PHYSICS 447: DEVELOPMEMT OF SCHROEDINGER’S EQUATION (1-D)

We will use:

1) deBroglie relations

(1)

(2)

2) Law of Conservation of Energy

(3)

3) Since the wave function is a linear combination of harmonic waves, the equation we are looking for must be linear in x,t).

4) Suppose we know the momentum and energy of a particle exactly. Then, according to the H.U.P., we can’t say anything about it’s position or about what it is doing in time. We’ll represent such a particle with a single harmonic wave that extends over all space and time. That is, its wave function is of the form

(4)

We know exactly the values of k and (momentum and energy). Recall that any harmonic wave can be written as a superposition of a cos and sin function. We choose to write the arbitrary constants multiplying the cos and sin as A and A. (We could have written them as A and B but we have chosen to factor out a common constant.)

Such a particle, represented by a single harmonic wave, is called a “free” particle. It can have a potential energy, but this potential energy must be independent of position and time since we know that the particle always has the same energy and momentum, regardless of x and t. We can represent its potential energy as V(x,t) = Vo.

------Let’s now see where these ideas lead us ------

Look at (4). Notice that taking the derivative of  with respect to time will pull out an . Taking the derivative of twice with respect to position x will pull out a k2. Now look at (3). The total energy involves  and the kinetic energy involves k2. Thus, we are tempted to try an equation of the form

(5)

where  and  are constants which keep our guess as general as possible. Note that the equation is linear in . Let’s now put our wave function for a free particle, (4), into (5) and see what happens.

so (5) becomes

(6)

How do we solve this? First, we can divide through by A. Now let’s equate the cos coefficients:

(7)

and now equate the sin coefficients

(8)

Let’s combine (7) and (8) since both must be satisfied. Solving (7) for V and substituting into (8) gives

To continue, we must make a choice of sign here. Let’s choose the “+” sign so that

(9)

Using this in (7) then gives

(10)

Now compare (10) to the conservation of energy expression in (3). By direct comparison, we can write that

(11)

(12)

We’re done! The equation for finding the wave function of a particle traveling in one-dimension (the 1-D Schroedinger Equation) is

(13)

Also note that we can write the wave function for a free particle as

or, using Euler’s theorem, as

(14)

OTHER NOTES:

1. If we had used (9) in (8) instead of (7), we would have obtained the same expressions for  and . You can try this and see it is so. This has to be since we combined (7) and (8) to obtain the value of .

2. We have assumed the particle is traveling to the right, hence the kx-t. The same results are obtained if we had started with a particle traveling to the left and used kx+t.