Shing

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Shing

Sep 16, 2024 15:16

In complex analysis, Jordan's lemma is a result frequently used in conjunction with the residue theorem to evaluate contour integrals and improper integrals. The lemma is named after the French mathematician Camille Jordan. == Statement == Consider a complex-valued, continuous function f, defined on a semicircular contour C R = { R e i θ ∣ θ ∈ [ 0 , π ]...

Shing

Sep 16, 2024 15:15

Shing

Sep 16, 2024 15:14

any intuitive way to see the Jordan's lemma?

Shing

Apr 1, 2024 14:52

@Slereah I have a feeling this somehow touch the measurement problem. if DNA counted as mac object (since it is so long), I am not so sure it can be predicted by Schrödinger equation / these is a wavefunction that can possibly represent a DNA.

Shing

Apr 1, 2024 12:07

I am wondering if Schrödinger equation can predict how DNA forms.

Shing

Dec 12, 2023 16:23

@ACuriousMind BG3? I share same feeling as you.

Shing

Dec 12, 2023 15:43

Have you guys played BG3?

Shing

Dec 12, 2023 15:24

@Slereah actually I think $a$ has eigenstate, and it is coherent state?

Shing

Dec 12, 2023 15:22

@ACuriousMind actually I think that's a cool idea lol

Shing

Dec 12, 2023 15:21

@ACuriousMind so pretty much we need this a|0> = 0 to make physics consistent?

Shing

Dec 12, 2023 15:19

so does a zero of a vector space has physical meaning?

Shing

Dec 12, 2023 15:19

I see, thanks.

Shing

Dec 12, 2023 15:18

because I forgot about this a|0> = 0 , hence all the contradictions came lol

Shing

Dec 12, 2023 15:17

so a zero vector?

Shing

Dec 12, 2023 15:17

@ACuriousMind I am more confused now :P

Shing

Dec 12, 2023 15:15

btw, what is the physics meaning of the right hand side 0 for $a|0> = 0$ ?

Shing

Dec 12, 2023 15:15

I think I get it. I made stupid miscalculation

Shing

Dec 12, 2023 14:04

I hope everyone is having a good time!

Shing

Dec 12, 2023 14:03

it is so good to see all the old faces are still here

Shing

Dec 12, 2023 14:02

$a = \frac{\partial}{\partial a^\dagger}$

Shing

Dec 12, 2023 13:57

I made some calculations with $[a,a^\dagger]=1$, all led to contradictions. I suspect creator operator is equal to partial derivative with respective to creation operator instead.

Shing

Dec 12, 2023 13:55

hey, I am studying QFT, a bit confused, would anyone be kind enough to explain to me: why the creator operator is equal to partial derivative with respective to annihilation operator? (in a lecture note)

Shing

Oct 20, 2023 03:43

I mean since dirac notation is pretty solid, and I did not touch anything I should not do (I think). then why the contradiction comes out?

Shing

Oct 20, 2023 03:42

I could be wrong, I have a bad memory.

Shing

Oct 20, 2023 03:41

I think I read something like this $(d/dx)^\dagger = -d/dx$ in Kip Throne's Modern Classical Physics somewhere

Shing

Oct 20, 2023 03:40

@naturallyInconsistent oh I knew that. I was confused why it raise if I only use dirac notation. I am thinking it has to do with tensor calculus

Shing

Oct 20, 2023 03:34

@naturallyInconsistent thanks! I did not know $\frac{d}{dx}$ will pick up a negative sign, may I know why?

Shing

Oct 20, 2023 03:19

@Relativisticcucumber yeah, that's why I tried to change it back to P at the end in order to avoid what you said. but I was kind of doing hand waving physics. thanks anyway

Shing

Oct 20, 2023 03:12

I know it is a silly question, but it bugs me...

Shing

Oct 20, 2023 03:12

then I get $<p| P |p> = - <p| P |p> $

Shing

Oct 20, 2023 03:11

it is like $<p| P |p>$ ~ $<p| i d/dx |p>$ (take dagger) -> $<p| i d/dx |p>$ ~ $ - <p| i d/dx |p>= -<p| P |p>$

Shing

Oct 20, 2023 03:09

@Relativisticcucumber thanks

Shing

Oct 20, 2023 03:06

I wonder what goes wrong

Shing

Oct 20, 2023 03:06

I will get wrong result. only if I calculate it explicitly like in the link you gave I will have the expected result.

Shing

Oct 20, 2023 03:05

@Relativisticcucumber thanks I am aware of that. But I am confused that if we only look at it with ket-bra notation

Shing

Oct 20, 2023 03:03

If I take dagger of P, then I will get silly result : $P \ineq P$

Shing

Oct 20, 2023 03:01

@Relativisticcucumber yes, I know, but $P$ ~ $i d/dx$

Shing

Oct 20, 2023 03:00

(I know how to solve this in probability distribution intergral)

Shing

Oct 20, 2023 02:59

where did I go wrong? (in ket-bra notation)

Shing

Oct 20, 2023 02:59

which is nonsense, since I am literally saying 4 = -4

Shing

Oct 20, 2023 02:59

I will get something $<p| P |p> = - <p| P |p>$

Shing

Oct 20, 2023 02:58

so if i take dagger

Shing

Oct 20, 2023 02:58

there is an $i$ in $P$

Shing

Oct 20, 2023 02:58

$|p>$ is eigenstate of $P$

Shing

Oct 20, 2023 02:58

In ket bra notation, I am a bit confused by say $<p| P |p>$

Shing

Oct 5, 2023 03:15

@naturallyInconsistent oh I didnt know there is a whole story behind it. I think I get it now! Thanks for your help.

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Shing

Sep 16, 2024 15:13

thanks guys!

Shing

Sep 16, 2024 14:34

I think $e^{iqx}$ is an "oscillation", why adding the contribution of these "oscillation" approaching infinity is nearly zero?

Shing

Sep 16, 2024 14:31

Shing

Sep 16, 2024 14:30

can anyone explain this to me?