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Arjun

05:54

Can the derivative of a delta function not be balanced by a delta function?I.e can't we have an equation in which we have delta function on one side and it's derivatives on the other?

My question stemmed from an answer by @mikestone to the following question on SE

A: Is continuity of the wavefunction "put in by hand" for the Dirac delta potentials?

Were $\psi(x)$ not continuous at $x=0$ then $\psi''(x)$ would contain the derivative of a $\delta$-function, and there is nothing else in the equation $H\psi=E\psi$ that could cancel it, so a discontinuity is not allowed.

naturallyInconsistent

@Arjun Can you give an example of what kind of absurdity you want us to consider? Also, I think Qmechanic's answer is much more illuminating than mike stone's

Arjun

06:21

@naturallyInconsistent I agree that qmechanic's answer makes much more sense..but I don't really understand what Mike stone is trying to say..even Barton zwiebach says the same thing in his book as to why $\psi(x)$ can't have a discontinuity

"The wave function must be continuous at all points. Assume ψ fails to be
continuous by having a set of finite discontinuities. Since the derivative of a step
function is a delta function (exercise 4.4), we conclude that the derivative ψ′ would
contain delta functions, and ψ′′, on the left-hand side of (6.2.4), would contain
derivatives of delta functions. This would require the right-hand side to have
derivatives of delta functions, and those would have to appear in the potential
since ψ only has finite discontinuities. Having declared that our potentials contain

6.2.4 is just TISE with ψ′′ on the lhs

Ryder Rude

06:36

@Arjun hi. the derivative of the delta function blows up to a bigger infinity than the delta function becuz $d/dx (1/a e^{-x^2/a^2})= -2x/a^3 e^{-x^2/a^2}$. This cant be cancelled by the delta function in the limit $a\rightarrow 0$

one can try these Gaussian potentials and try to cancel things out in the limit $a\rightarrow 0$. i dont think it works @Arjun

i mean that it's more singular than the delta function

Arjun

@RyderRude thanks for replying! Is there some way to make these arguments mathematically precise..i.e derivative of delta being more singular than delta and hence it being not able to balance its derivative?

Ryder Rude

try it with a Gaussian potential. $E\psi$ does not contain delta function, so you have to make sure $H\psi$ doesnt contain a singularity in the $a\rightarrow 0$ limit of the Gaussian potential @Arjun

i mean the lhs shud b a well defined function in the limit, becuz the rhs is well defined

maybe one can take $\psi(x)$ = theta function for simplicity

but it's not square integrable ofc

Arjun

06:55

Also why the heck does this equality,δ′(t)f(t)-f(0)δ′(t)=f′(0)δ(t) work? There is derivative of delta on lhs and delta on the rhs,doesn't this contradict what you've said?

I came across it in this answer..dsp.stackexchange.com/a/68736

Ryder Rude

@Arjun sorry i dont recognise this equality. is this related to $\int \delta ' f = -\int \delta f'$

Arjun

I think yes

I don't really have a problem with the integral on..but the op in the above linked answer has written the equality as one of the properties of δ′(t)

naturallyInconsistent

07:22

@Arjun Then I'm really wondering why you would want to expend so much effort to understand a clearly throwaway statement. Later, it seems like you have agreed that the derivative of a Dirac delta distribution diverges even faster than the Dirac delta distribution itself; is that sufficient for you to accept this as true and move on?

Ryder Rude

@Arjun sorry idk what that equality means. maybe someone else can help

Arjun

@naturallyInconsistent Sure,but what's not letting me move on is the the above equality that I've mentioned in the above comment..see the answer dsp.stackexchange.com/a/68736

Equality 2 says ,δ′(t)(f(t)-f(0))=f′(0)δ(t) ,why does this work if δ′(t) diverges off faster than ,δ(t)?

naturallyInconsistent

@Arjun This is some stupid formal manipulation result. By the integral definition of Dirac delta distribution and its derivatives, you get that $$\int\frac{(-t)^n}{n!}\frac{\mathrm d^n\ }{\mathrm dt^n}\delta(t)\,\mathrm dt=1$$ and $$\forall(m,n)\in(\mathbb Z^+)^2\wedge m>n\qquad\int t^m\frac{\mathrm d^n\ }{\mathrm dt^n}\delta(t)\,\mathrm dt=0$$ Now simply apply these relations to the formal Maclaurin's expansion of $f(t)$ and you will get that result.

Arjun

07:39

@naturallyInconsistent my point is if the above divergence argument is legit..why doesn't it lead to to a problem if the above equality were to hold? We have δ(t) on one side and δ'(t) on the other side of the equality after rearranging terms

naturallyInconsistent

07:49

@Arjun no, it does not. These are different equations. If you never started with the derivative of the Dirac delta distribution, then you will not have it appear after that.

Arjun

@naturallyInconsistent sorry but I don't really understand you ..could you elaborate a bit more?

naturallyInconsistent

@Arjun I am saying that you have to completely separate your jumbled questions. You asked "Why must $\psi(x)$ be continuous?" and the answer to that is what Qmechanic and what Barton Zweibach said. In a nutshell, what mike stone said is also similar. Then you also asked "How to get $\delta^\prime(t)f(t)=f(0)\delta^\prime(t)-f^\prime(0)\delta(t)$?" and the answer is what I have written. These are two completely different questions and deserve two different answers.

naturallyInconsistent

08:12

errm, did you get it?

Arjun

08:55

@naturallyInconsistent sorry for replying late..but I'm still note sure if I can use the gaussian function to show that δ'(t) diverges faster than δ(t)..Is it really correct to write dirac delta as a limit of the gaussian function..I think we can use the limit only under the integral..but then how does one show that δ'(t) faster than δ(t)?

@naturallyInconsistent * How do you show that δ'(t) goes to infinity faster than δ(t)? Was the last line of the above comment

@RyderRude i just realised that delta function isn't really the limit of the gaussian..it can only be said so under the integral..then how would you show that the derivative blows faster than the function?

naturallyInconsistent

@Arjun You can try out the many different expressions that have the same limit (as in, they all become the Dirac delta distribution in their appropriate limits), and see that all of them will give you the same result. It should suffice to convince you that this works.

@Arjun In a sense, the Dirac delta distribution only needs to go to infinity, and the come back. The derivative of the Dirac delta distribution has to go to negative infinity, come back to zero, overshoot towards positive infinity, and then come right back to zero. It should be obvious from such considerations that it diverges even faster.

Or the other way around; I cannot be bothered to get the sign correct right now.

Arjun

09:22

@naturallyInconsistent I just realised that the limits used to defined are not exact and they are defined in a "weak sense"..in any case I think it all makes sense to me now..but I'm slightly overwhelmed by the number of mathematical subtleties involved when it comes to distributions..

naturallyInconsistent

And the annoyance that is that, if you expend the effort to learn distribution theory from the mathematicians, all you will get out of it is a tremendously good understanding, but it will just justify all the sloppy trickery that physicists do with them.

1 hour later…

Ryder Rude

10:35

@Arjun the limit thing was the only argument i had in mind... it seems it's incorrect

1 hour later…

Arjun

11:54

@RyderRude I think as a physicist you can get away with it lol

2 hours later…

Ryder Rude

14:13

@Arjun lol