Hi Brant, thanks a lot for this very clear lecture. However, at 17:28, dont you think the final expression of the integral must be 1 + (e^-1)/2 + (e^1)/2 and not 1 + (e^-x)/2 + (e^x)/2 ?
awesome explanations of the properties. Thank you very much for doing this video. Made my concept more clear. I had been looking for this for a while. :)
25:30 I can see how the TOTAL "area under the curve" adds up to zero. Because you've got as much area above the x axis as you have below it. But TECHNICALLY, I can see how... you've got the Riemann integral. Then you've got Riemann integrals with the limit as the upper and lower limits of the integral tend to infinity - and THAT DOESN'T CONVERGE in this case. BUT.... I imagine that the LEBESGUE integral can come to the rescue here and fix things nicely! 🙂
This might be the best rapid intro to delta “functions” without all the distributional math. Well done, Brant! I loved watching every second of this lecture. I do have a minor question though. Do you have any recommendations for a mathy derivation of the delta decomposition into the sums by renormalizing the underlying distribution, as you implied visually? A picture as good as the one you gave, and some solid math, would go a really long way towards boosting my understanding. I’m working on a method to expand the types of equations you can solve via Fourier analysis and came across the need to reinterpret delta(g(x)) pretty much the same way you did it, but it was more of “intuition” for me, than anything else.
@1:20 the negative e power should make the e power term tend to 0 for small values of sigma and at the same time 1/(2pisigma) should tend to infinity. This doesn't make the distribution narrow for small values of sigma. Basically, what I'm saying is that the e power shouldn't be negative!!
At x=0 however, the large exponent doesn't matter anymore, and that part just becomes e^0 = 1. Everywhere else is heavily suppressed by a large negative exponent, so the thing peaks like a delta function. Blew my mind when I got it lol
You are wrong: it does make the distribution narrow, because e^(x^2) >> x, so the limit is 0 everywhere except at x = 0. You can verify this rigorously with L'Hôpital's theorem.
How did he evaluate delta(1)/2 as e^-x/2, and not e^-1, and the other answer too! Wouldn't it be e^2+e+1/e the integral? Does it mean evaluating at x for f(1) and so on?
Honestly not sure what sort of sociopath dislikes youtube videos like this. The man is literally teaching an entire QM course or two, free of charge, and you gonna give him shit?
Technically, the δ distribution is not even a limit of a sequence of probability distributions either, because probability distributions are still functions, whereas δ is not a function. The definition of a distribution, as a generalized function, actually requires some complicated details about topologies of linear functionals on a space of smooth functions of our choice. This is very much relevant to how the calculations happen to work here, and a heuristic demonstration is not sufficient.
Ultimately, I consider that this video had too much handwaving, but also too many mistakes. It was some nice effort and still informative, but I believe this could be done better by fixing those mistakes and trying to be more rigorous and derivation-oriented in the explanation.
Technicallities should be left to mathematicians (I'm one miself, by the way, and I've studied Distribution Theory). What a Physics student needs (and I consider myself one, at this point) is an operative understanding of the results, in order to apply them. Barring a couple of mistakes, I think the video achieves that. However a solid mathematical proof (or a precise deffinition) be convenient for the ease of mind (of said mathematicians), many advanced concepts were put forward, and used, PRIOR to rigorous deffinitions and proofs. Such is the case of the Heaviside step function, and the Dirac delta, now that we are talking of many things. en.wikipedia.org/wiki/Oliver_Heaviside
for d(x^3 - x) e^-x why couldn't we just use the symmetrical property and then have dirac simply be shifted by x^3 (g(x)) and have it return f(g(x)) which would be e^-(g(x)) ?
Well, that makes absolutely no sense according to the rules. What property are you using to get δ{x^3 - x} = x^3·δ{x}? There is no property that implies it. If you try deriving an integral for this, you will realize it does not work, because it is incorrect.
The linearization of g at x(i) is given by g[x(i)] + g'[x(i)]·[x - x(i)] = g'[x(i)]·[x - x(i)], and so you may consider δ{g(x)} as a sum δ{g'[x(i)]·[x - x(i)]} over i, because of the argument presented in the video. So the integrand you want to consider is δ{g'[x(i)]·[x - x(i)]}·f(x), and the result of the integration is the sum of these integrals over i. Here, you can do a shift x |-> x + x(i), changing the integrand to δ{g'[x(i)]·x}·f[x + x(i)], while the boundary of integration remains the same, as it is infinite. Now, it should be clear that δ{-y} = δ{y}, so δ{g'[x(i)]·x}·f[x + x(i)] = δ{|g'[x(i)]|·x}·f[x + x(i)]. Let x |-> x/|g'[x(i)]|. Again, the boundaries are unchanged, but the integrand becomes δ{x}·f{x/|g'[x(i)]| + x(i)}/|g'[x(i)]|. Evaluating the integral here results in f[x(i)]/|g'[x(i)]|. You can perform the summation here, but notice how the same integral is obtained by simply integrating δ{x - x(i)}·f(x)/|g'[x(i)]|. Therefore, summing these integrals over i should just give the integral δ{g(x)}·f(x). Therefore, δ{g(x)} is just the sum of δ{x - x(i)}/|g'[x(i)]| over i, since integration and multiplication by fixed f(x) are linear. Of course, this all assumes |g'[x(i)]| > 0.
Curious Bit it means the integral from neg to pos infinity equals 1 of your probability density function(in this case the wave function squared) since in QM the wave function squared gives a probability distribution, it can’t equal more than 1 over all space since you can’t have more than a 100% probability, so you put a constant out front which “normalizes” your wave function, e.g makes your integral from neg to pos infinity of your wave function squared equal 1
At the end of the lecture you explained why the integral = the delta function, but, as you wrote, it doesn't actually equal the delta function--it equals 2pi times the delta function. (The factor of 2pi emerges when deriving this equation using Plancherel's theorem (problem 2.26 in the textbook)).
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This is gold....Wish we had professors like you...
at 17:34 shouldn't you replace x with the values of x
Yes this is definitely an error. Should be 1+(e+(1/e))/2
Ok thank you for asking that I was confused why he said evaluate it at those points then proceeded to put variables.
When you did the example with x³-x g(x) and solved the integral, you forgot to sub in the values of x:
1 + 1/2e + e/2 as your xi ={0,1,-1}
Extraordinary explanation of this dirac delta , Professor. Thanks for your great efforts!
Hi Brant, thanks a lot for this very clear lecture. However, at 17:28, dont you think the final expression of the integral must be 1 + (e^-1)/2 + (e^1)/2 and not 1 + (e^-x)/2 + (e^x)/2 ?
Manu Kamin yeah it will be that... He didn't put those values of X
Yes this seens to have been a slip on thw lecturers part
Agreed ... so he does tutorial. Makes mistakes which have been corrected. He never comments. I guess that's public education.
Brant! I've been so confused about this for so long and this video helped me so much! Thank you!!!
These lectures are brilliantly constructed. Makes complex material, understandable.
Thank you about the last part in particular. It does seem quite fishy, but it is very necessary in QFT. That one part caught me by surprize.
Hey, I know this video is ages old, but at 21:30 isnt there a minus missing infront of the final term. correct me if i´m wrong. Might help others
Yeah, I think like that too
Yup, looks like it should have a negative sign there. en.wikipedia.org/wiki/Dirac_delta_function#Distributional_derivatives
yeah, thought so too. coming from the second term of integration by parts.
Thank you for your briliant explanation Meïr Krukowsky
17:25 Instead of putting e to the -x and e to the x, it should say e to the -1 and e to the 1, i.e. 1/e and e.
awesome explanations of the properties. Thank you very much for doing this video. Made my concept more clear. I had been looking for this for a while. :)
best video on dirac delta ive seen so far
Ur videos are crystal clear.Thank you very much sir 😊😊
It is best video of delta function i have discovered.
25:30 I can see how the TOTAL "area under the curve" adds up to zero. Because you've got as much area above the x axis as you have below it. But TECHNICALLY, I can see how... you've got the Riemann integral. Then you've got Riemann integrals with the limit as the upper and lower limits of the integral tend to infinity - and THAT DOESN'T CONVERGE in this case. BUT....
I imagine that the LEBESGUE integral can come to the rescue here and fix things nicely! 🙂
2,1 and 1/10
thank you, this really cleared some stuff up for me
Good explanation. thank you!
Awesome explanation, thank you very much!
I believe you missed a minus sign at 21:10.
im quite confused there.. its supposed to be minus sign rite?
+syed zack You probably already have the answer, but there is supposed to be a negative sign, as Dillon stated.
The deriative of dirac distribution multiplied with a function,f(x), integrated all over x should give the negative derivative of f(x) yes.
I think he just meant the integral calculation by itself then afterwards just forgot to connect with the rest of the equation.
I believe he misses a lot of things. Really makes me frustrated many times.
This might be the best rapid intro to delta “functions” without all the distributional math. Well done, Brant! I loved watching every second of this lecture.
I do have a minor question though. Do you have any recommendations for a mathy derivation of the delta decomposition into the sums by renormalizing the underlying distribution, as you implied visually? A picture as good as the one you gave, and some solid math, would go a really long way towards boosting my understanding.
I’m working on a method to expand the types of equations you can solve via Fourier analysis and came across the need to reinterpret delta(g(x)) pretty much the same way you did it, but it was more of “intuition” for me, than anything else.
Thank you very much for the effective lecture
Sir, you saved my day
Great examples
Absolutely mind blowing!!! Could you please tell me what software you are using for your slides and drawing ?
final i found the perfect video waster hours on youtube and online. God bless man!
Excellent video! Helped me a lot!
theanks for this great video....
I have a question
what will happen when whe have double root?
for example delta dirac for (x+1)^2 ?
Ill-defined?
thank you so much man , you are really great
Solid explanation
@1:20 the negative e power should make the e power term tend to 0 for small values of sigma and at the same time 1/(2pisigma) should tend to infinity. This doesn't make the distribution narrow for small values of sigma. Basically, what I'm saying is that the e power shouldn't be negative!!
If the sign of e power will be positive, the function would diverge for infinity x, and obviously that wouldn't be gaussian distribution.
At x=0 however, the large exponent doesn't matter anymore, and that part just becomes e^0 = 1. Everywhere else is heavily suppressed by a large negative exponent, so the thing peaks like a delta function. Blew my mind when I got it lol
You are wrong: it does make the distribution narrow, because e^(x^2) >> x, so the limit is 0 everywhere except at x = 0. You can verify this rigorously with L'Hôpital's theorem.
Thaannkksss you're brilliant!
How did he evaluate delta(1)/2 as e^-x/2, and not e^-1, and the other answer too! Wouldn't it be e^2+e+1/e the integral? Does it mean evaluating at x for f(1) and so on?
This is an error as others have already commented.
24:00, really nice.
at 12:51 the quadratic curve shall be upside down with maxima not minima :)
Honestly not sure what sort of sociopath dislikes youtube videos like this.
The man is literally teaching an entire QM course or two, free of charge, and you gonna give him shit?
At the very end, should it be delta(x - x0) instead of delta(x) ?
18:05 Check Your Understanding: The answers I got were 2, 1, 0. Should be correct, hoping it'd help.
can you please clarify why you omit the minus sign while integrating the derivative of delta(x) times f(x) at about 21'34'' in this video.
btw got great help from this video..
This was a mistake on his part.
Thank you very much for this.
very helpful! Thanks!!
Thank you so much....
What is the solution for the third example?
Technically, the δ distribution is not even a limit of a sequence of probability distributions either, because probability distributions are still functions, whereas δ is not a function. The definition of a distribution, as a generalized function, actually requires some complicated details about topologies of linear functionals on a space of smooth functions of our choice. This is very much relevant to how the calculations happen to work here, and a heuristic demonstration is not sufficient.
Ultimately, I consider that this video had too much handwaving, but also too many mistakes. It was some nice effort and still informative, but I believe this could be done better by fixing those mistakes and trying to be more rigorous and derivation-oriented in the explanation.
@Cephas Storm The confusion expressed by a relatively large proportion of people in the comments very much proves you wrong in that regard.
Technicallities should be left to mathematicians (I'm one miself, by the way, and I've studied Distribution Theory).
What a Physics student needs (and I consider myself one, at this point) is an operative understanding of the results, in order to apply them. Barring a couple of mistakes, I think the video achieves that.
However a solid mathematical proof (or a precise deffinition) be convenient for the ease of mind (of said mathematicians), many advanced concepts were put forward, and used, PRIOR to rigorous deffinitions and proofs. Such is the case of the Heaviside step function, and the Dirac delta, now that we are talking of many things. en.wikipedia.org/wiki/Oliver_Heaviside
Great learning
Thanks for the video. It helped me a lot ^. ^
for d(x^3 - x) e^-x why couldn't we just use the symmetrical property and then have dirac simply be shifted by x^3 (g(x)) and have it return f(g(x)) which would be
e^-(g(x)) ?
so e^-(g(x))
Well, that makes absolutely no sense according to the rules. What property are you using to get δ{x^3 - x} = x^3·δ{x}? There is no property that implies it. If you try deriving an integral for this, you will realize it does not work, because it is incorrect.
Hello sir,i am new to this chapter.why don't we use ∫δ(g(x))f(x)dx=Σf(k),where k=zeroes of g(x).integral limits from -∞ to ∞
Arun Kancharla no it can't be written exact like that... If you look at the |g'| in denominator... It also effect the overall result
@9:57 if it equals the sum of dirac(x-xi) then why did you do @15:28 as dirac(x+xi)?
He didn't do δ{x + x(i)}. Look carefully.
Helpful! thank you
How do we get here 10:32? Like how to to actually derive that?
The linearization of g at x(i) is given by g[x(i)] + g'[x(i)]·[x - x(i)] = g'[x(i)]·[x - x(i)], and so you may consider δ{g(x)} as a sum δ{g'[x(i)]·[x - x(i)]} over i, because of the argument presented in the video. So the integrand you want to consider is δ{g'[x(i)]·[x - x(i)]}·f(x), and the result of the integration is the sum of these integrals over i. Here, you can do a shift x |-> x + x(i), changing the integrand to δ{g'[x(i)]·x}·f[x + x(i)], while the boundary of integration remains the same, as it is infinite. Now, it should be clear that δ{-y} = δ{y}, so δ{g'[x(i)]·x}·f[x + x(i)] = δ{|g'[x(i)]|·x}·f[x + x(i)]. Let x |-> x/|g'[x(i)]|. Again, the boundaries are unchanged, but the integrand becomes δ{x}·f{x/|g'[x(i)]| + x(i)}/|g'[x(i)]|. Evaluating the integral here results in f[x(i)]/|g'[x(i)]|. You can perform the summation here, but notice how the same integral is obtained by simply integrating δ{x - x(i)}·f(x)/|g'[x(i)]|. Therefore, summing these integrals over i should just give the integral δ{g(x)}·f(x). Therefore, δ{g(x)} is just the sum of δ{x - x(i)}/|g'[x(i)]| over i, since integration and multiplication by fixed f(x) are linear. Of course, this all assumes |g'[x(i)]| > 0.
@@angelmendez-rivera351 thankyou for the detailed response 👍🏼
thanks
In case of complex root what happens?
Nothing, since this consideration of distributions specifically works with real valued functions of real variables.
Thanx
can someone explain what 'normalised' means??
Curious Bit it means the integral from neg to pos infinity equals 1 of your probability density function(in this case the wave function squared) since in QM the wave function squared gives a probability distribution, it can’t equal more than 1 over all space since you can’t have more than a 100% probability, so you put a constant out front which “normalizes” your wave function, e.g makes your integral from neg to pos infinity of your wave function squared equal 1
ThanX ........
ANSWERS TO THE QUESTIONS ARE : 2, 1, 1/65(SUSPICIOUS).
PLEASE CORRECT ME IF I AM WRONG!
Answer for 2nd is 3 and not 1.
awesome
THX THX THX
At the end of the lecture you explained why the integral = the delta function, but, as you wrote, it doesn't actually equal the delta function--it equals 2pi times the delta function. (The factor of 2pi emerges when deriving this equation using Plancherel's theorem (problem 2.26 in the textbook)).