Undergrad Dirac delta function integrated on a finite interval

Click For Summary
SUMMARY

The discussion centers on the integration of the Dirac delta function over a finite interval, specifically comparing two interpretations: one stating that $$\int_{-1}^1 \delta (x-x_0) \, dx = 1$$ for $$-1 \leq x_0 \leq 1$$ and the other for $$-1 < x_0 < 1$$. The consensus leans towards the first interpretation, with participants noting the implications of boundary conditions at $$x_0 = -1$$ and $$x_0 = 1$$. Additionally, references to Tobias Osborne's lectures on functional derivatives and the manipulation of the delta function in integration by parts were highlighted as valuable resources for further understanding.

PREREQUISITES
  • Understanding of the Dirac delta function and its properties
  • Familiarity with integration techniques in mathematical physics
  • Knowledge of functional derivatives and their applications
  • Basic concepts of distributions in mathematical analysis
NEXT STEPS
  • Study the properties of the Dirac delta function in detail
  • Learn about functional derivatives as presented in Tobias Osborne's lectures
  • Review mathematical texts on distributions for rigorous definitions and applications
  • Explore integration techniques involving the Dirac delta function in quantum field theory contexts
USEFUL FOR

Students and professionals in physics, particularly those studying quantum mechanics and field theory, as well as mathematicians interested in distributions and functional analysis.

Hill
Messages
737
Reaction score
577
TL;DR
How to account for integration limits with Dirac delta function?
Which is correct:
$$\int_{-1}^1 \delta (x-x_0) \, dx =\begin{cases} 1, -1\leq x_0 \leq 1 \\ 0, \text { otherwise} \end{cases}$$
or
$$\int_{-1}^1 \delta (x-x_0) \, dx =\begin{cases} 1, -1< x_0 < 1 \\ 0, \text { otherwise} \end{cases}$$
?
 
Physics news on Phys.org
OK, I know. The first one.
 
Sorry, not so sure, again. Because, e.g., $$\int_{-1}^2 \delta (x-1) \, dx = 1$$ but if the first case above is correct, then $$\int_{-1}^2 \delta (x-1) \, dx = \int_{-1}^1 \delta (x-1) \, dx +\int_1^2 \delta (x-1) \, dx =2$$ and if the second case is correct, then $$\int_{-1}^2 \delta (x-1) \, dx = \int_{-1}^1 \delta (x-1) \, dx +\int_1^2 \delta (x-1) \, dx =0$$ So, how does it work?
 
By symmetry, we must have:
$$\int_0^1 \delta(x) dx = \frac 1 2$$I use a rule of thumb that what holds for ##\delta (x)## must hold for the limit of appropriate Gaussians.
 
PeroK said:
By symmetry, we must have:
$$\int_0^1 \delta(x) dx = \frac 1 2$$I use a rule of thumb that what holds for ##\delta (x)## must hold for the limit of appropriate Gaussians.
I see. Then you disagree with the following example from the QFTftGA:

1700330922760.png


By your definition, it should rather be $$...=\begin{cases} 1, -1< x_0 < 1 \\ \frac 1 2, x_0=-1\\ \frac 1 2, x_0=1\\ 0, \text { otherwise} \end{cases}$$ Is it so?
 
I'd say so. I can check my notes, but it seems a minor detail. I told you the maths was wild and woolly.

As I recall, Tobias Osborne gives a good definition of the functional derivative somewhere in his lectures.
 
PeroK said:
I told you the maths was wild and woolly.
Yes, you did, and I keep it in mind. :smile:
 
PeroK said:
I'd say so. I can check my notes, but it seems a minor detail. I told you the maths was wild and woolly.

As I recall, Tobias Osborne gives a good definition of the functional derivative somewhere in his lectures.
This other example from the book is perhaps more "wild and woolly," but I might be wrong.

They use $$\frac {\partial} {\partial y}[f(y)+\epsilon\delta(y-x)]=f'+\epsilon\delta'(y-x)$$ Can we do this? Derivative of ##\delta()##?

A bit later, they integrate by parts:

1700340733766.png


taking ##\delta## outside. Is it OK?

P.S. There should be a typo: ##dy## is missing in the last integral.
 

Attachments

  • 1700340697751.png
    1700340697751.png
    2.9 KB · Views: 134
Finally, I'll go ahead and trust that these manipulations with Dirac delta are fine. I like this derivation:

1700343564515.png


They do not explain why the term in square brackets on the third line vanishes, but I understand that it is so because ##t_i<t<t_f##, and this is because we vary the trajectory ##x(t)## everywhere except the end points.
 
  • #10
This is Chapter 1.3. I studied from this book in 2020 so I can't remember everything. I have, however, found my notes.

Interesting, I have no notes on this section! This suggests there was something I didn't like about it! I guess I must have decided that L&B had produced the E-L equations by some exotic mathematics that I didn't understand or didn't like or both. But, the net result was just the E-L equations. Which I was already completey familair with. So, I moved on. I guess I was hoping that their particular derivation of E-L wouldn't turn out to be important. In the worst case, I'd have to come back to this section and try to figure it out.

If I'm studying advanced material like this, with no tutor, then I have to be pragmatic.

Also, when I later watched Tobias Osborne's videos, his approach to functional derivatives made much more sense.
 
  • #11
Here are two lecture notes (physics motivated, both under 6 pages) that helped me:
This one does go into the splitting bounds apart on page 5: http://jacobi.luc.edu/DiracDelta.pdf
This one is more pertaining to your HW thread (Higher order functional derivatives), and it may be more useful for you: https://www.physics.usu.edu/Wheeler/QFT2016/Notes/QFT09FunctionalDerivatives.pdf

However, you'll most likely need to go review a more mathematical oriented text on distributions if you want more generalizations and rigor.
 
  • Like
Likes PeroK and Hill

Similar threads

  • · Replies 25 ·
Replies
25
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
8
Views
2K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K