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8.04: Quantum Mechanics Professor Allan Adams
Massachusetts Institute of Technology 2013 March 5
Lecture 8
Quantum Harmonic Oscillator: Brute Force Methods
Assigned Reading:
E&R 5
all
1 , 2 , 8
Li. (^3) all, 41 , 51 , (^6) all
Ga. 2 4
all
Sh. 4 all
1 , 2
We will now continue our journey of exploring various systems in quantum mechanics for
which we have now laid down the rules.
Roughly speaking, there are two sorts of states in quantum mechanics:
- Bound states: the particle is somewhat localized and cannot escape the potential:
- Unbound states: the particle can escape the potential.
Note that for the same potential, whether something is a bound state or an unbound state
depends on the energy considered.
Figure 1: For the finite well, the energy represented by the lower black line is for a bound
state, while the energy represented by the upper black line is for an unbound state
But note that in qunatum mechanics, because of the possibility of tunneling as seen before,
the definition of whether a state is bound or not differs between classical and quantum
mechanics.
The point is that we need to compare E with lim x→±∞ V (x) to determine if a state is bound
or not.
Why do we split our cases that way? Why do we study bound and unbound states separately
if they obey the same equations? After all, in classical mechanics, both obey F = p˙. In
quantum mechanics, both obey
∂ψ(x, t) ˆ Eψ(x, t) = iJ.
∂t
Figure 2: The same energy denoted by the black line is a bound classical and quantum state
for the potential on the left, while the classical bound state is a quantum unbound state for
the potential on the right
The distinction is worth making because the bound and unbound states exhibit qualitatively
different behaviors:
Mechanics Bound States Unbound States
Classical periodic motion aperiodic motion
Quantum discrete energy spectrum continuous energy spectrum
For now, we will focus on bound states, with discussions of unbound states coming later.
Let us remind ourselves of some of the properties of bound states.
- Infinite square well:
- Ground state: no nodes
- nth excited state: n nodes (by the node theorem)
- Energy eigenfunctions chosen to be real
−
iEt
- Time evolution of energy eigenfunctions through complex phase e �^.
- Different states evolve at different rates
- Energy eigenstates have no time evolution in observables as p(x) for such
states is independent of t
- Time evolution of expectation values for observables comes only through in
terference terms between energy eigenfunctions
- Symmetric finite square well
- Node theorem still holds
- V (x) is symmetric
- Leads to symmetry or antisymmetry of φ(x; E)
- Antisymmetric φ(x; E) are fine as |φ(x; E)|
2 is symmetric
- Exponetial tails in classically forbidden regions leads to discrete energy spectrum
(picture of shooting for finite square well?)
- Asymmetric finite square well
�
is the energy eigenvalue equation for the harmonic oscillator. This is not an easy differential
equation to solve! For now, we will solve this through brute force methods; later, this will
be solved with more sophistication.
Before we dive into the brute force method, though, let us take a look at what we already
know:
- From dimensional analysis, we know that E ∝ Jω.
- From the uncertainty principle, we know that the ground state energy E 0 ∝ Jω.
- The energy operator is
pˆ
2 1 ˆ E = + mω
2 xˆ
2 .
2 m 2
There is a nice symmetry between xˆ and pˆ. Up to constants, our system looks the same
whether we view it in position space or in momentum space. That is to say, there will
be symmetries between φ(x; E) and φ
(k; E). Do we know of a function that looks the
same in both position space and momentum space? In other words, do we know of a
function that is functionally similar to its Fourier transform? We do, and that is the
Gaussian! We should expect to see some connection between the harmonic oscillator
eigenfunctions and the Gaussian function.
- By the node theorem, φ(x; n) should have n nodes.
- Here is a sneak preview of what the harmonic oscillator eigenfunctions look like: (pic
ture of harmonic oscillator eigenfunctions 0, 4, and 12?)
Our plan of attack is the following: non-dimensionalization → asymptotic analysis → series
method → profit! Let us tackle these one at a time.
We need to non-dimensionalize the equation
J
2 ∂
2 φ(x; E) 1
− + mω
2 x
2 φ(x; E) = Eφ(x; E). (0.4)
2 m ∂x
2 2
It is not good to carry around so many constants in that equation. Note that
m
J
2
2
ω
2
is
dimensionally the same as x
− 4
. This implies that we should define the constant
J
α ≡
mω
as our length scale. We can then define u ≡ α
x to non-dimensionalize position. Furthermore,
we know that E ∝ Jω, so non-dimensionalizing energy as ε =
E should work. This leaves ω
2
the energy eigenvalue equation in fully non-dimensional form as
2 φ
= (u
2 − ε)φ. (0.5)
∂u
2
ℏ
This is certainly much neater, but it is not any easier. To make things easier, we need to
develop some more mathematical techniques.
As a mathematical detour, we need to discuss asymptotic analysis. Let us suppose
that we are solving the equation
∂f (x) 1
∂x x
In asymptotic analysis, we look at the different limits. Let us consider that
lim 1 + ≈.
x→ (^0) x x
This means that
lim
x→ 0
∂f (x)
∂x
f (x)
x
which is solved by
f (x) = f 0 x
− 1 .
But this only considers the behavior as x → 0. Let us say that
f (x) = x
− 1 g(x)
where g(x) is finite and well-behaved as x → 0. Plugging this in yields
∂g(x)
= −g(x),
∂x
which is solved by
g(x) = g 0 e
−x .
This means that the full solution is
g 0 f (x) = (0.8)
xe
x
which is exact! The first approximation was just to make our lives easier, and
there ended up being no accuracy sacrificed anywhere. Furthermore, e
−x is indeed
relatively constant compared to x
− 1 as x → 0. To summarize, we looked at the
extreme behavior of the differential equation to peel off a part of the solution.
We then plugged that back in to get a simpler differential equation to solve.
∂g(x) But what if we did not know how to solve ∂x
- g(x) = 0? How would we figure
it out?
First, we apply asymptotic analysis:
lim (u
2 − ε) ≈ u
2 (0.17)
u→±∞
so
2 φ
lim ≈ u
2 φ. (0.18)
u→±∞ (^) ∂u 2
We can try
αu
2
2 φ(u) = φ 0 e.
We find that
∂φ
= αuφ
∂u
and
2 φ
= (α + α
2 u
2 )φ ≈ α
2 u
2 φ
∂u
2
in the same limit. Comparing this to our differential equation, we see that we need
α
2 = 1
so
α = ± 1.
We then get
2 2
2 2 φ(u) = φ A e
u
−
u
.
The first part is not normalizable, though, so we do not want that. We keep only the
second part and generalize the constant to a function of u that is relatively constant and
well-behaved as u → ±∞, so we say that
φ(u) = s(u)e^2
−
u 2
Now recall that the last step in asymptotic analysis is to plug this form into the original
differential equation to yield a differential equation that can be solved for s(u):
2 s ∂s
− 2 u + (ε − 1) s = 0. (0.19)
∂u
2 ∂u
Before we solve this, we need to figure out the behavior of s(u). We know that it should grow
2
less rapidly than e^2
u
as u → ±∞. Also, there should be multiple solutions corresponding
to discrete bound states. Finally, the nth solution should have n nodes.
Let us try the power series expansion
∞
s(u) = a j u
j (0.20)
j=
as our candidate solution. Plugging this into the differential equation yields
∞
((j + 1)(j + 2)a j+ − (2j + 1 − ε)a j )u
j = 0. (0.21)
j=
Again, as this is true for all u, the overall coefficients must all be identically zero. This
means that
2 j + 1 − ε
a j+ = a j
(j + 1)(j + 2)
is the recursion relation. This relates every other coefficient. This means that we need to
specify both a 0 and a 1 to find all of the coefficients. Why is this the case? It is because we
had a second-order differential equation. Does this meet our expectations? Let us consider
large j:
2 j a j lim a j+ ≈ a j
j→∞ (^) j 2 j
2
so
C
a j
j
2
and
j 2 n u u (^2)
s(u) ≈ C j
= C = Ce
u
. (0.25)
! n! j 2 n
Unfortunately, this is exactly what we do not want, as plugging this into φ(u) = s(u)e
−
u 2
2
recovers the non-normalizable solution.
Remember from before that to get a normalizable wavefunction, we had to impose a specific,
discrete set of energies. If we can do that here, can we get things to work?
The only way out of this conundrum is that the series must be finite. In particular, let us
suppose that that there exists some n such that when j = n, the numerator 2 j + 1 − ε = 0.
Then all of the subsequent a j = 0. Imposing that condition yields that ε = 2n + 1. But we
know that ε ≡ 2(Jω)
− 1 E. Therefore, the energy eigenvalues are
E
n = J n + ω. (0.26)
We now have our quantized energies! They are also evenly spaced as expected.
Note that imposing this condition only terminates either the odd series or the even series
because the recursion relation is spaced by two. We need to separately insist that the a j
for the other series. This is fine because it has been shown that if the potential is symmetric,
then the energy eigenfunctions can be taken to be either even or odd.
