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Two Dirac delta functions in an integral?
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For context, this is from a quantum mechanics lecture in which we were considering continuous eigenvalues of the position operator.
Starting with the position eigenvalue equation,
$$hatx,phi(x_m, x)=x_mphi(x_m,x)$$
where $x_m$ is the eigenvalue and $phi(x_m, x)$ are the continuous eigenfunctions of the position operator $hat x$.
The professor wrote that $phi(x_m, x)=delta(x-x_m)$ and stated that this is because the eigenbasis is continuous.
But then he wrote the following
$$beginalignint_-infty^inftyphi^*(x_m,x)phi(x_m',x),dx &=int_-infty^inftydelta(x_m-x)delta(x_m'-x),dx tag1\ &=colorreddelta(x_m-x_m')endalign$$
I don't understand how the expression in $(1)$ can be equal to the expression in red. I do understand that $$beginalignint_-infty^inftydelta(x_m-x)phi(x_m',x),dx &=phi(x_m',x_m)tag2\&=color#080delta(x_m-x_m')endalign$$
since the integral 'sifts' out the only value of $x$ where the argument of the Dirac delta function is zero (at $x_m$). I have applied this sifting property for one Dirac delta function (as in $(2)$).
But I don't understand how this works when there are two Dirac deltas in the integrand, $(1)$.
By my logic, I think it should sift out each value, one at a time, so I think that $(1)$ should be
$$int_-infty^inftydelta(x_m-x)delta(x_m'-x),dx =delta(x_m)+delta(x_m')$$
where the results of the integration are added, since the values $x_m$ and $x_m'$ are sifted out one after the other, depending on which of $x_m$ and $x_m'$ are larger (here I assumed $x_m'gt x_m$).
Could someone please derive or explain why equation $(1)$ is true, or give me any hints to help me understand it.
calculus multivariable-calculus mathematical-physics dirac-delta quantum-mechanics
asked Aug 9 at 5:00
BLAZE
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favorite
For context, this is from a quantum mechanics lecture in which we were considering continuous eigenvalues of the position operator.
Starting with the position eigenvalue equation,
$$hatx,phi(x_m, x)=x_mphi(x_m,x)$$
where $x_m$ is the eigenvalue and $phi(x_m, x)$ are the continuous eigenfunctions of the position operator $hat x$.
The professor wrote that $phi(x_m, x)=delta(x-x_m)$ and stated that this is because the eigenbasis is continuous.
But then he wrote the following
$$beginalignint_-infty^inftyphi^*(x_m,x)phi(x_m',x),dx &=int_-infty^inftydelta(x_m-x)delta(x_m'-x),dx tag1\ &=colorreddelta(x_m-x_m')endalign$$
I don't understand how the expression in $(1)$ can be equal to the expression in red. I do understand that $$beginalignint_-infty^inftydelta(x_m-x)phi(x_m',x),dx &=phi(x_m',x_m)tag2\&=color#080delta(x_m-x_m')endalign$$
since the integral 'sifts' out the only value of $x$ where the argument of the Dirac delta function is zero (at $x_m$). I have applied this sifting property for one Dirac delta function (as in $(2)$).
But I don't understand how this works when there are two Dirac deltas in the integrand, $(1)$.
By my logic, I think it should sift out each value, one at a time, so I think that $(1)$ should be
$$int_-infty^inftydelta(x_m-x)delta(x_m'-x),dx =delta(x_m)+delta(x_m')$$
where the results of the integration are added, since the values $x_m$ and $x_m'$ are sifted out one after the other, depending on which of $x_m$ and $x_m'$ are larger (here I assumed $x_m'gt x_m$).
Could someone please derive or explain why equation $(1)$ is true, or give me any hints to help me understand it.
calculus multivariable-calculus mathematical-physics dirac-delta quantum-mechanics
asked Aug 9 at 5:00
BLAZE
5,92092653
add a comment |Â
up vote
12
down vote
favorite
up vote
12
down vote
favorite
For context, this is from a quantum mechanics lecture in which we were considering continuous eigenvalues of the position operator.
Starting with the position eigenvalue equation,
$$hatx,phi(x_m, x)=x_mphi(x_m,x)$$
where $x_m$ is the eigenvalue and $phi(x_m, x)$ are the continuous eigenfunctions of the position operator $hat x$.
The professor wrote that $phi(x_m, x)=delta(x-x_m)$ and stated that this is because the eigenbasis is continuous.
But then he wrote the following
$$beginalignint_-infty^inftyphi^*(x_m,x)phi(x_m',x),dx &=int_-infty^inftydelta(x_m-x)delta(x_m'-x),dx tag1\ &=colorreddelta(x_m-x_m')endalign$$
I don't understand how the expression in $(1)$ can be equal to the expression in red. I do understand that $$beginalignint_-infty^inftydelta(x_m-x)phi(x_m',x),dx &=phi(x_m',x_m)tag2\&=color#080delta(x_m-x_m')endalign$$
since the integral 'sifts' out the only value of $x$ where the argument of the Dirac delta function is zero (at $x_m$). I have applied this sifting property for one Dirac delta function (as in $(2)$).
But I don't understand how this works when there are two Dirac deltas in the integrand, $(1)$.
By my logic, I think it should sift out each value, one at a time, so I think that $(1)$ should be
$$int_-infty^inftydelta(x_m-x)delta(x_m'-x),dx =delta(x_m)+delta(x_m')$$
where the results of the integration are added, since the values $x_m$ and $x_m'$ are sifted out one after the other, depending on which of $x_m$ and $x_m'$ are larger (here I assumed $x_m'gt x_m$).
Could someone please derive or explain why equation $(1)$ is true, or give me any hints to help me understand it.
calculus multivariable-calculus mathematical-physics dirac-delta quantum-mechanics
asked Aug 9 at 5:00
BLAZE
5,92092653
For context, this is from a quantum mechanics lecture in which we were considering continuous eigenvalues of the position operator.
Starting with the position eigenvalue equation,
$$hatx,phi(x_m, x)=x_mphi(x_m,x)$$
where $x_m$ is the eigenvalue and $phi(x_m, x)$ are the continuous eigenfunctions of the position operator $hat x$.
The professor wrote that $phi(x_m, x)=delta(x-x_m)$ and stated that this is because the eigenbasis is continuous.
But then he wrote the following
$$beginalignint_-infty^inftyphi^*(x_m,x)phi(x_m',x),dx &=int_-infty^inftydelta(x_m-x)delta(x_m'-x),dx tag1\ &=colorreddelta(x_m-x_m')endalign$$
I don't understand how the expression in $(1)$ can be equal to the expression in red. I do understand that $$beginalignint_-infty^inftydelta(x_m-x)phi(x_m',x),dx &=phi(x_m',x_m)tag2\&=color#080delta(x_m-x_m')endalign$$
since the integral 'sifts' out the only value of $x$ where the argument of the Dirac delta function is zero (at $x_m$). I have applied this sifting property for one Dirac delta function (as in $(2)$).
But I don't understand how this works when there are two Dirac deltas in the integrand, $(1)$.
By my logic, I think it should sift out each value, one at a time, so I think that $(1)$ should be
$$int_-infty^inftydelta(x_m-x)delta(x_m'-x),dx =delta(x_m)+delta(x_m')$$
where the results of the integration are added, since the values $x_m$ and $x_m'$ are sifted out one after the other, depending on which of $x_m$ and $x_m'$ are larger (here I assumed $x_m'gt x_m$).
Could someone please derive or explain why equation $(1)$ is true, or give me any hints to help me understand it.
calculus multivariable-calculus mathematical-physics dirac-delta quantum-mechanics
asked Aug 9 at 5:00
BLAZE
5,92092653
edited Aug 9 at 20:54
asked Aug 9 at 5:00
BLAZE
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asked Aug 9 at 5:00
BLAZE
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asked Aug 9 at 5:00
BLAZE
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Intuitively, $delta(x_m-x)$ is non-zero only when $x_m=x$. Likewise $delta(x_m'-x)$ is non-zero only when $x_m'=x$. So the factor $delta(x_m-x) , delta(x_m'-x)$ is non-zero only when $x_m=x=x_m'.$ After integration w.r.t. $x$ we still must have $x_m=x_m'$ which makes the result $delta(x_m-x_m')$ quite natural.
More formally, recall the formula $int_-infty^infty f(x) , delta(x_0-x) , dx = f(x_0)$. Apply this to the given integral with $x_0=x_m'$ and $f(x) = delta(x_m-x).$ Then the result $delta(x_m-x_m')$ falls out.
answered Aug 9 at 6:11
md2perpe
6,34811022
Thank you for a brilliant answer. I'm not sure why the argument of the Dirac delta function is $x_m-x_m'$, why not $delta(x_m+x_m')$ for instance? Or is it because from $x_m=x_m'$ such that $x_m-x_m'=0$? Hence, $delta(0)$?
– BLAZE
Aug 9 at 7:10
7
The second paragraph needs a more formal proof. Because the integral with $f$ holds for $f$ being a function, not a distribution.
– Yves Daoust
Aug 9 at 7:45
@md2perpe You write in your answer that "$x_m=x_m'$ makes the result $delta(x_m-x_m')$ quite natural". But I need to know why the 'result' takes that form.
– BLAZE
Aug 9 at 7:49
@BLAZE. The result should be $0$ when $x_m neq x_m',$ i.e. when $x_m - x_m' neq 0.$ That's why we don't have $delta(x_m+x_m').$ We then don't have many alternatives, according to a theorem actually only some linear combination of $delta^(k)(x_m-x_m')$ for $k=0,1,2,ldots$
– md2perpe
Aug 9 at 11:46
3
@YvesDaoust. I know. I didn't want to make it rigorous, just try to give some intuition why the result is as it is. The integral $int_-infty^inftydelta(x_m-x)delta(x_m'-x),dx$ itself isn't welldefined as a product of distributions isn't allowed, even less so the integral of them. For a rigorous treatment we need to give meaning to the integral which is not that difficult; it's basically a convolution. But I think that a rigorous treatment is not what BLAZE needs.
– md2perpe
Aug 9 at 12:24
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$delta(x)$ is never really well-defined by itself (at least not as a function). It's only defined when appearing in an integral (possibly, multiplied with another function). It is essentially syntactic sugar, and the “definition†is
$$ int_mathbbRmathrmdx f(x)cdot delta(x-x_0) := f(x_0) $$
So, what the identity in question actually means is
$$
int_mathbbRmathrmdy: f(y)cdot int_mathbbRmathrmdx: delta(y-x) cdot delta(x_m'-x)
= f(x_m')
$$
for any function $f$.
Why would that be? Well, it's basically just a matter of changing the order of integration:
$$beginalign
int_mathbbRmathrmdy: f(y)cdot int_mathbbRmathrmdx: delta(y-x) cdot delta(x_m'-x)
=& int_mathbbRmathrmdy int_mathbbRmathrmdx: f(y)cdot delta(y-x) cdot delta(x_m'-x)
\ =& int_mathbbRmathrmdx int_mathbbRmathrmdy: f(y)cdot delta(y-x) cdot delta(x_m'-x)
\ =& int_mathbbRmathrmdx: delta(x_m'-x) int_mathbbRmathrmdy: f(y)cdot delta(y-x)
\ =& int_mathbbRmathrmdx: delta(x_m'-x) int_mathbbRmathrmdy: f(y)cdot delta(y-x)
\ =& int_mathbbRmathrmdx: delta(x_m'-x) f(x)
\ =& int_mathbbRmathrmdx: delta(x-x_m') f(x)
\ =& f(x_m')
endalign$$
The above would be uncontroversial if $delta$ were just a smooth, odd function with compact support. It's not in fact, but you can approximate it with such functions, so at least for continuous $f$ it's unproblematic.
answered Aug 9 at 12:19
leftaroundabout
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The Dirac delta is a distribution, which means it acts on smooth functions. The Dirac delta itself is not a smooth function, which means that it cannot act on itself. The expression
$$
int_mathbb Rdelta(x-x_1)delta(x-x_2)mathrm dx=delta(x_1-x_2)tag1
$$
is meaningless from a rigorous point of view. While it's true that it makes intuitive sense, making it rigorous requires a bit of care.
To this end, we shall regard the Dirac delta as the limit (in the sense of measures) of a mollifier:
$$
delta_epsilon(x):=epsilon^-1eta(x/epsilon)tag2
$$
where $eta$ is an absolutely integrable function with unit integral.
With this,
$$
lim_epsilonto0^+int_mathbb Rf(x-y)delta_epsilon(x)mathrm dxequiv f(y)tag3
$$
for good enough $f$. Note that this is just a convolution: $fstardelta_epsilonto f$ as $epsilonto0^+$. This is a perfectly rigorous statement.
But note that $delta_epsilon$ is itself a smooth function, which means that we can apply this identity to $f=delta_epsilon'$:
$$
lim_epsilonto0^+int_mathbb Rdelta_epsilon'(x-y)delta_epsilon(x)mathrm dxequiv delta_epsilon'(y)tag4
$$
Finally, the formal limit $epsilon'to0^+$ yileds the identity in the OP (after the redefinition $y=x_2-x_1$ and the change of variables $xto x-x_1$). Needless to say, this latter limit only makes sense in the sense of measures, which means you should regard it as a statement that both sides are equal when integrated over good-enough functions.
answered Aug 9 at 14:37
AccidentalFourierTransform
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Distinguish two cases:
if $x_mne x'_m$, the product of the deltas is always zero and so is the integral;
if $x_m=x'_m$, you have a squared delta, which gives a divergent integral.
Hence, the value of the integral can be expressed as $delta(x_m-x'_m)$.
For rigor, one should show that the integral of a squared delta is the "same infinity" as a single delta. In other terms
$$int_yinmathbb Rint_xinmathbb Rdelta^2(x),dx,dy=1.$$
answered Aug 9 at 7:35
Yves Daoust
113k665207
4
The problem with this heuristic is that it doesn't explain why the integral isn't $2 delta(x_m - x'_m)$ or $frac12 delta(x_m - x'_m)$ or $- delta(x_m - x'_m)$ or $delta'(x_m - x'_m)$ or something else weird.
– Hurkyl
Aug 9 at 7:36
@Hurkyl: I was busy adding a comment on this.
– Yves Daoust
Aug 9 at 7:38
@Yves Please read the comment below md2perpe's answer, since that is the real question I have.
– BLAZE
Aug 9 at 7:38
@BLAZE: though this is not what you asked. I am answering why the form is $delta(x_m-x'_m)$.
– Yves Daoust
Aug 9 at 7:40
@Yves It wasn't my original question no. But it is the question that I have now in response to his answer. All I would like to know right now is; since $x_m-x_m'=0$ then is $delta(0)$ the result of the integration? Could you please answer it? A simple yes or no will suffice, thanks.
– BLAZE
Aug 9 at 7:45
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OP is asking:
Why is $ int_mathbbR! mathrmdy ~delta(x-y) delta(y-z)~=~delta(x-z)~?$
Answer: Both the lhs. and the rhs. are informal notations
$$ iiint_mathbbR^3! mathrmdx~mathrmdy~mathrmdz~ delta(x-y) delta(y-z) f(x,z)~=~u[f]~=~iint_mathbbR^2! mathrmdx~mathrmdz~ delta(x-z) f(x,z) $$
for the same distribution $uin D^prime(mathbbR^2)$ defined as
$$u[f]~:=~ int_mathbbR!mathrmdy~f(y,y), qquad f~in~D(mathbbR^2).$$
answered Aug 9 at 20:21
Qmechanic
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This might be the simplest way to think about it.
If we were working with discrete indices, $sum_xphi_x_m,,xdelta_x,,x_m=phi_x_m,,x_m'$ would be an equation in matrices, viz. $phimathbbI=phi$. The Dirac delta is an "identity matrix" on a vector space of uncountable dimension, where we integrate over $x$ instead of summing over it. The result you asked about is just $mathbbI^2=mathbbI$. In other words, the special case $phi=delta$ is legitimate; whether $phi$ is a true function or a measure doesn't really matter.
answered Aug 9 at 20:38
J.G.
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Here's a derivation using calculus
$$
int_-infty^infty delta(x_m - x)delta(x_m' - x)
$$
Integrating by parts
$$
= (delta(x_m' - x) int delta(x_m - x')dx') |^infty_-infty - int^infty_-infty fracddelta(x_m' - x)dx intdelta(x_m-x')dx' dx
$$
Note that $int delta(x_m -x')dx' (x)$ is the heaviside step function, $H(x_m - x)$, which is 0 below $x_m$, $1/2$ at $x_m$, and $1$ above $x_m$. Thus we have
$$
= lim_xrightarrow infty delta(x_m' - x)H(x_m - x) - lim_xrightarrow -infty delta(x_m' - x)H(x_m - x) -
int^infty_-infty fracddelta(x_m' - x)dx H(x_m - x) dx
$$
The limits go to $0$, because the heaviside function takes on a finite value and the delta functions are zero as $x rightarrow infty$ and as $x rightarrow -infty$. Thus
$$
= - int^infty_-infty fracddelta(x_m' - x)dx H(x_m - x) dx
$$
WLOG assume that $x_m leq x_m'$. We note that the heaviside function is $0$ below $x_m$ and $1$ above $x_m$ (we can ignore the value $frac12$ at $x_m$ in integrating because that is a content zero set of values). Thus we have that
$$
= - int^infty_x_m fracddelta(x_m' - x)dx dx
$$
$$
= - (delta(x_m' - x)|^infty_x_m) = delta(x_m' - x_m) = delta(x_m - x_m')
$$
To address the assumption $x_m leq x_m'$: we could have done integration by parts on the other delta function and gotten the same result with the assumption $x_m' leq x_m$, so we cover all cases.
answered Aug 9 at 15:10
Striker
631314
2
How is this more rigorous than a simple swapping of integration domains, as given in my answer? Strictly speaking, neither that nor integration of parts is “allowed†for something like $delta$, which is not a continuous (let alone differentiable) function.
– leftaroundabout
Aug 9 at 15:30
@leftaroundabout I think both are fine, considering that the only thing you can rigorously do with a delta function is 'sift' out a value on a function it's integrated with... but another delta function isn't a function anyway, so the above integral isn't well defined.
– Striker
Aug 9 at 15:46
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If you take $u=x_m'-x$, then you have $int_-infty^inftydelta(u+(x_m-x'_m))delta(u)$. Actually finding that rigorously is difficult, but if you use the heuristic that $int_-infty^inftyf(u)delta(u) = f(0)$, then taking $f(u)= delta(u+(x_m-x'_m))$ and setting $u=0$ gets you $delta(x_m-x'_m)$.
answered Aug 13 at 21:14
Acccumulation
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8 Answers
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Intuitively, $delta(x_m-x)$ is non-zero only when $x_m=x$. Likewise $delta(x_m'-x)$ is non-zero only when $x_m'=x$. So the factor $delta(x_m-x) , delta(x_m'-x)$ is non-zero only when $x_m=x=x_m'.$ After integration w.r.t. $x$ we still must have $x_m=x_m'$ which makes the result $delta(x_m-x_m')$ quite natural.
More formally, recall the formula $int_-infty^infty f(x) , delta(x_0-x) , dx = f(x_0)$. Apply this to the given integral with $x_0=x_m'$ and $f(x) = delta(x_m-x).$ Then the result $delta(x_m-x_m')$ falls out.
answered Aug 9 at 6:11
md2perpe
6,34811022
Thank you for a brilliant answer. I'm not sure why the argument of the Dirac delta function is $x_m-x_m'$, why not $delta(x_m+x_m')$ for instance? Or is it because from $x_m=x_m'$ such that $x_m-x_m'=0$? Hence, $delta(0)$?
– BLAZE
Aug 9 at 7:10
7
The second paragraph needs a more formal proof. Because the integral with $f$ holds for $f$ being a function, not a distribution.
– Yves Daoust
Aug 9 at 7:45
@md2perpe You write in your answer that "$x_m=x_m'$ makes the result $delta(x_m-x_m')$ quite natural". But I need to know why the 'result' takes that form.
– BLAZE
Aug 9 at 7:49
@BLAZE. The result should be $0$ when $x_m neq x_m',$ i.e. when $x_m - x_m' neq 0.$ That's why we don't have $delta(x_m+x_m').$ We then don't have many alternatives, according to a theorem actually only some linear combination of $delta^(k)(x_m-x_m')$ for $k=0,1,2,ldots$
– md2perpe
Aug 9 at 11:46
3
@YvesDaoust. I know. I didn't want to make it rigorous, just try to give some intuition why the result is as it is. The integral $int_-infty^inftydelta(x_m-x)delta(x_m'-x),dx$ itself isn't welldefined as a product of distributions isn't allowed, even less so the integral of them. For a rigorous treatment we need to give meaning to the integral which is not that difficult; it's basically a convolution. But I think that a rigorous treatment is not what BLAZE needs.
– md2perpe
Aug 9 at 12:24
add a comment |Â
up vote
14
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accepted
Intuitively, $delta(x_m-x)$ is non-zero only when $x_m=x$. Likewise $delta(x_m'-x)$ is non-zero only when $x_m'=x$. So the factor $delta(x_m-x) , delta(x_m'-x)$ is non-zero only when $x_m=x=x_m'.$ After integration w.r.t. $x$ we still must have $x_m=x_m'$ which makes the result $delta(x_m-x_m')$ quite natural.
More formally, recall the formula $int_-infty^infty f(x) , delta(x_0-x) , dx = f(x_0)$. Apply this to the given integral with $x_0=x_m'$ and $f(x) = delta(x_m-x).$ Then the result $delta(x_m-x_m')$ falls out.
answered Aug 9 at 6:11
md2perpe
6,34811022
Thank you for a brilliant answer. I'm not sure why the argument of the Dirac delta function is $x_m-x_m'$, why not $delta(x_m+x_m')$ for instance? Or is it because from $x_m=x_m'$ such that $x_m-x_m'=0$? Hence, $delta(0)$?
– BLAZE
Aug 9 at 7:10
7
The second paragraph needs a more formal proof. Because the integral with $f$ holds for $f$ being a function, not a distribution.
– Yves Daoust
Aug 9 at 7:45
@md2perpe You write in your answer that "$x_m=x_m'$ makes the result $delta(x_m-x_m')$ quite natural". But I need to know why the 'result' takes that form.
– BLAZE
Aug 9 at 7:49
@BLAZE. The result should be $0$ when $x_m neq x_m',$ i.e. when $x_m - x_m' neq 0.$ That's why we don't have $delta(x_m+x_m').$ We then don't have many alternatives, according to a theorem actually only some linear combination of $delta^(k)(x_m-x_m')$ for $k=0,1,2,ldots$
– md2perpe
Aug 9 at 11:46
3
@YvesDaoust. I know. I didn't want to make it rigorous, just try to give some intuition why the result is as it is. The integral $int_-infty^inftydelta(x_m-x)delta(x_m'-x),dx$ itself isn't welldefined as a product of distributions isn't allowed, even less so the integral of them. For a rigorous treatment we need to give meaning to the integral which is not that difficult; it's basically a convolution. But I think that a rigorous treatment is not what BLAZE needs.
– md2perpe
Aug 9 at 12:24
add a comment |Â
up vote
14
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up vote
14
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accepted
Intuitively, $delta(x_m-x)$ is non-zero only when $x_m=x$. Likewise $delta(x_m'-x)$ is non-zero only when $x_m'=x$. So the factor $delta(x_m-x) , delta(x_m'-x)$ is non-zero only when $x_m=x=x_m'.$ After integration w.r.t. $x$ we still must have $x_m=x_m'$ which makes the result $delta(x_m-x_m')$ quite natural.
More formally, recall the formula $int_-infty^infty f(x) , delta(x_0-x) , dx = f(x_0)$. Apply this to the given integral with $x_0=x_m'$ and $f(x) = delta(x_m-x).$ Then the result $delta(x_m-x_m')$ falls out.
answered Aug 9 at 6:11
md2perpe
6,34811022
Intuitively, $delta(x_m-x)$ is non-zero only when $x_m=x$. Likewise $delta(x_m'-x)$ is non-zero only when $x_m'=x$. So the factor $delta(x_m-x) , delta(x_m'-x)$ is non-zero only when $x_m=x=x_m'.$ After integration w.r.t. $x$ we still must have $x_m=x_m'$ which makes the result $delta(x_m-x_m')$ quite natural.
More formally, recall the formula $int_-infty^infty f(x) , delta(x_0-x) , dx = f(x_0)$. Apply this to the given integral with $x_0=x_m'$ and $f(x) = delta(x_m-x).$ Then the result $delta(x_m-x_m')$ falls out.
answered Aug 9 at 6:11
md2perpe
6,34811022
answered Aug 9 at 6:11
md2perpe
6,34811022
answered Aug 9 at 6:11
md2perpe
6,34811022
answered Aug 9 at 6:11
md2perpe
6,34811022
6,34811022
Thank you for a brilliant answer. I'm not sure why the argument of the Dirac delta function is $x_m-x_m'$, why not $delta(x_m+x_m')$ for instance? Or is it because from $x_m=x_m'$ such that $x_m-x_m'=0$? Hence, $delta(0)$?
– BLAZE
Aug 9 at 7:10
7
The second paragraph needs a more formal proof. Because the integral with $f$ holds for $f$ being a function, not a distribution.
– Yves Daoust
Aug 9 at 7:45
@md2perpe You write in your answer that "$x_m=x_m'$ makes the result $delta(x_m-x_m')$ quite natural". But I need to know why the 'result' takes that form.
– BLAZE
Aug 9 at 7:49
@BLAZE. The result should be $0$ when $x_m neq x_m',$ i.e. when $x_m - x_m' neq 0.$ That's why we don't have $delta(x_m+x_m').$ We then don't have many alternatives, according to a theorem actually only some linear combination of $delta^(k)(x_m-x_m')$ for $k=0,1,2,ldots$
– md2perpe
Aug 9 at 11:46
3
@YvesDaoust. I know. I didn't want to make it rigorous, just try to give some intuition why the result is as it is. The integral $int_-infty^inftydelta(x_m-x)delta(x_m'-x),dx$ itself isn't welldefined as a product of distributions isn't allowed, even less so the integral of them. For a rigorous treatment we need to give meaning to the integral which is not that difficult; it's basically a convolution. But I think that a rigorous treatment is not what BLAZE needs.
– md2perpe
Aug 9 at 12:24
add a comment |Â
Thank you for a brilliant answer. I'm not sure why the argument of the Dirac delta function is $x_m-x_m'$, why not $delta(x_m+x_m')$ for instance? Or is it because from $x_m=x_m'$ such that $x_m-x_m'=0$? Hence, $delta(0)$?
– BLAZE
Aug 9 at 7:10
7
The second paragraph needs a more formal proof. Because the integral with $f$ holds for $f$ being a function, not a distribution.
– Yves Daoust
Aug 9 at 7:45
@md2perpe You write in your answer that "$x_m=x_m'$ makes the result $delta(x_m-x_m')$ quite natural". But I need to know why the 'result' takes that form.
– BLAZE
Aug 9 at 7:49
@BLAZE. The result should be $0$ when $x_m neq x_m',$ i.e. when $x_m - x_m' neq 0.$ That's why we don't have $delta(x_m+x_m').$ We then don't have many alternatives, according to a theorem actually only some linear combination of $delta^(k)(x_m-x_m')$ for $k=0,1,2,ldots$
– md2perpe
Aug 9 at 11:46
3
@YvesDaoust. I know. I didn't want to make it rigorous, just try to give some intuition why the result is as it is. The integral $int_-infty^inftydelta(x_m-x)delta(x_m'-x),dx$ itself isn't welldefined as a product of distributions isn't allowed, even less so the integral of them. For a rigorous treatment we need to give meaning to the integral which is not that difficult; it's basically a convolution. But I think that a rigorous treatment is not what BLAZE needs.
– md2perpe
Aug 9 at 12:24
Thank you for a brilliant answer. I'm not sure why the argument of the Dirac delta function is $x_m-x_m'$, why not $delta(x_m+x_m')$ for instance? Or is it because from $x_m=x_m'$ such that $x_m-x_m'=0$? Hence, $delta(0)$?
– BLAZE
Aug 9 at 7:10
Thank you for a brilliant answer. I'm not sure why the argument of the Dirac delta function is $x_m-x_m'$, why not $delta(x_m+x_m')$ for instance? Or is it because from $x_m=x_m'$ such that $x_m-x_m'=0$? Hence, $delta(0)$?
– BLAZE
Aug 9 at 7:10
7
7
The second paragraph needs a more formal proof. Because the integral with $f$ holds for $f$ being a function, not a distribution.
– Yves Daoust
Aug 9 at 7:45
The second paragraph needs a more formal proof. Because the integral with $f$ holds for $f$ being a function, not a distribution.
– Yves Daoust
Aug 9 at 7:45
@md2perpe You write in your answer that "$x_m=x_m'$ makes the result $delta(x_m-x_m')$ quite natural". But I need to know why the 'result' takes that form.
– BLAZE
Aug 9 at 7:49
@md2perpe You write in your answer that "$x_m=x_m'$ makes the result $delta(x_m-x_m')$ quite natural". But I need to know why the 'result' takes that form.
– BLAZE
Aug 9 at 7:49
@BLAZE. The result should be $0$ when $x_m neq x_m',$ i.e. when $x_m - x_m' neq 0.$ That's why we don't have $delta(x_m+x_m').$ We then don't have many alternatives, according to a theorem actually only some linear combination of $delta^(k)(x_m-x_m')$ for $k=0,1,2,ldots$
– md2perpe
Aug 9 at 11:46
@BLAZE. The result should be $0$ when $x_m neq x_m',$ i.e. when $x_m - x_m' neq 0.$ That's why we don't have $delta(x_m+x_m').$ We then don't have many alternatives, according to a theorem actually only some linear combination of $delta^(k)(x_m-x_m')$ for $k=0,1,2,ldots$
– md2perpe
Aug 9 at 11:46
3
3
@YvesDaoust. I know. I didn't want to make it rigorous, just try to give some intuition why the result is as it is. The integral $int_-infty^inftydelta(x_m-x)delta(x_m'-x),dx$ itself isn't welldefined as a product of distributions isn't allowed, even less so the integral of them. For a rigorous treatment we need to give meaning to the integral which is not that difficult; it's basically a convolution. But I think that a rigorous treatment is not what BLAZE needs.
– md2perpe
Aug 9 at 12:24
@YvesDaoust. I know. I didn't want to make it rigorous, just try to give some intuition why the result is as it is. The integral $int_-infty^inftydelta(x_m-x)delta(x_m'-x),dx$ itself isn't welldefined as a product of distributions isn't allowed, even less so the integral of them. For a rigorous treatment we need to give meaning to the integral which is not that difficult; it's basically a convolution. But I think that a rigorous treatment is not what BLAZE needs.
– md2perpe
Aug 9 at 12:24
add a comment |Â
up vote
7
down vote
$delta(x)$ is never really well-defined by itself (at least not as a function). It's only defined when appearing in an integral (possibly, multiplied with another function). It is essentially syntactic sugar, and the “definition†is
$$ int_mathbbRmathrmdx f(x)cdot delta(x-x_0) := f(x_0) $$
So, what the identity in question actually means is
$$
int_mathbbRmathrmdy: f(y)cdot int_mathbbRmathrmdx: delta(y-x) cdot delta(x_m'-x)
= f(x_m')
$$
for any function $f$.
Why would that be? Well, it's basically just a matter of changing the order of integration:
$$beginalign
int_mathbbRmathrmdy: f(y)cdot int_mathbbRmathrmdx: delta(y-x) cdot delta(x_m'-x)
=& int_mathbbRmathrmdy int_mathbbRmathrmdx: f(y)cdot delta(y-x) cdot delta(x_m'-x)
\ =& int_mathbbRmathrmdx int_mathbbRmathrmdy: f(y)cdot delta(y-x) cdot delta(x_m'-x)
\ =& int_mathbbRmathrmdx: delta(x_m'-x) int_mathbbRmathrmdy: f(y)cdot delta(y-x)
\ =& int_mathbbRmathrmdx: delta(x_m'-x) int_mathbbRmathrmdy: f(y)cdot delta(y-x)
\ =& int_mathbbRmathrmdx: delta(x_m'-x) f(x)
\ =& int_mathbbRmathrmdx: delta(x-x_m') f(x)
\ =& f(x_m')
endalign$$
The above would be uncontroversial if $delta$ were just a smooth, odd function with compact support. It's not in fact, but you can approximate it with such functions, so at least for continuous $f$ it's unproblematic.
answered Aug 9 at 12:19
leftaroundabout
3,3411527
add a comment |Â
up vote
7
down vote
$delta(x)$ is never really well-defined by itself (at least not as a function). It's only defined when appearing in an integral (possibly, multiplied with another function). It is essentially syntactic sugar, and the “definition†is
$$ int_mathbbRmathrmdx f(x)cdot delta(x-x_0) := f(x_0) $$
So, what the identity in question actually means is
$$
int_mathbbRmathrmdy: f(y)cdot int_mathbbRmathrmdx: delta(y-x) cdot delta(x_m'-x)
= f(x_m')
$$
for any function $f$.
Why would that be? Well, it's basically just a matter of changing the order of integration:
$$beginalign
int_mathbbRmathrmdy: f(y)cdot int_mathbbRmathrmdx: delta(y-x) cdot delta(x_m'-x)
=& int_mathbbRmathrmdy int_mathbbRmathrmdx: f(y)cdot delta(y-x) cdot delta(x_m'-x)
\ =& int_mathbbRmathrmdx int_mathbbRmathrmdy: f(y)cdot delta(y-x) cdot delta(x_m'-x)
\ =& int_mathbbRmathrmdx: delta(x_m'-x) int_mathbbRmathrmdy: f(y)cdot delta(y-x)
\ =& int_mathbbRmathrmdx: delta(x_m'-x) int_mathbbRmathrmdy: f(y)cdot delta(y-x)
\ =& int_mathbbRmathrmdx: delta(x_m'-x) f(x)
\ =& int_mathbbRmathrmdx: delta(x-x_m') f(x)
\ =& f(x_m')
endalign$$
The above would be uncontroversial if $delta$ were just a smooth, odd function with compact support. It's not in fact, but you can approximate it with such functions, so at least for continuous $f$ it's unproblematic.
answered Aug 9 at 12:19
leftaroundabout
3,3411527
add a comment |Â
up vote
7
down vote
up vote
7
down vote
$delta(x)$ is never really well-defined by itself (at least not as a function). It's only defined when appearing in an integral (possibly, multiplied with another function). It is essentially syntactic sugar, and the “definition†is
$$ int_mathbbRmathrmdx f(x)cdot delta(x-x_0) := f(x_0) $$
So, what the identity in question actually means is
$$
int_mathbbRmathrmdy: f(y)cdot int_mathbbRmathrmdx: delta(y-x) cdot delta(x_m'-x)
= f(x_m')
$$
for any function $f$.
Why would that be? Well, it's basically just a matter of changing the order of integration:
$$beginalign
int_mathbbRmathrmdy: f(y)cdot int_mathbbRmathrmdx: delta(y-x) cdot delta(x_m'-x)
=& int_mathbbRmathrmdy int_mathbbRmathrmdx: f(y)cdot delta(y-x) cdot delta(x_m'-x)
\ =& int_mathbbRmathrmdx int_mathbbRmathrmdy: f(y)cdot delta(y-x) cdot delta(x_m'-x)
\ =& int_mathbbRmathrmdx: delta(x_m'-x) int_mathbbRmathrmdy: f(y)cdot delta(y-x)
\ =& int_mathbbRmathrmdx: delta(x_m'-x) int_mathbbRmathrmdy: f(y)cdot delta(y-x)
\ =& int_mathbbRmathrmdx: delta(x_m'-x) f(x)
\ =& int_mathbbRmathrmdx: delta(x-x_m') f(x)
\ =& f(x_m')
endalign$$
The above would be uncontroversial if $delta$ were just a smooth, odd function with compact support. It's not in fact, but you can approximate it with such functions, so at least for continuous $f$ it's unproblematic.
answered Aug 9 at 12:19
leftaroundabout
3,3411527
$delta(x)$ is never really well-defined by itself (at least not as a function). It's only defined when appearing in an integral (possibly, multiplied with another function). It is essentially syntactic sugar, and the “definition†is
$$ int_mathbbRmathrmdx f(x)cdot delta(x-x_0) := f(x_0) $$
So, what the identity in question actually means is
$$
int_mathbbRmathrmdy: f(y)cdot int_mathbbRmathrmdx: delta(y-x) cdot delta(x_m'-x)
= f(x_m')
$$
for any function $f$.
Why would that be? Well, it's basically just a matter of changing the order of integration:
$$beginalign
int_mathbbRmathrmdy: f(y)cdot int_mathbbRmathrmdx: delta(y-x) cdot delta(x_m'-x)
=& int_mathbbRmathrmdy int_mathbbRmathrmdx: f(y)cdot delta(y-x) cdot delta(x_m'-x)
\ =& int_mathbbRmathrmdx int_mathbbRmathrmdy: f(y)cdot delta(y-x) cdot delta(x_m'-x)
\ =& int_mathbbRmathrmdx: delta(x_m'-x) int_mathbbRmathrmdy: f(y)cdot delta(y-x)
\ =& int_mathbbRmathrmdx: delta(x_m'-x) int_mathbbRmathrmdy: f(y)cdot delta(y-x)
\ =& int_mathbbRmathrmdx: delta(x_m'-x) f(x)
\ =& int_mathbbRmathrmdx: delta(x-x_m') f(x)
\ =& f(x_m')
endalign$$
The above would be uncontroversial if $delta$ were just a smooth, odd function with compact support. It's not in fact, but you can approximate it with such functions, so at least for continuous $f$ it's unproblematic.
answered Aug 9 at 12:19
leftaroundabout
3,3411527
answered Aug 9 at 12:19
leftaroundabout
3,3411527
answered Aug 9 at 12:19
leftaroundabout
3,3411527
answered Aug 9 at 12:19
leftaroundabout
3,3411527
3,3411527
add a comment |Â
add a comment |Â
up vote
4
down vote
The Dirac delta is a distribution, which means it acts on smooth functions. The Dirac delta itself is not a smooth function, which means that it cannot act on itself. The expression
$$
int_mathbb Rdelta(x-x_1)delta(x-x_2)mathrm dx=delta(x_1-x_2)tag1
$$
is meaningless from a rigorous point of view. While it's true that it makes intuitive sense, making it rigorous requires a bit of care.
To this end, we shall regard the Dirac delta as the limit (in the sense of measures) of a mollifier:
$$
delta_epsilon(x):=epsilon^-1eta(x/epsilon)tag2
$$
where $eta$ is an absolutely integrable function with unit integral.
With this,
$$
lim_epsilonto0^+int_mathbb Rf(x-y)delta_epsilon(x)mathrm dxequiv f(y)tag3
$$
for good enough $f$. Note that this is just a convolution: $fstardelta_epsilonto f$ as $epsilonto0^+$. This is a perfectly rigorous statement.
But note that $delta_epsilon$ is itself a smooth function, which means that we can apply this identity to $f=delta_epsilon'$:
$$
lim_epsilonto0^+int_mathbb Rdelta_epsilon'(x-y)delta_epsilon(x)mathrm dxequiv delta_epsilon'(y)tag4
$$
Finally, the formal limit $epsilon'to0^+$ yileds the identity in the OP (after the redefinition $y=x_2-x_1$ and the change of variables $xto x-x_1$). Needless to say, this latter limit only makes sense in the sense of measures, which means you should regard it as a statement that both sides are equal when integrated over good-enough functions.
answered Aug 9 at 14:37
AccidentalFourierTransform
1,301627
add a comment |Â
up vote
4
down vote
The Dirac delta is a distribution, which means it acts on smooth functions. The Dirac delta itself is not a smooth function, which means that it cannot act on itself. The expression
$$
int_mathbb Rdelta(x-x_1)delta(x-x_2)mathrm dx=delta(x_1-x_2)tag1
$$
is meaningless from a rigorous point of view. While it's true that it makes intuitive sense, making it rigorous requires a bit of care.
To this end, we shall regard the Dirac delta as the limit (in the sense of measures) of a mollifier:
$$
delta_epsilon(x):=epsilon^-1eta(x/epsilon)tag2
$$
where $eta$ is an absolutely integrable function with unit integral.
With this,
$$
lim_epsilonto0^+int_mathbb Rf(x-y)delta_epsilon(x)mathrm dxequiv f(y)tag3
$$
for good enough $f$. Note that this is just a convolution: $fstardelta_epsilonto f$ as $epsilonto0^+$. This is a perfectly rigorous statement.
But note that $delta_epsilon$ is itself a smooth function, which means that we can apply this identity to $f=delta_epsilon'$:
$$
lim_epsilonto0^+int_mathbb Rdelta_epsilon'(x-y)delta_epsilon(x)mathrm dxequiv delta_epsilon'(y)tag4
$$
Finally, the formal limit $epsilon'to0^+$ yileds the identity in the OP (after the redefinition $y=x_2-x_1$ and the change of variables $xto x-x_1$). Needless to say, this latter limit only makes sense in the sense of measures, which means you should regard it as a statement that both sides are equal when integrated over good-enough functions.
answered Aug 9 at 14:37
AccidentalFourierTransform
1,301627
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The Dirac delta is a distribution, which means it acts on smooth functions. The Dirac delta itself is not a smooth function, which means that it cannot act on itself. The expression
$$
int_mathbb Rdelta(x-x_1)delta(x-x_2)mathrm dx=delta(x_1-x_2)tag1
$$
is meaningless from a rigorous point of view. While it's true that it makes intuitive sense, making it rigorous requires a bit of care.
To this end, we shall regard the Dirac delta as the limit (in the sense of measures) of a mollifier:
$$
delta_epsilon(x):=epsilon^-1eta(x/epsilon)tag2
$$
where $eta$ is an absolutely integrable function with unit integral.
With this,
$$
lim_epsilonto0^+int_mathbb Rf(x-y)delta_epsilon(x)mathrm dxequiv f(y)tag3
$$
for good enough $f$. Note that this is just a convolution: $fstardelta_epsilonto f$ as $epsilonto0^+$. This is a perfectly rigorous statement.
But note that $delta_epsilon$ is itself a smooth function, which means that we can apply this identity to $f=delta_epsilon'$:
$$
lim_epsilonto0^+int_mathbb Rdelta_epsilon'(x-y)delta_epsilon(x)mathrm dxequiv delta_epsilon'(y)tag4
$$
Finally, the formal limit $epsilon'to0^+$ yileds the identity in the OP (after the redefinition $y=x_2-x_1$ and the change of variables $xto x-x_1$). Needless to say, this latter limit only makes sense in the sense of measures, which means you should regard it as a statement that both sides are equal when integrated over good-enough functions.
answered Aug 9 at 14:37
AccidentalFourierTransform
1,301627
The Dirac delta is a distribution, which means it acts on smooth functions. The Dirac delta itself is not a smooth function, which means that it cannot act on itself. The expression
$$
int_mathbb Rdelta(x-x_1)delta(x-x_2)mathrm dx=delta(x_1-x_2)tag1
$$
is meaningless from a rigorous point of view. While it's true that it makes intuitive sense, making it rigorous requires a bit of care.
To this end, we shall regard the Dirac delta as the limit (in the sense of measures) of a mollifier:
$$
delta_epsilon(x):=epsilon^-1eta(x/epsilon)tag2
$$
where $eta$ is an absolutely integrable function with unit integral.
With this,
$$
lim_epsilonto0^+int_mathbb Rf(x-y)delta_epsilon(x)mathrm dxequiv f(y)tag3
$$
for good enough $f$. Note that this is just a convolution: $fstardelta_epsilonto f$ as $epsilonto0^+$. This is a perfectly rigorous statement.
But note that $delta_epsilon$ is itself a smooth function, which means that we can apply this identity to $f=delta_epsilon'$:
$$
lim_epsilonto0^+int_mathbb Rdelta_epsilon'(x-y)delta_epsilon(x)mathrm dxequiv delta_epsilon'(y)tag4
$$
Finally, the formal limit $epsilon'to0^+$ yileds the identity in the OP (after the redefinition $y=x_2-x_1$ and the change of variables $xto x-x_1$). Needless to say, this latter limit only makes sense in the sense of measures, which means you should regard it as a statement that both sides are equal when integrated over good-enough functions.
answered Aug 9 at 14:37
AccidentalFourierTransform
1,301627
answered Aug 9 at 14:37
AccidentalFourierTransform
1,301627
answered Aug 9 at 14:37
AccidentalFourierTransform
1,301627
answered Aug 9 at 14:37
AccidentalFourierTransform
1,301627
1,301627
add a comment |Â
add a comment |Â
up vote
2
down vote
Distinguish two cases:
if $x_mne x'_m$, the product of the deltas is always zero and so is the integral;
if $x_m=x'_m$, you have a squared delta, which gives a divergent integral.
Hence, the value of the integral can be expressed as $delta(x_m-x'_m)$.
For rigor, one should show that the integral of a squared delta is the "same infinity" as a single delta. In other terms
$$int_yinmathbb Rint_xinmathbb Rdelta^2(x),dx,dy=1.$$
answered Aug 9 at 7:35
Yves Daoust
113k665207
4
The problem with this heuristic is that it doesn't explain why the integral isn't $2 delta(x_m - x'_m)$ or $frac12 delta(x_m - x'_m)$ or $- delta(x_m - x'_m)$ or $delta'(x_m - x'_m)$ or something else weird.
– Hurkyl
Aug 9 at 7:36
@Hurkyl: I was busy adding a comment on this.
– Yves Daoust
Aug 9 at 7:38
@Yves Please read the comment below md2perpe's answer, since that is the real question I have.
– BLAZE
Aug 9 at 7:38
@BLAZE: though this is not what you asked. I am answering why the form is $delta(x_m-x'_m)$.
– Yves Daoust
Aug 9 at 7:40
@Yves It wasn't my original question no. But it is the question that I have now in response to his answer. All I would like to know right now is; since $x_m-x_m'=0$ then is $delta(0)$ the result of the integration? Could you please answer it? A simple yes or no will suffice, thanks.
– BLAZE
Aug 9 at 7:45
 |Â
show 4 more comments
up vote
2
down vote
Distinguish two cases:
if $x_mne x'_m$, the product of the deltas is always zero and so is the integral;
if $x_m=x'_m$, you have a squared delta, which gives a divergent integral.
Hence, the value of the integral can be expressed as $delta(x_m-x'_m)$.
For rigor, one should show that the integral of a squared delta is the "same infinity" as a single delta. In other terms
$$int_yinmathbb Rint_xinmathbb Rdelta^2(x),dx,dy=1.$$
answered Aug 9 at 7:35
Yves Daoust
113k665207
4
The problem with this heuristic is that it doesn't explain why the integral isn't $2 delta(x_m - x'_m)$ or $frac12 delta(x_m - x'_m)$ or $- delta(x_m - x'_m)$ or $delta'(x_m - x'_m)$ or something else weird.
– Hurkyl
Aug 9 at 7:36
@Hurkyl: I was busy adding a comment on this.
– Yves Daoust
Aug 9 at 7:38
@Yves Please read the comment below md2perpe's answer, since that is the real question I have.
– BLAZE
Aug 9 at 7:38
@BLAZE: though this is not what you asked. I am answering why the form is $delta(x_m-x'_m)$.
– Yves Daoust
Aug 9 at 7:40
@Yves It wasn't my original question no. But it is the question that I have now in response to his answer. All I would like to know right now is; since $x_m-x_m'=0$ then is $delta(0)$ the result of the integration? Could you please answer it? A simple yes or no will suffice, thanks.
– BLAZE
Aug 9 at 7:45
 |Â
show 4 more comments
up vote
2
down vote
up vote
2
down vote
Distinguish two cases:
if $x_mne x'_m$, the product of the deltas is always zero and so is the integral;
if $x_m=x'_m$, you have a squared delta, which gives a divergent integral.
Hence, the value of the integral can be expressed as $delta(x_m-x'_m)$.
For rigor, one should show that the integral of a squared delta is the "same infinity" as a single delta. In other terms
$$int_yinmathbb Rint_xinmathbb Rdelta^2(x),dx,dy=1.$$
answered Aug 9 at 7:35
Yves Daoust
113k665207
Distinguish two cases:
if $x_mne x'_m$, the product of the deltas is always zero and so is the integral;
if $x_m=x'_m$, you have a squared delta, which gives a divergent integral.
Hence, the value of the integral can be expressed as $delta(x_m-x'_m)$.
For rigor, one should show that the integral of a squared delta is the "same infinity" as a single delta. In other terms
$$int_yinmathbb Rint_xinmathbb Rdelta^2(x),dx,dy=1.$$
answered Aug 9 at 7:35
Yves Daoust
113k665207
edited Aug 9 at 7:38
answered Aug 9 at 7:35
Yves Daoust
113k665207
answered Aug 9 at 7:35
Yves Daoust
113k665207
answered Aug 9 at 7:35
Yves Daoust
113k665207
113k665207
4
The problem with this heuristic is that it doesn't explain why the integral isn't $2 delta(x_m - x'_m)$ or $frac12 delta(x_m - x'_m)$ or $- delta(x_m - x'_m)$ or $delta'(x_m - x'_m)$ or something else weird.
– Hurkyl
Aug 9 at 7:36
@Hurkyl: I was busy adding a comment on this.
– Yves Daoust
Aug 9 at 7:38
@Yves Please read the comment below md2perpe's answer, since that is the real question I have.
– BLAZE
Aug 9 at 7:38
@BLAZE: though this is not what you asked. I am answering why the form is $delta(x_m-x'_m)$.
– Yves Daoust
Aug 9 at 7:40
@Yves It wasn't my original question no. But it is the question that I have now in response to his answer. All I would like to know right now is; since $x_m-x_m'=0$ then is $delta(0)$ the result of the integration? Could you please answer it? A simple yes or no will suffice, thanks.
– BLAZE
Aug 9 at 7:45
 |Â
show 4 more comments
4
The problem with this heuristic is that it doesn't explain why the integral isn't $2 delta(x_m - x'_m)$ or $frac12 delta(x_m - x'_m)$ or $- delta(x_m - x'_m)$ or $delta'(x_m - x'_m)$ or something else weird.
– Hurkyl
Aug 9 at 7:36
@Hurkyl: I was busy adding a comment on this.
– Yves Daoust
Aug 9 at 7:38
@Yves Please read the comment below md2perpe's answer, since that is the real question I have.
– BLAZE
Aug 9 at 7:38
@BLAZE: though this is not what you asked. I am answering why the form is $delta(x_m-x'_m)$.
– Yves Daoust
Aug 9 at 7:40
@Yves It wasn't my original question no. But it is the question that I have now in response to his answer. All I would like to know right now is; since $x_m-x_m'=0$ then is $delta(0)$ the result of the integration? Could you please answer it? A simple yes or no will suffice, thanks.
– BLAZE
Aug 9 at 7:45
4
4
The problem with this heuristic is that it doesn't explain why the integral isn't $2 delta(x_m - x'_m)$ or $frac12 delta(x_m - x'_m)$ or $- delta(x_m - x'_m)$ or $delta'(x_m - x'_m)$ or something else weird.
– Hurkyl
Aug 9 at 7:36
The problem with this heuristic is that it doesn't explain why the integral isn't $2 delta(x_m - x'_m)$ or $frac12 delta(x_m - x'_m)$ or $- delta(x_m - x'_m)$ or $delta'(x_m - x'_m)$ or something else weird.
– Hurkyl
Aug 9 at 7:36
@Hurkyl: I was busy adding a comment on this.
– Yves Daoust
Aug 9 at 7:38
@Hurkyl: I was busy adding a comment on this.
– Yves Daoust
Aug 9 at 7:38
@Yves Please read the comment below md2perpe's answer, since that is the real question I have.
– BLAZE
Aug 9 at 7:38
@Yves Please read the comment below md2perpe's answer, since that is the real question I have.
– BLAZE
Aug 9 at 7:38
@BLAZE: though this is not what you asked. I am answering why the form is $delta(x_m-x'_m)$.
– Yves Daoust
Aug 9 at 7:40
@BLAZE: though this is not what you asked. I am answering why the form is $delta(x_m-x'_m)$.
– Yves Daoust
Aug 9 at 7:40
@Yves It wasn't my original question no. But it is the question that I have now in response to his answer. All I would like to know right now is; since $x_m-x_m'=0$ then is $delta(0)$ the result of the integration? Could you please answer it? A simple yes or no will suffice, thanks.
– BLAZE
Aug 9 at 7:45
@Yves It wasn't my original question no. But it is the question that I have now in response to his answer. All I would like to know right now is; since $x_m-x_m'=0$ then is $delta(0)$ the result of the integration? Could you please answer it? A simple yes or no will suffice, thanks.
– BLAZE
Aug 9 at 7:45
 |Â
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up vote
2
down vote
OP is asking:
Why is $ int_mathbbR! mathrmdy ~delta(x-y) delta(y-z)~=~delta(x-z)~?$
Answer: Both the lhs. and the rhs. are informal notations
$$ iiint_mathbbR^3! mathrmdx~mathrmdy~mathrmdz~ delta(x-y) delta(y-z) f(x,z)~=~u[f]~=~iint_mathbbR^2! mathrmdx~mathrmdz~ delta(x-z) f(x,z) $$
for the same distribution $uin D^prime(mathbbR^2)$ defined as
$$u[f]~:=~ int_mathbbR!mathrmdy~f(y,y), qquad f~in~D(mathbbR^2).$$
answered Aug 9 at 20:21
Qmechanic
4,47811746
add a comment |Â
up vote
2
down vote
OP is asking:
Why is $ int_mathbbR! mathrmdy ~delta(x-y) delta(y-z)~=~delta(x-z)~?$
Answer: Both the lhs. and the rhs. are informal notations
$$ iiint_mathbbR^3! mathrmdx~mathrmdy~mathrmdz~ delta(x-y) delta(y-z) f(x,z)~=~u[f]~=~iint_mathbbR^2! mathrmdx~mathrmdz~ delta(x-z) f(x,z) $$
for the same distribution $uin D^prime(mathbbR^2)$ defined as
$$u[f]~:=~ int_mathbbR!mathrmdy~f(y,y), qquad f~in~D(mathbbR^2).$$
answered Aug 9 at 20:21
Qmechanic
4,47811746
add a comment |Â
up vote
2
down vote
up vote
2
down vote
OP is asking:
Why is $ int_mathbbR! mathrmdy ~delta(x-y) delta(y-z)~=~delta(x-z)~?$
Answer: Both the lhs. and the rhs. are informal notations
$$ iiint_mathbbR^3! mathrmdx~mathrmdy~mathrmdz~ delta(x-y) delta(y-z) f(x,z)~=~u[f]~=~iint_mathbbR^2! mathrmdx~mathrmdz~ delta(x-z) f(x,z) $$
for the same distribution $uin D^prime(mathbbR^2)$ defined as
$$u[f]~:=~ int_mathbbR!mathrmdy~f(y,y), qquad f~in~D(mathbbR^2).$$
answered Aug 9 at 20:21
Qmechanic
4,47811746
OP is asking:
Why is $ int_mathbbR! mathrmdy ~delta(x-y) delta(y-z)~=~delta(x-z)~?$
Answer: Both the lhs. and the rhs. are informal notations
$$ iiint_mathbbR^3! mathrmdx~mathrmdy~mathrmdz~ delta(x-y) delta(y-z) f(x,z)~=~u[f]~=~iint_mathbbR^2! mathrmdx~mathrmdz~ delta(x-z) f(x,z) $$
for the same distribution $uin D^prime(mathbbR^2)$ defined as
$$u[f]~:=~ int_mathbbR!mathrmdy~f(y,y), qquad f~in~D(mathbbR^2).$$
answered Aug 9 at 20:21
Qmechanic
4,47811746
answered Aug 9 at 20:21
Qmechanic
4,47811746
answered Aug 9 at 20:21
Qmechanic
4,47811746
answered Aug 9 at 20:21
Qmechanic
4,47811746
4,47811746
add a comment |Â
add a comment |Â
up vote
2
down vote
This might be the simplest way to think about it.
If we were working with discrete indices, $sum_xphi_x_m,,xdelta_x,,x_m=phi_x_m,,x_m'$ would be an equation in matrices, viz. $phimathbbI=phi$. The Dirac delta is an "identity matrix" on a vector space of uncountable dimension, where we integrate over $x$ instead of summing over it. The result you asked about is just $mathbbI^2=mathbbI$. In other words, the special case $phi=delta$ is legitimate; whether $phi$ is a true function or a measure doesn't really matter.
answered Aug 9 at 20:38
J.G.
14k11425
add a comment |Â
up vote
2
down vote
This might be the simplest way to think about it.
If we were working with discrete indices, $sum_xphi_x_m,,xdelta_x,,x_m=phi_x_m,,x_m'$ would be an equation in matrices, viz. $phimathbbI=phi$. The Dirac delta is an "identity matrix" on a vector space of uncountable dimension, where we integrate over $x$ instead of summing over it. The result you asked about is just $mathbbI^2=mathbbI$. In other words, the special case $phi=delta$ is legitimate; whether $phi$ is a true function or a measure doesn't really matter.
answered Aug 9 at 20:38
J.G.
14k11425
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This might be the simplest way to think about it.
If we were working with discrete indices, $sum_xphi_x_m,,xdelta_x,,x_m=phi_x_m,,x_m'$ would be an equation in matrices, viz. $phimathbbI=phi$. The Dirac delta is an "identity matrix" on a vector space of uncountable dimension, where we integrate over $x$ instead of summing over it. The result you asked about is just $mathbbI^2=mathbbI$. In other words, the special case $phi=delta$ is legitimate; whether $phi$ is a true function or a measure doesn't really matter.
answered Aug 9 at 20:38
J.G.
14k11425
This might be the simplest way to think about it.
If we were working with discrete indices, $sum_xphi_x_m,,xdelta_x,,x_m=phi_x_m,,x_m'$ would be an equation in matrices, viz. $phimathbbI=phi$. The Dirac delta is an "identity matrix" on a vector space of uncountable dimension, where we integrate over $x$ instead of summing over it. The result you asked about is just $mathbbI^2=mathbbI$. In other words, the special case $phi=delta$ is legitimate; whether $phi$ is a true function or a measure doesn't really matter.
answered Aug 9 at 20:38
J.G.
14k11425
answered Aug 9 at 20:38
J.G.
14k11425
answered Aug 9 at 20:38
J.G.
14k11425
answered Aug 9 at 20:38
J.G.
14k11425
14k11425
add a comment |Â
add a comment |Â
up vote
1
down vote
Here's a derivation using calculus
$$
int_-infty^infty delta(x_m - x)delta(x_m' - x)
$$
Integrating by parts
$$
= (delta(x_m' - x) int delta(x_m - x')dx') |^infty_-infty - int^infty_-infty fracddelta(x_m' - x)dx intdelta(x_m-x')dx' dx
$$
Note that $int delta(x_m -x')dx' (x)$ is the heaviside step function, $H(x_m - x)$, which is 0 below $x_m$, $1/2$ at $x_m$, and $1$ above $x_m$. Thus we have
$$
= lim_xrightarrow infty delta(x_m' - x)H(x_m - x) - lim_xrightarrow -infty delta(x_m' - x)H(x_m - x) -
int^infty_-infty fracddelta(x_m' - x)dx H(x_m - x) dx
$$
The limits go to $0$, because the heaviside function takes on a finite value and the delta functions are zero as $x rightarrow infty$ and as $x rightarrow -infty$. Thus
$$
= - int^infty_-infty fracddelta(x_m' - x)dx H(x_m - x) dx
$$
WLOG assume that $x_m leq x_m'$. We note that the heaviside function is $0$ below $x_m$ and $1$ above $x_m$ (we can ignore the value $frac12$ at $x_m$ in integrating because that is a content zero set of values). Thus we have that
$$
= - int^infty_x_m fracddelta(x_m' - x)dx dx
$$
$$
= - (delta(x_m' - x)|^infty_x_m) = delta(x_m' - x_m) = delta(x_m - x_m')
$$
To address the assumption $x_m leq x_m'$: we could have done integration by parts on the other delta function and gotten the same result with the assumption $x_m' leq x_m$, so we cover all cases.
answered Aug 9 at 15:10
Striker
631314
2
How is this more rigorous than a simple swapping of integration domains, as given in my answer? Strictly speaking, neither that nor integration of parts is “allowed†for something like $delta$, which is not a continuous (let alone differentiable) function.
– leftaroundabout
Aug 9 at 15:30
@leftaroundabout I think both are fine, considering that the only thing you can rigorously do with a delta function is 'sift' out a value on a function it's integrated with... but another delta function isn't a function anyway, so the above integral isn't well defined.
– Striker
Aug 9 at 15:46
add a comment |Â
up vote
1
down vote
Here's a derivation using calculus
$$
int_-infty^infty delta(x_m - x)delta(x_m' - x)
$$
Integrating by parts
$$
= (delta(x_m' - x) int delta(x_m - x')dx') |^infty_-infty - int^infty_-infty fracddelta(x_m' - x)dx intdelta(x_m-x')dx' dx
$$
Note that $int delta(x_m -x')dx' (x)$ is the heaviside step function, $H(x_m - x)$, which is 0 below $x_m$, $1/2$ at $x_m$, and $1$ above $x_m$. Thus we have
$$
= lim_xrightarrow infty delta(x_m' - x)H(x_m - x) - lim_xrightarrow -infty delta(x_m' - x)H(x_m - x) -
int^infty_-infty fracddelta(x_m' - x)dx H(x_m - x) dx
$$
The limits go to $0$, because the heaviside function takes on a finite value and the delta functions are zero as $x rightarrow infty$ and as $x rightarrow -infty$. Thus
$$
= - int^infty_-infty fracddelta(x_m' - x)dx H(x_m - x) dx
$$
WLOG assume that $x_m leq x_m'$. We note that the heaviside function is $0$ below $x_m$ and $1$ above $x_m$ (we can ignore the value $frac12$ at $x_m$ in integrating because that is a content zero set of values). Thus we have that
$$
= - int^infty_x_m fracddelta(x_m' - x)dx dx
$$
$$
= - (delta(x_m' - x)|^infty_x_m) = delta(x_m' - x_m) = delta(x_m - x_m')
$$
To address the assumption $x_m leq x_m'$: we could have done integration by parts on the other delta function and gotten the same result with the assumption $x_m' leq x_m$, so we cover all cases.
answered Aug 9 at 15:10
Striker
631314
2
How is this more rigorous than a simple swapping of integration domains, as given in my answer? Strictly speaking, neither that nor integration of parts is “allowed†for something like $delta$, which is not a continuous (let alone differentiable) function.
– leftaroundabout
Aug 9 at 15:30
@leftaroundabout I think both are fine, considering that the only thing you can rigorously do with a delta function is 'sift' out a value on a function it's integrated with... but another delta function isn't a function anyway, so the above integral isn't well defined.
– Striker
Aug 9 at 15:46
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here's a derivation using calculus
$$
int_-infty^infty delta(x_m - x)delta(x_m' - x)
$$
Integrating by parts
$$
= (delta(x_m' - x) int delta(x_m - x')dx') |^infty_-infty - int^infty_-infty fracddelta(x_m' - x)dx intdelta(x_m-x')dx' dx
$$
Note that $int delta(x_m -x')dx' (x)$ is the heaviside step function, $H(x_m - x)$, which is 0 below $x_m$, $1/2$ at $x_m$, and $1$ above $x_m$. Thus we have
$$
= lim_xrightarrow infty delta(x_m' - x)H(x_m - x) - lim_xrightarrow -infty delta(x_m' - x)H(x_m - x) -
int^infty_-infty fracddelta(x_m' - x)dx H(x_m - x) dx
$$
The limits go to $0$, because the heaviside function takes on a finite value and the delta functions are zero as $x rightarrow infty$ and as $x rightarrow -infty$. Thus
$$
= - int^infty_-infty fracddelta(x_m' - x)dx H(x_m - x) dx
$$
WLOG assume that $x_m leq x_m'$. We note that the heaviside function is $0$ below $x_m$ and $1$ above $x_m$ (we can ignore the value $frac12$ at $x_m$ in integrating because that is a content zero set of values). Thus we have that
$$
= - int^infty_x_m fracddelta(x_m' - x)dx dx
$$
$$
= - (delta(x_m' - x)|^infty_x_m) = delta(x_m' - x_m) = delta(x_m - x_m')
$$
To address the assumption $x_m leq x_m'$: we could have done integration by parts on the other delta function and gotten the same result with the assumption $x_m' leq x_m$, so we cover all cases.
answered Aug 9 at 15:10
Striker
631314
Here's a derivation using calculus
$$
int_-infty^infty delta(x_m - x)delta(x_m' - x)
$$
Integrating by parts
$$
= (delta(x_m' - x) int delta(x_m - x')dx') |^infty_-infty - int^infty_-infty fracddelta(x_m' - x)dx intdelta(x_m-x')dx' dx
$$
Note that $int delta(x_m -x')dx' (x)$ is the heaviside step function, $H(x_m - x)$, which is 0 below $x_m$, $1/2$ at $x_m$, and $1$ above $x_m$. Thus we have
$$
= lim_xrightarrow infty delta(x_m' - x)H(x_m - x) - lim_xrightarrow -infty delta(x_m' - x)H(x_m - x) -
int^infty_-infty fracddelta(x_m' - x)dx H(x_m - x) dx
$$
The limits go to $0$, because the heaviside function takes on a finite value and the delta functions are zero as $x rightarrow infty$ and as $x rightarrow -infty$. Thus
$$
= - int^infty_-infty fracddelta(x_m' - x)dx H(x_m - x) dx
$$
WLOG assume that $x_m leq x_m'$. We note that the heaviside function is $0$ below $x_m$ and $1$ above $x_m$ (we can ignore the value $frac12$ at $x_m$ in integrating because that is a content zero set of values). Thus we have that
$$
= - int^infty_x_m fracddelta(x_m' - x)dx dx
$$
$$
= - (delta(x_m' - x)|^infty_x_m) = delta(x_m' - x_m) = delta(x_m - x_m')
$$
To address the assumption $x_m leq x_m'$: we could have done integration by parts on the other delta function and gotten the same result with the assumption $x_m' leq x_m$, so we cover all cases.
answered Aug 9 at 15:10
Striker
631314
edited Aug 10 at 15:32
answered Aug 9 at 15:10
Striker
631314
answered Aug 9 at 15:10
Striker
631314
answered Aug 9 at 15:10
Striker
631314
631314
2
How is this more rigorous than a simple swapping of integration domains, as given in my answer? Strictly speaking, neither that nor integration of parts is “allowed†for something like $delta$, which is not a continuous (let alone differentiable) function.
– leftaroundabout
Aug 9 at 15:30
@leftaroundabout I think both are fine, considering that the only thing you can rigorously do with a delta function is 'sift' out a value on a function it's integrated with... but another delta function isn't a function anyway, so the above integral isn't well defined.
– Striker
Aug 9 at 15:46
add a comment |Â
2
How is this more rigorous than a simple swapping of integration domains, as given in my answer? Strictly speaking, neither that nor integration of parts is “allowed†for something like $delta$, which is not a continuous (let alone differentiable) function.
– leftaroundabout
Aug 9 at 15:30
@leftaroundabout I think both are fine, considering that the only thing you can rigorously do with a delta function is 'sift' out a value on a function it's integrated with... but another delta function isn't a function anyway, so the above integral isn't well defined.
– Striker
Aug 9 at 15:46
2
2
How is this more rigorous than a simple swapping of integration domains, as given in my answer? Strictly speaking, neither that nor integration of parts is “allowed†for something like $delta$, which is not a continuous (let alone differentiable) function.
– leftaroundabout
Aug 9 at 15:30
How is this more rigorous than a simple swapping of integration domains, as given in my answer? Strictly speaking, neither that nor integration of parts is “allowed†for something like $delta$, which is not a continuous (let alone differentiable) function.
– leftaroundabout
Aug 9 at 15:30
@leftaroundabout I think both are fine, considering that the only thing you can rigorously do with a delta function is 'sift' out a value on a function it's integrated with... but another delta function isn't a function anyway, so the above integral isn't well defined.
– Striker
Aug 9 at 15:46
@leftaroundabout I think both are fine, considering that the only thing you can rigorously do with a delta function is 'sift' out a value on a function it's integrated with... but another delta function isn't a function anyway, so the above integral isn't well defined.
– Striker
Aug 9 at 15:46
add a comment |Â
up vote
1
down vote
If you take $u=x_m'-x$, then you have $int_-infty^inftydelta(u+(x_m-x'_m))delta(u)$. Actually finding that rigorously is difficult, but if you use the heuristic that $int_-infty^inftyf(u)delta(u) = f(0)$, then taking $f(u)= delta(u+(x_m-x'_m))$ and setting $u=0$ gets you $delta(x_m-x'_m)$.
answered Aug 13 at 21:14
Acccumulation
5,2342515
add a comment |Â
up vote
1
down vote
If you take $u=x_m'-x$, then you have $int_-infty^inftydelta(u+(x_m-x'_m))delta(u)$. Actually finding that rigorously is difficult, but if you use the heuristic that $int_-infty^inftyf(u)delta(u) = f(0)$, then taking $f(u)= delta(u+(x_m-x'_m))$ and setting $u=0$ gets you $delta(x_m-x'_m)$.
answered Aug 13 at 21:14
Acccumulation
5,2342515
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If you take $u=x_m'-x$, then you have $int_-infty^inftydelta(u+(x_m-x'_m))delta(u)$. Actually finding that rigorously is difficult, but if you use the heuristic that $int_-infty^inftyf(u)delta(u) = f(0)$, then taking $f(u)= delta(u+(x_m-x'_m))$ and setting $u=0$ gets you $delta(x_m-x'_m)$.
answered Aug 13 at 21:14
Acccumulation
5,2342515
If you take $u=x_m'-x$, then you have $int_-infty^inftydelta(u+(x_m-x'_m))delta(u)$. Actually finding that rigorously is difficult, but if you use the heuristic that $int_-infty^inftyf(u)delta(u) = f(0)$, then taking $f(u)= delta(u+(x_m-x'_m))$ and setting $u=0$ gets you $delta(x_m-x'_m)$.
answered Aug 13 at 21:14
Acccumulation
5,2342515
answered Aug 13 at 21:14
Acccumulation
5,2342515
answered Aug 13 at 21:14
Acccumulation
5,2342515
answered Aug 13 at 21:14
Acccumulation
5,2342515
5,2342515
add a comment |Â
add a comment |Â
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