A harmonic oscillator is in the state ?(y) = N (y²

A harmonic oscillator is in the state ?(y) = N (y² – 3y + 1) exp\left(-\frac{y²}{2}\right) (a) Expressed the state function in terms of the oscillator energy eigenstates, and determine the normalization. (b) If energy is measured, what would its predicted value be? (c) What is the chance that the system is in the ground state?

Transcript

0:00 Hi there.

00:01 So for this problem, we are told that the ground state wave function for a harmonic oscillator is equal to that depends on adds and is equal to a 2 times the exponential of minus adds to the square divided by 2 times b to the square.

00:30 So for part a of this problem, we need to determine the normalization constant a2.

00:54 So we know that we are given the oscillator wave function, this one right here.

01:03 So what we are going to use to obtain the value for the normalization constant is, well, we know that the normalization of the, the wave function requires that the integral from minus infinite to positive infinity is equal to the wave function to the square, the absolute value of the wave function to the square should be equal to one.

01:43 So with that set, what we are going to have is that this is, this is, also equal to the, well, we put this expression for the wavelength, so we'll have minus infinity to positive infinity.

02:03 So we have the a2 times the exponential of minus x to the square divided by two times b to the square.

02:20 And this times the x, l in this to the square.

02:27 So taking out what is common in this integral, we know that a 2 to the square is a constant in this case, and we will have that this is minus infinity to infinity.

02:41 The integral of the s penumptial of minus, since this is elevated to the square, we can get rid of the absolute value.

02:56 So we're going to have minus x to the square divided by 2 times b to the square the x.

03:02 Now we know this integral.

03:05 This integral, it has a known solution.

03:09 And in this case, because we have b in there, we're going to have that the solution is, well, we will have a to the square, a2 to the square times b times the square root of pi.

03:25 So remember that this is equal to 1 by the normalization of the wave function.

03:32 We know that we need to require that.

03:35 So if we solve for the a2, we're going to have that that is 1 overt, the square root of b times the fourth root root of pi.

03:52 So that's a solution for part a of this problem...

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