LECTURES ON QUANTUM MECHANICS

A line can be drawn and quantum theories coming before Schrodinger equation are called the old quantum theory. The old quantum theory is a variety of results spanning the years 1900–1925.

These attempts went as far as they could with the classical notion of trajectories and added quantum postulates to them. The uncertainty principle had not been enunciated .

Stated by the inequality ∆x∆p ≥ h /(4π) we see that zero uncertainty in position implies infinite uncertainty in the momentum.

In classical mechanics one has, in one dimension, Newton’law ,a second order differential equation. For example suppose that on a particle of mass m acts the total force F= F(x,v). The equation of motion is

d²x/dt² = (1/m) F(x,v) . (1)

A second order DE require two initial conditions , position and velocity , x₀ and dx₀/dt . They are assumed to be known exactly x₀ and v₀ or x₀ and p₀/m. An numerical solution is provided by a Taylor series about t₀=0 and has of the form,

x(t) =x₀ + v₀t + (1/2) t² F(x₀ ,v₀) /m + ( 1/6) t³ d{ F(x₀ ,v₀) /m}/dt … ( 2 )

But the uncertainty principle set limitations on the simultaneous knowledge of x₀ and v₀. Therefore the first two terms in the series , eq (2) , can not be known with absolute precision.

That the classical notion of trajectory had to be done away with as a starting point in quantum mechanics was not apparent to the pioneers.

One of the early approaches to quantum mechanics consisted in applying a t postulate that quantized the integral

J = ∫ p dq = nh where the variables p and q form a conjugate pair. J is called the action . For example if q=x then p=m dx/dt. But more important if p is the angular momentum of a particle going around in a circle of radius –r- , around the Z axis , p_φ = m r² (dφ/dt)=mr²ω ~ units of angular momentum ~ J-s.

WARNING THIS APPROACH IS SUPERSEDED BY SCHRODINGER’S THEORY. The quantization of angular momentum is shown there to be given by another rule.

hydrogen atom

Going bak to the action J , dq = dφ= ωdt and

J = mr²ω (2π) = n h . (3)

Another equation is that of mechanical equilibrium (in CGS units), in the case of the hydrogen atom ,

e²/r² = m r ω² . (4)

One can view (3)-(4) as two equations for the variables r and ω or alternativelyfro the variables r and v ( v= ωr) .

From (3)

m r v = n (h/2π) = n h’ , (h’=h/2π) . (5)

And from(4)

m r v² = e² . (6)

Dividing eq (6) by eq. ( 5) one obtains a quantized orbital velocity

v_n = e² /(n h’) = 2.18E8/n² ~ cm/s = 2.18E6 /n² ~ m/s (7)

≈ 7.27E-3 c /n² , where c is the velocity of light.

This in turn leads to quantized orbit radius

r_n = (h’²/(e²m) ) n² = 5.30E-9 n² (cm) =5.30E-11 n² (m) . (8)

Writing down the energy in the n-th quantum state ,

E_n = T + U = (1/2)m v_n² + (-e²/r_n)

= (1/2) m e⁴ /(n² h’²) - m e⁴ /( n² h’²)

= -(1/2) m e⁴ /( n² h’²) , (9)

with n=1,2,3 … .

We give the energy in various systems of units;

E_n = - 2.17163076E-011 /n² ~ ergs

= -2.17 E-18 /n² ~ joules

= -13.6 /n² ~ eV (10)

This result was anticipated in the Introduction.

Particle in the box

Another elementary example is that of the particle in a one dimensional box.

A particle of mass m moves between the impenetrable walls at x=0 and x=L .

The kinetic energy T equals the total energy E ( i.e. U(x)=0)

E= p²/(2m) ; p=(2mE)^1/2 . (11)

The action integral is J= 2∫ p dx = 2∫ (2mE)^1/2 dx = 2L(2mE)^1/2 = n h

and

E_n = { h²/(8mL²) } n² , (11)

n=1,2,3…..

A more complete treatment with quantum mechanics arrives at the same result of the energy of the particle in a box.

Comparing (11) with (10) we see two different dependence on the quantum number n. In (10) the coulomb attractive potential gives rise to a negative energies with a dependence 1/n² . In (11) the eigenvalues are positive and grow as ~ n² .

Quantum harmonic oscillator

The total energy for an oscillator is

E= p²/(2m) + (1/2) k x² (12)

and p(x) = [ 2m(E-(1/2)kx²)] ^1/2 . Letting m=1 ,k=1 , h^’ =1 ,l gives the unit (scale) of energy E_scale = h’ ( k/m)^1/2 = h’ω .

So p = [ 2(E-(1/2)x²)] ^1/2 and the action integral takes the form

J= 2 ∫[ 2(E-(1/2)x²)] ^1/2 dx = nh= n(2π)h’

= n(2π) , -(2E)^1/2 ≤ x ≤ (2E)^1/2 , (13)

with x₁=-(2E)^1/2 , x₂ = +(2E)^1/2 .

Define a² = 2E , X = (a² –x²) we find the integral

2 ∫ X^1/2 dx = (x X^1/2 +a² arcsin (x/a) ) │^x2 _x1

= 2E( arcsin(1) –arcsin(-1) )=2E( π/2 – (-π/2) )

= 2πE = n(2π)

Thus in our units E = n (n=1,2,3 …..)

or E _n=n h’ω . (14)

Eq (14) is Planck formula.

It is shown later that the qunatum harmonic oscillator has a spectrum given by E _n=(n +1/2)h’ω n= 0,1,2,3… (15)

The minima E ₀ = (1/2) h’ ω is called the zero point energy and can have physical significance , it is not an arbitrary constant to be swept away.

Now we introduce an algorithm to find the eigenvalues by integrating

∫ p dx for different values of E. The eigenvalues are those for which the numerical integral equals n(2π) = 2π , 4π, 6π ,….

FORTRAN code

c Wilson Somnerfeld Quantization rule

real k ,m

data k,m,nstep,ne/1.,1.,1000,50/

data epsi/1.e-5/

v(x)=(1./2.)*k*x**2

p(x)=sqrt(2.*m*(E-v(x)))

pi=2.*asin(1.)

ei=.001

ef=5.

de=(ef-ei)/float(ne)

e=ei

do 10 ie=1,ne

x1=-sqrt(2.*e/k)

x2=-x1

dx=(x2-x1)/float(nstep)

sum=0.

do 20 ix=1,nstep

x=x1+dx*float(ix)

sum=sum+2.*(dx/2.)*(p(x-epsi)+p(x-dx+epsi))

20 continue

print 100,e, sum, 2.*Pi,4.*pi

e=e+de

10 continue

100 format('e,sum,2.*Pi,4.*pi=',4(2x,e10.3))

stop

end

Fig 2.1 Eigenvalues of the harmonic oscillator. The first two values are

E= 1 and E=2.

Another example is the linear potential on the positive halfline, U = kx , x > 0 , U= ∞ , x ≤ 0.

The result of differs from the full quantum mechanical treatment but for large quantum numbers (n) they converge.

Dimensionally p²/m ~ k L where L is the length scale.

Using p~ h’/L one gets

h’²/(mL²) = k L , from which follows that

L = {h’²/(mk)}^1/3 (15)

The scale of energy is ,

E_scale = kL = { h’²k²/m}^1/3 . (16)

Letting m=1, k=1, h’=1 ,h = 2π , the quantum condition is

2∫ { 2(E-x) }^1/2 dx = n(2π) , 0 ≤ x ≤ E . (17)

The integral is

2^5/2E ^3/2 /3 = 2π n , (18)

thus E = {(3π)^2/3/2 } n^2/3 . (19)

This simple linear potential , with proper values of k and m , can serve to estimate among other things such exotic themes as the nature of quark confinement.

In the three example given we have found different dependencies of the energy E on the quantum number n.

We have E ~ 1/n² for the coulombic potential , E ~ n for the harmonic oscillator , U= (1/2)kx 2 and E ~ n^2/3 for the linear potential U = kx.

The dependency of E on the quantum number n is related to dimensional analysis. Suppose the potential energy is of the form U = kx^α (α≠0).Introducing a length scale L, dimensional analysis says that,

p²/m ~ h² /mL² ~ k L^α ~ E . (20)

Then L ~ (h² /mk)^1/(2+α) and E ~ k (h² /mk)^2α/(2+α)

or E ~ factor1 * h ^{2α / (2+α)} or vice versa

h ~ factor2 * E^(2+α)/2α . (21)

We need not ascertain at this point what factor1 and factor 2 are.

What matters is that ∫ p dx has dimensions of h , thus

∫ p dx ~factor* E^(2+α)/2α ~ n (from the quantization condition).

Hence the energy dependence on thequantum number n has the form

E_n ~ n^2α/(2+α) . (22)

α U E ~

-1	U= -1/r	~ - n^-2
+ 1	U = x	~ n^2/3
+ 2	U = x²	~ n

For the particle in the box one has U=0 inside the box and

h² /mL² ~ E or h ~ E^1/2 .

Thus ∫ p dx ~ E^1/2 ~ nh and E ~ n² .

A numerical integration of ∫p dx for the first two values of n (n=1 and n=2) can provide the energy expression in the form

E_n = constant ( n ^power) .

Fig 2. 2 The action J/(2π) vs energy .

The plot shows that E_n=1 ≈ 2.26 and E_n=2= 2.26 (2)^γ ≈ 3.59 . Therefore γ = ln(1.59)/ln(2) = 0.669

This leads to

E_n = 2.26 n^0.669 , (23)

which matches the analytical result given by eq(19).

Fortran code for Fig 2.

c Wilson Somnerfeld Quantization rule
real k ,m
data k,m,nstep,ne/1.,1.,1000,60/
data epsi/1.e-5/
c v(x)=(1./2.)*k*x**2
v(x)=k*x
p(x)=sqrt(2.*m*(E-v(x)))
pi=2.*asin(1.)
ei=.01
ef=5.
de=(ef-ei)/float(ne)
e=ei
do 10 ie=1,ne
x1=0.
x2=e/k
dx=(x2-x1)/float(nstep)
sum=0.
do 20 ix=1,nstep
x=x1+dx*float(ix)
sum=sum+2.*(dx/2.)*(p(x-epsi)+p(x-dx+epsi))
20 continue
print 100,e, sum/(2.*pi)
e=e+de
10 continue
100 format('e,sum =',2(2x,e10.3))
stop
end