Delta property, integration by parts, heaviside simple property proof

[binbagsss]

Homework Statement

I am trying to show that $\int \delta (x-a) \delta (x-c) dx = \delta (-a-c)$ via integeration by parts, but instead I am getting $\delta (c-a)$ (or $\delta (a-c)$ depending how I go...).
Can someone please help me out where I've gone wrong: struggling to spot it:

Homework Equations

above.

The Attempt at a Solution

I am trying to show that $\int \delta (x-a) \delta (x-c) dx = \delta (-a-c)$ via integeration by parts, but instead I am getting $\delta (c-a)$ (or $\delta (a-c)$ depending how I go...).
Can someone please help me out where I've gone wrong: struggling to spot it:

By parts $dv/dx= \delta (x-c)$
$v= H(x-c)$, $H(x)$ the Heaviside function
$u=\delta(x-a)$
$du/dx=d/dx ( \delta (x-a)$

Then

$\delta (x-a) \delta (x-c) dx = ^\infty _{-\infty} [ \delta (x-a) H (x-c) ] - \int H(x-c) d/dx (\delta (x-a)) dx$
First term vanishes,
=$- ^{\infty}_{c} [\delta (x-a) ] = \delta (c-a)$, if $c \geq a$
or $=0$ if $c<a$

The limit change from $H(x-c) \neq 0$ for $x-c > 0$.

I can't spot what's gone wrong.

thanks a lot.

 

[PeroK]

Homework Statement

I am trying to show that $\int \delta (x-a) \delta (x-c) dx = \delta (-a-c)$

Are you sure about this? The answer you are getting of $\delta (a-c) = \delta(c-a)$ looks right.

 

[PeroK]

As as aside, here's something I use to take some of the mystery out of the Dirac delta function. You replace the delta function with a finite spiked function. Any function will do, but two obvious candidates are:

$S_1(x) = \frac{1}{2\alpha} \ (-\alpha < x < \alpha) \ ; \ S_1(x) = 0,$ otherwise. For some small, positive $\alpha$

$S_2(x) = \frac{1}{\sqrt{2\pi}\sigma}exp(-\frac{x^2}{2\sigma^2})$. For some small, positive $\sigma$

Loosely, the Dirac delta function is the limit of either of these functions as $\alpha$ or $\sigma$ tends to $0$.

Now, in this example, you could replace the second Delta function with one of these finite spiked functions:

$\int \delta(x-a)S(x-c)dx = S(a-c) \approx \delta(a-c)$

You know this from the standard, defining property of the Delta function. $S(a-c)$ is just your spiked function $S(x)$ evaluated at $a-c$. In the finite case, this is effectively $0$ when $a$ is not close to $c$ and large when $a$ is close to $c$. In any case, it's a approximately $\delta(a-c)$.

This loosely shows that the standard property of the delta function can be applied, even when the second function is also a delta function - which wasn't immediately obvious. And that confirms that you are trying to prove an incorrect identity!

 

[Ray Vickson]

Homework Statement

I am trying to show that $\int \delta (x-a) \delta (x-c) dx = \delta (-a-c)$ via integeration by parts, but instead I am getting $\delta (c-a)$ (or $\delta (a-c)$ depending how I go...).
Can someone please help me out where I've gone wrong: struggling to spot it:

Homework Equations

above.

The Attempt at a Solution

I am trying to show that $\int \delta (x-a) \delta (x-c) dx = \delta (-a-c)$ via integeration by parts, but instead I am getting $\delta (c-a)$ (or $\delta (a-c)$ depending how I go...).
Can someone please help me out where I've gone wrong: struggling to spot it:

By parts $dv/dx= \delta (x-c)$
$v= H(x-c)$, $H(x)$ the Heaviside function
$u=\delta(x-a)$
$du/dx=d/dx ( \delta (x-a)$

Then

$\delta (x-a) \delta (x-c) dx = ^\infty _{-\infty} [ \delta (x-a) H (x-c) ] - \int H(x-c) d/dx (\delta (x-a)) dx$
First term vanishes,
=$- ^{\infty}_{c} [\delta (x-a) ] = \delta (c-a)$, if $c \geq a$
or $=0$ if $c<a$

The limit change from $H(x-c) \neq 0$ for $x-c > 0$.

I can't spot what's gone wrong.

thanks a lot.

You should have $\delta(c-a)$, not $\delta(-a-c)$.

Besides the method of Perok in post #3, you can instead look at the effect as a generalized function, applied to a test function. That is, let $F(c) = \int \delta(x-a) \delta(x-c) \, dx$, with $a$ considered as an input constant. We have
$$\int F(c) f(c) \, dc = \int_c \int_x f(c) \delta(x-a) \delta(x-c) \, dx \, dc\\
= \int_x \delta(x-a) \left(\int f(c) \delta(x-c) \, dc\right) \, dx = \int f(x) \delta(x-a) \, dx = f(a),$$
so considered as an operator on functions we have $F(c) = \delta(c-a)$.

 

[binbagsss]

See attached - quantum theory. first line.
Setting $a=k_1+k_2$ and $c=-k_3-k_4$
$\delta ( -a -c )$ gives me the correct expression of $\delta(k_3+k_4-k_1-k_2) = \delta ( -k_3 - k_4 + k_1 + k_2)$
whereas
$\delta(a-c)$ gives me $\delta(k_1+k_2+k_3+k_4)$

 

[Orodruin]

See attached - quantum theory. first line.

Well, that step is wrong. Without seeing the rest of the argumentation it is difficult to tell you what is going on.

 

[Ray Vickson]

See attached - quantum theory. first line.
Setting $a=k_1+k_2$ and $c=-k_3-k_4$
$\delta ( -a -c )$ gives me the correct expression of $\delta(k_3+k_4-k_1-k_2) = \delta ( -k_3 - k_4 + k_1 + k_2)$
whereas
$\delta(a-c)$ gives me $\delta(k_1+k_2+k_3+k_4)$

Fact:
$$\int \delta(x-a) \delta(x-c) \, dx = \delta(a-c) = \delta(c-a). $$
You have been shown at least two correct, independent proofs of that fact. If you are reading an article that says something else then that article is wrong!

 

[binbagsss]

Homework Statement

I am trying to show that $\int \delta (x-a) \delta (x-c) dx = \delta (-a-c)$ via integeration by parts, but instead I am getting $\delta (c-a)$ (or $\delta (a-c)$ depending how I go...).
Can someone please help me out where I've gone wrong: struggling to spot it:

Homework Equations

above.

The Attempt at a Solution

I am trying to show that $\int \delta (x-a) \delta (x-c) dx = \delta (-a-c)$ via integeration by parts, but instead I am getting $\delta (c-a)$ (or $\delta (a-c)$ depending how I go...).
Can someone please help me out where I've gone wrong: struggling to spot it:

By parts $dv/dx= \delta (x-c)$
$v= H(x-c)$, $H(x)$ the Heaviside function
$u=\delta(x-a)$
$du/dx=d/dx ( \delta (x-a)$

Then

$\delta (x-a) \delta (x-c) dx = ^\infty _{-\infty} [ \delta (x-a) H (x-c) ] - \int H(x-c) d/dx (\delta (x-a)) dx$
First term vanishes,
=$- ^{\infty}_{c} [\delta (x-a) ] = \delta (c-a)$, if $c \geq a$
or $=0$ if $c<a$

The limit change from $H(x-c) \neq 0$ for $x-c > 0$.

I can't spot what's gone wrong.

thanks a lot.

Thank you for your replies.
However can someone please help me see what is wrong with the step above where I have
=$- ^{\infty}_{c} [\delta (x-a) ] = \delta (c-a)$, if $c \geq a$
or $=0$ if $c<a$

I have $c \geq a$ since $a$ must be $\in$ the domain I am integrating over, however this doesn't seem to make sense because $\delta (c-a) = \delta (a-c)$, so if I do the above derivaiton the other way around I'd instead get the condition $=0$ if $a<c$ ... ?

Many thanks.

 

[Ray Vickson]

Thank you for your replies.
However can someone please help me see what is wrong with the step above where I have
=$- ^{\infty}_{c} [\delta (x-a) ] = \delta (c-a)$, if $c \geq a$
or $=0$ if $c<a$

I have $c \geq a$ since $a$ must be $\in$ the domain I am integrating over, however this doesn't seem to make sense because $\delta (c-a) = \delta (a-c)$, so if I do the above derivaiton the other way around I'd instead get the condition $=0$ if $a<c$ ... ?

Many thanks.

Derivatives of $\delta$ do not behave the way you think they should. Formally, we may regard the derivative as
$$\delta'(x) = -\frac{1}{x} \delta(x).$$
This does not make much sense as a statement about a function; it really means that $x \delta'(x) = -\delta(x)$.

$\delta(x)$ is not really an "ordinary" function, so using ordinary methods on it may lead to disaster, as they have done in your case.

For more on this, see, eg., http://mathworld.wolfram.com/DeltaFunction.html
or https://en.wikipedia.org/wiki/Dirac_delta_function .

 

[binbagsss]

Thank you for your reply,

so rather above where i wrote:

=$- ^{\infty}_{c} [\delta (x-a) ] = \delta (c-a)$, if $c \geq a$
or $=0$ if $c<a$

The limit change from $H(x-c) \neq 0$ for $x-c > 0$.

It should have been $= - ^{\infty}_{c} [\delta (x-a) ] = \delta (c-a)$ for all $c$,$a$, i.e. without any condition?
what's the proper way I should have wrote this?

thanks a lot.