Near Normalization calculation for given wavefunction

[JamesJames]

Near "Normalization" calculation for given wavefunction

Homework Statement

A wave function is given by Y(E) = CEexp(-E/kt)

1. Find C so that Y(E) becomes Y0 where Y0 is a constant.

2. Calculate the mean energy with respect to Y(E).

3. Find Y(t) as a function of wavelength and calculate the mean wavelength.

The Attempt at a Solution

1. Ok, I'm a bit conufsed by this "normalization" concept as it applies here. I understand normalization requiring Y*Y = 1 etc. but that would require the outcome to be 1. Here, the outcome is a constant. I could just say

$$\int_{-\infty}^{\infty}Y\left(t\right)Y^*\left(t\right)dt=1$$

but the "Y0" part is throwing me off.

2. Shouldn't it just be

<E> = $$\int_{-\infty}^{\infty}Y\left(t\right)EY^*\left(t\right)dt$$

where the exponentialterms would cancel leaving E and other constants? I would presumably calculate this AFTER having solved for C above.

3. I will get to this a bit later and post my attempt here a little later.

 

[Jhero]

Hello,

Normalization is a concept that is commonly used in quantum mechanics to ensure that the wave function is properly normalized. In this case, the constant Y0 represents the probability of finding the particle in the given energy state, so it is important to ensure that the total probability is equal to 1. This is why we need to find the value of C that will make Y(E) equal to Y0.

To find C, we can use the normalization condition:

\int_{-\infty}^{\infty}Y\left(E\right)Y^*\left(E\right)dE = 1

Substituting the given wave function, we get:

C^2\int_{-\infty}^{\infty}exp\left(-\frac{2E}{kt}\right)dE = 1

Solving this integral, we get:

C^2\left(-\frac{kt}{2}\right)exp\left(-\frac{2E}{kt}\right)\Big|_{-\infty}^{\infty} = 1

Since we want Y(E) to be equal to Y0, we can substitute Y0 for Y(E) and solve for C:

C = \sqrt{\frac{2}{ktY0}}

Now, to calculate the mean energy, we can use the formula you mentioned:

<E> = \int_{-\infty}^{\infty}Y\left(E\right)EY^*\left(E\right)dE

Substituting the given wave function and solving the integral, we get:

<E> = \frac{3}{2}\frac{kt}{2}

Finally, to find Y(t) as a function of wavelength, we can use the relation:

E = \frac{hc}{\lambda}

Substituting this into the given wave function, we get:

Y(\lambda) = \frac{2}{kt\lambda}exp\left(-\frac{hc}{kt\lambda}\right)

To calculate the mean wavelength, we can use the formula:

<\lambda> = \int_{-\infty}^{\infty}Y\left(\lambda\right)\lambda Y^*\left(\lambda\right)d\lambda

Substituting the calculated Y(\lambda) and solving the integral, we get:

<\lambda> = \frac{hc}{4\pi k