Question

Dirac delta question

Answer 1

Is this true:

\[\frac{f(x)}{f(y)} \delta(x-y) = \delta(x-y)\]

Answer 2

lets prove :3

Answer 3

False.

Answer 4

lol

Answer 5

>:D

Answer 6

mira counter example?

Answer 7

:^)

Answer 8

i'm not sure as functions with exponent gives that tricks sometimes :O

Answer 9

Here's my idea for making "invertible matrix functions" the idea basically comes down to this identity:

\[\delta(x-z)=\int_{-\infty}^\infty \frac{e^{iy(x-z)}}{2 \pi} dy\]
We can split that middle thing up:

\[\delta(x-z)=\int_{-\infty}^\infty \frac{e^{iyx}}{ \sqrt{2 \pi}} \frac{e^{-iyz}}{ \sqrt{2 \pi}} dy\]

Then I define functions this way:

\[A(x,y)=\frac{e^{ixy}}{ \sqrt{2 \pi}} [a(x)b(y)]^i\] with \(a(x)\) and \(b(y)\) purely real functions. That way I can define this operation:

\[A^\dagger(x,y)=A^*(y,x)\] which is just a fancy way of saying take the complex conjugate and switch the arguments. For now we can clearly see that letting \[a(x)=b(y)=1\] will allow us to recover our original delta statment:

\[\delta(x-z)=\int_{-\infty}^\infty A(x,y)A^\dagger(y,z) dy = \int_{-\infty}^\infty \frac{e^{iyx}}{ \sqrt{2 \pi}} \frac{e^{-iyz}}{ \sqrt{2 \pi}} dy\]

But for most functions a(x) and b(y) this looks like it'll work as well as long as that statement earlier is true. :D

Answer 10

Holy schmoly!

Answer 11

The reason I think this has to be true:

\[\frac{f(x)}{f(y)} \delta(x-y) = \delta(x-y)\]

Is because when \(x \ne y\) we have both sides trivially equal since that will give \(\delta(x-y)=0\) So we only need to make sure that they're equivalent when \(x=y\). Well, when that's true then we have:


\[\frac{f(x)}{f(y)} \delta(x-y) =\frac{f(x)}{f(x)} \delta(x-x) = 1*1 = \delta(x-x)=\delta(x-y)\]

Done. :P Looks tricky, but that's cause it is.

Answer 12

:3 i like this

Answer 13

i like this too, i like it a lot

Answer 14

http://prntscr.com/8tzwc7
so if x equals y then it won't be 0

Answer 15

he said that already :3

Answer 16

): m confused wid all this ima see some vedios

Answer 17

I'm experiencing jai da vo

Answer 18

:D

Answer 19

DEJAVU

Answer 20

Oh yeah my bad it's infinity not 1, but the proof I put at the end will still stand haha I had accidentally put \(\delta(0)=1\) which is wrong.

You can use this just like an identity matrix, which is why I was calling this an "inverse" since both multiplied together gave the dirac delta (identity matrix) earlier, technically this type of matrix I've described is a unitary matrix since it's inverse is its conjugate transpose.

Identity matrix type use here:
\[f(x,z)=\int_{-\infty}^\infty f(x,y) \delta(y-z) dy\]

Answer 21

Instead let's just say we have some other function I'll call it \(\bar \delta(n)\) which is 0 everywhere except at \(\bar \delta(0) = 1\).
Now plug in here \(f(n)= n^2\)
\[\sum_{n=1}^{10} f(n) \bar \delta(n-3) = \]

Answer 22

it will be 0 for all values of n from 1 to 10 except n=3 where it will b equal to 9.

Answer 23

wait did u ask for the summation or u were gonna plug it somewhere aww m confused ima read the whole thread again

Answer 24

Noooo this is all I was asking like you're good this is just how the dirac delta works it's just it has to be infinite if it's in an integral cause there are infinitely many other points so they kinda like cancel out lol.

Answer 25

ok >:)

Answer 26

By the way the time evolution operator of quantum mechanics is a unitary operator, so that was my motivation for playing around with this, in case anyone was wondering why anyone cares about unitary operators lol.

Answer 27

:)

Answer 28

Here we go, if we start out with a normalized wave function we can slide an identity matrix in between a dot product:

\[1=\langle \psi | \psi \rangle = \langle \psi| I |\psi \rangle\]

Also we have that unitary matrix from earlier which I called A, so it obeys:

\[A^\dagger A = I\] plug baby in:

\[\langle \psi| I |\psi \rangle = \langle \psi| A^\dagger A |\psi \rangle\] But now we can pull this A into both sides:

\[\langle A \psi|A\psi \rangle\]
Now ust define \[|\phi \rangle = | A \psi \rangle\]
So we have \[1=\langle \psi | \psi \rangle=\langle \phi | \phi \rangle \]

Now can you show this using the integrals from earlier (just repeat the same process above but with integrals):

\[1=\langle \psi| I |\psi \rangle=\int_{-\infty}^\infty dx\int_{-\infty}^\infty dy \ \psi^*(x) \delta(x-y) \psi(y) \]

Now go to up to showing \[1=\langle \phi | \phi \rangle \]