The h Bar

6:00 PM

Although non-commutation does not imply that there is no simultaneous eigenstates, it merely implies that there is no basis of simultaneous eigenstates, but that is enough to show that there is no observable corresponding to this process.

enumaris

well it's Semiclassical's question heh, I'm just pondering :P

Semiclassical

Yeah, I could well believe that there's no observable. But there should still be a statistic.

ACuriousMind

@Semiclassical I don't know what it means for a statistic to "be there"

Semiclassical

In the sense that I should still be able to compute probabilities for each of the various

enumaris

yeah it's not apparent how you would make that correspondence

ACuriousMind

6:02 PM

@Semiclassical Sure - just apply the measurements in succession, computing the probabilties step-by-step

Semiclassical

yeah

enumaris

I think there's also an issue that you can only calculate probabilities at this point

not wave functions anymore

Semiclassical

meh

Anonymous

@Semiclassical You mean something like a density matrix for the outcome states?

enumaris

so it goes back to @dmckee's point of maybe having to introduce a density matrix

Semiclassical

6:03 PM

I'm fine with working in the density matrix formalism here, so that's no big deal to me

@Blue I guess so?

Anonymous

@enumaris Yeah, but that would just involve computing the probabilities at each step to get the final mixture I guess

Anonymous

as ACM says

enumaris

I mean, conceptually it's easy to imagine calculating those probabilities

Novalium Company

"When you connect two identical batteries in parallel, you double the output capacity while keep the output voltage the same as either battery". Is output capacity the watt-hour or the amp-hour measurement?

Semiclassical

yeah, it's just a bit tedious.

enumaris

6:05 PM

and starting from a pure state $\Psi$, you can calculate all those probabilities out to the end.

ACuriousMind

@Semiclassical Everything that's not the SHO is tedious :P

Semiclassical

lol

enumaris

that's why we transform everything we can into SHO

Semiclassical

Here's the original motivation I had, just for context

enumaris

even if the potential doesn't look like $x^2$, we can still make some taylor expansion to do it

Semiclassical

6:07 PM

If I measure the spin components along axes $\hat{a},\hat{b}$, then the corresponding observables are $A=\hat{a}\cdot \vec{S}$ and $B=\hat{b}\cdot\vec{S}$

enumaris

I'm getting a little annoyed that L+L's square root signs don't extend over the object being square-rooted

Semiclassical

or just $A=a_i S_i,B=b_i S_i$ using the summation convention

enumaris

yes indeedy

Semiclassical

One then has $\frac12(AB+BA) = \frac12 a_j b_k(S_j S_k+S_k S_j) = a_j b_k \delta_{jk} I_2 = (\hat{a}\cdot \hat{b})I_2$

In which case $\langle \frac12(AB+BA)\rangle = \hat{a}\cdot \hat{b}$

What I was trying to figure out is if one could actually extract that from spin measurements.

e.g. repeatedly prepare the same initial state and measure A, measure B, then take the product

Novalium Company

I like how I have no idea what you are guys talking about xD

Semiclassical

6:13 PM

and average over many such trials to get $\langle \frac12(AB+BA)\rangle$.

by contrast, btw, $\langle AB\rangle = \hat{a}\cdot\hat{b}+i(\hat{a}\times \hat{b})\times \langle \vec{S}\rangle$

which is obviously not going to be an observable quantity when the second term fails to vanish.

Anonymous

@NovaliumCompany You'll get there if you keep at it :)

enumaris

@NovaliumCompany if you double the amp-hour and keep the voltage the same, you also double the watt-hour

so both the total charge and total energy are doubled in that case

which should be clear since you have double the batteries - 2 batteries hold twice as much charge and twice as much energy as 1

Novalium Company

@enumaris Got it, thanks.

@Blue Hopefully, one day. I'm motivated so... I'm currently reading a book on electronics and calculus and they are exciting as hell.

Anonymous

@NovaliumCompany Great. As long as you're enjoying what you're learning, you're on the right path. Fwiw I didn't know anything about QM around 8 months back....but I understand a teeny-weeny bit now

enumaris

if you can follow all of Semiclassical's ramblings, you know more than a teeny-weeny bit :P

Semiclassical

6:21 PM

lol

if my ramblings are semi-understandable, you either know a lot or have gone semi-insane

ACuriousMind

(not an exclusive or there)

Semiclassical

^

well

in the case of Sx, Sy

For whatever state you start off with, there's some probability p that you measure Sx=up=+1 (in units where hbar=2 lol) and some probability 1-p that you measure Sx=down=-1

in either case, though, you have fifty-fifty odds of getting up vs. down for Sy

so the probabilities for the measurement sequences are P(++) = P(+-) = p/2 and P(-+)=P(--) = (1-p)/2

as such, the probability of getting a positive product is p/2 + (1-p)/2 = 1/2

so 50-50 odds

ACuriousMind

For every state? That's neat

Semiclassical

should be, yeah

ACuriousMind

I didn't realize you could make a perfect quantum coin-flip by measuring two orthogonal spins in succession without any state preparation whatsoever

Semiclassical

6:33 PM

well, once you measure Sx you're definitely in a pure state

ACuriousMind

Yeah, thinking it through it's "obviously" correct

Semiclassical

yeah. doesn't feel that way though.

ACuriousMind

Doesn't matter which of the states I landed in, they both have 50-50 odds of measuring either of the outcomes of the second operator

Semiclassical

Right.

Full disclosure: I worked out stuff in the density matrix formalism (with mathematica doing the matrix multiplication) before I noticed how simple the result was :P

What I should do now is consider a case like $S_z$ and $a S_z+b S_x$

in which case one doesn't have Sz=up implying 50-50 odds for aSz+bSx

enumaris

but I mean...you could just use Sx to put the state into a system of getting 50-50 for Sy...there's no need to multiply with the result form Sx if you just want to get a coin flip :P

Semiclassical

6:40 PM

yeah. i'm interested in the product for other reasons

but you definitely can just prepare prepare a pure state by measuring Sx and then get Sy as a coin flip.

Novalium Company

@enumaris Why would you usually (in a practical case) want to hook up batteries in parallel? I mean, why would you want to have less voltage delivered but more energy and charge and what does this actually mean?

enumaris

you hook up batteries in parallel when you don't need a voltage boost and you just want the thing you're hooking them up to to last longer

Semiclassical

less load per battery

Anonymous

@NovaliumCompany Parallel connections have some advantages like if one of the components (in one of the branches) get damaged then the others will still work fine as they're directly connected to the power source

Anonymous

Oh, but you're talking about batteries in parallel

Anonymous

6:46 PM

Even that has some advantages

Semiclassical

though I feel like, if you had batteries in parallel, then any differences in voltage could lead to some weirdness

Novalium Company

So less voltage but will run for more time, how can that be proven mathematically?

Semiclassical

i guess if they're all connected to some load then it shouldn't be an issue though

Anonymous

For such systems you have $$E_{\text{parallel}}=\frac{E_1/r_1+E_2/r_2+....}{1/r_1+1/r_2+...}$$

enumaris

less voltage will run for more time...is not what I said

Anonymous

6:48 PM

Where $E$ stands for EMF/cell-volatage

Novalium Company

same voltage*?

enumaris

better to use $\varepsilon$ for emf I think...

E feels too much like energy

Semiclassical

$\mathcal{E}$ is what I usually do

enumaris

look at mr. fancy pants over here

Anonymous

I'm too lazy to type so many epsilons :P

Semiclassical

6:49 PM

lol

Novalium Company

So if we have 2 batteries in parallel (each 5V), the voltage will be equal to 5V (zero increase in voltage) and Ah and Wh will increase twice?

Anonymous

@NovaliumCompany Use the formula I gave :P

Anonymous

Remember that when you connect cells in parallel you always consider some internal resistance

Anonymous

@NovaliumCompany That's true though

Novalium Company

I thought we could use the Wh = Ah * V formula or something?

Maybe not?

Anonymous

6:51 PM

If the cells have the same internal resistance

Anonymous

You consider $r_1=r_2=r$

Anonymous

@NovaliumCompany Yeah, power is VI

Anonymous

emf multiplied by current outflow

Anonymous

Wh is an unit of energy

Novalium Company

I mean I want to prove that keeping the voltage the same (5V) increases the lifespan of the total battery system, can I do that?

Anonymous

6:53 PM

First of all, what determines lifespan of a battery system?

Novalium Company

The load... :D

How much current/energy the load draws per second?

Anonymous

@NovaliumCompany From each cell

Anonymous

Right

Novalium Company

What do you mean by cell?

Anonymous

A battery is a collection of cells, technically speaking

dmckee

6:55 PM

@NovaliumCompany A "battery" is a collection of cells.

Anonymous

But some people consider cell=battery

Novalium Company

Can I consider it?

Cuz it confuses me a bit.

dmckee

Most cells in current use produce 1.5 V or less, so a 6 V batter has at least 4 cells in series.

Anonymous

Doesn't matter, you can, as long as the person you're speaking to understands what you're saying :P

Novalium Company

Oh, so batteries are made up of cells. Are these the electrolytes and the electrode or?

dmckee

6:56 PM

If you look at an automotive lead-acid battery you can see that it has 6 separate parts, and each of those cells produces a shade over 2 V, so that the whole thing produces a bit over 12 V.

Anonymous

@NovaliumCompany Have you ever seen the AAA batteries used in your TV remote?

Anonymous

Each of those batteries is technically a "cell"

Novalium Company

pencil?

Oh, got it.

I googled.

Ok, for now I'll just consider cell a battery and vice-versa.

Anonymous

See @dmckee's example though ^

Semiclassical

@ACuriousMind well, I'm ending up with: If I measure $A=S_z=\uparrow$, then the probability of having $B=\hat{b}\cdot \vec{S}=\uparrow$ is $\cos^2(\theta/2)$ where $\theta$ is the polar angle, and $\sin^2(\theta/2)$ for $B=\downarrow$

Novalium Company

6:59 PM

@Blue Yep, I got it.

Semiclassical

whereas if I measure $S_z=\downarrow$, then these probabilities get swapped

dmckee

In the demo closet at my school, they have actual 6V camping batteries and 12 V lead acid bateries that have been drained and cut open to expose the workings. I wish I had pictures.

Anonymous

@NovaliumCompany So, for what you want to prove, draw a circuit having two cells (having some internal resistances) connected across an external resistor

Anonymous

And find the current through each branch

Anonymous

Compare it to the case when just a single cell of emf E is connected

ACuriousMind

7:00 PM

@Semiclassical sounds legit

Novalium Company

@Blue Ok gimme 1 min.

Semiclassical

so, using the same p as before, I have Pr(up-up) = $p \cos^2(\theta/2)$, Pr(up-down) = $p \sin^2(\theta/2)$,
Pr(down-up) = $(1-p) \sin^2(\theta/2)$, and Pr(down-down)=$(1-p)\cos^2(\theta/2)$

which, assigning $\pm$ values to up/down, means that the product is positive with probability $\cos^2\theta/2$ and negative with probability $\sin^2(\theta/2)$

and therefore the expected value of this produce is $\cos^2(\theta/2)-\sin^2(\theta/2)=\cos\theta$

by comparison again, one can show that $\frac12(AB+BA)=(\hat{a}\cdot\hat{b})I_2 = (\cos\theta) I_2$

and so $\langle \frac12(AB+BA)\rangle = \cos \theta$

Novalium Company

@Blue In the case of a single cell, the current will be constant throughout the circuit (Kirchhoff's current law) and in the parallel circuit with 2 cell connected, the current going through the external resistor should be equal to the sum of the currents that the two batteries combined produce?

Semiclassical

Which makes it seem like $\frac12(AB+BA)$ is indeed the observable.

this, of course, would be entirely consistent with this being special to the $n=2$ case

Anonymous

@NovaliumCompany Sounds right. So the current each cell produces in the second case will be half of the current a single cell produces in the first case, since total current in both cases will obviously be same (consider all the cells are identical)

Semiclassical

7:06 PM

so I'm not going to try to draw any conclusions for $n\geq 3$

ACuriousMind

@Semiclassical Does iterating the process change the probabilities? If so, it's dangerous/wrong to say that is "the observable" because the outcomes of measurement of an observable need to be eigenstates

Semiclassical

hmmm! that's a very good point.

Novalium Company

@Blue Yep, also the voltage in both circuits should be equal?

Semiclassical

though, after one such A,B measurement you'll definitely be in a pure state ($B = \pm 1$)

Anonymous

@NovaliumCompany Right, since two cells of emf E, in parallel will again produce a net emf E

Novalium Company

7:08 PM

@Blue Yep.

Semiclassical

ah, but if you measure $A,B$ and get $1,1$, it is not the case that measuring A,B again will guarantee getting a positive product

ACuriousMind

Also, your observable gives the correct expected value, but the wrong prescription for the states post-measurement - since it's the identity, measuring it won't change the state at all

Semiclassical

you could get 1,1, and then 1,-1

ACuriousMind

I.e. the expected value is correct, but the variance is not.

Semiclassical

yeah

Good call.

I think the first objection is decisive for me, though.

If you measure A,B and take the product, there's no guarantee that you'll get the same product if you repeat this

As such, there's no way that this can correspond to an observable.

It's interesting that this gives the correct expected value, though. I wonder why that is.

That might be worth a question on its own.

Novalium Company

7:15 PM

@Blue So in the parallel example, the load spreads the current draw between the batteries and that's why they will die after more time?

Anonymous

@NovaliumCompany Right

Novalium Company

@Blue So in the serial example, if the circuit draws from a single cell let's say 3A, in the parallel example, where the cells are two, 1.5A of current will be drawn from each one of them, that's why they will live (around) twice as long?

Semiclassical

@ACuriousMind This does raise a question of how one might actually have $\frac12(AB+BA)$ show up in a quantum system. But it evidently can't be "measure A and then measure B".

(assuming non-commuting of course)

Anonymous

@NovaliumCompany Exactly

ACuriousMind

@Semiclassical Is that a question? Do you ask the same for the commutator $[A,B]$?

(note that your expression is just the anti-commutator)

Novalium Company

7:18 PM

@Blue Ok, thanks so much. See you for now :)

Semiclassical

Hmm.

That's fair.

ACuriousMind

I mean, (anti-)commutators are important in QM but I wouldn't expect them to necessarily show up naturally when talking about measuring the individual observables involved

Semiclassical

Hmm

I guess here's the question, for which I won't be surprised if the answer is no (or simply ill-defined)

Suppose I've got a quantum system which I can probe in two different ways, one measuring observable $A$ and another measuring observable $B$. Must there be an experiment I can do which measures the observable $\frac12(AB+BA)$ ?

I'm not sure how one would falsify that, as I think through it.

Main reason I wonder is the following historical fact: Jordan introduced the concept of a Jordan algebra (which you can build from an associative algebra by introducing the symmetrization product) in order to formalize the notion of an algebra of observables in QM. (i'm more or less quoting the Wiki page for that)

Anonymous

Does the anti-commutator (observable) have any physical significance? (sounds like a good question)

Semiclassical

I guess you don't actually need to be able to measure $\frac12(AB+BA)$ in order to conclude that it's an observable though.

Yeah.

Start with two measurable observables, and construct a third by symmetrization. Must this new observable be measurable as well?

ACuriousMind

7:27 PM

@Semiclassical An observable is not necessarily something you can straightforwardly measure!

Semiclassical

The tricky thing in this is that, while 'observable' has a clear cut definition, a 'measurable quantity' doesn't seem to be so simple

@ACuriousMind Yeah, I guess that's the point

If you can measure it repeatedly and always get the same value, you have an observable. But an observable is observable even if you don't actually have a device which implements that.

ACuriousMind

@Blue @Semiclassical Depending on your quantization prescription, the (anti-)commutator will be the quantum observable corresponding to the classical product of the two observables.

Semiclassical

"I can perform experiment A repeatedly on the same system and always the same result after the first measurement" is sufficient for A to be an observable. But it evidently isn't necessary.

ACuriousMind

E.g. $\{x,p\}$ is quantum orbital angular momentum in the standard (Weyl) prescription

@Semiclassical I think you have to formalize "measurement" for that to even be a meaningful statement

Semiclassical

hmm

ACuriousMind

7:32 PM

An observable is simply an element of the algebra of observables, which is axiomatically part of the definition of an abstract quantum system.

The relation to a physical measurement procedure is not short or straightforward.

I think Ozawa (possibly wrong spelling) has written a lot on that.

Semiclassical

ugh, measurement is such a pain

ACuriousMind

Yes, physics would be much easier if we didn't measure things :)

Semiclassical

or at least if we could stick to measuring observables that commute

stupid [x,p] = i hbar

enumaris

it's why your name is Semiclassical

and not Semiquantum

Semiclassical

ikr

ACuriousMind

7:35 PM

@Semiclassical Isn't it wonderful how much flows from so simple a statement? ;)

Semiclassical

hmm. all of that said, I wouldn't be surprised if there is some way to measure $\frac12(AB+BA)$ in the spin context

e.g. run the electrons through a quadrupole field rather than a dipole field

ACuriousMind

@Semiclassical In the spin-1/2 context. it's just the identity!

Measuring the identity is trivial - do nothing.

I'm measuring the identity all the time

Semiclassical

eh, it's $\frac12(AB+BA)=(\hat{a}\cdot\hat{b})I_2$

that dot product doesn't count for nothing

ACuriousMind

@Semiclassical Measuring a multiple of the identity is still trivial.

Semiclassical

It is pretty boring.

ACuriousMind

7:38 PM

Just use a device with a pointer permanently taped to $(\hat{a}\cdot\hat{b})$.

Semiclassical

lol

enumaris

all my experimental devices have pointers permanently taped to different numbers

lılostafa

enumaris

the results of my experiments involve me choosing which pointer I like best at the moment

lılostafa

how is a plot like this created? ^

enumaris

7:39 PM

can't view images sorry

lılostafa

@enumaris why

Semiclassical

or, y'know, include a protractor

ACuriousMind

@lılostafa His workplace is worried he will slack off if he can

enumaris

indeed

lılostafa

Use Psiphon or another VPN service

ACuriousMind

7:41 PM

Everyone knows you can't slack off on the internet without viewing images!

@lılostafa I have no idea what that plot is showing

Semiclassical

@ACuriousMind it does remain interesting to me that the expected value of that A,B product is just cos(theta)

enumaris

yesterday I discovered Github is blocked on my desktop's ghrome

chrome*

Semiclassical

it's just a 3D bar graph I guess?

enumaris

kind of hilarious that github would be blocked for me...probably the only person in this company that actually needs github...

ACuriousMind

@enumaris wat

No code for you!

enumaris

7:42 PM

I can still access github on my IE

so I guess I have to do that lol

lılostafa

@ACuriousMind It is an array of circles, each of them having a number assigned to it. This value is shown by the height of each circle (now cylinder).

Semiclassical

@ACuriousMind to be more precise on this point (since I found myself momentarily questioning it): one has $$\langle [\frac12(AB+BA)]^2\rangle = \langle (\cos\theta\, I_2)^2\rangle = \cos^2\theta$$.

but the only possible products are $\pm 1$, so the expected value of the squared product is just 1.

ACuriousMind

@lılostafa I'm sure I could produce something like that graph if you gave me a few hours and an incentive to do so :P

Semiclassical

So yeah, they're definitely not the same.

enumaris

LOL - I just heard someone quote Stalin about why the Germans lost in Russia today...

Anonymous

7:47 PM

@lılostafa I think even Excel can do it

ACuriousMind

"even"

Anonymous

I mean it doesn't seem anything extra-ordinary :P

ACuriousMind

Do you know how many software developer hours are in that piece of software called Excel? :P

enumaris

I wonder how much longer my pipeline is gonna take...been going for almost 3 hours now

@ACuriousMind are we counting all the countless hours those developers spent waiting for code to compile?

Anonymous

@ACuriousMind Okay, okay, we sort of take Excel for granted ;)

Semiclassical

7:49 PM

In particular, Sx and Sy anticommute. so the symmetrized product in that case is just the 0 operator, with mean 0 and variance zero. by contrast, the product of the measured Sx, Sy values will have mean zero but variance one.

so yeah, that doesn't work.

Anonymous

(proprietary malware after all)

enumaris

(Bernardo is leaking)

Anonymous

Bernardo has vanished for a few months now

Anonymous

Apparently

Anonymous

Can't ping him anymore :/

enumaris

7:51 PM

hmmm

maybe he's preparing for his war on microsoft

Anonymous

Incidentally his favourite GitHub got sold to Microsoft in the meanwhile

Anonymous

Which maybe one reason we're not seeing him anymore :P

Semiclassical

i somehow read that as "his war on minecraft" at first

ACuriousMind

He's probably busy after getting that job he wanted. Daniel (who's also not been here in a while, but a shorter while) said he's doing fine.

@Blue You lose the ability to ping people after 2 days :P

Anonymous

@ACuriousMind Oh, that's good :)

Anonymous

7:53 PM

@ACuriousMind SE sucks!

ACuriousMind

@Blue ::hovers over mod action button:: Say what? ;)

dmckee

@Blue Moderators have a extra-secret super-special ping power which works even on users who've never visited chat.

So obviously what you need to join our cabal.

ACuriousMind

@dmckee I'm not sure I want to use that power now I know it's extra-secrete. Ew.

dmckee

Fair, fair. Now I'm sorry I fixed it.

ACuriousMind

Hehe

Anonymous

7:56 PM

@dmckee And yet they don't seem to use it when the need arises :P Pinging AFT and Bernardo from time to time is absolutely necessary to keep h-bar alive ;)

Anonymous

(anyhow, just joking)

ACuriousMind

@Blue AFT has made a conscious decision to not hang around here idly anymore (as evidence by him dropping in when he has something specific to discuss) and we should respect that.

Anonymous

I wonder how the hbar will be, say around ~5 years from now

Anonymous

All of us will grow older and boring-er (and busier of course) :)

dmckee

@Blue But new young-and-exciting people will pop in. The interesting question is will they join/change/re-form the community here?

Anonymous

8:01 PM

@dmckee That's true though. BTW in your eyes, how has hbar and PSE (as a whole) changed in the last 5-6 years? (I've only been around for the past 1.5 years or so)

ACuriousMind

@Blue I don't plan to grow any busier :P

Anonymous

@ACuriousMind Just wait till you get married and get a family ;)

dmckee

At the start it was pretty dead and only picked up slowly. I would describe the last few years as "vibrant", but the change came on slowly so it is hard to say when it started.

Anonymous

@dmckee I see. From what I could deduce, in the middle, I guess there was a phase when PSE was popular as "Maimon's battlefield" :P

Anonymous

And oh, of course Motl's too

enumaris

8:06 PM

time to go read some code I wrote 2 months ago

danielunderwood

Has there always been the occasional "I have a secret theory of everything" person or is that more of a recent thing? More generally, were those people around much in physics before the internet?

enumaris

hopefully I won't hate myself after this

ACuriousMind

@danielunderwood Yes and yes.

enumaris

dives

Semiclassical

cranks are as old as science, I suspect

ACuriousMind

8:06 PM

Before the internet, they just sent their ideas unbidden to semi-random physicists/faculties

Anonymous

lol

dmckee

@danielunderwood These folks are always present where they think they might get a hearing.

ACuriousMind

(They still do, but emails are easier to delete than manuscripts)

dmckee

I got my first email from one of them less than a week after getting a staff email address when I started grad school.

For years, deleting such emails was something I did every week (until the filters started catching most of them).

Semiclassical

I've somehow avoided getting crank emails

enumaris

8:09 PM

I never got crank emails

I do get a couple emails asking me to publish in random obscure journals

Anonymous

@enumaris Potential start-up idea: Work on making a neural net based on this. It will sell like hot cakes among the physics faculties :P

enumaris

hmmm...

you could build a neural net to detect "crackpot theories" perhaps

if you gave me enough data

labeled data

Semiclassical

some sort of supervised learning, perhaps? give it a bunch of message board posts, rated by crackpottery

Naveen Balaji

Hey guys! Please check this out when free

enumaris

sets of (paper, crackpot_or_not)

Anonymous

8:12 PM

@enumaris You can just contact vzn or JD for the data

Naveen Balaji

physics.stackexchange.com/questions/413614/…

ACuriousMind

@enumaris Quality test: Make it good at arXiv vs. snarXiv

Apr 21 at 15:41, by ACuriousMind

@AlexKChen Please don't do that - if you think the question is of particular interest to someone here, ping them with it, but don't simply post your question to draw attention, especially not periodically. Our site is not meant to deliver answers in real-time and you need to be a little patient.

enumaris

lol

Semiclassical

following up on the A,B stuff from earlier. Let $\mathcal{A},\mathcal{B}$ be observables with eigenvalues $a_j,b_k$ and projection operators $A_j,B_k$ respectively. (so $\mathcal{A}=\sum_j a_j A_j$ and $\mathcal{B}=\sum_k b_k B_k$)

ACuriousMind

Oooh, \mathcal observables. It's getting serious now

Semiclassical

8:15 PM

lol

I guess I could call the projection operators something like $P_j^A$

yeah, think I'll do that. observables $A,B$ with eigenvalues $a_j,b_k$ and projection operators $P_j^A,P_k^B$

I'll take the initial state to have density matrix $\rho$

then the probability of measuring $A=a_j$ is $p_j^A=\text{tr}(\rho P_j^A)$, yielding the new state as $\rho_j^A = (P_j^A \rho P_j^A)/p_j^A$

and the probability to subsequently measure $B=B_k$ after having measured $A=a_j$ is $\text{tr}(\rho_j^A P_k^B) = \text{tr}(P_j^A \rho P_j^A P_k^B)/p_j^A$

hence the probability to measure $A=A_j$ and then $B=B_k$ is just $p_j^A \text{tr}(\rho_j^A P_k^B) =\text{tr}(\rho P_j^A P_k^B P_j^A)$

But $(P_j^A P_k^B P_j^A)^2 = P_j^A P_k^B P_j^A P_k^B P_j^A \neq P_j^A P_k^B P_j^A$ in general

(It'll be true if the projections commute for all j,k; but that'd imply $A$ and $B$ themselves commute)

danielunderwood

Oh neat I hadn't heard of snarxiv before, but it's something that I thought would be cool back a couple years ago

Semiclassical

so in general the probability of getting $A=A_j$ and then $B=B_k$ isn't of the form $\text{tr}(\rho P)$ for a projection operator $P$.

hence this isn't the probability of an observable etc. etc.

for a further comparison, the expected value of that statistic will be $\sum_{j,k} a_j b_k \,\text{tr}(\rho P_j^A P_k^B P_j^A)$

whereas $\langle AB\rangle = \text{tr}(\rho A B) = \sum_{j,k} a_j b_k \,\text{tr}(\rho P_j^A P_k^B)$

with that extra projection operator in the former being the difference.

Bob van de Voort

8:38 PM

I was wondering if there's anyone around with knowledge of QED and if they could hopefully have a look at the following question

0

Q: The Purcell effect, it's influence on the lifetime and quantum yield (of fluorophores)

So I've been looking into the Purcell effect and how it interacts with fluorophores (fluorescent molecules). The Purcell arises when you have a dipole in a cavity or even just near a dielectric or metal interface. It can then enhance the emission of the dipole in comparison to free space. Now fo...

Anonymous

@BobvandeVoort Put a bounty on it maybe

David Z

8:52 PM

@dmckee I think it took about a year after we started regular chat sessions for activity to build up in the chat room.

danielunderwood

9:08 PM

Cool I didn't know that there were chat sessions. Are they set topics or just anything? And for anyone like me who didn't know: chat.stackexchange.com/rooms/info/71/the-h-bar?tab=schedule

ACuriousMind

@danielunderwood They were originally introduced to just bring traffic to the room at set times, then we tried for a while to make them more structured with set agendas...but nowadays they aren't really special

If you have an idea for what we could do with them it would be very welcome!

lılostafa

@ACuriousMind I need the plot for a paper I'm going to publish (hopefully in an APS or Nature journal). If you show me how to do it, I will thank you in the acknowledgments section of the paper. Is this a good enough incentive? ;)

Anonymous

@lılostafa acknowledgment doesn't help to buy beer ;)

Anonymous

@lılostafa BTW do you need something exactly like this:

Anonymous

9:17 PM

Or just any generic 3D bar graph will do?

lılostafa

@Blue Okay what about this: I can mine a coin of your choice on my PC ( a core i7 6670K with a decent GPU) for one week to your wallet.

@Blue Yeah similar to this

Anonymous

@lılostafa Where did you find it though?

lılostafa

@Blue nature.com/articles/nphoton.2013.274

ACuriousMind

@lılostafa If I find the time I'm gonna look into it, but no promises - pretty busy this week

lılostafa

@ACuriousMind Okay, thanks

BTW I read here a few days ago that you've become a COBOL programmer.

Did you just give up doing string theory in favor of programming in COBOL? :o

Anonymous

9:23 PM

Oh, not that again

Anonymous

XD

lılostafa

what

Anonymous

59 secs ago, by lılostafa

Did you just give up doing string theory in favor of programming in COBOL? :o

ACuriousMind

@lılostafa The COBOL thing is a joke - work for SAP now mainly programming in their own programming language ABAP, which was originally sorta inspired by COBOL but is now an entirely different language.

lılostafa

@Blue But, going with this logic, doing string theory usually doesn't buy you a beer; programming in COBOL does.

Bob van de Voort

10:48 PM

@Blue I will do that once I have above 175/180 reputation

frogeyedpeas

11:46 PM

howdy would anyone like to come on a physics adventure with me

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