hello, i am wondering why exactly the WKB approximation is needed? i get the process - using it to deal with a spatially varying potential, taking a slowly varying potential, and and solving for a wave function under these conditions. however, i am confused why this is needed to piece solutions together around "turning points"? i see that turning points are points where the energy is equal to the potential energy, [...]

[...] but i am not seeing the connection between these and needing a semiclassical approximation to solve the schrodinger equation in these regions

is f(x) a constnat?

@SillyGoose in what context?

@SillyGoose what is a constnat

honk

it's a blueb

honk

Hi All...

hi

I found this oldie but goldie: Stephen Hawking reveals formula for World Cup success RIP.

Hello @JohnRennie Sir

Hi All...

@Relativisticcucumber The WKB is an approximation method used to solve 1D TISE. The approximation can be applied only in some given limits. The approximation works well in the classically allowed region and it turns out it can be used also in the classically forbidden region. In such regions one finds $\psi(x)\propto\frac{1}{{\lvert\sqrt{2m[E-V(x)]}\lvert}^{\frac{1}{2}}}\exp\left\{\pm\frac{i}{\hbar}\int^x\{\lvert\sqrt{2m[E-V(x')]}\lvert dx'\right\}$, which as you can see breaks at turning points $E=V(x)$

The reason why it breaks is that the approximation ceases to be legit around such points (the denominator is zero), so around such point you need to find the wavefunction using some other method and the patch the wavefunction at such points

Why is $G$ given the same importance as $c$ and $\hbar$? $c$ and $\hbar$ are tied to the geometry of the universe. $G$ just seems to be an interaction strength like the fine structure constant

@RyderRude $G$ is not like the fine-structure constant because it is dimensionful

in GR terms, it's just a proportionality factor between the curvature and the stress-energy, since the Einstein equations (with $c=1$) are $R_{\mu\nu} = 8\pi G T_{\mu\nu}$

This isn't an interaction strength, it's just saying that $R$ and $T$ are measured in different units and we need a conversion factor

setting $G=1$ means we measure space/time (in $R$) and mass/energy (in $T$) with the same units

much like $c=1$ means measuring time and space in the same units

@ACuriousMind isn't it a co incidence of 4 four dimensions that the fine structure constant is a unit-less quantity? Otherwise, the fine structure constant would also represent a unit-conversion factor between acceleration and electromagnetic field?

@RyderRude I don't know why the fine structure constant would be a conversion constant between acceleration and the EM field

in QED terms, it's just $\frac{e^2}{4\pi}$ where $e$ is the dimensionless coupling between the EM 4-potential and charged fermion fields

and there are plenty of ratios it's equals to that are manifestly independent of dimension, see e.g. Wiki - all these energy ratios work in any dimension

@ACuriousMind is e still dimensionless even in a theory with, say, 5 space dimensions?

yes

@ACuriousMind oh, i misread this somewhere

what changes between dimensions is the mass dimension of the fermion field since the kinetic term $\bar\psi\gamma^\mu D_\mu\psi$ needs to have the right mass dimension

Yeah

So $G$ is special in this way? What about the $\lambda$ from $\phi ^4$ theory?

Is $\lambda$ also dimensionless

the $\lambda$ changes its mass dimension

It doesn't have a fixed dimension?

But dimension gets fixed from the Lagrangian

no - that's also the reason why the renormalization of $\phi^4$ in 4d is different from other dimensions (triviality and all that)

@RyderRude its dimension is determined by the dimension of $\phi^4$, and the dimension of $\phi^4$ varies with the spacetime dimension

Oh

So, we can that $\lambda$ is also kind of a unit-conversion factor like $G$

no, not exactly

but you would need to construct a different dimensionless constant to properly represent the coupling strength there

$G$ is not a coupling constant, it's just a pre-factor of the Einstein-Hilbert action

Oh, so coupling constants are dimensionless

I wouldn't say that.

coupling constants are just things that stand in front of coupling terms in the Lagrangian

@ACuriousMind you said that we'd have to construct a different dimensionless constant to properly represent the coupling

by "properly" I mean in the sense so that we can really talk about the value and changes in it

as long as you just do theory it's fine to not worry about that

but when you want to talk in terms of measurements and changes in values, dimensionless quantities are kind of necessary, see physics.stackexchange.com/a/176296/50583

I used to think that $G$ was a coupling constant. But now I think it's also tied to geometry of the universe

I mean G is a coupling constant in some sense

You can set it to 0 and just not have gravity

The difference between G and the fine structure constant is that there is no (as far as we know) discrete charge for gravity

Otherwise $\epsilon_0$ is a dimensionful constant, too

Also there is something about how gravity's coupling is like a mass term which is another thing

@Slereah if we focus on classical field theory, then the charge for electromagnetism is also not discrete, right?

Sure, you can just have any old charge

Unless you impose it by hand

The fine structure constant being dimensionless relies on the fact that we have a specific charge that sets the strength of the interaction for everything

In introductory physics, $G$ and $k$ from Coulomb's law play a similar role, as unit conversion factors

Although the fact that gravity's coupling is to energy makes things more complicated because those kinds of units are fundamentally different from other charges

@Slereah what specific charge? You mean the fact that all charge is multiples of the electron charge?

yeah

I will have to read more about this. I have paused my qft study for months

i.imgur.com/OEGyVed.png this wording is confusing me. Isn't the diameter of the spherical mirror necessarily twice the radius of curvature..?

@Obliv I think "diameter" is a poor choice of words there - I'm pretty sure they mean that the size of the mirror is supposed to be small against the radius of curvature (the size is a "diameter" if your spherical mirror is a circular piece of the surface of a sphere...)

Ok that makes more sense. Two sources now that I've come across giving a vague description of that. I wonder what the actual size relation is called

If it's related to the radius of curvature then I'm guessing it's a percentage of the spherical mirror

the key point of the approximation they want you to make is that the incident rays form small angles with the normals of the mirror surface

the exercise is really worded terribly because "large" spherical mirrors in this sense don't even have a unique focal point

yeah it's a focal axis

depending on the orientation of the mirror to the rays.. but I'm having trouble understanding how the size of the mirror changes its ability to focus the rays

like can't you change the angle of incidence to the normal by orienting the rays differently from being parallel to the principal axis

generally we define the focal point to be where rays parallel to the principal axis get focused, no?

so like even if you orient the rays like this, you have a "focal point"

if you orient it to a more exaggerated angle does it continue to focus on this axis of focal points

in that case the approximation they want you to make involves both the size of the mirror and the direction of the incident rays

this is not evident from the exercise text but in order to derive the $f = R/2$ you need that the rays make small angles with the mirror normal over the whole mirror. The usual assumption for this is to make the incident rays parallel to the principal axis and the size of the mirror small against the radius of curvature

that's what "the focal length" of the mirror refers to

Oh I see

Would it suffice to prove that two rays meet at the same point through some geometry

nvm probably have to prove for n rays

@Feynman_00 ah i see, and can you clarify what the "classically forbidden region" refers to? + thanks so much for the response !

@Slereah may i ask what your specific focus/interest is in gravitational theory? this is what you research, right?

@Relativisticcucumber Mostly differential geometry stuff

do you have a specific focus within differential geometry @Slereah

Lately looking into structures and such

in the context of relativity or just purely maths ? @Slereah

in the context of relativity yeah

@Relativisticcucumber A region inside which $E>V(x)\quad\forall x$. The turning points are the boundary of such region and beyond them you are in the classically forbidden region

@Feynman_00 you mean $E<V(x)$ - the forbidden regions are the regions where the particle has less energy than it would need classical potential energy to get there

Oh ok, I read again and I thought they were asking for the allowed region :P

In our lecture today we were dealing with the shapes of different orbitals. The lecturer said the following: If we have the bound state(eigenstate of H,L^2,L_z)$\phi_{n,l,m}(r,\theta,\phi)$ and we have a fixed r, then the shape is dependent from the spherical harmonics, particularly the $\theta$ angle.

If I remember correctly the l=1 orbital, in one direction has the form of a sand clock

how exactly is the distance r, here constant?

@imbAF you need to read these visualizations together with their definition. See e.g. the caption of the spherical harmonic pictures on Wiki - the radial distance in these pictures is a measure of the value of the function $Y(\theta,\phi)$ in the $\theta,\phi$ direction.

That is what I have been looking at actually

but how is the distance r constant in let's say l=1 in z axis orientation

that shape is like a bottleneck in the nucleus

this is not a picture in 3d space

what?

btw which section should I look at in the link you provided?

as the caption says: The distance of the surface from the origin indicates the absolute value of ${\displaystyle Y_{\ell }^{m}(\theta ,\varphi )}$ in angular direction $(\theta,\varphi)$

i.e. the "radial direction" here is not the radial direction in 3d space, it's the axis on which the value of the spherical harmonic is represented

@imbAF right at the caption of the picture with the "shapes" of the harmonics at the top of the article

I am trying to understand it currectly

why tf is how difficult

to understand

One thing r, \phi and \theta

work like how they normally work in spherical coordinates, right?

there is no r in the picture for the spherical harmonics

the spherical harmonics are functions on the sphere, i.e. only of $\theta$ and $\phi$

yes

hold on i gotta read what you wrote again

The distance of the surface from the origin indicates the absolute value of {\displaystyle Y_{\ell }^{m}(\theta ,\varphi )}{\displaystyle Y_{\ell }^{m}(\theta ,\varphi )} in angular direction {\displaystyle (\theta ,\varphi )}(\theta,\varphi).

I kinda get it but one thing confuses me

So because I am incapable to put it into words, I'll try to explain it with an example

for n=2, you have l=1, and m=-1,0,1, these are the 3 orientations of the 3 orbitals p

but the statement above is

in angular direction...

we can have whatever phi and theta angle

but the p_x,p_y,p_z are only in 3 particular direction

isn't that somehow not the same?

@ACuriousMind This is a statement I found in a book: To visualize their angular dependence (of the eigenfunctions),we can measure a distance on the axis characterized by the polar angles theta and phi which is prop to the absolute value of the eigenstate for any fixed r, that is prop. to the spherical harmonics

Now I understand that particular values of theta and phi, characterize an axis

but why it's important to specify that r is fixed>

?

again, there is no r when we're just talking about spherical harmonics

I mean

the "constant r" you read comes from the part where we write a solution to the Schrödinger equation in 3d space $f(r,\theta,\phi)$ as the product of some radial function and a spherical part $f_r(r)Y(\theta,\phi)$

now at constant r this is a function on the sphere and can be drawn like the spherical harmonics

$\phi_{n,l,m}(r,\theta,\phi)=R_{n,l}(r)Y_l^m(\theta,\phi)$ in the square amplitude the radial component is there

" this is a function..." this=?

the $f(r,\theta,\phi) = f_r(r)Y(\theta,\phi)$, there is no other function in that message :P

Maybe the problem is that you don't understand why we're trying to visualize only parts of this function: Do you understand why we can't really draw the full wavefunction $f(r,\theta,\phi)$? It's a function in 3d space, so its graph would have to be drawn in 4 dimensions, which we can't draw.

why can't you? I am sorry, but when i read r,theta,phi I simply think of spherical coordinates

is it because we have 3 variables

@imbAF they are spherical coordinates

but how are you going to draw a function that assigns a value to every point in 3d space?

which give us a value

as a forth quantity

this isn't a problem with "spherical coordinates", you would have the same difficulty with $f(x,y,z)$ in cartesian coordinates

so basically you have x, and you get an y

you have x and y and you get a z in 3d

so because we can't draw this function, we try to draw its component functions $f_r(r)$ and $Y(\theta,\phi)$. Drawing $f_r(r)$ is easy, it's just a function on $\mathbb{R}$

you have a x,y,z and you get a value in 4d?

@imbAF yes, that is what I mean when I say the graph would have to be drawn in 4 dimensions

I get why iss hard

I can't visually understand why is hard, and what i mean by that

I can easily picture

the process of going x to y, x and y to z, and I can understand that you cannot picture going from x,y,z to w

I can't graphically/visually picture that with the bound states,

but what does r in the eigenstates represent physically?

but drawing $Y(\theta,\phi)$ is still a bit hard, because $(\theta,\phi)$ is a point on the 2-sphere or equivalently a direction in 3d space (the direction of the vector pointing at that point on the 2-sphere). So the way we draw this as a 3d graph is by drawing a point in the direction of $(\theta,\phi)$ at distance $\lvert Y(\theta,\phi)\rvert$. These points then form the surfaces you see in the pictures

@imbAF I don't understand the question. What does $x$ in a 1d wavefunction $\psi(x)$ "represent physically"?

Can't put it into words

then how am I supposed to answer the question :P

cuz I thought you would be able to put it into words lol

I didn't know it was impossible

But for example, theta and phi, represent a point in space, for fixed r

it's easy to be put into words

what

I mean, yes, those are words, but I don't really get what they're supposed to mean :P

and btw I get it now, the distance from the nucleus depends on the square of the spherical harmonics

@ACuriousMind let's dumb it down, with an example

you have n=1,l=0, the region the electron can be found has a spherical shape

is that a correct statement?

no, the region the electron can be found (support of the wave function) is the whole space

the probability density has spherical symmetry

ahaaa

@fqq that would be for the electron $|phi_{1,0,0}|^2$ right?

@imbAF That's what I meant when I said the pictures for the spherical harmonics are not pictures in 3d space!

They are not regions in 3d space

this is not a picture of where you'll find the electron

but it is?

probability density shape?

this is a picture of in which directions you are more likely to find the electron, but it is only the directional part

again, this does not depict the radial dependence $f_r(r)$

one sec

you use the word picture

what do you mean

the shape i see in wikipedia :P ?

for the spherical harmonics>

Ok, what?

isn't this the shape of the region in which the electron might be

with a certain probability

no

dependent from theta and phi, for fixed radial component ?

but fqq said that (n=1,l=0) the probability has spherical shape

no they didn't

ah I said region,that's why you said no ?

What should i understand with this?

saying the probability density has spherical symmetry is not the same as saying that the region where you can find the electron is a sphere

so what it;s meant with probability density has spherical symmetry ?

it's just saying that there is no particular direction in which you are more likely to find the electron than in any other

the density extends over the whole of space, it doesn't have a boundary

and for l=1, you say what

maybe this plot (also from Wiki) helps bring the point across:

the density extends over the whole space,it doesnt have a boundary

but what about the direction?

this is plotting the same functions as the 3d picture with the dumbbell shapes above (just for higher $\ell$

it's just plotting the value of the function as color instead of distance from the origin

would saying the probability density is isotropic make more sense

I am trying to understand the picture

we have no info about l here right?

it's only n and m

the captions are bad, the "n" here is $\ell$

could you elaborate what is happening, I am not that good with this

What is n and m here? Maybe if you explain it to me it'll help you @imbaF

you mean l and m

yeah

l would be the 5th orbital (meaning n=6 or more) and m are the mangetic q. numbers, which are the possible orientations of the orbital l=5

@Obliv we're plotting spherical harmonics $Y_\ell^m(\theta,\phi)$. Since imbAF keeps misinterpreting the representation as 3d shapes, I posted a color plot instead (the "n" in the color plot is $\ell$)

in order to drive the point home that the 3d shapes plot is not to be interpreted as literal regions in three-dimensional space

I really don't know what else to say, the shapes plot plots the value of $Y$ as distance from the origin, the color plot plots the value as color

neither of these are directly related to any shape in $r,\theta,\phi$-space

so color is used to determine how positive/negative the angular momentum values are

no, the value of $Y$ is not an "angular momentum value"

oh idk what the function represents, but the color is representing how positive/negative it is

the $Y_\ell^m$ are eigenfunctions of the angular momentum operator, but that doesn't mean their values as functions are angular momenta

that goes back to what I initially said: The shapes are not a picture in normal 3d space

they're a picture in "function value space", where the radius is about the value of a function, not about spatial distance in the physical space we use the $r,\theta,\phi$ coordinates for

It's a representation of how the function changes w.r.t. phi/theta

ahaaa

function value space

no idea what that means, but at least is a start

that's not a technical term, I just made that up :P

if i misinterpret the representation as 3d shapes of the harmonics, what is the correct interpretation, possible directions where we might find the particle

the larger the shape in a certain direction is, the more likely you are to find the particle when you look along that direction

but the distance at which you are more or less likely to find the particle has nothing to do with this shape, since that information is encoded in the radial part $f_r(r)$ of the wavefunction $\psi(r,\theta,\phi) = f_r(r)Y(\theta,\phi)$ and we're not considering that at all here

the distance from what?

nucleus?

the distance from the nucleus (or whatever the particle is orbiting)

on wiki it says you can denote this as $Y_\ell^m(\theta,\phi)$ or $Y_\ell^m(r)$ in the second one r is a vector?

oh it's the unit vector makes sense.

So I can probably imagine this: for a theta and phi, there is a particular direction (if I can express myself like this) related to the probability density. then the radial component "enlarges "this, like for example n=1 l=0 , and n=5 l=0, what is one thing that is different from these? phi and theta are same, it should be the radial part no?

I believe no. Maybe the radial part has to do with some exponential decrease in probability density, I am not sure

I neither know what you mean by "there is a particular direction" nor what "enlarging" means here, but yes the radial part is the only difference between $n=1$ and $n=5$ when both have $\ell = 0$

and this difference in r how does it affect the probability density ?

or does it affect at all?

I mean, the full probability density is just $f_r(r)Y(\theta,\phi)$

the two $f_r(r)$ are different (that's what "the radial part" means"), so the probability density is different

you'll have to look at plots for $f_r(r)$ for $n=1$ and $n=5$ to see how they're different

well in the directional sense there should be no difference

as you said

@ACuriousMind this

@imbAF sure - at $\ell = 0$ this just mean in both cases you are equally likely to find the particle when you look along a line from the origin no matter the direction from the line

what differs is - again - the distance at which you are likely to find it

the distance as in a physical sense

or in the function value space?

yes, I mean the physical distance from the center of the orbit, that is what the radial $f_r(r)$ describes - the probability to find the particle at a certain distance from the center

so r helps you narrow the radial region and the angles help you with which direction there is a higher probability of finding the particle,if i can say so

I'm not sure what that means

these are just probability densities, they don't "help" me with anything

in theory we said that the particle can be at any point in space

@ACuriousMind ok

the goal here, in general, is not to maximize the probability to detect the particle or anything like that

we're just looking at a quantum mechanical wavefunction

Different harmonics, represent different cluster of points/section/region of the 2d sphere?

in a 3d representation?

nope

each harmonic is a function on the sphere

I don't know what you mean by them "representing a region" of the sphere

they're just functions on it

you said : " So the way we draw this as a 3d graph is by drawing a point in the direction of $(\theta,\phi)$ at distance $\lvert Y(\theta,\phi)\rvert$. These points then form the surfaces you see in the pictures"

$(\theta,\phi)$ is a point on the 2-sphere or equivalently a direction in 3d space

th 2 sphere is the sphere in 3D

right?

yes

ah

I forgot

distance $\lvert Y(\theta,\phi)\rvert$

the distance varies

One last thing because I want to be able to express myself correctly

correct me if I'm wrong but there is no distance component in this

it's just a sphere of arbitrary radius that's color coded by positive or negative values of the function

youtube.com/watch?v=5PMqf3Hj-Aw this seems to be a good resource

@Obliv we're now talking about the "shapes plot" again, not the color plot :P

oh lol

(or at least I hope we are because the stuff imbAF cited was what I said about the shapes plot :P)

the shapes plot are those electron orbital pictures from before?

@Obliv yes

And I said: so r helps you narrow the radial region and the angles help you with which direction there is a higher probability of finding the particle,if i can say so

How is my statement different then your 2 statements

it's different in that I have no idea what you mean by "help" or "narrow"

maybe you mean the same as what I said, but I can't understand that just from what you wrote

wait, so we can be on the same page about it

@ACuriousMind is the probability described by the radial component at a certain distnace from the center, the one with the highest value

I don't understand what you mean by "the one with the highest value", either

the highest value among what?

among other values of the probability for different values of the distance from the center

if the radius starts from 0 to infinity

for every distance value a probability should be associated with

the radial part is a function $f_r(r)$

$r$ can have any value to want - you put a value in, it spits a value out

$\int_{r_1}^{r_2} f^2_r(r) r^2\mathrm{d}r$ is the probability to find the particle at a distance between $r_1$ and $r_2$

but if you have n=1,l=0 and n=5 and l=0, I don't think you'd get the same value for the radial component

for a given r value

of course not, if you got the same value they'd be the same functions!

I don't understand why we're discussing the elementary notion that two different functions should have different values at at least one point

so that for a given r value

we get two different values, meaning two different probabilities

radial part inputs a direction and outputs a value where the larger it is the higher the probability?

@imbAF probability densities

I'm really not in the mood for this to digress into another discussion about what the difference between a probability density and a probability is :P

to put it simply one can with certainty say that the electron in the H atom is found at a certain region/distance from the protone

somehow in some way, there should exist a mathematical function/process idk the name, that would showcase that

to put it simply that is wrong and I have no idea how you arrived at that conclusion

the probability to find an electron inside the nucleus is non-zero, at least for some orbitals!

so the electron of a H atom can be in the moon while the proton is in lab? there is a probability for this to hold true?

@ACuriousMind ah yes okay got it thanks

you should input those values @imbaF

@imbAF nothing we've said so far implies it can't!

distance component being the moon's distance from earth, and the direction as well

no nothing implies it can't, but somehow there must be a most probable region where the electron is

and I assume there should exist a way to tell where this region is

isn't it described by the functions you're talking about?

sure, usually it's more probabie to find it closer to the nucleus than the moon :P

yes

I can say that, because....

@imbAF That'S what the probability density is for!

you integrate it over a region and it tells you how likely it is to find the particle inside that region

Well that is what I am trying to say

@ACuriousMind you said

@ACuriousMind this one

I stand by that

I still don't know what you meant by "help"

aren't you using the pdf to draw informations?

a tool cannot help you, only you can. - obliv 2022

if you just wanted to state that the probability density tells us how likely it is to find the particle somewhere, then...yes, that's the definition of the probability density

I'm confused why you think this needs to be explicitly stated in this context

and you didn't say "the probability density helps me to find the particle", you said " r helps you narrow the radial region" which I still don't understand

@ACuriousMind I was trying to say with this that there must be some physical difference between different radial functions i.e n=1/l=0 and n=5/l=0

physical difference= in spatial terms

But since I don't have a clear understanding of all of this, I might be off and don't understand even why

@ImbaF what are n and l?

energy level and sub energy level

$Y_\ell^m(\theta, \phi)$ gives you the eigenstate of the angular momentum operator (whatever that means) which I get, but what you're referring to as radial functions I do not.

Are you referring to this same function?

I know it's wrong to say, but in that example I picture two spheres with different radis

can you tell me the function or give me a link im curious

that takes in the n,l values

@Obliv eigenstates of an electron in central potential have a radial component too, an eigenstate looks like $\phi_{n,m.l}=R_{n,l}(r)Y_l^m(\theta, \phi)$

okay, let's try one final visualization:

(from en.wikipedia.org/wiki/…)

this is a section through the x-z plane

ok

orbitals and the levels too

the center is the nucleus, and the brighter the pixel is, the more likely you are to find the electron at that position

the left column is $\ell = 0$

ok

so top left is n=1/l=0 and bottom left is n=3/l=0

yes