Mixing
Are these distributions the same?
$begingroup$
Consider the following distribution, where $delta$ is the Dirac delta:
$$f(x,y)=delta(x)+delta(y).tag1$$
This can be viewed as a limit of the following sequence of smooth functions:
$$newcommand{sech}{operatorname{sech}}
%
f_a(x,y)=frac a2left(sech(ax)^2+sech(ay)^2right),tag2$$
with $atoinfty$.
Looking at the vicinity of $x=y=0$, we can see that this function always has some "extra" peak there due to the sum of the wavefronts, regardless of $a$. Suppose we remove this central peak, getting the following sequence:
$$g_a(x,y)=frac a2left(sech(ax)^2+sech(ay)^2-sech(ax)^2sech(ay)^2right).tag3$$
Will this still result in the same distribution $(1)$, or will it be different in some regards? In particular, if $(3)$ is used as (minus) potential in a Schrödinger equation, will the eigenfunctions be the same as those with $(1)$ used instead?
functional-analysis distribution-theory dirac-delta
asked Jan 18 at 11:38
RuslanRuslan
3,72721633
$endgroup$
add a comment |
$begingroup$
Consider the following distribution, where $delta$ is the Dirac delta:
$$f(x,y)=delta(x)+delta(y).tag1$$
This can be viewed as a limit of the following sequence of smooth functions:
$$newcommand{sech}{operatorname{sech}}
%
f_a(x,y)=frac a2left(sech(ax)^2+sech(ay)^2right),tag2$$
with $atoinfty$.
Looking at the vicinity of $x=y=0$, we can see that this function always has some "extra" peak there due to the sum of the wavefronts, regardless of $a$. Suppose we remove this central peak, getting the following sequence:
$$g_a(x,y)=frac a2left(sech(ax)^2+sech(ay)^2-sech(ax)^2sech(ay)^2right).tag3$$
Will this still result in the same distribution $(1)$, or will it be different in some regards? In particular, if $(3)$ is used as (minus) potential in a Schrödinger equation, will the eigenfunctions be the same as those with $(1)$ used instead?
functional-analysis distribution-theory dirac-delta
asked Jan 18 at 11:38
RuslanRuslan
3,72721633
$endgroup$
add a comment |
$begingroup$
Consider the following distribution, where $delta$ is the Dirac delta:
$$f(x,y)=delta(x)+delta(y).tag1$$
This can be viewed as a limit of the following sequence of smooth functions:
$$newcommand{sech}{operatorname{sech}}
%
f_a(x,y)=frac a2left(sech(ax)^2+sech(ay)^2right),tag2$$
with $atoinfty$.
Looking at the vicinity of $x=y=0$, we can see that this function always has some "extra" peak there due to the sum of the wavefronts, regardless of $a$. Suppose we remove this central peak, getting the following sequence:
$$g_a(x,y)=frac a2left(sech(ax)^2+sech(ay)^2-sech(ax)^2sech(ay)^2right).tag3$$
Will this still result in the same distribution $(1)$, or will it be different in some regards? In particular, if $(3)$ is used as (minus) potential in a Schrödinger equation, will the eigenfunctions be the same as those with $(1)$ used instead?
functional-analysis distribution-theory dirac-delta
asked Jan 18 at 11:38
RuslanRuslan
3,72721633
$endgroup$
Consider the following distribution, where $delta$ is the Dirac delta:
$$f(x,y)=delta(x)+delta(y).tag1$$
This can be viewed as a limit of the following sequence of smooth functions:
$$newcommand{sech}{operatorname{sech}}
%
f_a(x,y)=frac a2left(sech(ax)^2+sech(ay)^2right),tag2$$
with $atoinfty$.
Looking at the vicinity of $x=y=0$, we can see that this function always has some "extra" peak there due to the sum of the wavefronts, regardless of $a$. Suppose we remove this central peak, getting the following sequence:
$$g_a(x,y)=frac a2left(sech(ax)^2+sech(ay)^2-sech(ax)^2sech(ay)^2right).tag3$$
Will this still result in the same distribution $(1)$, or will it be different in some regards? In particular, if $(3)$ is used as (minus) potential in a Schrödinger equation, will the eigenfunctions be the same as those with $(1)$ used instead?
functional-analysis distribution-theory dirac-delta
functional-analysis distribution-theory dirac-delta
asked Jan 18 at 11:38
RuslanRuslan
3,72721633
asked Jan 18 at 11:38
RuslanRuslan
3,72721633
asked Jan 18 at 11:38
RuslanRuslan
3,72721633
asked Jan 18 at 11:38
RuslanRuslan
3,72721633
asked Jan 18 at 11:38
RuslanRuslan
3,72721633
3,72721633
add a comment |
add a comment |
1 Answer
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$begingroup$
That subtraction indeed doesn't matter, because the term $frac{a}{2}operatorname{sech}(ax^2)operatorname{sech}(ay^2)$ goes to zero (weakly). If we multiplied it by $frac{a^2}{4}$ instead of $frac a2$ to properly compensate for two dimensions of stretching, we would get the identity $delta(x)delta(y)$ in the limit. It doesn't matter whether we have a double-height peak at the origin, a peak to match the ridge lines, or even a valley at that scale; all that matters is that we don't have something infinitely larger there.
The thing is, when we're talking about Dirac deltas? It just doesn't matter how we got to them. Work with the $delta$s directly, not some arbitrary choice of functionals that converge to them.
answered Jan 18 at 12:03
jmerryjmerry
10.8k1225
$endgroup$
$begingroup$
It's not possible to work with $delta$s directly when doing numerics, which is where my question originates.
$endgroup$
– Ruslan
Jan 18 at 12:19
$begingroup$
$delta$ isn't a function - it's an operator. It only really becomes meaningful when we convolve it with something; here, $f*h(x,y)=int_{mathbb{R}}h(x,v),dv+int_{mathbb{R}}h(u,y),du$. That's something you can work with.
$endgroup$
– jmerry
Jan 18 at 12:24
$begingroup$
You do have to approximate it somehow e.g. when it's a potential in a Schrödinger equation you're solving numerically.
$endgroup$
– Ruslan
Jan 18 at 12:30
add a comment |
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1 Answer
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$begingroup$
That subtraction indeed doesn't matter, because the term $frac{a}{2}operatorname{sech}(ax^2)operatorname{sech}(ay^2)$ goes to zero (weakly). If we multiplied it by $frac{a^2}{4}$ instead of $frac a2$ to properly compensate for two dimensions of stretching, we would get the identity $delta(x)delta(y)$ in the limit. It doesn't matter whether we have a double-height peak at the origin, a peak to match the ridge lines, or even a valley at that scale; all that matters is that we don't have something infinitely larger there.
The thing is, when we're talking about Dirac deltas? It just doesn't matter how we got to them. Work with the $delta$s directly, not some arbitrary choice of functionals that converge to them.
answered Jan 18 at 12:03
jmerryjmerry
10.8k1225
$endgroup$
$begingroup$
It's not possible to work with $delta$s directly when doing numerics, which is where my question originates.
$endgroup$
– Ruslan
Jan 18 at 12:19
$begingroup$
$delta$ isn't a function - it's an operator. It only really becomes meaningful when we convolve it with something; here, $f*h(x,y)=int_{mathbb{R}}h(x,v),dv+int_{mathbb{R}}h(u,y),du$. That's something you can work with.
$endgroup$
– jmerry
Jan 18 at 12:24
$begingroup$
You do have to approximate it somehow e.g. when it's a potential in a Schrödinger equation you're solving numerically.
$endgroup$
– Ruslan
Jan 18 at 12:30
add a comment |
$begingroup$
That subtraction indeed doesn't matter, because the term $frac{a}{2}operatorname{sech}(ax^2)operatorname{sech}(ay^2)$ goes to zero (weakly). If we multiplied it by $frac{a^2}{4}$ instead of $frac a2$ to properly compensate for two dimensions of stretching, we would get the identity $delta(x)delta(y)$ in the limit. It doesn't matter whether we have a double-height peak at the origin, a peak to match the ridge lines, or even a valley at that scale; all that matters is that we don't have something infinitely larger there.
The thing is, when we're talking about Dirac deltas? It just doesn't matter how we got to them. Work with the $delta$s directly, not some arbitrary choice of functionals that converge to them.
answered Jan 18 at 12:03
jmerryjmerry
10.8k1225
$endgroup$
$begingroup$
It's not possible to work with $delta$s directly when doing numerics, which is where my question originates.
$endgroup$
– Ruslan
Jan 18 at 12:19
$begingroup$
$delta$ isn't a function - it's an operator. It only really becomes meaningful when we convolve it with something; here, $f*h(x,y)=int_{mathbb{R}}h(x,v),dv+int_{mathbb{R}}h(u,y),du$. That's something you can work with.
$endgroup$
– jmerry
Jan 18 at 12:24
$begingroup$
You do have to approximate it somehow e.g. when it's a potential in a Schrödinger equation you're solving numerically.
$endgroup$
– Ruslan
Jan 18 at 12:30
add a comment |
$begingroup$
That subtraction indeed doesn't matter, because the term $frac{a}{2}operatorname{sech}(ax^2)operatorname{sech}(ay^2)$ goes to zero (weakly). If we multiplied it by $frac{a^2}{4}$ instead of $frac a2$ to properly compensate for two dimensions of stretching, we would get the identity $delta(x)delta(y)$ in the limit. It doesn't matter whether we have a double-height peak at the origin, a peak to match the ridge lines, or even a valley at that scale; all that matters is that we don't have something infinitely larger there.
The thing is, when we're talking about Dirac deltas? It just doesn't matter how we got to them. Work with the $delta$s directly, not some arbitrary choice of functionals that converge to them.
answered Jan 18 at 12:03
jmerryjmerry
10.8k1225
$endgroup$
That subtraction indeed doesn't matter, because the term $frac{a}{2}operatorname{sech}(ax^2)operatorname{sech}(ay^2)$ goes to zero (weakly). If we multiplied it by $frac{a^2}{4}$ instead of $frac a2$ to properly compensate for two dimensions of stretching, we would get the identity $delta(x)delta(y)$ in the limit. It doesn't matter whether we have a double-height peak at the origin, a peak to match the ridge lines, or even a valley at that scale; all that matters is that we don't have something infinitely larger there.
The thing is, when we're talking about Dirac deltas? It just doesn't matter how we got to them. Work with the $delta$s directly, not some arbitrary choice of functionals that converge to them.
answered Jan 18 at 12:03
jmerryjmerry
10.8k1225
answered Jan 18 at 12:03
jmerryjmerry
10.8k1225
answered Jan 18 at 12:03
jmerryjmerry
10.8k1225
answered Jan 18 at 12:03
jmerryjmerry
10.8k1225
10.8k1225
$begingroup$
It's not possible to work with $delta$s directly when doing numerics, which is where my question originates.
$endgroup$
– Ruslan
Jan 18 at 12:19
$begingroup$
$delta$ isn't a function - it's an operator. It only really becomes meaningful when we convolve it with something; here, $f*h(x,y)=int_{mathbb{R}}h(x,v),dv+int_{mathbb{R}}h(u,y),du$. That's something you can work with.
$endgroup$
– jmerry
Jan 18 at 12:24
$begingroup$
You do have to approximate it somehow e.g. when it's a potential in a Schrödinger equation you're solving numerically.
$endgroup$
– Ruslan
Jan 18 at 12:30
add a comment |
$begingroup$
It's not possible to work with $delta$s directly when doing numerics, which is where my question originates.
$endgroup$
– Ruslan
Jan 18 at 12:19
$begingroup$
$delta$ isn't a function - it's an operator. It only really becomes meaningful when we convolve it with something; here, $f*h(x,y)=int_{mathbb{R}}h(x,v),dv+int_{mathbb{R}}h(u,y),du$. That's something you can work with.
$endgroup$
– jmerry
Jan 18 at 12:24
$begingroup$
You do have to approximate it somehow e.g. when it's a potential in a Schrödinger equation you're solving numerically.
$endgroup$
– Ruslan
Jan 18 at 12:30
$begingroup$
It's not possible to work with $delta$s directly when doing numerics, which is where my question originates.
$endgroup$
– Ruslan
Jan 18 at 12:19
$begingroup$
It's not possible to work with $delta$s directly when doing numerics, which is where my question originates.
$endgroup$
– Ruslan
Jan 18 at 12:19
$begingroup$
$delta$ isn't a function - it's an operator. It only really becomes meaningful when we convolve it with something; here, $f*h(x,y)=int_{mathbb{R}}h(x,v),dv+int_{mathbb{R}}h(u,y),du$. That's something you can work with.
$endgroup$
– jmerry
Jan 18 at 12:24
$begingroup$
$delta$ isn't a function - it's an operator. It only really becomes meaningful when we convolve it with something; here, $f*h(x,y)=int_{mathbb{R}}h(x,v),dv+int_{mathbb{R}}h(u,y),du$. That's something you can work with.
$endgroup$
– jmerry
Jan 18 at 12:24
$begingroup$
You do have to approximate it somehow e.g. when it's a potential in a Schrödinger equation you're solving numerically.
$endgroup$
– Ruslan
Jan 18 at 12:30
$begingroup$
You do have to approximate it somehow e.g. when it's a potential in a Schrödinger equation you're solving numerically.
$endgroup$
– Ruslan
Jan 18 at 12:30
add a comment |
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