Hmm... Schrodinger's Equation for 7th graders... Okay, let's do it!
First, let's have a look at the ol' Schrodinger equation:
H [psi] = E [psi].
It doesn't look so scary, does it?
Before we jump into solving for [psi], let's address the issue of what this equation actually represents anyway. In simplest terms, the Schrodinger equation is a probabilistic interpretation of the energy conservation law. For now, don't worry about what this means, let's take a closer look at the different parts of the equation.
The Schroinger wavefunction [psi] gives us information about the existence of an object in a system. By itself, it doesn't make much sense to us physically. But when we square the absolute value of this function we get a probability density function.
At this point, you might be asking yourself, what is a probability density function? To answer this question, I pose the following problem.
There is an electron sitting on top of a ruler. Just for fun, let's say the hand of God moves this electron around the ruler from time to time, but God is kind of anal retentive so he like place the electron exactly on top of integer markings... in other words, the 1 inch mark, the 2 inch mark, etc. Let's further suppose that God has told us that 1/16 of the time he puts the electron on top of the 4, 3/16 of the time he puts it on the 5, 1/2 of the time he puts it on the 6, 3/16 of the time he puts it on the 7, and 1/16 of the time he puts it on the 8. Now, I ask you to calculate the average the position of the electron along the ruler. I'm sure you readily see that answer is 6, but to calculate this statistic rigorously, maybe you would write something on paper that looks like this:
(1/16)*4 + (3/16)*5 + (1/2)*6 + (3/16)*7 + (1/16)*8 = 6
Notice the fractional factor in front of each term. These numbers actually describe a probability density function. Let's call this density function P and say that it is a function of the distance x measured along the ruler in units of inches:
P(x < 4) = 0,
P(x = 4) = 1/16,
P(x = 5) = 3/16,
P(x = 6) = 1/2,
P(x = 7) = 3/16,
P(x = 8) = 1/16,
P(x > 8) = 0.
Remember what we said before. The square of the absolute value of the wavefunction [psi] gives us a probability density function. Symbolically, this means:
P(x) = (Abs[[psi](x)])^2.
There are two important characteristics to notice about our function P(x). The first, as we're already mentioned, is that we can find the average distance along the ruler where the electron exists by multiplying each distance point by that points corresponding probability P(x) and summing these products.
Average[x] = Sum x * P(x).
In this sense, we can think of the value x as being it's own operator. When you want to find the average value of a quantity, multiply the operator by the probability density and some over all points. Here are some examples of common physical values and their corresponding quantum mechanical operators:
x --> x, x = distance;
p --> -i hbar d/dx (in one dimension), p = momentum; and
E --> i hbar d/dt, E = energy.
Don't worry about what these operators really mean right now.
The second important characteristic of P(x) is the fact that it is normalized (remember this key word). When we say it is normalized, we mean that if we sum P(x) for all values of x we get unity.
Sum P(x) = Sum (Abs[[psi](x)])^2 = 1
Let's get back to describing the Schrodinger equation. The factor H in this equation is called the Hamiltonian. It is also an operator. To really understand where the Hamiltonian comes from, you need to understand variational calculus. But to understand it's function, you don't need to know jack sh:t! I jest... you need to some sh:t... but just a little. Actually, the Hamiltonian describes the allocation of the energy in the system. In other words it is the kinetic energy plus the potential energy.
At this point, let's recap and fill in as much information about Schrodinger's equation that we have thus far established.
H [psi] = E [psi], where
H = T + V,
T = the kinetic energy of the system,
V = the potential energy of the system,
E = the total energy of the system, and
[psi] = a wavefunction which gives information about the existence of the object in our system.
Now, here comes the part you've been waiting for. We're going to solve the Schrodinger equation for the case of an unbound electron flying through the void of space! Does that sound exciting or what?
First, let's get rid of the time dependency that shows up in the energy operator.
E --> i hbar d/dt
This term only applies to cases where total energy of the system is changing. There's no external force being applied to our electron; so, we're dealing a closed system, and we can allow E to be a constant. This assumption leaves us with:
(T + V) [psi] = E [psi].
Of course, we have already stated that this electron is unbound. Therefore, we can set the potential energy to zero... leaving us with:
T [psi] = E [psi].
As we've stated before, T is the kinetic energy term. Maybe you've heard in your science class that classically kinetic energy is equal to:
T = 1/2 mv^2, where
m = mass, and
v = velocity.
Perhaps, you've also heard that classically, momentum is:
p = mv.
Well, putting these two equations together, we get:
T = p^2 / 2m.
In quantum mechanics, we must use the momentum operator for p, so:
T = (hbar^2 / 2m) * d^2/dx^2.
Our equation is now:
(d^2 [psi] /dx^2) + k^2 * [psi] = 0, where
k = (2mE / hbar^2)^1/2.
This expression is a second order differential equation. You probably won't really learn how to solve these equations until high school, but you can easily find the solution to this form of equation by simply referencing the CRC Handbook for Physics and Chemistry. The solution will give you:
[psi](x) = A*Exp[+ i k * x] + B*Exp[- i k * x].
There are two terms in this equation. The first represents a wave traveling in the positive x direction. The second represents a wave traveling in the negative x direction. We stated before that our electron is flying through space; so, let's just arbitrarily say that our electron is moving in the positive direction, and let's get rid of the second term.
[psi](x) = A*Exp[i k * x]
Now, all we have to do is solve for A. We can do this by normalizing the wavefunction.
Sum (Abs[[psi](x)])^2 = 1
In this case, we have to sum over a continuum of points. Therefore, we must actually integrate our probability density over all of space:
Integral (Abs[[psi](x)])^2 = 1.
Fortunately though, we have a shortcut because the squared absolute value of [psi](x) is:
Abs[[psi](x)]^2 = (A*Exp[+i k * x]) * (A*Exp[-i k*x]) = A^2.
So we only have to integrate over one:
Integral (A^2) = (A^2) * Integrate[1] = 2pi * A^2 = 1.
Solving for A, we get:
A = (2pi)^(-1/2).
Putting A back into our solution for [psi], we have:
[psi](x) = (2pi)^(-1/2) * Exp[i k * x], where
k = (2mE / hbar^2)^1/2.
eNtRopY
