@ACuriousMind what does an "excited bound state" mean?

Nontrivial bound state?

More than one bound state?

It's a state of a bound system that's not the ground state of the system.

obbiously

What's the ground state?

The one with the lowest energy

@0celo7:hahah..thanx for your valuabel comments....I can't stop to laugh upon myself..what the irony..I was giving a 'sermon'to @MartianCactus and that sermon came back 2 me indirectly through your comments

*The cursed sermon :P

@ACuriousMind ok well I'm looking at an attractive delta potential

So the only two solutions of the SE are one with zero energy and one with negative energy

The zero energy one is the trivial solution

So does that negative energy one count as an excited ground state or not

no, the delta potential has exactly one bound state, it's not meaningful to speak of any excited bound state in it.

Note that the "trivial" solution (I'm guessing you mean $\psi = 0$) is not a solution that corresponds to a quantum state, btw.

Yeah, I am basically reading nearly every single article on group theory in wikipedia just to make a recipe of all known types of groups

because the idea of orbits in group theory interested me and I want to have a recipe book to cook up and test some groups

@ACuriousMind o right

I forgot this was physics

This include sampling all important theorems and properties of groups and other algebraic structure and do a meta analysis

But I managed to solve the SE all by myself without looking at notes or Wiki. I'm proud of that

hey guys I have a confusion here: in my book there' question to find the value of 'A' for which

y = (x^2 - x)/(1-ax) is real

but

Ultimately, the primary trigger of all of this reading is really because of my divison by zero project. But readig these things will be useful for other things too. In particular, some of the group theoric stuff mentioned by ACM started to make sense such as Lie group and representations

for what value of a ca y be unreal?(or any other field)

I have tried to graph it using desmos but still couldn't arrive at any reslut?

Any ideas?

well you cannot plug ax=1 else you will be on an asymptote

A =Real number

so there will be a line of values ax such that y is not defined

(i.e. an obilque asymptote)

How do you invite a user into a custom chat room?

@MAFIA36790, thanks! One more quick question: is there a way to transfer relevant messages into that custom chat room?

@MAFIA36790, okay.

@heather Do you mean transfer or copy? Because moderators can move messages from one room to the other, effectively deleting them in the original room

@ACuriousMind, just copy.

@Secret :but books is saying ; a = [1,infty), for which oblique asymptotes exist

yes, point is, a can be anything $\geq 1$ and then y=ax becomes an obiquoe asymptote

if a=0, then there is no asymptote

what the heck is an oblique asymptote

I had a very poor high school education @bloo

I grew up in a suburb of DC

@bloo: There are high chances he read the same thing under a different name

In fact, a can be any real number $\neq 0$

and you still get the oblique asymptote

but it's asking values of a for which function is defined everpoint

what the heck is an oblique asymptote?

and athese values there is atleast some point where y is not define

@0celo7:purplemath.com/modules/asymtote3.htm

h@Secret:But if a is very very large ther e is no asymptote

@bloo that sounds familiar

but my mind from that time of my life is messed up

^open desmose and enter a = 10000000

too much crack

l0l

that was the LSD

$y = \frac{x^2 - x}{1+ax}=\frac{x^2}{1+ax}-\frac{x}{1+ax}$ If a is large, then ax will dominate 1 thus $y\sim\ \frac{x^2}{ax}-\frac{x}{ax}=\frac{x}{a}-\frac{1}{a}=\frac{x-1}{a}$ which is why the asymptote seemed to disappear as a gets really large

@ACuriousMind ok, I need conceptual help

What is a good book on complexity theory?

This is kinda confirmed here as note that for very large a, the asymptote tends towards the line with slope 1

@ACuriousMind So I start with a plane wave in the I region, then I use boundary conditions on the wave function + derivatives to solve the SE in all three regions

And what are some requirements for learning complexity theory?

then I compute the currents in regions I and III

@ACuriousMind then I compute ratios of these currents to get the reflection and transmission coefficients?

@ACuriousMind But won't the answer also depend on the initial energy of the wave?

@Secret:so the conclusion is books's soultuion is right for a = [1,infty) for very large values..?

we did basic step function scattering in class and the answer depended on the energy of the initial wave

in fact, I will argue that $a\neq 0$ and oblique asymptote exists

so for what values oblique asymptote doesn't exist that is y is defined at every point..hence book is wrong ain't it?

only when a=0 there is no oblique asymotote

because only then y is defined for all values of x

if the book excludes $a\in (0,1)$ I suppose there might be a reason for it that I don't know, perhaps a has some extra assumptions built on it?

like a is required to be an integer?

@MAFIA36790 I lost, badly

No contest

a is real(assume it..cause no info is given)..secondly book asks for complete setof A for which y is defined..but the graphs contradicts books answer aA = [1,infty)

@Secret:HEre take a look at book's solution

y = (x^2 -x )/(1-ax)

=>x^2 -x = y -axy

x^2 + x(ay-1) - y = 0

Since x is real:Discriminant should be >= 0

=> (ay-1)^2+ 4y >= 0

=> a^2y^2 + 2y(2-a) + 1 >=0 where y is REAL

SO, as a^2 >0

=> 4(2-a)^2 -4a^2 <= 0

or 4 - 4a <= 0

=> A = [1,infty)

Any ideas?

totally confused :(

I don't think the discriminant is the correct approach here Note when you plug a=0 you still get discriminant > 0, thus x is still real

@ACuriousMind I'm so stupid

To get an asymptote you need to find where x is undefined. Here, because there are no square roots in that expression of y, there is no way for x to become complex

why? because both of us were over reacting

it was in my best interest to take a dive

Typo: I mean, even if a=0 is plugged, there are values of y for any values of x such that the discriminant is > 0

@Secret Well before heading furthur can we confirm the conclusion that A = [1,infty) is wrong since asymptotes exist for small values althogh for large x it vanishes

it actually never vanishes, because for any finite a, 1-ax can always become zero for some x

but it looks like it vanishes to first order because ax dominates 1 in thsi case

sorry I meant for very large A...soo book's solution is wrong that for A= [1,infty) y isdefined(attains real values)

@bloo wtf

Hi, everybody.

That's what I figured, but (suppose this is from a school exercise), I will consider double checking with your teacher to ensure there is nothing else I overlooked cause that might change the conclusion

\o

@DanielSank, hello. Anything new and cool happening with quantum computing?

@heather Yes!

Always.

@DanielSank How the heck do I remember which wave is left/right moving :/

@DanielSank =)

I've been reading about Shor's algorithm

And I got it for about thirty seconds

Well, one thing that's exciting for me personally is that the paper describing an experiment I did last year was accepted for publication in Physical Review Letters!

Congrats!

@heather That's about how long I understood it too.

I know a lot about the physics of qubits, but less about the actual algorithms.

@DanielSank, okay, then I probably only understood it for five seconds =)

@0celo7 I had trouble with that too. Simple trick: look at $f(x - v t)$, if $t$ increases what do you have to do to $x$ to keep the argument constant?

You increase $x$. Therefore, as time goes on, the position increases --> right moving wave.

@heather lol

@DanielSank ahhhh what

I understood this when I took PDE but I don't any more

@DanielSank, I watched a bunch of YouTube and read many a website but in the end I found this the most helpful.

@heather Ah yes, Scott Aaronson's blog.

@DanielSank, a handy resource. That particular article was recommended by Shor himself (::insert pure awe here::).

:O

@0celo7 Yes, it probably will

^ "Pure awe" face

@0celo7 What part don't you understand?

@DanielSank All of it. I'll go back and read my notes on the wave eq. / characteristic solution methods.

If I increase $t$ by $\delta t$, but I also shift where I'm looking by $v \delta t$, then I see the same waveform. Therefore, the wave has moved $+ v \delta t$ to the right.

Solving linear equations right now, why is this so damn hard??

@0celo7, what is the system of equations?

(not that I'll be able to solve it or anything)

@0celo7 Uh, because you have a lot of variables?

I can solve it

@DanielSank there's only three variables

I'm bad at linear algebra

@0celo7, my chances of solving this system went from 0% to 0.00000000000000000001%

Okay, maybe a little more than that.

there's typos in my lecture notes, which isn't helping

probably my typos, the prof was scribbling equations pretty quickly and I can't write that fast

In middle school they actually ask if you're done writing down the notes =)

@MAFIA36790 at least she's talking to me again ;/

??

@heather I miss that :(

my nuclear physics prof stands right in front of what he's writing, then moves to the podium to lower the screen

you have maybe 10 seconds to copy whatever he was writing

also his hand writing is awful

@0celo7, oh, I'm sorry - ick, that's way too little time.

Oh, even worse.

I don't suppose you could take a photo of the board...?

@MAFIA36790 no, people would make fun of me

Hippies?

@0celo7, I'm sorry, that's really unfortunate

@MAFIA36790, what do you mean "hippies"?

Yes I know the proof.

@ACuriousMind dropbox.com/s/b0atjgtg5g8y01t/qmhw6.pdf?dl=0

is what I have on page 5 right so far?

when I work with that I end up with a contradiction

so I'm assuming something must be wrong with my setup

Are you scattering on a rectangular potential?

Yes

It's the problem from the picture earlier

Then that appears to be the correct ansatz

Ok, I'll type of the rest of what I did...

@ACuriousMind With $A_2'=A_3'=0$, right?

@0celo7 I'm not convinced about $A'_2 = 0$.

In fact, I think you shouldn't assume that

@ACuriousMind Why wold there be left moving waves in the block?

Because the wave back-scatters from the second step

hmm

ok, back to the drawing board then

@Xasel Some corrections: sorry I mentioned the wrong way to calculate the oblique asymptote. The oblique asymptote is $y=-\frac{1}{a}\left(x+\frac{1}{a}-1\right)$. However the same conclusion of $a\neq 0$ is unaffected Details on how to calculate oblique asymptotes here, the same thing taught back in my high school

Note now that when a gets large $-\frac{1}{a}\left(x+\frac{1}{a}-1\right) \rightarrow y=+1$ as expected

thus the asymptote tends towards a horizontal one as a gets large

The method I mentioned previosuly, are used for finding vertical asymptotes

@ACuriousMind How do I know what to calculate

Stupid question, I know

what the heck do I do :/

Should I compute the TR coeffs first generally

then see what I need to extract from the boundary conditions?

You get four equations from the boundary conditions, and you are free to choose $A_1$. With $A_3'= 0$, this means you have four equations and four unknowns, and you can solve for the unknowns.

@ACuriousMind :/ you know I can't solve linear equations...

@ACuriousMind Seriously, how do I solve a 4x4 system by hand?

Do you know gaussian elimiantion?

no

do you know how to invert a matrix?

no

one of those two would be useful if you want to work on differential equations

I tried understanding linear algebra and gave up

Algebra is too hard

@MAFIA36790 What part is unclear?

no

u r doing diff geo and think linear algebra is too hard? :O

yes, why is this hard to believe?

0celo7 so you already tried reading those notes I referred you earlier and cannot work out what happened?

@Secret no

@Sanya Yes, this has been a source for puzzlement in this chat for quite a while now :P

I do not believe anyone thinks linear algebra is intuitive/makes sense

pretty sure that's graduate level @MAFIA36790 ...

cause in linear algebra I always just had to prove what was instinctively clear - diff geo held a few surprises for me and well .... LA was my first year of studies, diff geo was later