# Relativistic Momentum Operator

• Yitzach
In summary, according to Yitzach, the relativistic momentum operator is still p = -ih∇, but never mind that. You don't need special relativity to solve the square well! The usual thing to do when your integrals diverge is to put everything in a box of side L.Perhaps I wasn't quite clear. I have the wavefunction in "configuration" space for the infinite square well which does a very nice job of giving a probability distribution for where the particle should be. I want the probability distribution for the momentum of the particle. Getting the wavefunction in momentum space is reasonable enough. The problem is that it is not square-integrable on the real number line. The following is the momentum
Yitzach
I have in mind to build a game to help teach or demonstrate some concepts in QM and I thought it would be nice to be able to measure momentum. So as a proof of concept before I get too many man hours burned on the project I thought it would be good to do the infinite square well. I managed to get the un-normalized momentum probability out of my math. The only problem with it was that the integral on the real number line did not converge due to it being sinusoidal with a constant additive that gave periodic zeros in the trough. So then I think to integrate between -c and +c. But as this is a classical system, c is not a necessary universal speed limit. So now I would like the relativistic momentum operator. I'm not seeing that available, yet. So using Griffths's book on QM as my resource:
$\hat{p}=\frac{mv}{\sqrt{1-(\frac{v}{c})^2}}$6.47
$p=-i\hbar\nabla\Rightarrow v=-\frac{i\hbar}{m}\nabla$cover
$\hat{p}=\frac{-i\hbar\nabla}{\sqrt{1-(\frac{-i\hbar\nabla}{mc})^2}}$
$\hat{p}=\frac{-i\hbar\nabla}{\sqrt{1+(\frac{\hbar}{mc})^2\nabla^2}}$
Use:
$\int\Psi*\frac{-i\hbar\nabla}{\sqrt{1+(\frac{\hbar}{mc})^2\nabla^2}}\Psi dx=\left<p\right>$
$\int\Psi*\frac{-i\hbar\nabla\Psi}{\sqrt{1+(\frac{\hbar}{mc})^2\nabla^2\Psi}} dx=\left<p\right>$
This doesn't look quite intelligent to me. Does anyone have a better solution to the problem? It would be wonderful if you could provide the eigenfuction for it as well. If this is it, the next trick will be to solve the following equation or an approximation of it:
$\frac{-i\hbar\nabla\Psi}{\sqrt{1+(\frac{\hbar}{mc})^2\nabla^2\Psi}}=p\Psi$

The relativistic momentum operator is still p = -ih∇, but never mind that. You don't need special relativity to solve the square well! The usual thing to do when your integrals diverge is to put everything in a box of side L.

Perhaps I wasn't quite clear. I have the wavefunction in "configuration" space for the infinite square well which does a very nice job of giving a probability distribution for where the particle should be. I want the probability distribution for the momentum of the particle. Getting the wavefunction in momentum space is reasonable enough. The problem is that it is not square-integrable on the real number line. The following is the momentum probability distribution that I derived 3.54 and 3.56 of Griffths. $|A|^2(-2\cos(\pi n)\cos(\frac{p}{\hbar})+\cos^2(\pi n)+1))$ As it is periodic, the integral over all real numbers does not converge. I would be very surprised if classical QM required that any particle with mass can exceed any speed greater than or equal to that of light. I can make it integrate to a number by requiring that the momentum does not exceed 2.730924×10^-22 Js/m for an electron. But that is relativistic with a gamma of square root of 2. If the momentum operator is unchanged due to relativistic effects, then something else must be added somewhere else to make the distribution square-integrable over the real numbers. Momentum in both classical mechanics and relativity is confined to the real numbers, but relativity restricts the speed to less than or equal to c. Some work with paper and pencil may be required.

The problem is that it is not square-integrable on the real number line.
That's why you put it in a box of side L.
If the momentum operator is unchanged due to relativistic effects, then something else must be added somewhere else to make the distribution square-integrable over the real numbers.
Box of side L, Yitzach.

Let's (mostly me) try this again. I have the following:
$\Psi_n(x)=\sqrt2\sin(n\pi x)\ x\in[0,1]$
$|\Psi_n(x)|^2=2\sin^2(n\pi x)\ x\in[0,1]$
$\Phi_n(p)=\frac{1}{\sqrt{2\pi\hbar}}\int^\infty_{-\infty}e^{-ipx/\hbar}\Psi(x)dx\ x\in[0,1],p\in\Re$3.54
$\Phi_n(p)=\frac{\pi n\hbar^2}{\sqrt{\pi\hbar}}\frac{e^{-ip/\hbar}(e^{ip/\hbar}-\cos(\pi n))}{\pi^2\hbar^2n^2-p^2}\ p\in\Re$
$|\Phi_n(p)|^2=\pi n^2\hbar^3\frac{-4\cos(\pi n)\cos(p/\hbar)+4}{2(p-\pi\hbar n)^2(p+\pi\hbar n)^2}\ p\in\Re$3.56 filled in for this system.
That looks much better. It should already be normalized, ie that last one should integrate over the real numbers to 1. This time I dropped the sin(n pi) as soon as I saw it and I think it solved a few problems. The other problem was that I had a variable in what I thought was my constant. That should be my solution. Wolframalpha was used for a number of leaps. I would recommend you check my handiwork.

Last edited:
Good, I'm glad you got it to work out. Actually from your original description I thought you were doing it numerically. Is that where you're headed with this?

I think that analytically will be easier on the processor(s) and so I would like as many analytical solutions as possible. But I suspect that eventually, there will be a need for numerical solutions. Numerical is just a little hard on CPU's if you are lacking with that in your computer.

## 1. What is the Relativistic Momentum Operator?

The Relativistic Momentum Operator is a mathematical operator used in the theory of relativity to describe the momentum of a particle. It takes into account the effects of both special relativity and quantum mechanics.

## 2. How is the Relativistic Momentum Operator different from the classical momentum operator?

The Relativistic Momentum Operator is different from the classical momentum operator in that it includes the effects of special relativity, such as time dilation and length contraction. It also takes into account the particle's energy and rest mass, rather than just its velocity.

## 3. What is the equation for the Relativistic Momentum Operator?

The equation for the Relativistic Momentum Operator is P = γm0v, where P is the momentum operator, γ is the Lorentz factor, m0 is the rest mass of the particle, and v is its velocity.

## 4. How is the Relativistic Momentum Operator used in quantum mechanics?

The Relativistic Momentum Operator is used in quantum mechanics to describe the momentum of particles at high speeds, where the effects of special relativity cannot be ignored. It is an important component in the Dirac equation, which describes the behavior of relativistic particles.

## 5. What is the significance of the Relativistic Momentum Operator in modern physics?

The Relativistic Momentum Operator is significant in modern physics because it allows for a more accurate description of the momentum of particles, particularly at high speeds. It is also a key component in the theory of relativity, which has revolutionized our understanding of space and time.

Replies
56
Views
3K
Replies
9
Views
845
Replies
7
Views
953
Replies
6
Views
1K
Replies
9
Views
2K
Replies
5
Views
1K
Replies
3
Views
1K
Replies
3
Views
871
Replies
15
Views
957
Replies
4
Views
2K