Important because:

It illustrates quantum mechanical principals

It illustrates the use of differential eqns. & boundary conditions to solve for y

It shows how discrete energy levels arise when a small particle is confined to a region of space

It can predict the absorption spectrum of some linear conjugated molecules by treating the p electrons as free particles in a 1-dimensional box with infinite walls

Involves only 1 independent variable

x - independent variable

y(x) - dependent variable

y’(x) = dy/dx, y"(x) = d²y/dx², ... y⁽ⁿ⁾(x) = dⁿy/dxⁿ

A differential eqn. expresses a functional relationship between x & the derivatives of y with respect to x:

f(x, y’(x), y"(x), ... y⁽ⁿ⁾(x)) = 0

The order of the eq. is the order of the highest derivative of y with respect to x.

An n-th order differential eq. has n independent solutions (i.e. solutions that are not multiples of each other)

Examples: y⁽⁴⁾ + (y’)² + sin x cos y = 3 e^x, order = 4

x (y’) ² + sin x cos y = 3 e^x, order = 1

A_n(x) y⁽ⁿ⁾ + A_n-1(x) y^(n-1) + ... + A₀(x) y = g(x)

g(x) = 0: homogeneous

g(x) not =0: inhomogeneous

Example: Schrödinger Eq.

d^2y (x)/dx² + (2m/h²) (E - V(x)) y (x) = 0

is a 2nd order homogeneous linear differential eq. with

A₀(x) = (2m/h²) (E - V(x))

A₁(x) = 0

A₂(x) = 1

To solve a 2nd order homogeneous linear differential eq.,

A₂(x) y" + A₁(x) y’ + A₀(x) y = 0,

divide by A₂(x) to give:

y" + P(x) y’ + Q(x) y = 0, (1)

where P(x) = A₁(x)/A₂(x) and Q(x) = A₀(x)/A₂(x).

y₁" + P(x) y₁’ + Q(x) y₁ = 0

and

y₂" + P(x) y₂’ + Q(x) y₂ = 0,

then the general solution is

y = c₁ y₁ + c₂ y₂

(Prove this yourself by plugging the general solution into Eq. (1).)

In order to solve for c₁ & c₂, must have two eqs. for y. Get the 2nd Eq. by using boundary conditions &/or normalization conditions. This will be illustrated by the Particle in a Box.

But first, consider a special case--a 2nd order linear differential eq. with constant coefficients:

P(x) = p, Q(x) = q, p& q are constants

Then,

y" + p y’ + q y = 0

and the solution must be a function which has the same functional form for the 1st & 2nd derivatives as the function itself, i.e.

y = e^sx.

Then y’ = s e^sx = s y

and y" = s² e^sx = s² y

So that s² e^sx + ps e^sx + q e^sx = 0

and the auxiliary eq. is

s² + ps + q = 0

or s = -p/2 + (1/2) Ö (p² - 4q) = s₁, s₂

and y = c₁ e^s1x + c₂ e^s2x

Now we will use this knowledge of 2nd order linear differential eqs. to solve the Schrödinger Eq. for a model problem--

The Particle in a 1-DimensionalBox

The particle is constrained to move on the x-axis & is subject to an infinite potential outside the box & a zero potential inside. The box stretches from x = 0 to x = L (see Fig. 2.1)

Must solve the Schrödinger Eq.

d^2y (x)/dx² + (2m/h²) (E - V(x)) y (x) = 0

in three regions:

I & III: V=¥ , (E-V) blows up, so y (x) must be taken as 0.

II: V = 0, d^2y (x)/dx² + (2m/h²) E y (x) = 0

This is a 2nd order homogeneous linear differential eq. with constant coefficients: p = 0, q = (2m/h²)E

So the auxiliary eq. is: s² + q = 0,

or s = + Ö (-q) = + Ö -(2m/h²)E

E = kinetic energy + potential energy

kinetic energy is always > 0

here, potenial energy = V(x) = 0

so E > 0 ans s = + iÖ (2m/h²)E = s₁, s₂

So y _II = c₁ e^s1x + c₂ e^s2x = c₁ e^iq + c₂ e^-iq

where q = x Ö (2mE/h²)

y _II = c₁(cos q + isin q ) + c₂(cos q - isin q )

= (c₁ + c₂) cos q + i(c₁ - c₂) sin q

= A cos q + B sin q , where A = (c₁ + c₂) & B = i(c₁ - c₂)

Solve for A & B by applying the Boundary Condition: y must be continuous at the boundaries of the different regions:

lim y _I = lim y _II

x® 0 x® 0

0 = lim (A cos q + B sin q )

_{x® 0}

As x® 0, q ® 0, cos q ® 1, & sin q ® 0

To make A cos q = 0 at x = 0, must choose A = 0.

Then y _II = B sin q

lim y _II = lim y _III

x® L x® L

B sin q = 0

B can’t be zero because then y would be zero everywhere & the box would be empty. So must have

sin q = 0, or q = 0, +p , +2p , +3p ,...

At x=L, q = L Ö (2mE/h²) = +np , n = 0, 1, 2, ...

This leads to a quantum condition on the energy:

E = n^{2p 2}h²/(L²2m) = n²h² /(L²8m), n = 1,2,3, ...

Then y _II = B sin q = B sin [x Ö (2mE/h²)]

= B sin (+ np x/L)

Since sin x = sin (-x), +np x/L gives the same solution as

-np x/L. So

y _II = B sin (np x/L)

ò - ¥ ¥ ½ y ½ ² dx = 1

= ò - ¥ ^{0½ y} _I½ ² dx + ò ₀ ^{L½ y} _II½ ² dx + ò _L ¥ ½ y _III½ ² dx

= ò ₀ ^{L½ y} _II½ ² dx

= ½ B½ ^2ò ₀ ^Lsin²(np x/L) dx

= ½ B½ ^2ò ₀ ^L[1/2 - 1/2 cos (2np x/L)] dx

= ½ B½ ² [L/2 - 0]

B = Ö (2/L)

y _II = Ö (2/L) sin (np x/L), n = 1, 2, 3, ...

wehere n is called the quantum number. A node is a point at which the wavefunction equals zero. For each increment in n, the number of nodes inceases by 1 (see Fig. 2.3) so that there are (n+1) nodes.

Note that at n=2, there is zero probability for the particle to be in the center of the box. But how does it get from one side to another? The situation is different than that of a classical particle. A classical particle of constant energy would have an equal probability of being found anywhere in the box since constant energy means constant speed. A quantum mechanical particle of constant energy would have a maximum probability of being found in the center for n=1. But as n increases, the number of maxima & number of nodes increase and the quantum mechanical behavior approaches the classical limit, i.e. the particle would have equal probability of being found anywhere in the box. This is an example of the Bohr Correspondence Principle: In the limit of large quantum number, quantum mechanics approaches classical mechanics.

"Free" means not subject to any force, so V is a constant (independent of x). Arbirarily set V = 0. Then region II is spread out over the whole x-axis; no regions I & III.

d^2y (x)/dx² + (2m/h²) E y (x) = 0

y = c₁ e^iAx + c₂ e^-iAx, where A = (1/h)Ö (2mE)

But how can one solve for the constants in y ? Previously they were determined by the boundary conditions between regions.

Since ½ y ½ ² is the probability density, y must be finite as x® ¥ .

For E>0, y = c₁ e^iAx + c₂ e^-iAx

As x® ¥ , e^iAx & e^-iAx oscillate as sin (Ax) &

cos (Ax), so y is finite

For E<0, y = c₁ e^iAx + c₂ e^-iAx

Let A’ = (1/h)Ö (2m½ E½ )

Then e^iAx = e^-iA’x, which ® 0 as x® ¥ .

And e ^-iAx = e^A’x, which ® ¥ as x® ¥ .

So y ® ¥ as x® ¥ .

So, in order to have a well-behaved wavefunction, one must choose E>0. There is no quantum condition on the energy. It can take on a continuous range of values.

But note, for E>0, y can’t be normalized because

ò - ¥ ¥ ½ y ½ ² dx is not finite.

A free particle is an unphysical situation because there is no partcle in the physical world that is not acted on by any forces.

Particle in a 1-Dimensional Well (See Fig. 2.5a)

V(x) = V₀ for x < 0

= 0 0 < x < L

= V₀ for x > L

In Regions I & III, have

d^2y (x)/dx² + (2m/h²) (E - V₀) y (x) = 0

This has the form of a homogeneous linear 2nd order differential eq. with constant coefficients

p = 0, q = (2m/h²) (E - V₀)

Setting y = e^sx leads to the auxiliary eq.

s² + ps + q = 0 = s² + q

Or s = + Ö (-q) = + Ö (2m/h²) (V₀ - E)

Therefore, the general form of y for Regions I & III is

y _I,III = c₁ e^s1x + c₂ e^s2x

with s₁ = + Ö (-q) = Ö (2m/h²) (V₀ - E)

and s₂ = - Ö (-q) = Ö (2m/h²) (E - V₀)

In Region II, have the same form of the solution as in the case with infinite walls:

y _II = c₁ e^{ixÖ q’} + c₂ e^{-ixÖ q’}

But c₁ & c₂ will be different for the regions I, II, & III. They are determined, as before, by choosing them so that the wavefunction is well-behaved as x® +¥ (For Regions I & III), by matching the solutions at the boundaries, and by normalization. Since the wavefunction is not zero in Regions I & III, this is more difficult to solve mathematically. Matching Regions II & II results in the relation