Question

anyone good at quantum here?

Answer 1

Depends on the question

Answer 2

just some concept questions.

Answer 3

Do you understand what the dirac delta function is?

Answer 4

and its applications?

Answer 5

I haven't made it to that part yet, but it's something to do with spikes I can't entirely remember, though I've heard of it, maybe @Kainui knows.

Answer 6

yeah you're right, it has to do with spikes or impulses but strangely its area is always 1

Answer 7

I didn't really get why

Answer 8

Mhm sort of reminds me of normalizing, is it related to that at all?

Answer 9

The dirac delta function?

Answer 10

yes

Answer 11

Hey look, a great starting point: http://en.wikipedia.org/wiki/Dirac_delta_function you can look at references for more input

Answer 12

The dirac delta function has many different interpretations, it's sort of more hyped up than it should probably be. You can think of it as "sampling" a function at a point, so if you have a bunch of them, they will sample a function at different intervals.

Ok, so what is the thing? It's simply this: \[\Large \delta(x)\] so when x=0 then delta=1. Otherwise for all other values it equals 0. You can kind of think of it as a sort of continuous kronecker delta.

So if you want to sample a function at one spot, what do you do? This: \[\Large \int\limits_{-\infty}^\infty f(x) \delta(x-a)dx=f(a)\]

Answer 13

Kronecker delta?

Answer 14

yes the kronecker delta is the discrete analog of the dirac delta function

Answer 15

See, the point is the entire integral is 0 except at one spot, when x=a since it will give you the answer: f(a)*1. Since it is infinitely high but also multiplied by dx which is infinitely small they cancel each other out in a sense and give you a finite number.

Kronecker's delta is essentially a way of talking about the identity matrix in linear algebra. \[\Large \delta_{ij}\] when i=j delta = 1, i=/=j, delta=0. So the entries are all 0 except the diagonal where the rows and columns are the same.

Answer 16

so what do mean exactly by sample function

Answer 17

Suppose you want to take a discrete sampling of a function, you can use a "comb" function which is really just defined as a sum of dirac delta functions at separate points. Check this out, it's pretty straight forward I think: http://en.wikipedia.org/wiki/Dirac_comb

Answer 18

so its a model of some infinitely high and infinitely narrow pulse?

Answer 19

I don't really see why its needed if you can always just take the definite integral from any two points to analyze the pulse

Answer 20

Yes, you can also look at it as a gaussian that has been squeezed infinitely tightly together since they are normalized to have an area of 1 under them, (think 100%) just like you've done with 65, 95.7,99.5 or whatever the rules were for standard deviations.

Or it can be looked at as the derivative of a stepfunction, you might have heard them called Heaviside functions, and they are very simple, essentially a single square wave.

Answer 21

What do you mean?

Answer 22

okay that makes sense. so the primary application is to analyze spikes in wave functions or whatever

Answer 23

It is an infinitely high and infinitely narrow pulse that integrates to 1 over it. But it has other uses and applications outside of that. It's like saying when you were in 3rd grade what's the point of having a calculator when I can just count how many stones are laying in the pile? There's more to it than what you see.

Answer 24

yeah im just curious as to what some of the applications are so I better understand it

Answer 25

I agree

Answer 26

but I will say I get it a lot better now though

Answer 27

Nice explanation kai :D

Answer 28

yeah

Answer 29

I also have a question about bound and scattered states

Answer 30

if you don't mind?

Answer 31

still there?

Answer 32

well thanks for the help

Answer 33

Sorry I was distracted by another question.

Answer 34

Sure, what do you want to know? It would help me answer you better if you had a more specific question. The last question you asked was pretty broad, so I kind of answered it generally but I can probably help you more if I can see where you're struggling. Plus it saves me time from reexplaining stuff you already know. Also, while you're at it, can you just ask a new question and close this one? OS tends to lag when these get too long and it sucks lol.

Answer 35

okay i have a question about bound and scattering states

Answer 36

namely: if a particle has higher E than the v(x) potential barrier, then that is considered a scattered state right?

Answer 37

because the particle is free essentially

Answer 38

Yes exactly. It would be bound if it is below that, kind of like how water is "bound" by the cup it is held in when the water level is below the top of the cup. If you overfill the cup "too much energy" than the potential energy barrier "cup walls" then it spills "scatters".

Of course this isn't an exact metaphor because you would have to cut holes in the cup to allow tunnelling =P

Answer 39

okay cool. just one more question: v(x) represents potential barriers right? like walls sort of ?

Answer 40

Not walls, just potential energy. Check this out https://www.youtube.com/watch?v=HFu3nbowletMA&list=PL65jGfVh1ilueHVVsuCxNXoxrLI3OZAPI

Answer 41

says its unavailable

Answer 42

Weird https://www.youtube.com/channel/UCNIEAv633WRg4ubBIhOcwMg

Answer 43

brant carlson is the man! watch his videos all the time.

Answer 44

thanks again for your help

Answer 45

haha nice, np