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:

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

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

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

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

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

 

[member 553083]

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.