"Particle" in a 1D Box
In this section we will revisit the particle in a one-dimensional box as we previously calculated with the Wilson-Sommerfeld quantization condition. This time we will derive the wave functions using Schrödinger's equation. We set up the box by describing the potential energy that confines the particle. The bottom of the box has potential energy of zero, and the walls are assumed infinitely tall. This just guarantees that the particle is trapped in the box, since it'd take infinite energy to escape the box. The box will start at x=0 and end at x=L. Thus the potential energy function to describe the box is
Once we have the potential energy, we know that the wave function can only be non-zero within the region where U=0 since it would take infinite energy to have a wave function outside the box. Therefore, not only do we know that the wave function is uniformly zero outside the box, but at the boundaries at x=0 and x=L the wave function, which must be continuous, must be zero valued. Therefore we know that
Solving Schrödinger's Equation
When we write out Schrödinger's equation for this particle inside the box, U(x)=0. Therefore we get
You have seen equations of this form before. It is the exact same mathematical form as the simple harmonic motion of a mass on a spring or a pendulum at small angles. Therefore it should not surprise you that the solution will be a linear combination of a sine and a cosine, or We can use our boundary conditions to find values for A and B, and by plugging this solution into the differential equation we can also find E, the energy of the system. Let's do this now.
Before discussing that, I should emphasize that this is NOT the same solution as the one used in the last section for the classical wave equation. That function had both temporal (t) and spatial dependence (x) while this one only has spatial dependence. If you looked at the linked article on the derivation of Schrödinger's equation , you saw that in its general form the equation DOES have time dependence, but even so it is not in the same mathematical form as the classical wave equation.
Let's now consider the boundary conditions and how we use them. If our function needs to return then we know that B=0, since the value of the wave function will be equal to B at x=0. So now we know that the function must be and using the boundary condition at x=L and knowledge of the sine function we can find k. The sine function only has zeros at so we know that at the right side boundary where x=L, we can write I can't use N=0 since then there would be no wave and no wave function. Solving for the wave number gives us So now we may rewrite our wave function as
Plugging that solution into Schrödinger's equation gives us:
This result should look familiar since we got the exact same result using the Wilson-Sommerfeld quantization condition in the last chapter. However, looking at our wave function we still have an undetermined constant A.
Normalization of the Wave Function
The last step to get the wave function is to determine the constant A. The squared wave function is the probability density p(x) of finding a particle. By "finding" we mean measuring. So what we mean is that where the probability of "finding" the particle is large, there is a high probability of wave function collapse taking place at that location - which is analogous to field collapse. Any measurement of this electron must search within a finite range. The probability of finding the particle between x=A and x=B is written:
Since the squared wave function represents probability, the area under the squared function must be unity. That way if we search everywhere we have 100% chance of detecting the electron. Mathematically this is stated:
Notice that while the integral is written with infinite ranges, that in our particle in a box our wave function is only non-zero between x=0 and x=L. Therefore that is the only range over which we must integrate. This process of integrating the squared wave function over the entire applicable range and setting equal to unity allows us to normalize the wave function which will allow us to find the amplitude of the wave function such that we are guaranteed a proper probability estimate.
Given our function , we write and solve for A. This naturally requires the substitution Plugging in and solving the integral in which the cosine term has no contribution, gives us
So we now have a properly normalized wave function for the 1D particle in a box: