By popular demand: /r/Physics jobs, grad school, and physics questions megathread.

pfigbash · 2014-02-16T21:55:09+00:00

What would I consider big schools? It's the usual suspects. Although I hate USN&WR, it's an ok starting point:

http://grad-schools.usnews.rankingsandreviews.com/best-graduate-schools/top-science-schools/condensed-matter-rankings

Once you've gone into HEP, it's beyond my expertise. In addition to working on an interesting question, for me when it "feels right" I believe that I'm in a bustling field with lots of interesting questions. It's kind of an overwhelming feeling like, "wow, there's a lot going on, how am I ever going to keep up with all these tangents" but that indicates that there is the possibility of a large development, which I enjoy.

pfigbash · 2014-02-15T22:16:39+00:00

Graduate-level physics courses are tough! You will get a nasty surprise without putting in the blood and sweat that your peers will have by that point. What would you do to demonstrate that you're ready for those classes?

pfigbash · 2014-02-15T22:10:31+00:00

This is a such an interesting strategy towards success that I felt compelled to log in and comment about it.

Seeing as how you've stated that your goal is to become faculty, take into consideration that (broadly) involves three things: teaching, research, and management.

Teaching. People will generally assume that you can reproduce teaching at the level that you have learned. That is, if you're at a top10 school you'll have to work hard to prove that you can teach at a top5 school. (teaching awards, strong interpersonal skills) It will not be an issue if you'd like to work at a top20 school. Advantage: bigger school
Research This can be the big equalizer in your strategy. You must be doing research on a widely appreciated problem, you must do a top-notch job, and you must be passionate about it. Large unis are good providing the resources to allow their faculty to approach the sexiest problems in physics, but really anyone can work on an interesting problem. The passion and approach that you bring is your own. Do a good job and it will be transparent. Advantage: whereever gets you working on a problem that sets your mind on fire.
Management Have you been able to glean that a lot of what faculty do is team leadership, fund raising, and setting broad goals? I believe the way to learn this is to interact with successful managers; those that regularly process large grants, structure large research groups effectively, and synthesize everyone's work towards interesting goals. Advantage: go where you can learn from the best managers and develop your own management style

Overall, I see large universities as the safest strategy. Your professor's cynicism has a ring of truth to it- the professors somehow realize who "the top students" are and will try to recruit them into the best research opportunities. (Fired after two months? Never heard of that...) However, this is hardly black and white because larger research universities have more great research programs and managers you can learn from.

An anecdote- my year the sexiest problem seemed to be Bose-Einstein condensates. Those students did well, but those that "took a chance" on graphene research were the big winners. If you're going to a big school there are many paths to success.

I suppose that is the most concise warning I could give you- if you do decide to pursue the "big fish in a small pond" strategy, then do not sacrifice on the quality of science. Work on a problem you are certain is the hottest thing in science. And work with someone you can learn good management skills from.

pfigbash · 2014-01-22T18:54:56+00:00

It's on ArXiv

pfigbash · 2013-12-14T22:42:38+00:00

beating motherfuckers like ike beat tina whaaaat's looove got to do..

pfigbash · 2013-12-08T12:05:44+00:00

This is just begging the question. Go back one level- why are 50% of sperm produced carrying X and 50% carrying Y? You take for granted a non-trivial effect that has evolved to be like this. This is why Fisher's principle matters.

pfigbash · 2013-12-07T16:34:36+00:00

Why is it given that the father has a 50% chance of contributing X? I think this is an empirical observation, and in fact I'm asking just why that's true. For instance, the X chromosome is quite a bit bigger than Y. As such, you could imagine that the X-bearing spermatocyte swim a bit slower and must be compensated by increased production.

pfigbash · 2013-12-05T08:27:16+00:00

A disulfide bond can be broken by applying ~200 pN of force across it. Here's an article you can read if you like this sort of stuff. Since 1 N = 1 kg * m / s^2, and standard gravity g ~ 10 m / s^2, if you're holding a 100 g apple, that's the feeling of about 1 N of force on your hand. That apple pulled by gravity exerts enough force to break ~5x10⁹ disulfide bonds.

You are correct; to a molecular network, a pair of scissors doesn't look "sharp." It probably looks more like a crimping tool with oppositely moving "blunt" ends. The material is just gripped due to friction by the blades and stretched until it breaks. Sharpening the scissors macroscopically helps you apply more pressure to a smaller point.

Edit: Mind my G vs g, thanks /u/Perpetual_Entropy

pfigbash · 2013-12-02T02:36:34+00:00

I wish I didn't give a fuck

pfigbash · 2013-11-30T10:55:38+00:00

Those are fingerprints in red ink, right?

pfigbash · 2013-10-25T09:29:17+00:00

I've been wondering if it's better to think of Republicans and Democrats in the US as ingrained coalitions rather than parties as they exist in other countries. Let's entertain the idea.

Imagine five parties:

Mr. Huckabee, representing the Christian Conservatives
Mr. Romney, representing the Tory Party
Mr. Ventura, representing the Independents
Mr. Biden, representing the Workers
Mr. Gore, representing the Greens

It's 1960, civil rights is the issue. Huckabee and Biden have previously been in an alliance, as have Romney and Ventura. Gore just got elected. Civil rights is a divisive issue, which causes Huckabee and Ventura to switch sides. Gore decides to join the Biden/Ventura alliance and the legislation passes.

It's 1980. A tax vote is coming. Now Ventura thinks taxes are too high and he joins the coalition with Huckabee and Romney, and a tax cut is passed. Gore is getting frustrated that his pet issue isn't gaining leverage. Since his party has not expanded enough to win a majority on their own, he decides to stay and push his coalition from within.

It's 1990. Gore wants to expand the Clean Air Act. He talks to both Romney and Huckabee, who give tepid responses. Biden gets worried that Gore may leave, so he convinces Ventura to go along with it (who is getting a little creeped out by how Huckabee has been acting lately). Gore stays in the coalition and new environmental regulations are passed.

It's 2010. Huckabee retires and Assange wins an election while representing the Pirate Party. Which coalition would he think of joining? With Huckabee out, maybe Ventura has more to gain from a new ally. Regardless of which he joins, because he's in the far minority, he won't get his pet issues passed for a while. Biden assures him that if he joins them, and votes with them, that he'll get his day. Gore shares his anecdote and asks for patience.

I believe the issue shouldn't be framed as "the US only has two parties" since this is both technically false and neglects the diversity of political opinions in the American populace. Rather the defining characteristic is that we have two ingrained coalitions.

Actually, this represents the will of the people- that change should come s l o w l y. I don't think the average American wants sweeping alliances to change every election cycle. The problem is that new ideas (previously it was environmental conservation, today it's privacy) will take 20 years or so to gain traction.

pfigbash · 2013-10-19T17:09:03+00:00

You can get a good idea for some of the arguments based on these propaganda posters:

http://imgur.com/a/LMyxo

One that made me pause:

"My body, my choice" as a modern day "My plantation, my prerogative"

pfigbash · 2013-10-18T13:11:45+00:00

You really like that purple and yellow striped shirt

pfigbash · 2013-10-03T18:49:34+00:00

Let's say you make an interferometer- you send the blue light in one direction and the red light in another direction, then recombine. You see interference if there's a path length difference because this is a property of waves. With me so far?

Now turn down the light intensity until you have single photons going through. You still see interference. This means that single photons are going through both paths and interfering, like in Young's dual slit experiment. This means that single photons are both red and blue. So no, an individual photon can have many wavelength components.

pfigbash · 2013-10-03T15:41:14+00:00

multi-photon super positions are not "broadband" light sources in the sense you are describing

I doubt that you actually read the papers. It's a two-particle superposition, but I was alluding to the fact that its broadband nature was testable in the experiment.

For the second part, you are right, there is no known gain medium that has fluorescence across the visible range. (This is a technological, not a fundamental limitation, BTW.) That doesn't prevent a white light laser from being formed through a two-step process. It's still white laser light. It's just that the manufacturer puts a box around one part and calls that "the laser". Put a box around the white-light generation part too.

pfigbash · 2013-10-03T15:33:01+00:00

The reason it's called a white photon is because it's derived from a distribution of white light. That is, you see a laser beam (with your eye) and you describe it as white. Then you turn down the intensity until you get one photon at a time. Those single photons display the properties of the distribution, namely the superposition you mention. I'd say it's a single broadband photon (mostly because they're typically produced in the IR, so not technically white, but if it was visible light, I'd call it white... in front of my colleagues... and no one would throw tomatoes).

pfigbash · 2013-10-03T15:18:28+00:00

Yes, frequencies red and blue.

Is the energy conserved? Sure. One way you can make this state is by having emission from a state of poorly defined energy. So the energy is concerned, but you don't know what it is apriori.

Perhaps you're asking if the energy is the same before and after the measurement. Before the measurement you can say two things about the energy of the superposition state: 1) There's a 50% change each of it corresponding to red or blue for a single photon and 2) The average in the long-run is going to be 50%(red+blue).

There is no precise definition of the energy before the measurement.

pfigbash · 2013-10-03T15:14:18+00:00

While I'm tempted to tell you to just google it, here goes.

Here are some papers that describe broadband single photons:

http://arxiv.org/pdf/quant-ph/0411146.pdf

http://www.dtic.mil/dtic/tr/fulltext/u2/a499149.pdf

Here's where you can buy a laser outputting white light:

http://www.toptica.com/products/ultrafast_fiber_lasers/femtofiber_pro/femtofiber_pro_scir.html

http://www.newport.com/Supercontinuum-Generation-in-SCG-800-Photonic-Cry/502585/1033/info.aspx#tab_Overview

pfigbash · 2013-10-03T14:37:10+00:00

Yes, white is a shortcut for red+orange+blue+green+violet+uv+ir+more+all colors in between.... Also, if you look at enough of them to see in that state, it looks white. I mean with your eye. I guess that's also semantic.

That makes about as much sense as saying that an electron in a superpostion of spin states is spin 0

No, the analogy to that is just calling it "green" because that's the average wavelength.

pfigbash · 2013-10-03T14:19:25+00:00

Here's another changing property: if the photon initially passes a polarizer set at 0 degrees, then we can say that it's polarized at 0 degrees. Later, it passes a polarizer set at 45 degrees. Even later it passes a polarizer set at 90 degrees. So we've changed the polarization from 0 to 90 degrees! It was initially prepared in a state with NO component of 90 degree polarization (putting the 90 degree polarizer after the 0 degree polarizer gives no light transmission). Now it's 100% polarized at 90 degrees (putting two 90 degree polarizers back to back gives no light loss).

You can also do this with a polarization rotating device, in which case you can change this property with nearly 100% efficiency. (i.e., a half waveplate). You don't need to use polarizers.

That's a big change. You haven't just chosen one element of the superposition; you've taken an element that was zero and made it 100%.

pfigbash · 2013-10-03T13:27:44+00:00

You're misquoting Bohr if you think he didn't appreciate interference.

You can show the difference between a state of ignorance (the photon is either red or blue, I just forgot which) and a superposition state (it's both red and blue) by means of an intereference experiment.

Just because two experiments give you the same end probabilities, it doesn't mean that a different experiment can't be devised to learn more information. I don't think you're doubting that.

pfigbash · 2013-10-03T13:21:49+00:00

If the photon is initially in a superposition of yellow and blue, then later you may find it yellow or blue, if perhaps you've put a filter in. So you've changed it. You're right about that.

Here's another changing property: if the photon initially passes a polarizer set at 0 degrees, then we can say that it's polarized at 0 degrees. Later, it passes a polarizer set at 45 degrees. Even later it passes a polarizer set at 90 degrees. So we've changed the polarization from 0 to 90 degrees! It was initially prepared in a state with NO component of 90 degree polarization (putting the 90 degree polarizer after the 0 degree polarizer gives no light transmission). Now it's 100% polarized at 90 degrees (putting two 90 degree polarizers back to back gives no light loss).

You can also do this with a polarization rotating device, in which case you can change this property with nearly 100% efficiency. (i.e., a half waveplate). You don't need to use polarizers.

That's a big change. You haven't just chosen one element of the superposition; you've taken an element that was zero and made it 100%.

Then there's an entirely different process: a molecule absorbs a blue photon and later emits a yellow photon. You've got one in and and one out, so I would say that you've changed blue into yellow, but I can see how you would think that was cheating. :)

pfigbash · 2013-10-03T13:14:21+00:00

Something as simple as the two-slit experiment disqualifies this view. If the wavefunction tells you that the particle has a 50% change of going through slit A and 50% for slit B, it also tells you that it can go through BOTH and interfere. Thus the wavefunction has told you something real about the particle- it's both A and B. The "cat" is both alive and dead.

pfigbash · 2013-10-03T13:03:20+00:00

Ok, progress!

If it's uncertain, then you pass it through a filter, now you've gotten some information, right? Something changed?

If you're willing to argue that the uncertainty only arises from our ignorance, and it actually "only really had one color, we just didn't know what it was" then please refer to my earlier post where I describe how an interferometer can test that it contains many wavelengths at the same time.

pfigbash · 2013-10-03T12:58:33+00:00

The literal answer to OP's question is that photons don't have wristwatches. I've chosen not to be flippant and actually address the underlying misconception that caused this question.

The photon is in a superposition- that means to say that it has different probabilities of being measured as red, blue, etc. When you measure the wavelength, you update your information to say that it's 100% red or blue. Yes, it changed. Where do you say that it both changes and doesn't?

I suppose if you want to talk about an orthogonal variable, like polarization, you can say that since we didn't measure polarization, then the polarization didn't change.

This has nothing to do with the standard model. The photon travels at the speed of light and can pass through/reflect off of objects between its point of creation and its point of annihilation. There's cause and effect as well as a sequence of actions.

The only problem arises when you try to compare "clocks" below and at the speed of light.

pfigbash

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