I Derivation of Eigenfunctions/Eigenvalues of the Momentum Operator

TRB8985
Messages
74
Reaction score
15
TL;DR Summary
Just trying to understand the derivation of the eigenvalues & eigenfunctions of the momentum operator.
Good afternoon all,

In David Griffiths' "Intro to Quantum Mechanics", I'm looking through Example 3.2 on page 115 that shows how to get the eigenfunctions and eigenvalues of the momentum operator.

I completely understand everything up until this part:

##\int_{-\infty}^{\infty} f_p'^*(x) f_p(x) dx = |A|^2 \int_{-\infty}^{\infty} e^{(i(p-p')x/\hbar)}dx = |A|^2 2\pi \hbar \delta(p-p')##

Where ##f_p(x) = Ae^{ipx/\hbar}.##

I'm not really understanding how the first parts follow into the last. When I try to do the integration explicitly of the middle part, I end up with the following:

## |A|^2 \frac {\hbar} {i(p-p')} \int_{-\infty}^{\infty}e^{(i(p-p')x/\hbar)}dx ##

I can see how ## \hbar ## ends up in the numerator on the end, but the factor of ##2 \pi## seemingly comes out of nowhere, and the delta function just appears without any factor of ##i## attached somehow.

Could anyone explain the missing link between the middle part of the equation and the last? Thank you so very much.
 
Physics news on Phys.org
I think I've figured this out, so I thought I'd post what I found so if anyone else gets stuck they can reference back here. Please feel free to point out if this is incorrect!

Referring back to the definition of the Dirac delta function on page 89:

$$ \delta(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ikx} dk $$

If I pull up the expression that was giving me trouble in the middle, we basically have the same kind of form, but with integration to x instead of k:

$$ |A|^2 \int_{-\infty}^{\infty} e^{i(p-p')x/\hbar} dx$$

Thus, I'm not going have have a delta in the form of ## \delta(x) ## like when the integration variable was with respect to k, but now, since the variable has switched to x, my delta function essentially "switches feet" in the exponent to ## \delta(\frac{p-p'}{\hbar}) ##.

Additionally, we have the following quality for the delta function:

$$ \delta(cx) = \frac{1}{|c|}\delta(x) $$

Which means,

$$ \delta(\frac{p-p'}{\hbar}) = \frac{1}{|1/\hbar|}\delta(p-p') = \hbar \delta(p-p')$$

Recall the hidden ##1/2\pi## buried in the definition at the top of this post. To keep the equation balanced (not off by a factor of ##1/2\pi##), we'll need to multiply by ## 2\pi##. In doing so, we match the right-hand side of the equation at the top of my original post:

$$ \implies |A|^2 \int_{-\infty}^{\infty} e^{i(p-p')x/\hbar} dx = |A|^2 2\pi \hbar \delta(p-p')$$

Thank you for reading!
 
  • Like
Likes vanhees71 and stevendaryl
I am not sure if this belongs in the biology section, but it appears more of a quantum physics question. Mike Wiest, Associate Professor of Neuroscience at Wellesley College in the US. In 2024 he published the results of an experiment on anaesthesia which purported to point to a role of quantum processes in consciousness; here is a popular exposition: https://neurosciencenews.com/quantum-process-consciousness-27624/ As my expertise in neuroscience doesn't reach up to an ant's ear...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
I am reading WHAT IS A QUANTUM FIELD THEORY?" A First Introduction for Mathematicians. The author states (2.4 Finite versus Continuous Models) that the use of continuity causes the infinities in QFT: 'Mathematicians are trained to think of physical space as R3. But our continuous model of physical space as R3 is of course an idealization, both at the scale of the very large and at the scale of the very small. This idealization has proved to be very powerful, but in the case of Quantum...
Back
Top