Degeneracy CHEM 614
MODULE 2
MOTION, FREE AND CONFINED
In this module we begin our examination of whether the Schrödinger equation that we have
arrived at is capable of being of any practical value to us. We begin this with a look at the
motion of particles, large and small. Note that the concept of “smallness” is not absolute; it
depends on whether our method of observation imparts any significant perturbation on the state
of the body being measured.
Free Motion
(a) Macroscopic particles (grains of sand, bullets, balls, planets, galaxies …)
Consider a ball that can move in the x-dimension, without potential energy barriers, i.e., V(x) =
0. Like this:
x
and not like this, where potential barriers exist:
V2
V1
Our ball has a mass m and velocity v in either direction (x or -x). Because the object is a macroentity we have no problem in determining its instantaneous position and velocity. From the latter
we can arrive at the momentum
p  mv  m
dx
dt
(2.1)
Then we can calculate the kinetic energy
T  mv 2 / 2  p 2 / 2m
(2.2)
Experience tells us that there are no theoretical restrictions on the values of p or T. They can be
provided with any value we (or Mother Nature) choose.
1
(b) Nano-object
Suppose the entity we are considering is of submicroscopic size, for example a carbon atom, and
let us use our newly formulated Quantum Mechanics to analyze the motion. Our first two
postulates tell us that there will be a wave function that describes the system and that there will
be an operator that we can use to extract the total energy (in this case it is all kinetic energy since
V(x) = 0) and the momentum. In fact we can use the Schrödinger equation
Ĥ  E
(2.3)
where Ĥ is the operator for total energy and E is its eigenvalue. What we need to do is to solve
this eigenvalue equation to evaluate E.
Since we are setting V(x) = 0
Hˆ  Tˆ
(2.4)
and after substitution (2.3) becomes
Tˆ  Ekin
(2.5)
where Ekin is the eigenvalue for kinetic energy. We saw in Module 1 that in one-dimension
Tˆ  
2
d2
2m dx 2
(2.6)
Then our equation for the motion of our quantum body in one dimension is

d 2 ( x )
 Ekin ( x )
2m dx 2
2
(2.7)
or
d 2 ( x )
 k 2 ( x )
dx 2
where k 2  2mEkin /
(2.8)
2
We know (Math Module A or Barrante: 6) how to solve linear second order differential
equations of the kind shown in (2.8). A general solution is
 ( x)  Aeikx  Beikx
(2.9)
and
Ekin 
2
k 2 / 2m
(2.10)
You can verify that this is a solution by substituting in to (2.8) and differentiating twice.
2
Note that equation (2.9) is a linear combination of particular solutions of equation (2.8). The
linear combination is a more general solution.
Our carbon atom is a free particle and it can continue unhindered in motion between x = infinity
and x = - infinity. Thus our general solution cannot be made more specific and therefore Ekin
can have any value we wish.
Thus a carbon atom that moves without restriction along the x dimension (or any other) in an
unbounded fashion (except constrained to x) is not subject to any energy restrictions.
An interesting point is that if we put Ekin = p2/2m (the classical energy-momentum connection)
into (2.10), we obtain
p2 
2 2
(2.11)
k
or
p k
(2.12)
This is identical to the de Broglie equation if k = 2/Thus we can find the parameters of the
matter waves thatmove along with the body in motion. Thus our carbon atom (or electron, or
neutron, or HCl molecule, etc) when in unbounded motion behaves exactly as it would if
Newton's mechanics held.
Thus, size itself is not a determining factor.
Confined Motion
Now we consider the same particle (carbon atom) on the same one-dimensional track, but now
we place two totally reflective walls on the track to keep the particle confined within a fixed
region of one-dimensional space. Again the potential energy between the walls is zero.
TO INFINITY
TO INFINITY
L
X
3
This enclosure can be constructed, at least in our imaginations, by making the reflective walls
from potential energy barriers of infinite height.
Thus Vx = 0 = Vx =L = ∞
and V(x) = 0 when 0 < x > L
This problem is referred to as the particle in a one-dimensional box, or trap, or square well.
We follow the same procedure as above, writing down the Schrödinger equation
Ĥ  E
(2.13)
and expanding the hamiltonian operator (since Vx = 0) as
Hˆ  Tˆ  Vˆ  Tˆ
Proceeding as above our eigenvalue equation becomes

d 2 ( x )
 Ekin ( x )
2m dx 2
2
(2.14)
This is again familiar to us and we are in a position to write down the general solution as either
 ( x)  Aeikx  Beikx
(2.15)
 ( x )  C sin(kx )  D cos(kx )
(2.16)
or
where A,B,C, and D are constants.
We can use either of these but here let us choose (2.16) and recognize that we can make this
general solution more specific by applying boundary conditions. But first we must decide what
the boundary conditions are. To do this let us see what is happening outside the box.
We posed the problem by calling it the particle confined in a one-dimensional trap and we might
assume that because of this the probability of finding the particle in some element dx outside the
walls is zero. Hence *dx = 0, or  = 0, since dx is a real and finite space element and cannot
be zero.
However we have no way of knowing that we are permitted to make this assumption, so we must
inquire further.
Outside the trap V(x) = ∞, and by writing down the Schrödinger equation for this case we obtain

d 2 ( x )
 ( Ekin  V ) ( x )
2m dx 2
2
4
(2.17)

whence
d 2 ( x )
  ( x )
dx 2
(2.18)
The implication is that the curvature of the function (given by the second derivative) is infinite.
This is certainly possible, but only for an infinitesimal range of x, and it cannot be infinite
everywhere outside the box. Thus it is most likely that (out) = 0
And thus *dx = 0 outside the box, i.e. there is no probability of finding the particle outside the
walls, as per our assumption above.
Note that this is only valid when the potential at the walls goes to infinity.
Now, since (out) = 0, and our postulates require the wavefunction to be continuous, so as
(inside) approaches the walls at x = 0 and x = L, it must asymptotically approach zero. Thus
our boundary conditions are
 ( x  0)   ( x  L )  0
Now let us see what these boundary conditions do to equation (2.16).
Putting (0) = 0, we see that C sin(k 0)  D cos(k 0)  0 , but cos 0 = 1 hence D = 0, and our
general solution reduces to

 ( x )  C sin(kx ) 
which satisfies the x = 0 boundary condition. At the other boundary, (L) = 0, i.e.
 ( L)  0  C sin(kL)
This is acceptable if C = 0, but in this case the full solution becomes trivial.
However, at x = L, let sin(kL) = 0
Sine functions become zero when the argument is zero, i.e., at integral multiples of ,
that is, when kL = 0, , 2, 3,…, or kL = n, where n = 1, 2, 3,…
and k = n/L
[the n = 0 possibility is discarded because it leads to another trivial solution.]
Thus the application of the boundary conditions to  leads to (x) = Csin (kx)
with kL = n, and n =1,2,3,…
or
 ( x )  C sin(n x / L)
5
(2.19)
and we see that our wavefunctions form a set of discrete sine equations, governed by integral
multiplier in the argument.
We saw above that the original eigenvalue equation that we solved had the kinetic energy as the
eigenvalue, thus
E  2 k 2 / 2m
this is an identical situation (where V = 0) and we can now use this and substitute for k  n / L
whence
En  n 2h 2 / 8mL2
[note 2.20 has h and not
(2.20)
]
Thus, restricting the motion of the particle to a defined region of space imposes restrictions upon
the value that the kinetic energy can take. Applying the boundary conditions has removed all the
non-integral solutions, there remains a very large number, because there are a large number of
integers! The energy is quantized-it can take discrete values only (even though there is a large
number of allowed values corresponding to the integers). We can now imagine that whenever
we confine a particle, by whatever means, we can expect a similar result, viz., that the allowed
energy states are restricted. This is typical of quantum behavior-it is the true natural state of
motion of matter when it is confined, but it only reveals itself when m and L are small compared
to h. The above treatment has been truly theoretical, arguing from the postulates, and using
operators that we have constructed. We have been able to arrive at the form of the wave function
of a particle in a box, and at values for the eigenstates of the system. Now we can see clearly
why transitions between states are quantized.
Zero point energy
From equation (2.20) we see that the lowest energy that the particle can have is when n = 1. This
leads us to the concept of zero point energy
E1  h 2 / 8mL2
The zero point energy is a purely quantum mechanical concept that has no counterpart in
classical physics. We can think of this in terms of curvature. We argued above in setting up the
boundary conditions, that  must be zero at the inner walls in order to maintain continuity with
the zero function outside the walls. A function that is zero at two points can be zero or non-zero
between the points. If = 0 between 0 and L, then we have no particle in the box (* must be
finite), and thus we have to conclude that  must always be greater than zero within the box.
6
Any function that has non-zero amplitude between points at which it is zero must have curvature,
and thus some kinetic energy. As the length of the box is increased for constant m, E1 decreases,
and the limit is when L is infinite, E1  0 . In this case the particle is without restriction and the
quantization of its kinetic energy is removed and, in consonance with a classical particle, the
zero point energy vanishes.
Normalization
To complete our wavefunction we need to find the value of the constant C that precedes the sine
function in (2.19). This can be achieved by normalization, i.e., what must be the value of C that

 * dx  1
all space
gives the probability of finding the particle somewhere in the well equal to unity. In our case "all
space" is the x-dimension between 0 and L, thus
 n x 
C 2  sin 2 
 dx  1
 L 
0
the integral is a standard form and integration leads to C = (2/L)1/2
L
Thus the full wave function is
2
 n x 
(2.21)
sin 

L
 L 
Figure (2.1) shows a plot of equation (2.21) for an electron (m = 9.31x10-31kg) in a well with L =
 ( x) 
1000 pm, for n = 1,2,3,4 and 5. The vertical scale is energy (in eV) as a function of n.
11





x 1
x 2
11
8.8
6.6
x 3
x 4
x 5
0
4.4
2.2
0
0
200
400
0
600
x
Figure 2.1
7
800
1000
1000
Figure 2.2 shows plots of 2 as a function of n.
11.5
z
x 1
z
x 2
11.5
9.2
6.9
z
x 3
z
x 4
4.6
z
x 5
2.3
0.376
0
0
0
200
400
600
x
800
1000
1 10
3
Figure 2.2
Remember that En is a set of eigenvalues of the hamiltonian operator and has a definite value for
each value of n, i.e. E is a constant for all the different stationary states of the particle. In our
derivation we have forced V to be zero and thus H = T . However, V need not be zero for other
systems (see later), moreover, V can vary within the confines of the system. In such cases, T
will also vary in such a way as to keep E constant.
At this point I recommend that you familiarize yourselves with the physics of the stretched
string, as discussed in Lowe 1-3 and 1-4. You find there that the motion of a violin, e.g., string (
a macro object) when plucked or bowed can be exactly analyzed by Newton’s second law. The
equation of motion is a second order differential equation in length and time which can be
separated into a pair of eigenvalue equations, one spatial and one temporal. Solving the spatial
equation and applying the boundary conditions that the vibration must have zero amplitude at the
points where it is clamped leads to the result that the string can only oscillate with frequencies
that are integral multiples of some fundamental frequency, exactly analogous to the particle in
the box, described above. Thus the application of Newton’s laws to a macroscopic system leads
to exactly the same kind of quantum behavior as the matter waves of the particle confined within
8
the infinite well. In both cases we see that the functions that are allowed fit an integral number
of half sine waves into the space delineated.
Orthogonality
The different wavefunctions of the particle are orthogonal, which means that
L
 
n
m
dx  0
0
when n ≠ m.
First we use a symmetry argument to show this is true in certain cases. Observe that the n = 2
function in Figure 2.1 is antisymmetric (equal amounts of positive and negative area) and its
integral between 0 and L vanishes (equals zero). All the even-numbered wavefunctions are
antisymmetric about the centerline (x = L/2) and have vanishing intervals; the odd-numbered
wavefunctions are symmetric about the centerline and do not vanish on integration. In general
the product of a pair of symmetric functions is symmetric (  x    ) as is the product of a pair of
antisymmetric functions (  x    ) .
The product of an anti-symmetric function with a
symmetric one yields an antisymmetric function (  x   ) . Thus we see that the orthogonality
relationship is true for the product of symmetric (n is odd) and antisymmetric (n is even)
functions, because the product will be odd and its integral will therefore vanish. We cannot see
this so straightforwardly for even products but substituting for the wavefunctions and evaluating
the integrals shows that all products with n ≠ m obey the orthogonality condition.
Later we shall see a proof of a theorem that the eigenfunctions of a Hermitian operator
corresponding to different eigenvalues are orthogonal.
Earlier we saw that the normalization condition was
*
   dx  1
all space
and since  and * refer to the same quantum state (say m) we can write

 m* mdx  1
all space
The conditions of normality and orthogonality can be combined into a single orthonormality
relation

 n* m d   n ,m
(2.22)
all space
and  n ,m  1 when n = m (normal), and  n ,m  0 otherwise. This is the Kronecker "delta" and is
nothing more than convenient shorthand.
9
The orthonormality condition can be expressed very succinctly with the use of the Dirac bra-ket
notation:
(2.23)
n m   n,m
here the combination of the bra and the ket with the vertical line separating implies the all space
integration of the product of the pair of functions. Using this notation we have normality as
m m 1
(2.24)
n m 0
(2.25)
and orthogonality as
Integrals over operators
In quantum chemistry we seek to make contact between calculations done using operators and
the actual outcome of experiments. This usually requires us to evaluate certain integrals, all of
which have the form
ˆ g d
I   f *
(2.26)
where f and g are wavefunctions and ̂ is an operator. The integration is carried out over the
volume element d, which in one dimension is simply dx. In Dirac notation the integral would
be symbolized as
ˆ g
I f 
(2.27)
An even briefer symbolism would be
ˆ
I 
(2.28)
f ,g
here, the RHS has the appearance of an element of a matrix placed at the intersection of row f
and column g. For this reason the integral is referred to as a "matrix element". We shall meet
many of these in the next few months.
In many cases, the operator ̂ is simply multiplication by 1and then the matrix element has a
special symbol
(2.29)
f 1 g  f g S
The parameter S is given the name "overlap integral" and is a measure of the similarity of a pair
of functions. S can take values from 0 to 1. We have already seen the limiting values of S in the
f g   n ,m
orthonormality condition. 
10
The two-dimensional well
The two-dimensional well relates to the one dimensional well in exactly the same way as a drum
skin relates to a guitar string. Here the potential is infinite at all points outside the rectangular
well of sides L1 (on x) and L2 (on y), and zero within. Inside the walls the eigenvalue equation is
 2  2
2mE
 2  2 
2
x
y
(2.30)
After separation of the x and y variables, finding the general solutions, and applying the
boundary conditions and normalizing we find
 k ,l 
 k x   l y 
2
sin 
 sin 

L1L2
 L1   L2 
(2.31)
and
Ek ,l 
h2  k 2 l 2 
  
8m  L12 L22 
(2.32)
where k and l are independent quantum numbers and equal 1,2,3,…
Degeneracy
When the box is square then L1 = L2 = L and the energies are given by
h2
Ek ,l 
(k 2  l 2 )
(2.33)
2
8mL
Thus the states represented by 1, 2 and 2,1 will have the same energies even though their wave
functions are different.
The degenerate behavior comes about because of the symmetry that arises from the square. With
a rectangular box, the symmetry is lost, as is the degeneracy.
11