The h Bar

12:59 AM

Most funny comment on the main site the last 24 hours:

Are you accusing OP of having a hole in his head? — Ross Millikan 7 hours ago

1:32 AM

@Qmechanic I just left a flag on physics.stackexchange.com/a/101998/22494 but I've realized that the issue was actually a mistake on my end. Sorry :-/

Brandon Enright

1:43 AM

Is it true that the orbital angular momentum of an electron in a hydrogen atom is zero? I thought there was both an electron spin angular momentum and orbital angular momentum.

dmckee's answer here makes me think otherwise:

14

A: Why do electrons occupy the space around nuclei, and not collide with them?

In fact the electrons (at least those in s-shells) do spend some non-trivial time inside the nucleus. The reason they spend a lot of time outside the nucleus is essentially quantum mechanical. To use too simple an explanation their momentum is restricted to a range consistent with begin captured...

Is that because the 'orbit' isn't really like a planetary orbit but rather due to position uncertainty?

user54412

1:58 AM

@BrandonEnright the ground state is 1s^1, and s-orbitals have 0 angular momentum

user54412

excited hydrogen can have orbital angular momentum

Brandon Enright

@ChrisWhite Does that imply their path / position is pretty much random an not orbital in any real sense of a classical idea of an orbit?

Or am I thinking about the electron being localized to a point and even that view is just wrong?

user54412

you can view the wavefunction in position space, where the probability of finding the electron in a region V is $\int_V |\psi(x)^2| dx$

user54412

for the ground state, $\psi(x) = (1/(\pi a_0^3))^{1/2} e^{-r/a_0}$

user54412

so it's position (or rather the probability distribution of its position) is spherically symmetric, simply decaying exponentially away from the origin

Brandon Enright

2:05 AM

And it's best to view that as a uniform distribution where the electron is in all places at once weighted by some probability rather than a localized electron zooming around so fast that it seems like it could be anywhere? Because if it were zooming around fast you'd expect some angular momentum but if it were really spread out it some probability distribution then you wouldn't

user54412

the first view is the best one

user54412

but if you want the second view, realize that an electron that has been localized (via measurement or theoretical fiat) is not in the energy ground state anymore

user54412

so sure, there is some angular momentum in a sense once you localize it, but the angular momentum of the undisturbed ground state must be a fair weighted average of all those potential angular momenta you get after localizing the electron

user54412

and spherical symmetry tells you this average must come out to 0

Brandon Enright

Alright so allow me to continue to make a fool of myself with a follow up question :-) You said an excited electron in a hydrogen atom can have orbital angular momentum. Can that be viewed as a non-uniform probability distribution where the peek (most likely place to find the electron) is moving around the nucleus?

user54412

2:10 AM

hmm, I may be digging myself into a hole here...

user54412

consider the 2p orbital

Brandon Enright

Well I'm trying to shed my classical views of these things but I never quite know if my mental models are convenient lies or they actually somewhat accurately describe things.

is 2p where there are two lobes like dumbbells?

user54412

yeah - but be careful, many of those pictures are misleading

user54412

wikipedia has a decent representation

user54412

Brandon Enright

2:14 AM

Something like:

user54412

the lobes are rather broader than most textbook cartoons make them out to be

user54412

yeah, that works

Brandon Enright

So that's clearly mirror symmetrical and rotationally symmetrical about a specific axis

but not spherically symmetrical

So it's that asymmetry alone that explains an angular momentum?

user54412

ah, so that's where the representations break down again

user54412

fix a spherical coordinate system, with $\phi$ going around the equator and $\theta$ measuring latitude

user54412

2:18 AM

you can write the m=0 (call it 2p_z) part as something proportional to $r e^{-r/2a_0} \cos(\theta)$

user54412

then the m=+/-1 parts are something like $r e^{-r/2a_0} \sin(\theta) e^{+/-i\theta}$

user54412

note the complex exponential

user54412

if you take the square magnitude of any of these functions, call it the probability density, and plot a surface of (1) constant probability density that (2) encloses, say, 90% of the total probability, then you get the lobes plotted

Brandon Enright

For my image. For yours you just assign a color to the p-value

user54412

but this was arbitrarily imposing a choice of how to decompose the full l=2 wavefunction - we chose to break it up into the eigenfunctions of $L_z$

user54412

2:21 AM

@BrandonEnright yes

user54412

if you add all three wavefunctions together (with the proper normalizations), you get a spherically symmetric wavefunction

user54412

in this sense: depending on you choice of overall phase and coordinates, different parts might be real or imaginary, but the square magnitude of this wavefunction gives a spherically symmetric probability distribution in space

Brandon Enright

And this is for 2p? Because those images aren't spherically symmetric.

Sorry, I'm trying to get it. This is out of my comfort zone though.

user54412

yeah - we took a spherically symmetric thing and split it into 3 parts, none of which are spherically symmetric

user54412

the trick is, the 1s orbital is an eigenstate of not only the Hamiltonian (i.e. energy) and total angular momentum (L^2), but also angular momentum along any fixed axis (typically L_z)

user54412

2:32 AM

2p as a whole, while also an eigenstate of H and L^2, is not an eigenstate of L_z

Brandon Enright

Yikes. Despite your efforts I'm pretty lost. When I look at the image I posted and I imagine adding the $2p_x$ with $2p_y$ and $2p_z$ the result is definitely not spherically symmetric. I must have missed something.

user54412

for any fixed definition of z-axis, you can measure an electron initially in 2p and you will find it's z-angular momentum will be either +1, -1, or 0

user54412

@BrandonEnright that's because those images took the square magnitude already - you have to work from the complex formulas I gave, add them, and then take the square magnitude

Brandon Enright

@ChrisWhite Ah that makes sense. I hadn't thought about that.

user54412

but of course, visualizing a complex function of 3D space is pretty challenging, so no one makes pictures like that

Brandon Enright

2:36 AM

So in the 2p orbital is the electron in some superposition of all three of those probability distributions?

And when you measure it, you find it to be in some specific one with either +1, 0, -1 angular momentum?

user54412

yes - if you (thought)-experimentally place an electron in 2p without preference for its "substates" I'll call them, then you might measure it to be in 2p_x, 2p_y, or 2p_z (or whatever names you give them)

user54412

hmm, forget my specific connections to which wavefunctions go to which values - I don't want to state the wrong thing

user54412

by analogy with light polarization, my guess is 0 corresponds to 2p_z in your diagram, while +/- 1 correspond to things like 2p_x +/- 2p_y

user54412

that is, measuring with L_z only guarantees you'll collapse to an eigenstate of L_z, and I can't remember whether 2p_x/2p_y are themselves such eigenstates or only the right linear combinations are - but that's just details

Brandon Enright

Details that are mostly over my head.

But I think I've understood at least some of this :-)

user54412

2:56 AM

news.sciencemag.org/funding/2014/03/…

user54412

I guess it's good I'm doing computational work and not fusion

user54412

actually, the budget explains the flood of papers and press releases I've seen recently from worried areas - atmospheric observatories, fusion projects, etc

Brandon Enright

I have a friend working for General Atomics in San Diego. He does plasma fusion flow / stability computations. He was furloughed when in the last government shutdown and when that ended, they started gutting his department due to much lower funding.

Another physics quack giveaway: quacks seem obsessed with the history of discoveries and the drama of the people behind those discoveries. It's like they feel that because they've tuned into some soap opera drama about the people behind the science, they've somehow gained some insight into which science is right and which is wrong.

user54412

shame about fusion - such good science and engineering there - but I guess being a limitless source of clean power isn't sufficient to warrant the "green" label

user54412

but I guess we learned that lesson from fission already

Brandon Enright

3:09 AM

Ignorance and Cost are the Captain Planet of energy research. With their powers combined pretty much anything fission / fusion is screwed.

►

user54412

A few years ago, Obama mentioned Caltech in the State of the Union. Everyone was so excited, because he might be the first president to know the school existed. And since his comment was something about leading the way in environmental stuff (that really confused all of us), the school immediately built 2 or 3 new buildings dedicated to green power.

3

user54412

This is on a 1-square block campus, mind you. And as far as I can tell there were about 2 professors working on green energy, so I guess they each have a building now.

dmckee

4:00 AM

@BrandonEnright Yes, the S orbitals are all spherically symmetric (zero orbital angular momentum, and the electron's spin does not much affect the shape of the distribution) and have a anti-node at the center.

And no, you should not think of a localized electron moving along a path...the orbital is not a eigenstate of either position of momentum, so the particle does not have a fixed value of either \vec{x} or \vec{p}, just distributions.

Physics Meta

9:16 AM

0

Q: Whats Worse: Posting a HW Question or Answering a HW Question

I have very recently started participating in SE, and I have (embarrassingly) answered some the HW questions. After seeing some the meta posts of the more experienced users and mods, I now feel pretty bad about answering HW questions (Sorry!!). I have never posted any HW questions either. So my ...

Arafat

3:45 PM

Does every quantum state is a minimum uncertainty state?

ManishEarth

4:33 PM

@Arafat Uh, no, in the harmonic oscillator for example only the ground state is minimum uncertainity

Physics Meta

5:18 PM

1

Q: Are we clear on what the "not an answer" flag is for?

I've noticed a bunch of disputed not-an-answer flags in the queue recently. While it's great that people are actively flagging things, this trend does suggest some disagreement on or misunderstanding of what these flags should be used for. Is that really the case? Are people clear on the purpose ...

discussion flagging

Waffle's Crazy Peanut

5:45 PM

@BrandonEnright Now, I have an evil Steward for CVs... :D

Physics Meta

9:18 PM

0

Q: Why can't I see rendered MathJax/LaTeX when reviewing edits?

Posts from new users often requiring edits that add LaTeX code. When I review edits from users that add such code, all I see on the LHS and RHS are the raw LaTeX codes, such as $a_y=g$. This is the case even after I select the "rendered" view button. I'd rather just compare the two rendered outpu...

feature-request mathjax reviews

Brandon Enright

9:50 PM

@ChrisWhite per our type 1a supernova discussion, this is in the news recently: today.lbl.gov/2014/03/04/…

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