |ezEz |1si = 0
h2py |ezEz |1si = 0
h2pz |ezEz |1si 6= 0
(39)
(40)
(41)
(42)
In the last integral, the integrand is even in x and y, and since z occurs as z 2 (one z from 2pz and one from ezEz ) it
is even in z as well: the integrand is non-zero. To write out this integral in full we need to go to spherical coordinates,
and write z as r cos θ:
3/2
5/2
ZZZ
1
1
1
1
−r/2a0
√
re
e−r/a0 r2 sin θdrdθdφ (43)
cos
θ|eE
r
cos
θ|
h2pz |ezEz |1si =
z
π a0
4(2π)1/2 a0
Z 2π
Z π
Z ∞
eEz
2
dφ
sin θ cos θdθ
r4 e−3r/2a0 dr
(44)
= √ 4
4π 2a0 0
0
0
5
In evaluating the radial integral we use
R∞
0
xn e−qx dx =
n!
, n > −1, q > 0. The integral evaluates to
q n+1
a0
h2pz |ezEz |1si = 12 × (2/3)6 × √ eEz
2
According to eq 30 we need to square this and divide by E1 − E2 = −
(45)
3Rhc0
. We then get, using eq 33
4
1
122 × (2/3)12 a20 e2 Ez2
− αEz2 = −
2
2 × (3Rhc0 )/4
(46)
This enables us to get an expression for the polarizability of the H atom in the ground state.
In the same way you could calculate the polarizability of the H atom when it is in the n = 2 state. The diagram shows
how the energy levels are changed when the electric field is switched on; we have neglected the effect of n = 3 and
higher levels.
Sfrag replacements
(0)
|2pz i + cs |1si
mixed state
|2si, |2pxi, |2py i
pure states
2s, 2p
E2
perturbation “pushes apart” energies
(0)
E1
1s
Ĥ
Ĥ (0)
6
(0)
|1si + cp |2pz i
+ Ĥ
(1)
mixed state