00:01
To be dealing with a particle in a symmetric box in which we have the boundary conditions that at minus a over 2 and a over 2, what is going to happen to the wave function is that the wave function needs to be equal to 0, because we're going to have an infinite potential outside of those positions.
00:20
Recall that over here the potential is equal to 0, and outside of the box the potential is going to be infinite.
00:25
This infinite potential is going to ensure us that the movement of the particle is going to be confined inside of the box.
00:37
So we want to determine the energy for the lowest energy state for this particle in a box.
00:41
We know that this needs to satisfy schrodinger's equation.
00:45
In this case, the hamiltonian is going to be only the kinetic energy part.
00:49
Part, so what we're going to have is that our wave function applied to our schrodinger equation is going to look like this.
01:01
The second derivative plus, in this case, this is going to be equal to the energy times our wave function.
01:13
Recall that this is nothing but hamiltonian applied to the wave function is equal to energy applied to the wave function.
01:20
It's going to be the time -independent schrodinger equation.
01:24
And in this case, the second derivative refers refers to the second derivative of the wave function with respect to position.
01:31
So we can reduce this equation into a more familiar formula in which we're going to have the second derivative of the wave function with respect to x plus a constant squared times my wave function is going to be equal to zero and this constant is going to be equal to k squared equal to 2me divided by h bar squared.
01:50
So from here we know that the value for k, let me put it over here, here, the value for k is the square root of this quantity, 2me divided by h -bar squared.
02:02
Now, notice that this is a very familiar function that is actually just a harmonic oscillator, so we know that the general solution for the wave function is going to look like this, a cosine of kx plus b sine of kx, and upon applying the binary conditions, definitions, meaning when we apply a wave function at a over 2, we're going to get a cosine kl over 2 plus b sine kl over 2 is going to generate two unique solutions.
02:43
Recall that this needs to be equal to 0.
02:45
At the same time, with a negative sign, this is also going to be equal to 0...