The Harmonic Oscillator

The harmonic oscillator for which the restoring force F is proportional to the deviation of x from the equilibrium position can be described by the potential V(x). Since F ~ −d/dxV(x) the potential can be written as follows:

V(x) = ½ kx²

And the Hamiltonian operator and the Schroedinger equation:

(−h²/2µd²/dx² + ½ kx²) y = E y

where µ is the reduced mass. Since we wanna be shortly we will introduce new variables z² = x²(kµ/h²)½ and ε = E/h(µ/k)½ = E/h w with [w = (k/µ)½] and finally obtain (d²/dx² = (kµ/h²)½d²/dz²):

½ (−d²/dz² + z²) y(z) = E/h w = ε · y(z) = H y(z)

One who is not interested in the mathematical solution can follow here for the final result.

Now we want to factorize the term (−d²/dz² + z²) as for instance −a² + b² = (−a + b) (a + b). It's not so easy for operator because we must pay attention to the precise consequence. And we produce for check

1/Ö2(−d/dz + z) | 1/Ö2(d/dz + z)
|¾¾¯¾¾¾| | |¾¾¾¯¾¾|
a+ | a−

a+a− y(z) = ½ (−d²/dz² + z²) y(z) + ½(−d/dz z + z d/dz)y(z)

where ½ (−d²/dz² + z²) = H and H y = e y one can obtain:

a+a− y = ey + ½ (-y(z) − z d/dzy + z d/dzy) = (e-½)y

Now it's our equation for the harmonic oscillator. And now let's try to calculate the permutation relation between a− and a+:

[a−, a+] = a−a+− a+a− = ½[(d/dz + z) (−d/dz + z) − (−d/dz + z) (d/dz + z)] y

= ½ [−d²/dz²− z d/dz + d/dz z + z² + d²/dz²² − z d/dz + d/dz z − z²] y = [−z d/dzy + d/dz (z y)] = y

a−a+ − a+a− = 1

Applying a− to our equation a+a− y = (ε −½)y :

a−a+ a-y = (e- ½) a-y

(a−a+ )(a-y) = (e- ½) (a-y)

(a+a− + 1) (a-y) = (e- ½)(a-y)

a+a− (a-y) = (e- 3/2)(a-y)

We obtain by applying the operator a− to our wavefunction y the wavefunction (a−y) that is enough for our harmonic oscillator equation a+a-y = (ε − ½) y, for the eigenvalue which is smaller than about 1: (e- ½) → (e-3/2). The operator a− decreases the eigenvalue on 1 and therefore it's also called the annihilation operator.

Since ε is the measure of energy E (ε = E/h w) then it should be the lowest state yo. If we apply the annihilation operator to yo then we will obtain more deeper state that is not possible! And finally it's considered for the lowest state

a− yo = 0 and εo-½ = 0

εo = ½→ Eo = ½ h w

The lowest energy state has the energy ½ h w. And calculating the corresponding wavefunction one can obtain:

a− yo = (d/dz + z) yo = 0 → dyo/dz = −z yo → dyo/yo = −zdz → ln yo = Co' e−z²/2

yo = Co e−z²/2

Co can be obtained from normalization condition -¥ò+∞yo* yo dz = 1 in the following way:

Co = p-¼

And now applying a+ to our equation a+a-y = (ε − ½) y:

a+ (a+a-)y = (e- ½) a+y

a+ (a−a+− 1) y = (e-½) a+y

a+a− (a+y) = (ε + ½)(a+y)

The application of a+ to y increases the eigenvalue on about 1. Therefore a+ is also called the production operator. If we apply a+ to the lowest state yo then we will obtain the wavefunction y1 that is the eigenfunction closer to the higher eigenvalue:

y1 = a+yo

y1 = C1 (−d/dz + z) · e−z²/2

y1 = C1 2z · e−z²/2

The corresponding energy value is as follows:
E1 = 3/2h w

y2 = a+y1 = C2 (−d/dz + z) z · e−z²/2

y2 = C2(2z² − 1) e−z²/2

E2 = 5/2h w

and so on.

Generally one can obtain for any n-wavefunction yn

yn = Cn (a+)n yo

Result:

In general one can write down the following formula for possible energy values:

En = h w (n + ½) n = 0, 1, 2, ....

The ground state energy level Eo = ½ h w is the zero point of energy. Generally one can obtain for the wavefunction of n-state

yn(z) = Cn Hn(z) e−z²/2

where the constant Cn can be obtained from the normalization condition -¥ò+∞yn* yn dz = 1:

Cn = 1/(π½n!2n )½

and the function Hn(z) is the Hermitian polynomial:

H0(z) = 1
H1(z) = 2z
H2(z) = 4z2 − 2
H3(z) = 8z3 − 12z
H4(z) = 16z4 − 48z2 + 12
H5(z) = 32z5 − 160z3 + 120z
H6(z) = 64z6 - 480z4 + 720z2 − 120

All other polynomials can be obtained from the recursion formula

Hn(z) = 2z Hn−1(z) − 2(n−1) Hn−2(z)

for instance in order to obtain H3(z) one must do the following Hn−1(z) ® H2(z) and Hn−2(z) ® H1(z):

H3(z) = 2z · (4z2−2)− 2(2) (2z) = 8z3 − 12z

Fig. 1: Four lowest wavefunctions and its probability density for the harmonic oscillator.

The highest detection probability in the ground state is in the equilibrium position (see Fig. 1 for n=0). When increasing vibrations the quantity |yn|2 would be higher for classical turning points E = V(x); see figure to the left for n=5, 10, 20.

How much does the probability density for the classical oscillator in the range [x, x+dx]?

Classical oscillator: x = xm sin wt xm = maximum deviation

In the time interval dt (in which one observes the particle) the deviation dx will be as follows:

dx/dt = xmw cos wt

dt = dx/xmw cos wt = dx/xmw (1−sin²wt)½

dt = dx/w (xm²−x²)½

The quantity 1/w(xm²−x²)½ shows the probability density to find particle in the range [x, x+dx]; i.e. the highest chance to detect particle is in the classical turning points xm.