Download The Quantum Mechanical Harmonic Oscillator

The Quantum Mechanical Harmonic Oscillator
In this class, (Math 364A), we have already spent some time discussing the (onedimensional) classical harmonic oscillator (e.g., mass-spring systems). This problem is a
mainstay of 18th-century (Newtonian) physics. I would now like to give a brief description of an
important 20th-century successor to this problem: the (one-dimensional) quantum mechanical
harmonic oscillator. Although there’s no such physical thing as an ideal harmonic oscillator,
either classical or quantum mechanical, this provides a good beginning for a discussion of
molecular vibrations.
First, let’s review the classical problem. Put a mass on a spring. I would like to have as
few physical constants as possible to distract us from the essential mathematics, so I assume that
the mass m equals 1 and the spring constant k also equals 1. (An honest physicist would feel
obliged to return to this assumption and see what we would have for more arbitrary values of m
and k). Let this be the ideal mass-spring system with no damping or resistance forces at all.
Denote the displacement of the mass from its equilibrium position by x = x(t). Then this system is
governed by the differential equation x'' + x = 0, which has as its solutions simple harmonic
motion of angular frequency 1: x = C1cos t + C2sin t, which can be rewritten as x = R cos (t – ).
(These forms of the motion are connected by the equations C1 = R cos  and C2 = R sin , so that
1
C12 + C2 2 . R is called the amplitude of the motion.) Since kinetic energy = 2 ( x') 2 and
1 2
1
1 2
1 2
2
potential energy = 2 x , then the total energy = E = 2 ( x') + 2 x = 2 R . If we solve this
for R in terms of E, we have that the amplitude of the motion equals 2E .
To put ourselves in a sufficiently modern frame of mind, let’s assume that we don’t know
the exact initial conditions, but we do know the energy E. Let us suppose that the device is
oscillating in the dark until we set off a flash photograph at some randomly chosen instant. What
will the photograph say about the position? In the language of Math 380, the position is a random
variable which I will refer to as X. Furthermore, if we assume that the time at which we took the
picture is itself a random variable uniformly distributed on an interval equal to one period of the
motion, the we can compute the cumulative distribution function of X:
R=
P(X ≤ x) =





0
if
1 1
 x 
+
arcsin


2 π
 2E
if
1
if
x≤–
–
1
2E
1
1
<x<
2E
2E
x≥
1
2E
As in Math 380, we can now differentiate this to get the probability density function for x:
ƒ(x) =



1
2E – x 2
if
–
1
1
<x<
2E
2E
0
(1)
otherwise
In the context of classical mechanics, this probability density appears as an oversight. If
we could just take a few more measurements, we would discover the exact equation of the
motion as a function of time - at least that’s what all of the physicists for several generations after
Newton would have thought. In a wrenching change of attitude that took place during the first
third of the twentieth century, physicists learned a new way of thinking: the way of quantum
mechanics. Quantum mechanics gives up on ever being able to tell exactly where a particle is all we can learn about that particle’s position is a probability density similar to that of formula
(1). But (1) is not the correct formula for the quantum mechanical harmonic oscillator. No such
formula derived from classical considerations will ever be correct. Instead, we are faced with a
differential equation called Schrödinger’s equation for a function (x) whose solution we
interpret by making  2(x) the probability density function for the random variable X(the position
of the particle). If I choose a system of units in which Planck’s constant h equals 2π and if the
mass of the particle equals 1, then Schrödinger’s equation (one-dimensional, time-independent)
is
1 d 2
– 2 2 + V(x)  = E 
dx
(2)
2
with the condition that  be a probability density function:
∞
2

  (x) dx = 1
(3)
–∞
The function V(x) in equation (2) is the potential energy function. In the harmonic
1
oscillator case (and with the spring constant equal to 1) it is V(x) = 2 x 2. The manner in which
E appears in (2) is as an eigenvalue.
Now I will do four things to modify (2): bring the term on the right over to the left
1
(emphasizing that this is a homogeneous equation), multiply through by –2, substitute 2 x 2 for
V(x), and rename the function as u(x) instead of (x) (so I won’t have to keep changing fonts
while typing this). The result is the following differential equation:
u'' + (2E – x 2) u = 0
(4)
The integral condition (2) requires forces us to have lim u(x) = 0. This in turn means
x ± ∞
that we won’t be able to accept just any solution of the differential equation as having physical
meaning. A qualitative look at (4) reveals this: if |x| < 2E , then u'' = (some negative number)u,
so the graph of u curves toward the axis and thus has a chance to oscillate back an forth. If
|x| > 2E , then u'' = (some positive number)u, so the graph of u curves away from the axis.
Such curvature is exactly what you need to have the axis be a horizontal asymptote, but it takes
hitting a knife-edge of an exact set of conditions for this to happen; more likely, the curve swings
far away form the axis and keeps on going. Looking more closely, we see that for large x, the
differential equation is “almost” u'' – x 2u = 0. I can’t solve this at a glance, so I turn to the
2
equation u'' –  2u = 0, which has e  x and e –  x as solutions. If I now let  = x, I have e x and
e–x
2
as proposed so “almost” solutions, but they both seem to introduce unwanted powers of 2
2
2
into their derivatives. Finally, I settle on e x /2 and e – x /2 as my “almost” solutions.
You could be forgiven for not finding any point to this last calculation - I’ve found two
functions which aren’t exactly solutions to a differential equation which isn’t exactly my original
equation. But look at the method of variation of parameters: if we know the solutions of a
homogeneous equation (“almost” but not exactly the same equation), them by merely setting the
solution equal to a “linear combination” of the known solutions with arbitrary functions as
coefficients, we discover a soluble differential equation for the unknown coefficients. And
consider the method of reduction of order: if we know one solution of a second order equation, I
can set the solution equal to an unknown function times this known solution and obtain a soluble
differential equation for the unknown coefficient. In that same spirit, we suppose that the
solutions of (4) are arbitrary functions times the “almost” solutions of the “almost” equation.
Since I’m interested in solutions that go to 0 at ∞, I restrict my attention to functions times the
“almost” solution that tends to 0. That is, I assume that (4) has solutions of the form u =
ye
– x 2 /2
. Let’s substitute this into (4) to get an equation for y.
2
2
2
u' = y'e – x /2 + ye – x /2(–x) = e – x /2(y' – xy )
[
]
2
2
2
2
2
u'' = y''e – x /2 + y'e – x /2(–x) + y'e – x /2(–x) + y e – x /2(–1) + e – x /2(x 2)
(
2
u'' = e – x /2 y'' – 2xy' + (x 2 – 1)y
(
2
e – x /2 y'' – 2xy' + (x 2 – 1)y
so that (4) becomes
)
)
2
+ (2E – x 2) e – x /2 y = 0
y'' – 2xy' + (2E – 1) y = 0
(5)
If we now write the constant 2E – 1 as , we have exactly the Hermite equation given as
(21) at the top of page 205 of Boyce & DiPrima and solved on the remainder of that page. This is
a parity-preserving equation, so its solutions split into an even solution and an odd solution. If we
∞
assume that y(x) =
 anx n , then we arrive at the general recurrence relation
n =0
2n – 
an +2 = (n + 2)(n + 1) ,
n≥0
(6)
Two things are apparent from this recurrence relation:
a
x n +2
n
+2
lim 
anx n
n∞ 

2n – 
 = lim
|x| 2 = 0
(n
+
2)(n
+
1)

n∞

(7)
so for all x, the series converges (its radius of convergence is infinite), and

for n > 2 , if an ≠ 0, then
an +2
an
>0
(8)
so, after a certain point, all of the coefficients of the power series for y(x) have the same sign and for an entire function like y, that means that as |x|  ∞, |y(x)| grows faster than any
polynomial function of x. This last item presents us with a problem, since to have a physically
2
meaningful solution, we must have u(x)  0 as |x|  ∞. With u(x) being equal to y(x)e – x /2 ,
we can still have that for some possible y’s that grow faster than any polynomial - but we’re very
suspicious. We remember the idea that with the curve concave away from the axis it is very
difficult to get it to tend to that axis, and we recall that our two “almost” solutions for u(x) were
2
2
e – x /2 and e x /2 . It would be easy for all the solutions we are getting to resemble the latter for
large x. Is there any way out of this predicament? Equation (8) contains the key: consecutive
coefficients are the same sign unless they are zero. If any coefficient is zero, then all subsequent
coefficients of the same parity are zero - that is, one of the two independent solutions is a
polynomial. Without going through all of the argument necessary, let me state this: the only
physically meaningful solutions of this equation are the polynomial solutions.
Under what circumstances are there polynomial solutions? This can only happen if the
numerator in (6) is zero for some possible value of n, and that only happens if  is a nonnegative
even integer. So the allowable values of  are 0, 2, 4, 6, 8, ... , and the corresponding values of
1 3 5 7
the total energy are (substituting  = 2E + 1) E = 2 , 2 , 2 , 2 , ... . In the following table, we
go through the first few allowable even integer values of , set a0 = 1 and a1 = 0 if  is divisible
by 4 and a0 = 0 and a1 = 1 otherwise. Starting from that point, we compute the polynomial
2
solution for y(x). Finally, we set u(x) = Cy(x)e – x /2 , square this and integrate this from –∞ to ∞
to determine what C should be for this to be a probability distribution.

E
0
1
2
3
2
2
y(x)
u(x)
1
2
π– 1 /4 e – x /2
x
2
π– 1 /4 2 xe – x /2
4
6
8
10
5
2
7
2
9
2
11
2
1 – 2x
2
2
x – 3 x3
4
1 – 4x 2 + 3 x 4
4
4
x – 3 x 3 + 15 x 5
1
2
– x 2 /2
2 (1 – 2x ) e
2
 2 
π– 1 /4 3 x – 3 x 3 e – x /2


2
3
4 
π– 1 /4 8 1 – 4x 2 + 3 x 4 e – x /2


15  4 3 4 5 – x 2 /2
π– 1 /4
4 x – 3 x + 15 x  e
π– 1 /4