Schrodinger's Equation

The wave description of de Broglie is not adequate to explain the properties of the electron under different types of external potential fields. Schrödinger derived an equation that conserves energy using the wave function to describe the electron. Then he solved the equation for different potentials applied to the electron. His description has proven to be correct and is the bases for the modern understanding of the electron.

Conservation of Energy

The energy of an electron acted on by a conservative force is

Equation 1

where E is the total energy, U is the potential energy due to the force, and K is the kinetic energy of the electron. The nonrelativistic relation between kinetic energy and momentum, p, is used. The electron is described by a summation of many different waves according to,

Equation 2

To relate the two equation, first take derivative of the wave function with respect to time and position. Remember, the derivative of the sum is equal to the sum of derivates (a linear relation). The derivative of each term in the above sum is:

According to the relation, E = hu = hw/(2p) = ħw, or w = E/ħ.The first equation above is then

Also, according to de Broglie, k = 2p/l = 2p(p/h) = p/ħ so that

Multiply the conservation of energy equation by y to get

The final version of Schrödinger’s equation is:

Equation 3

This very important equation is a result of conservation of energy and the relations between total energy and w and between the momentum and the wave number, k. Remember that the wave function in equation 3 is the sum of waves in equation 2. Schrödinger’s equation is a 2^nd order partial differential equation that is linear. Each term in the summation in equation 2 is differentiated independent of the other terms. We can write a Schrödinger equation for each term of the sum of waves.

Time-Independent Equation

A very powerful method of solving Schrödinger’s equation is with separation of variables. If the function, y(x,t) is written as a product of two independent functions, then the partial differential equation (equation 3) can be reduced to ordinary differential equations in each variable. For example, Schrödinger’s equation as written in equation 3 is a function of two variables, x and t. Write the wave function y(x,t) as:

Equation 4

A plane wave (one term in equation 2) can be written in this form (e^ikx e^-i^wt ). Substitution of Equation 4 into Equation 3 yields:

Divide this equation by y(x)f(t) the get,

Equation 5

The left side of equation 5 depends on x only while the right side of the equation depends only on t. The only way these two independent parts of the equation can equal each other is when they are equal to a constant. For practical reasons shown below, call the constant E.

so that now we have two ordinary differential equations:

The time dependent equation is easy to solve. Rewrite the equation as

which leads to,

where as before E = ħw. The time independent Schrödinger equation is,

Equation 6

where the full solution of the wave equation is given by

Equation 7

We will solve Equation 6 for different physically interesting situations.

Free Particle

First assume that the particle has no forces acting on it which means the potential energy is zero. Schrödinger’s equation reduces to

Equation 8

To solve this equation, first let k² = 2mE/ħ². The equation now reduces to the standard form of the wave equation. The general solution to this equation is,

where A and B are the two integration constant required for a 2^nd order differential equation. If you are given the solution to a differential equation, it is simple to test if it is correct. Take derivatives of the function and see if it satisfies the equation. For example, two derivatives of y(x) above yields,

which satisfies Schrödinger’s equation. The mathematicians tell us that when you find on solution to a differential equation, you have found all the solutions (the uniqueness of the solution theorem). But to have a normalizable, localized function we must sum over an infinite number of waves.

Solutions for Constant Potential

If the potential energy, U(x), is constant then Schrödinger’s equation becomes:

which is very similar to Equation 8. To solve this equation, consider two cases:

1) k₁ = 2m/ħ² (E-U) > 0 E > U

2) k₂ = 2m/ħ² (U-E) > 0 U > E

In these two cases, Schrodinger’s equation becomes:

Equation 9

The solution to the first case was given above:

Equation 10

The solution to the second case is (check and see!):

Equation 11

These solutions have very important interpretations. The first case has solutions that oscillates (sin and cos). Since E > U, then E = K + U implies that the kinetic energy (K) > 0. In classical physics, this solution for the kinetic energy is the only possibility. The second case has solutions that exponentially rise or fall (sinh and cosh). Since E < U, K <0, impossible in classical physics. Negative kinetic energy does occur in quantum physics and is the principle behind the phenomenon of tunneling.

Particle in a Box

We can solve a particle trapped in a box of width L in two cases. First assume a nonrealistic case that the walls are infinitely tall. The particle has to be in the box so that the wavefunction must go to zero at the walls. Inside the box, U = 0 so the solution is equation 10 where k²₁ = 2m/ħ² E > 0 and E is the energy of the particle.

In this one case, it is easier to see the solution if instead of using equation 10, use a completely equivalent form,

Since this function must go to zero at the boundaries, choose y(x=0) = 0 and y(x = L) = 0. Or,

The last equation requires that k₁ L = n p so that

The constant B is determined by the normalization:

The energy for the particle is given by k²₁ = (n p/L)² = 2m/ħ² E , or

The energy for a particle in a box is quantized – the energy cannot have any value.

Real Particle in a Box

A more accurate model for a particle trapped in a box is instead of the walls having infinite potential (i.e. the wave function is zero at the walls), let the walls have a finite potential, U₀.

To solve this problem, find the solution to Schrödinger’s equation in each of the regions:

In region I, k²₁ = 2m/ħ² (U₀- E) and the solution is given by equation 11.

In region II, k²₂ = 2m/ħ² (E) and the solution is given by equation 10.

In region III, k²₃ = 2m/ħ² (U₀- E) and the solution is again given by equation 11.

The wavefunction must be normalizable over all space so that the function must go to zero as x approaches ∞. In region 1 where x < 0, e^-kx must be zero. Similarly, in region III where x > 0, e^kx must be zero so the solutions in region I and III reduces to

where A, B, C, and D are integration constants. To solve for these constants we have to force the wavefunction to be continuous and smooth. We require that the function and its first derivative are continuous at each boundary.

These four equations might allow us to solve for the four integration constants. However, as we will see, these equations allow us to constrain the energy.