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value) E . Thus, the physical interpretation of the eigenvalue equation
n
Hu (x)=E u (x),
n n n
is that the eigenvalues corresponding to the Hamiltonian operator H are
the energies of the accessible states for the system. This makes sense, since
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the Hamiltonian itself is the quantum operator corresponding to the classical
energy.
We havethus seenthatthe states ofthe systemformanenumerableinfinite
set. The first stage of abstraction in quantum mechanics is to note that, al-
thoughwemightneedspecialpropertiesofthesesolutions,itisnotbenecessary
to work with the explicit representation in terms of sine and cosine functions.
Our next example system, the harmonic oscillator, will illustrate this.
4.2 Linear harmonic oscillator
The harmonic oscillatoris a simple and extremely useful model of physicalsys-
tems both in classical physics and quantum physics. Classically, it can be used
tomodelmechanicalvibrations. Inquantumphysicsitismodelforthemodesof
electromagneticwaves. Itsusefulnessstemsfromthefactthatevencomplicated
many-particle systems or continuous media can often be analyzed in terms of
normal modes of vibration, perhaps after linearization, and furthermore, that
it is a completely solvable model.
The model is also very useful in that it has enough features in order to
develop large portions of elementary classical and quantum mechanics within
it. This is precisely what will be done in this section.
Thesimpleclassicalharmonicoscillatorconsistsofaparticleconnectedwith
aspringto rigidwall. The forcefromthe springis proportionalto the displace-
mentofthespringfromitsnaturallength. Ifotherforceslikeair-resistanceand
friction are neglected, the particle will oscillate forever once it is set in motion.
Duringthisoscillation,therewillbeanoscillationofenergybetweenkineticen-
ergy of motion and potential energyin the spring. The total energy is constant
during the motion. The total energy is said to be a constant of the motion.
When setting up a modelfor this system,physicists normallyabstractaway
from the wall, instead considering a particle of mass m attracted to a fixed
centerby a forcethat is proportionalto the displacementfromthe center. This
gives a more symmetrical formulation, and the displacement can take negative
values. Letting x denote the displacement from the center, the force acting on
the particle is F = kx. The negative sign makes the force attracting. The
−
equilibrium position is in the center where x = 0. The so called conserva-
tive forces, i.e forces that conserves total energy, can always be derived from a
potential by the relation
d
F = V(x).
−dx
It is thus easy to see that the potential energy for a harmonic oscillator is
1
V(x)= kx2.
2
The total classical energy of the oscillator is
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p2 1
E =K+V = + kx2. (4.20)
2m 2
Since, as noted, E is a constant during the motion, we can read off the
oscillation of the energy between kinetic energy and potential energy. When
x = 0, corresponding to the particle passing through the center of motion, all
energy is kinetic and the momentum p takes its maximum value. On the other
hand, when p = 0, the kinetic energy is zero and all energy is potential. This
corresponds to the turning points at the maximum distance from the center.
We will not analyze the classical model further, but turn directly to the
quantum harmonic oscillator.
4.2.1 Quantization of the oscillator
Referring back to the quantization rules of the previous section we can now
easily write down the Hamiltonian for the harmonic oscillator. Applying the
quantization rules to the energy (4.20) we get
¯h2 ∂2 1
E H = + kx2.
−→ −2m∂x2 2
The Schr¨odinger equation becomes
∂ ¯h2 ∂2 1
i¯h ψ = ψ+ kx2ψ. (4.21)
∂t −2m∂x2 2
Theseparationofvariablesandthesolutionforthetime-dependence,i.e. the
stepsrecordedinsection4.2.1,areexactlythe samefortheharmonicoscillator.
We need only concentrate on the space-dependence. In fact, this is true for all
systems for which the forces are time-independent. Thus, after separation of
thevariableswealwaysendupwithequation(4.9)ofthe previoussection,with
the appropriate potential.3 Therefore, the eigenvalue equation to solve is
¯h2 ∂2 1
u+ kx2u=Eu. (4.22)
− 2m∂x2 2
At this stage, one can proceed as in the previous section, and solve this
differential equation to obtain an infinite set of basis functions. The set of
orthonormalbasis functions can be written as
1
u (x)=N H (αx)exp( α2x2).
n n n
−2
Here, α is a constant and N is a normalization constant
n
α4 = m ¯h2k N n =( √πα 2nn!)1 2
3Inthreedimensionsofspace,andinothercoordinatesystemsthanrectangular,theequa-
tion is more complicated - but in principle it is always the same equation with the relevant
potential.
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and H are Hermite polynomials, the first few of which are
n
H (x)=1, H (x)=2x, H (x)=4x2 2.
0 1 2