How Do You Calculate the X-Component Probability Distribution in Momentum Space?

[BLaH!]

Hey,
We are given the 1s spatial wave function for the hydrogen atom:

[tex]\psi(\vec{r}) = \frac{1}{\sqrt{a_{0}^3r}}e^{-r/a_{0}[/tex]

We are asked to find the momentum space wave function [tex]\phi(\vec{p})[/tex]. Obviously this is just the Fourier transform of the spatial wave function. In calculating [tex]\phi(\vec{p})[/tex] I used the following theorem:

[tex]For f(\vec{r}) = f(r), \rightarrow F(\vec{q}) = \frac{4\pi}{q}\int_{0}^{\infty} sin(qr) f(r) r dr[/tex]

Here [tex]F(\vec{q})[/tex] is simply the Fourier transform of [tex]f(\vec{r})[/tex]Anyway, this will give you the momentum space wave function in terms of the magnitude of momentum [tex]p[/tex]. After we find this, how do we find what the probability distribution is for the x-component of momentum [tex]p_{x}[/tex].

What should I do? Insert a complete set? Do another transformation? Any help would be appreciated.

 

[Nate810]

Thanks!Yes, you should insert a complete set in order to find the probability distribution of the x-component of the momentum. Specifically, you need to calculate the following: \phi(p_x) = \int \phi(\vec{p}) e^{i p_x x} dp_x where \phi(\vec{p}) is the momentum space wave function you obtained from the Fourier transform. The result of this integration will be the probability distribution of the x-component of momentum.

 

[wais]

Hi there,

To find the momentum space wave function \phi(\vec{p}), we can use the Fourier transform as you mentioned. However, we need to take into account the three-dimensional nature of the problem. This means that we need to use the three-dimensional Fourier transform, which is given by:

\phi(\vec{p}) = \frac{1}{(2\pi)^{3/2}} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \psi(\vec{r}) e^{-i\vec{p}\cdot\vec{r}} dxdydz

This will give us the momentum space wave function in terms of the three components of momentum (p_x, p_y, p_z). To find the probability distribution for the x-component of momentum p_x, we can use the Born rule, which states that the probability of finding a particle in a particular state is proportional to the square of the magnitude of the wave function in that state.

In this case, we can find the probability distribution for p_x by taking the square of the magnitude of the momentum space wave function with p_y and p_z integrated out:

P(p_x) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |\phi(\vec{p})|^2 dp_y dp_z

This will give us the probability distribution for the x-component of momentum p_x. Keep in mind that this is only for the 1s state, and for other states, the probability distribution may be different. I hope this helps! Let me know if you have any other questions.

 

[M98Ranger]

Hi there,

Yes, you are correct in using the Fourier transform to find the momentum space wave function. However, in order to find the probability distribution for the x-component of momentum, we need to use the momentum operator \hat{p_{x}}. This operator acts on the momentum space wave function \phi(\vec{p}) to give us the probability distribution for the x-component of momentum.

To do this, we can use the following formula:

\langle p_{x} \rangle = \int \phi(\vec{p})^* \hat{p_{x}} \phi(\vec{p}) d\vec{p}

Where the integral is over all possible values of momentum p. This will give us the expectation value for the x-component of momentum, which is related to the probability distribution.

Alternatively, we can also use the position operator \hat{x} to find the probability distribution for the x-component of momentum. This can be done using the following formula:

P(p_{x}) = \int |\phi(\vec{p})|^2 |\psi(\vec{x})|^2 d\vec{x}

Where \psi(\vec{x}) is the position space wave function, given by the inverse Fourier transform of \phi(\vec{p}). This method gives us the probability distribution directly, without having to calculate the expectation value first.

I hope this helps clarify the process for finding the probability distribution for the x-component of momentum using the momentum space wave function. Let me know if you have any further questions.