- #1

Kara386

- 208

- 2

## Homework Statement

The operator for the momentum of a particle in 1D is ##\hat{p}_x = -i\hbar\frac{d}{dx}## has the eigenfunction ##\psi(x) = e^{ikx}##.

The Schrodinger equation for a free particle in 1D is

##\frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi##

Which has the solution ##\psi(x) = cos(kx)##.

a) Find the corresponding energy eigenvalue E.

b)What are possible outcomes of measuring ##\hat{p}_x## and the relative probabilities of these outcomes?

c)Is it true that the radial probability density for the electron in the lowest-energy atomic state of hydrogen is

##P(r) = \frac{1}{\pi a_0^3}e^{-2r/a_0}##, if the normalised wavefunction is

##\psi(r) = \sqrt{\frac{1}{\pi a_0^3}}e^{-r/a_0}##

And what's the most probable value of ##r##?

## Homework Equations

## The Attempt at a Solution

For a) I subbed ##\psi(x)## into the LHS and set it equal to the RHS, which gave

##\frac{\hbar^2k^2}{2m} \psi(x) = E \psi(x)##

So the eigenvalue is ##\frac{\hbar^2k^2}{2m}##.

b) The possible outcomes are the eigenvalues of the momentum operator which are ##\hbar k##. For their relative probabilities I think you have to calculate coefficients of something, can't remember what, which are given by

##c_n = \int_{-\infty}^{\infty} e^{ikx}\cos{(kx)} dx##

Which is a tricky integral and in any case I don't know why those are the probabilities, I'm just basing it on an example we were given. I'd have to repeat the integral again with ##e^{-ikx}## afterwards. So is this the right method and if so how do I integrate that? And why are those the probabilities?

c) I think the probability density function is correct, calculated using ##P(r) = \psi^{*}\psi##. Is the most probable value of ##r## given by finding the maximum of the probability density function? I'm not sure differentiating and setting = 0 really gets me anywhere because it's an exponential and only has a turning point not a maximum, right?

Thanks for any help!