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Delta integration property result/proof

I have that

\int dk \delta (k-a) \delta (k-c) f(k) = \delta (a-c) f(k)

is this correct and as simple form as I can get it?

How do I go about proving it?

Many thanks

Reply 1

atsruser

Original post by xfootiecrazeesarax

I have that

\int dk \delta (k-a) \delta (k-c) f(k) = \delta (a-c) f(k)

is this correct and as simple form as I can get it?

How do I go about proving it?

Many thanks

Didn't you ask the same question some time ago? I replied that it looked like something that only makes sense inside a double integral, and that Dirac proves something vaguely similar in the Principles of Quantum Mechanics.

Reply 2

RichE

Original post by xfootiecrazeesarax

I have that

\int dk \delta (k-a) \delta (k-c) f(k) = \delta (a-c) f(k)

is this correct and as simple form as I can get it?

How do I go about proving it?

Many thanks

That can't be right as the dummy variable k is in the answer in the right hand side.

Reply 3

atsruser

Original post by xfootiecrazeesarax

I have that

\int dk \delta (k-a) \delta (k-c) f(k) = \delta (a-c) f(k)

is this correct and as simple form as I can get it?

How do I go about proving it?

Many thanks

1. As RichE pointed out, your variables look screwed up. In general, bear in mind that delta functions only make sense inside integrals, so given that your final expression has a delta function, it must be integrated w.r.t some variable. But:

\int \delta(a-c) f(k) \ dk

makes no sense (or at least, doesn't look useful), and neither do the other possibilities with da or dc.

2. Take a look at page 70 of this:

https://archive.org/stream/DiracPrinciplesOfQuantumMechanics/Dirac%20-%20Principles%20of%20quantum%20mechanics#page/n69/mode/1up

Reply 4

xfootiecrazeesarax

Original post by RichE

That can't be right as the dummy variable k is in the answer in the right hand side.

ahh my apologies !

that should be f(a) , or f(c) on the RHS with delta (a-c) multiplying it so it doesn't matter which !

Reply 5

RichE

Original post by atsruser

\int \delta(a-c) f(k) \ dk

makes no sense (or at least, doesn't look useful), and neither do the other possibilities with da or dc.

It makes sense to write the delta function outside of an integral. The delta function can be given rigorous meaning as distribution and in that context I think what the OP writes if ok (when f(k) is changed to f(c)).

One could for example write

H' =\delta

to convey that the delta function is the derivative of the Heaviside function.

Reply 6

atsruser

Original post by RichE

You've confused me here. In this case, the OP seems to using an informal "behaviour inside an integral" approach to the delta function, so given that, I think that it's accurate to say that whatever expression she ends up with must work inside an integral.

Anyway, she has confirmed that she made a typo - I'm still not sure if it's the correct result though, even with the fix.

Reply 7

RichE

Original post by atsruser

I don't think there's anything in what the OP's done that means a delta functions can only appear within integrals.

A special of the desired identity in fact appears on the wiki page

https://en.wikipedia.org/wiki/Dirac_delta_function

at the end of the "translation" section.

Reply 8

atsruser

Original post by RichE

A special of the desired identity in fact appears on the wiki page

https://en.wikipedia.org/wiki/Dirac_delta_function

at the end of the "translation" section.

That's the not quite same identity as she has quoted - there's a sign change, and I'm not sure if we can trivially ignore that due to evenness of the delta function - I tend to confuse myself easily with delta functions so I'm very cautious using them.

The page from Dirac's book that I quoted above in fact proves almost the exact form of the identity that she wants, but with the same sign change, by means of integration rather than any fancy functional stuff, though.

Reply 9

RichE

Original post by atsruser

I've now looked at some of the dirac book. I'm now confused even more confused why you say why you say delta function only make sense in integrals. There are plenty of identities on page 60 (of the book. 70 of the pdf) where identities are given that don't involve integrals.

Reply 10

atsruser

Original post by RichE

I suspect that our confusion is mutual - I'm not really sure where you are envisaging using deltas outside of integrals - could you give an example? Or are you merely pointing out that a delta can be represented as a linear functional, and that the whole integral thing is hand-waving?

Anyway, maybe I didn't word things properly - by saying that deltas only "make sense" inside integral, I'm saying that e.g. all of the identities that Dirac quotes in his book are meant to be used as integrands, and he proves the results from that POV. The page that I showed above in fact says: "The meaning of any of these equations is that its two sides give equivalent results as factors in an integrand". Does that clarify things?

Reply 11

atsruser

Original post by RichE

Just as a final point: are you referring to cases where delta functions appear in DEs to represent impulsive forces, or point charges, or the like? If so, then yes, I'd agree that you can see "bare" delta functions, but even then, we end up immediately integrating them via Laplace transforms or whatever, no?

Reply 12

RichE

Original post by atsruser

I was only concerned with your phrasing

"In general, bear in mind that delta functions only make sense inside integrals, so given that your final expression has a delta function, it must be integrated w.r.t some variable."

I would at best say this is confusing for the OP. We've since discussed lots of examples where delta functions make sense in the absence of any integral and are commonly used in this way.