# Measuring momentum using position wavefunction

• Joao Victor
In summary, the author was trying to solve an exercise from Cohen-Tannoudji's book, but then got stuck in this question. He wrote and expression for the probability in terms of the wave function in momentum space Φ(p), and then i wrote Φ(p) in terms of ψ(r) (since they are related by a Fourier transform). However, the computation of the probability uses the modulus squared of Φ(p), which is Φ(p)Φ*(p). He gets, then, a product of integrals, which is a real mess. Is there a way to clean it up a bit?
Joao Victor
I was solving an exercise from Cohen's textbook, but then I got stuck in this question.

"Let ψ(x,y,z) = ψ(r) the normalized wave function of a particle. Express in terms of ψ(r) the probability for:

b) a measurement of the component Px of the momentum, to yield a result included between p1 and p2.
c) simultaneous measurements of X and Pz to yield
x1 ≤ x ≤ x2
pz ≥ 0"

I've tried to work it out, but using the wave function in position space instead of using in momentum space really got me in trouble. I do now that they are related to each other by a Fourier Transform, but the expressions in terms of ψ(x,y,z) are a mess! I hope you guys can help me soon, so I can proceed on my study.

Joao Victor said:
I was solving an exercise from Cohen's textbook, but then I got stuck in this question.

Hmm ... I can't find this exercise in the book (on my shelf) "An Introduction to Hilbert Space and Quantum Logic" by Cohen. Or maybe Canadian singer/songwriter/poet Leonard Cohen wrote a a quantum book of which I am unaware. I can find this exercise in one of my books by an author who has a completely different name. Sorry for being harsh, but it is disrespectful to get an author's name so wrong.

Can you show your attempt for b?

@George Jones : in defence of the poster: in his first post he used the full regalia
Joao Victor said:
Yesterday, I was solving an exercise from Cohen-Tannoudji's book - Quantum Mechanics

I fully subscribe your request to let the OP present his attempt at solution. In the thread above he did that correctly and already had good pointers from @nrqed

It is also long overdue to say: Hello Joao, ##\quad##

and: use of the template is mandatory -- see guidelines

Hello!

As BvU already pointed out, the complete name of the author is Claude Cohen-Tannoudji, one of the nobel prize winners on 1997. I also appreciate you guys reception, for it is only the second thread I create here on Physics Forums.
When I tried to solve b, I wrote and expression for the probability in terms of the wave function in momentum space Φ(p), and then i wrote Φ(p) in terms of ψ(r) (since they are related by a Fourier transform). However, the computation of the probability uses the modulus squared of Φ(p), which is Φ(p)Φ*(p). We get, then, a product of integrals, which is a real mess. Is there a way to clean it up a bit?

It would really help if you showed us your actual work rather than simply describing what you did because often the problem arises in the details of your work.

BvU
Here is what I did for b.

We can express this probability using the wave function Φ(p) in momentum space. It is straightfoward - Px has to be between p1 and p2, while Py and Pz can be anything. So:

Prob = ∫∫∫ Φ(p) Φ*(p) dpx dpy dpz, where the last integral (in px) is from p1 to p2 and the other two integrals are from -∞ to ∞.

Now, we have to express that in terms of the wave function in position space: Φ(p) is related to ψ(r) by a Fourier Transform:

Φ(p) = (2πħ)-3/2 ∫∫∫ e-ip.r/ħ ψ(r) dxdydz, with the limits of integration of the three integrals being from -∞ to ∞.

Similarly, we can write an expression for Φ*(p):

Φ*(p) = (2πħ)-3/2 ∫∫∫ eip.r/ħ ψ*(r) dxdydz.

If we substitute this expressions on the equation for Prob, we get:

Prob = (2πħ)-3 ∫∫∫ [(∫∫∫ e-ip.r/ħ ψ(r) dxdydz) * (∫∫∫ eip.r/ħ ψ*(r) dxdydz)] dpx dpy dpz.

I'm sorry for the terrible format above. The equation for Prob is really a mess. Is there a way to "clean it up"?

It's worth spending a little bit of time to learn LaTeX. PF has a primer here: https://www.physicsforums.com/help/latexhelp/.

Your work looks good so far, except you shouldn't be using ##\vec{r}## in both Fourier transform integrals in your expression for the probability. One should be ##\vec{r}'##.

## 1. What is the concept of momentum in physics?

Momentum is a fundamental concept in physics that refers to the quantity of motion possessed by an object. It is defined as the product of an object's mass and velocity, and is a vector quantity with both magnitude and direction.

## 2. How is momentum measured using the position wavefunction?

In quantum mechanics, momentum can be measured using the position wavefunction, which is a mathematical function that describes the probability of finding a particle at a certain position. The momentum of a particle can be obtained by taking the derivative of the position wavefunction with respect to time.

## 3. What is the relationship between momentum and uncertainty principle?

The uncertainty principle, also known as Heisenberg's uncertainty principle, states that it is impossible to simultaneously measure the exact position and momentum of a particle. This is because the act of measuring one quantity will affect the other, and the more accurately we know one, the less accurately we can know the other.

## 4. How do you calculate the average momentum of a particle using the position wavefunction?

The average momentum of a particle can be calculated by finding the expectation value of its momentum operator, which is represented by the derivative of its position wavefunction. This expectation value is obtained by integrating the position wavefunction multiplied by the momentum operator over all possible positions.

## 5. Can momentum be negative when measured using the position wavefunction?

Yes, momentum can be negative when measured using the position wavefunction. This is because the position wavefunction is a complex-valued function, and the imaginary part of the function can result in a negative momentum value. However, the magnitude of momentum is always positive.

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