©Fernando Caracena 2013
The harmonic oscillator is an important model in physics because so much of the oscillatory motion in the universe is Simple Harmonic (SH) Motion (SHM). In classical physics, the one dimensional harmonic oscillator is a favorite class room piece, which is easy to solve. The simplest laboratory experiment for investigating SHM in one dimension is that of a mass on the end of a spring. To solve for its motion, apply Newton's laws of motion. When the mass hangs motionless, the net force on the mass is zero, and it is at an equilibrium position (x=0). A spring that is not overly stretched obeys Hooke's law
F = -k x , (1a)
where k is the spring constant, x, the displacement from equilibrium and F the force generated by stretching or compressing the spring, which note is in the opposite direction to the displacement.
Displacing the mass to ether side of equilibrium x causes a restoring force to accelerate the mass back toward equilibrium, where it overshoots travelling to the other side of equilibrium. the result is that the mass oscillates about its equilibrium position.
m dx2/dt2 = -k x (1b)
or
dx2/dt2 = -(k/m) x . (1c)
The solution to the above is found by inspection
x=A e-ωt . (1d)
By plugging this solution (1d) into (1c), we find
ω2 = k/m , (1e)
where
ω = 2 π ν (1f)
is the angular frequency associated with the oscillation and νis the frequency in cycles per second, or Hertz (Hz).
The quantum mechanical counter part of the harmonic oscillator is important in modern physics as well; but in this case, the problem is not approached through Newton's laws, but through Schrödinger's equation, when we are dealing with non-relativistic motions.
In this presentation, we use the quantum mechanical treatment of the harmonic oscillator as an occasion to demonstrate the elegant power of operator methods, which allow us to avoid carrying out superfluous calculations and allow us to solve some serious problems with algebraic methods. In doing so, we need to be careful in respecting the order of operators that act on the state vectors because the have specified commutative properties that ordinary numbers do not have.
To begin, write Schrödinger's equation for a particle moving in one dimension
iħ∂/∂t ψ(x, t) = H ψ(x, t) (2a)
The time operator, iħ∂/∂t, has an energy eigenvalue, E. The Hamiltonian
H = p2/2m+½ k x2 (2b)
where the spring constant can be replaced by the mass and classical angular frequency by inverting (1e)
k= mω2 , (1g)
in which case, the Hamiltonian becomes
H = p2/2m+½ mω2 x2. (2c)
The time dependant Schrödinger equation
iħ∂/∂t ψ(x, t) = E e-itE/ħΦ(x) (2d)
can be separated into a time part and a coordinate part by the substitution
ψ(x, t) = e-itE/ħ Φ(x), (2e)
which results in the time-independent Schrödinger equation
H Φ(x) = EΦ(x). (3)
Divide through by the energy unit ħω, constructed from the classical frequency of the harmonic oscillator and you get the following dimensionless values and operators:
E= E/(ħω) (4a)
H Φ(x) = E Φ(x) (4b)
P = p/√(mħω) (4c)
Q =√(mω/ħ) x (4d)
H=½(P2+Q2) (4e)
We explore the commutation relations between the dimensionless momentum and position by using the commutation relation for x and p
[x,p] = iħ (4f)
[Q, P]= (1/ħ) [x, p] (4g)
[Q,P]=iħ(1/ħ) (4h)
[Q,P]=i . (4i)
Show that the operator P is the following in the coordinate representation of the wave function
P=-i ∂/∂Q . (4j)
Factor H by using the operators
A= (Q+iP)/√2 (5a)
and
A†= (Q-iP)/√2 , (5b)
which have the commutation relation
[A†,A] =i [Q, P] (6a)
[A†,A] =-1 . (6b)
The product of these operators is related to the dimensionless Hamiltonian as follows:
A†A= ½(P2 + Q2) + i/2 [Q, P] (6c)
A†A=H-½ (6d)
H=A†A+½. (6e)
To save writing this product repeatedly, define a new operator
N=A†A , (7)
which relates directly to the dimensionless Hamiltonian
H=N+½. (8a)
Next, we derive the commutation relations between this operator and
[A† ,A]A = (A†A) A - A(A†A) (8b)
[N,A]=-A . (8c)
Show that
[N,A†]=+A† (8d)
[N,A†] Φ=+A†Φ (8e)
[N,A] Φ= -AΦ . (8f)
Now, we are ready to solve for the system of energy levels implied by the modified Schrödinger equation
HΦ=EΦ . (9a)
Substitute for H using (8a) and (8e)
HA†Φ=A†NΦ+½Φ+A†Φ (9b)
HA†Φ=EA†Φ+A†Φ , (9c)
since E is a real number, it commutes freely with any operator. Collecting terms on the RHS of (9c) it follows that
HA†Φ=(E+1)A†Φ . (9d)
It is left to the reader to show that
HAΦ=(E-1)AΦ . (9e)
It is clear from (9d) and (9e) that A†Φ is an eigenfunction that has an energy level increased by one unit over that of Φ; and from, (9e) that AΦ is an eigenfunction that has an energy level decreased by one unit over that of Φ. Also, it is clear from (4e) that the energy levels are positive definite so that one cannot indefinitely decrease the energy eigenvalues by the procedure in (9e). Therefore, there must be a lowest energy state for which the following rule applies
AΦ0 =0 , (10a)
and also because of (10a)
NΦ0 =0 (10b)
or
HΦ0 =½. (10c)
By multiplying both sides of (10c), we have the beginning of the solution to our original problem
H Φ0 =½ħω. (10d)
Note that the lowest energy level of a harmonic oscillator is not zero, but is one half of an energy unit, viz., ½ħω. This an important result in physics because it turns out that the energy the field of electromagnetic radiation in space can be evaluated as a very large number of harmonic oscillators, each of which contains at least one half quantum of energy (½ħω) which they cannot give up. This is called the zero point energy of otherwise empty space. Although our reasoning seems very abstract, and for that reason,we would tend to take this result not very seriously, current observations support the existence of this zero point energy. This has happened several ties in the history of physics, that abstract quantities that emerge in the process of deriving wanted, physical results are dismissed as inconvenient abstractions that somehow must be removed from the calculations. Zero point energy at first appeared as infinite, but which could be ignored because it was constant and unobservable. Today, cosmologists have realized that the zero point energy is an important component of the universe, called dark energy, which causes the Hubble expansion of the universe to accelerate.
The power of the operator method is realized when we use the algebra of non-commuting variables to solve for all of the energy levels and even the wave functions. First repeat (10b)
NΦ0 = 0 Φ0, (10b)
noting that A†Φ0 has an eigenvalue that is increased over Φ0 by one
NA†Φ0 =(0+1)A†Φ0 (11a)
NA†Φ0 = 1 A†Φ0. (11b)
This means that the next eigenfunction up Φ1is proportional to A†Φ0. Therefore we write
Φ1 = c1A†Φ0 , (11c)
where c1 is the constant of proportionality that we are free to chose in such a way that the eigenfunction is normalized to unity through the integral
<Φ1 |Φ1 > = |c1|2 ∫ dx Φ*1(x) Φ1(x), (12a)
where the integral is written on the LHS of (12a) as an equivalent scalar product of an eigenvector with its dual, which is a more convenient form of writing it. Note that the above integral written without limits stands for a definite integral that extends over the full range of of integration(-∞, ∞).
Carrying out the calculation
<Φ1 |Φ1 > =|c1|2 <Φ0|AA†|Φ0> (12b)
<Φ1 |Φ1 > =|c1|2 <Φ0|(N+1)|Φ0> (12c)
<Φ1 |Φ1 > =|c1|2 <Φ0|1|Φ0> (12d)
we find that
<Φ1 |Φ1 > = |c1|2, (12e)
which in this case results in
c1=1. (12e)
Repeating the calculation for the next eigenvector
<Φ2|Φ2 > =|c2 |2 <Φ1|(N+1)|Φ1> (12f)
results in a different value for the constant of proportionality
<Φ2|Φ2 > =|c2 |2 2 (12g)
or
c2 =1/√2,
which results in the relation
|Φ2 > =√(½) A†|Φ1>. (12h)
It is left as an exercise for the reader to show that
|Φ3> =√(1/3) A†|Φ2> (12i)
or
|Φ2 > =√(1/3!) (A† )2 |Φ0>. (12j)
Given that
|Φn > =(n!)-½(A† )n|Φ0>, (13a)
show that
|Φn+1 > =[(n+1)!]-½(A† )n+1|Φn+1>. (13b)
Since (13a) is true for n=1 and n=2, showing that (13b) follows from (13a)
constitutes a proof of (13a) by mathematical induction.
Note that
n!=1*2*3*...*n
defines what is called the factorial of a number.
Show that 3!=6 and 4!=24 and 5!=120.
The operator N is called the number operator because its eigenvalue (n) relates to the energy number of the wave function
N Φn=nΦn. (14a)
Because of this property (9a), we can write the energy eigenvalues for the whole series
HΦn=(n+½)Φn, (14b)
where n = 0,1,2,3, ....,∞ ; and by multiplying both sides byħω we have solved for the full range of energy eigenvalues of the harmonic oscillator
H Φn=(n+½) ħω Φn. (14c)
One loose thread remains; there are some people that want to see the explicit form of the wave functions. This we can do by using (10a) to evaluate the lowest order wave function and (5a) to evaluate the operator A , which expands to
(Q+∂/∂Q)Φ0 = 0, (15a)
which has the solution
Φ0 = π-¼exp(-Q2/2). (15b)
The 1/4th power of π in the above expression comes from nomalizing the integral
<Φ0|Φ0>=1/√π ∫exp(-Q2)dQ . (15c)
Note that the Gaussian integral integrated over the full range of x evaluates as follows:
∫exp(-x2)dx =√π ,
which means that the normalizing constant is the square root of the reciprocal of √π , or
π-¼.
Note that once one obtains the explicit for Φ0 in (15a) it is just a matter of applying the following, explicite form of the operator A† repeatedly in (13a) to compute any desired wave function Φn :
A†= (Q-∂/∂Q). (15d)
Next, we will compute the next in the series from Φ0 explicitly to show how this is done. If anyone desires to compute any member of the set of harmonic oscillator wave functions, be advised that it is much simpler to use a symbolic manipulation package, such as MathematicaTM or MapleTM. The calculation runs as follows
Φ1 =1/√(1!) A†Φ0 (15e)
A†Φ0 =(Q-∂/∂Q) [ π-¼exp(-Q2/2)]/√2, (15f)
which evaluates to
Φ1 = π-¼ √2 Qexp(-Q2/2) . (15g)
Note that these functions are written in terms of dimensionless variables, and the solutions should be converted to the variable x to be complete using the following expression
x =√(ħ/mω) Q . (15h)