The h Bar

in particular there are dark bands at higher $n$ where the electron is less likely to be found at a particular distance and then becomes more likely again

Obliv

that is interesting..

ACuriousMind

and in principle this probability is non-zero even as $r\to\infty$, it's just hard to draw because it falls off exponentially and we can't see arbitrarily low brightness

there is no bounded region inside which you'd find the electron with 100% probability

imbAF

the exponential fall is a result of the radial part?

ACuriousMind

yes

but what is true is that at higher n - except for the bands - the region where it's "pretty likely" to find the electron is larger

imbAF

12:04 AM

does it mean that the pdf decreases with increasing n?

ACuriousMind

I don't know what that means

Obliv

n can only be natural numbers correct?

imbAF

@Obliv yes

ACuriousMind

the total integral over all of space is always 1, that's the definition of a density

imbAF

Right

just so i can confirm for 2s we see one black band

and for 3s at least 2?

or am i wrong?

ACuriousMind

12:06 AM

yes

these are called "nodes" of the radial part and there are always $n-\ell - 1$ nodes

imbAF

hmm

is this similar to

n=k+l ?

ACuriousMind

what

imbAF

I don't know any link to this

but forget it

I wasn't aware of these nodes

but that is what it is actually

But at least I understood what the the spherical part represents

@ACuriousMind the direction in which it is most likely to find the particle,as you said here

@ACuriousMind for l=0 orbits right?

Obliv

12:29 AM

@imbaF i'm not sure if the notation you're using is the same as in wiki., but the azimuthal quantum number $\ell$ is given by the letters at the top of each column in that last picture he linked.

imbAF

I was thinking something else

Obliv

which he specified to be $\ell = 0$ which I don't see why.

imbAF

I call it total orbital angular momentum quantum number :P

Obliv

I guess $\ell = 0$ is the $s$ orbital.

imbAF

yes I am aware

Obliv

12:31 AM

that last picture he linked depicts the probability of finding an electron based on the brightness of the pixel.

imbAF

It's just that he also said that there are orbitals where there is a non-zero probability that the electron is in the nucleus, and these are the l=0 orbitals. But I wonder how that doesn't collapse the atom?

Obliv

Why would that "collapse" the atom?

imbAF

Because the positive nucleus and the negative electron would attract each other

I don't know actually what would happen/happens when the electron is located in the nucleus

Obliv

I don't know enough about QM to tell you :P

ACuriousMind

@imbAF quantum mechanics is not classical mechanics - just because the probability is non-zero doesn't mean the electron "is" there in the classical sense

and no, it's not just for $\ell = 0$

imbAF

12:36 AM

oh not only for l=0

Obliv

if anything, it'd be even weirder if an electron couldn't exist in the nucleus (from a laymann's point of view)

but I don't know how you'd prove or disprove anything to do with electron orbitals via experiment.

imbAF

One more thing that i noticed $|Y_l^m|^2$ we know that we have the term $e^{i\phi..}$ in the spherical harmonics, and as a result the square amplitude is not a function of $\phi$. Is there any intuitive meaning to this @ACuriousMind ?

@Obliv well that's what the orbitals are supposed to represent, regions where the electron might be, if it's correct to say so

ACuriousMind

@imbAF The $Y_\ell^m$ are eigenfunctions of the $L_z$ operator, and the $L_z$ operator generates rotations in $\phi$. So that means that under a rotation by $\phi_0$, $Y_\ell^m(\theta,\phi + \phi_0) = \mathrm{e}^{\mathrm{i}m\phi_0}Y_\ell^m(\theta,\phi)$, so they can't have any other dependence on $\phi$ except through a phase

imbAF

but can one also say that the distance |Y| does not depend from phi?

since we said that $\lvert Y(\theta,\phi)\rvert$ is the distance

ACuriousMind

12:56 AM

the picture is actually drawing the real spherical harmonics with $\cos(m\phi)$ instead of $\mathrm{e}^{\mathrm{i}m\phi}$

imbAF

I somehow thought there was something off about this

so if we would draw the $\mathrm{e}^{\mathrm{i}m\phi}$

would that be an intersection

or am overcomplicating the things

ACuriousMind

if you just draw $\lvert Y\rvert$ for the $\mathrm{e}^{\mathrm{i}m\phi}$ they just all look like the $m=0$ version :P

well, no, not exactly

but they would all be symmetric around the z-axis

imbAF

so an intersection of the 3d picture by either of the xoz or yoz planes

And i guess this symmetry around the z axis would always be the case, since always regardless of phi value $\lvert Y\rvert$ proportional to some function of Theta

Anyway thank you for your time. I definitely understand more then before

PM 2Ring

3:34 AM

@JohnRennie There's a fairly new Greg Egan short story, available free online. gregegan.net/BIBLIOGRAPHY/Online.html clarkesworldmagazine.com/egan_04_22 (although financial contributions to Clarkesworld are welcome). Dream Factory is set in the very near future. It involves YouTube, Bluetooth, and cats.

John Rennie

5:20 AM

@PM2Ring Thanks, I'll take a look :-)

123

6:22 AM

Hello @JohnRennie Sir

John Rennie

Hi :-)

123

@JohnRennie Sir If your time permit. Can we discuss further?

John Rennie

Let's switch to the problem solving room

123

@JohnRennie Aye Sir

Shiki Ryougi

10:19 AM

This post: https://physics.stackexchange.com/questions/558238/killing-vectors-of-schwarzschild-metric
Doesn't seem to have an answer that addresses the main point, which is how those constants were specifically chosen to find the 3 killing vectors. I run into a similar problem while trying to evaluate the killing vectors of 2d AdS, where I found the expressions of the vectors but with 3 undetermined coefficients A,B,C (after solving the Killing equation). The author then states that there are 3 independent solutions for such and such choices of A,B,C, without explaining how you can come a…

In other words, what is the condition that leads to the specific choice of the constants?

ACuriousMind

12:14 PM

@ShikiRyougi the Killing vectors form a vector space

so in your solution with three coefficients, any choice of A, B, C is a Killing vector

and since there are three coefficients, you know there are at most three independent solutions

so when you have three different choices of A, B, C that are all independent, you know you've found all of them

there is no "need" to start with (1, 0, 0), (0, 1, 0), (0, 0, 1) as the three choices, but before trying complicated combinations, why not start with these?

Shiki Ryougi

That helped a lot, thanks. I will now try solving it with this in mind, be right back.

imbAF

is there a reason why $|Y_l^m|^2$ is interpreted as the distance of the surface from the origin?

ACuriousMind

it's just how the plot is drawn

there is no more reason for it than for it to be associated with color in the color plot I showed

like, when you draw the graph of a function $f(x)$ in the normal 2d way, there is no "reason" for $f(x)$ to be associated with the distance from the x-axis either

that's just how we draw graphs

imbAF

And when we try to draw by considering all 4 components, we do it by using colors?

Like in the last picture you showed

ACuriousMind

since that picture was a section through the x-z plane, it's not drawing "all 4 components"

imbAF

12:28 PM

ah

ACuriousMind

if you wanted to draw the full probability density at once, you'd have to assign a value (e.g. color) to every point in 3d space

but you can't draw that on a flat screen

imbAF

and in the x-z plane, the coloring is the drawing of the radial component ?

ACuriousMind

13 hours ago, by ACuriousMind

12 hours ago, by ACuriousMind

this is a combination of the radial and directional information

12 hours ago, by ACuriousMind

so this actually shows the probability density

imbAF

@ACuriousMind the directional info should, I assume be shapes we see

ACuriousMind

???

imbAF

12:31 PM

well from n=2 from s to p you can clearly see two different shapes

ACuriousMind

The directional information is literally just that the brightness at a point really does correspond to the probability to detect the particle there

the entire graph is "directional + radial information" because the points in it correspond to the actual spatial points $(r,\theta,\phi)$ in the x-z plane

imbAF

ah

ACuriousMind

in less confusing words, this really is drawing $f_r(r)Y(\theta,\phi)$ instead of just $Y(\theta,\phi)$ like the other pictures did

don't interpret anything more into it

imbAF

ok

the thing is that I never considered the idea of "drawing" a wave function

Never

ACuriousMind

I doubt that

imbAF

12:34 PM

All I considered was the square amplitude of it

that is the pdf

and for an observable

ACuriousMind

don't be like that, the picture is drawing $\lvert f_r(r)Y(\theta,\phi)\rvert^2$, I'm just too lazy to type the bars every time

imbAF

integrating, we will get a interval in which the value of the physical quantity represented by the observable can be

ACuriousMind

of course we're not drawing the complex wavefunction itself, how would a scalar value like brightness or distance correspond to a complex number?

imbAF

it felt natural to talk about the square amplitude when we were in one dimension and plot it, and the graph was a line in the 2d plane. Somehow I find it confusing in 3D

Maybe it had to do with misinterpreting the spherical harmonics as pictures in 3d space

Shiki Ryougi

1:47 PM

It worked (almost) flawlessly. I did find the killing vectors, but there is sign difference on the formulae. For example, I found $K_{(2)}=t\partial_t + a\partial_r$, while the author found $K_{(2)}=t\partial_t - a\partial_r$. This is because I chose (0,0,a) for (A,B,C) while they used (0,0,-a). It seems like I've absorbed the minus sign into the constant a, which was given in the definition of the metric: $ds^2=e^{\frac{2r}{a}}\mathrm{d}t^2+\mathrm{d}r^2. $

If that minus sign is arbitrary, why bother putting it there in the first place?

So I'm thinking that I must have made a mistake in my reasoning

I did check that with $a$ and $-a$ they are both solutions

I.e. (a,0,0), (0,a,0), (0,0,a) and (-a,0,0), (0,-a,0), (0,0,-a) are both linearly independent vectors

ACuriousMind

2:06 PM

@ShikiRyougi the form with the minus might be convenient for some later calculation, or match with some "conventional" way in the literature

I wouldn't worry too much about it

Shiki Ryougi

Great, thank you! ;D

imbAF

chem.libretexts.org/Bookshelves/… Under which conditions or when does the radial component becomes zero? I thought it exponentially decreases

ACuriousMind

@imbAF the page you link already answers that, since they write the radial part as $R(r) = Np(r)\mathrm{e}^{-kr}$. This clearly "generally" exponentially decreases but has zeros where the polynomial $p(r)$ is zero. What more do you need?

imbAF

I see

Tanner - reinstate LGBT people

3:26 PM

I'm curious about something. Did physicists ever attempt to model electrons and protons as spheres with small but non-zero radius, behaving classically?

Obviously, such a model would not have produced accurate predictions.

But I feel like it would be interesting to read about hypotheses like that and exactly where they break down.

ACuriousMind

@Tanner-reinstateLGBTpeople sure, see e.g. Bohr model, plum pudding model

imbAF

5:42 PM

for an orbital p is it correct to say that the probability density has $\pi$ symmetry ?