00:01
Okay, so in this problem, the first thing we need to calculate is the expectation value for the potential energy of the quantum harmonic oscillator.
00:11
First of all, the expectation value for the potential energy is this quantity here.
00:19
We must remember that the ground state wave function of the harmonic oscillator is n0, exponential of minus 0.
00:31
Beta square x squared divided by two what else and we must remember that the potential energy in the harmonic oscillator is just half of m omega square position square okay well so let's begin to calculate this quantity here expectation value so the expectation value or the potential energy is going to be the integral from minus infinity to infinity of si of zero v, si star of zero, d x, okay? therefore this is going to be equal let's put in here.
01:32
This is going to be equal n0 squared divided by 2, m omega square of the integral from minus infinity to infinity of x square exponential of minus beta square x square dx okay well we have two problems in here first the solution to this integral second the value of the normalization constant.
02:15
Let's begin with the normalization constant.
02:18
So to find the normalization constant as we just make in the problem 64 let's put in here we need to calculate the integral from minus infinity to infinity of psi 0 psi star 0 d x equals 1 and that's what gives us n0 sorry n0 square of the integral from minus infinity to infinity of the exponential of minus beta square x square dx and this and here is just the gaussian integral but instead we have alpha we have beta square well the solution to the gaussian integral we already know i just forgot to put here equals one sorry the solution to the gaussian integral, we know the solution is just pi divided by alpha and the square root.
03:34
Okay, therefore the normalization constant n0 square multiplies pi square divided by beta equals 1.
03:49
And because of that, we can say that n0 square is going to be equal, beta divided by the square root of pi.
04:02
So keep this solution here.
04:05
Now we can go back in the expectation value of the potential energy.
04:09
And just see that this integral here is a second derivative of the gaussian integral.
04:21
Actually not even a second derivative a first derivative so imagine let's put here observation imagine that we have the gaussian integral here which is the exponential of minus beta square x square d x equals pi square root of pi divided by so if we make, actually let's put in this way, instead of putting the square root here, we're going to put the beta square and the square root in here.
05:06
So if we make a derivative, a partial derivative with respect of beta square of the gaussian integral, we actually doing a partial derivative with respect of beta square and the solution.
05:24
So if we apply this derivative in here, what we have is the integral from minus infinity to infinity of x square minus exponential of minus beta square, x square dx.
05:46
And in the right side, what we have is, let's put in here the square root of pi...