The h Bar

bolbteppa

00:00

mathcha.io/editor

You can draw a picture and get the tikz code with that

enumaris

I'll just draw on powerpoint...

danielunderwood

hey that looks pretty neat

bolbteppa

Yeah it's a life saver for tikz

danielunderwood

it looks like it'd be good for quick tex notes too

bolbteppa

'TIL Terry Tao thought he didn't have to study for a quantum physics class in his undergrad. He ended up failing the final and the class because he was unprepared'

(He was 12)

web.math.princeton.edu/generals/tao_terence

danielunderwood

00:18

I thought I didn't have to study when I started college. Then I went through college with an ever-increasing amount of studying

Semiclassical

00:43

@bolbteppa looool

@bolbteppa If you're going to start with a quote that basically treats anyone with sympathy for the deBB as a fool, than you can basically [CENSORED]

bolbteppa

@Semiclassical point of the quote was context for the Von Neumann comments on hidden variables

Semiclassical

I really don't care what you were attempting to support with it.

bolbteppa

In reference to your earlier (starred) comment on Von Neumann's proof on hidden variables being silly

Semiclassical

@bolbteppa yeah, and I specifically indicated later in that conversation that that "conventional wisdom" is probably wrong. i even cited a paper to that effect.

bolbteppa

It's debatable whether Von Neumann was right or wrong was all I was saying

Semiclassical

00:49

I don't think there's any qualms about whether von Neumann proved what he did. The issue is whether it has the significance that was initially attached to it.

But again, I don't really care what point you were trying to make. It might have been a fine point. But the source you quote is so permeated with venom as to make me entirely disinterested.

bolbteppa

The source is not venom, he knows more about this stuff than any of us and I have yet to find errors in his postings

My only point was to point out your comment that what Von Neumann did was not silly, it's debatable whether what he did was ok the way he did it, but silly is a very unfair thing to say

Semiclassical

@bolbteppa Sure. I'll agree with that. "not false but foolish" is Bell's words, not mine, and I think it's fair to dispute them

But again, I don't really care what point you're making. You walk in here with a source that begins with "the whole attempt is a silly caricature...parasitic, ideologically driven"

And you expect me to give a damn about what you say next?

bolbteppa

You said "But then Bell comes along and points out that the assumptions von Neumann made are silly", when it's actually more likely the case that von Neumann's assumptions were the most correct part of what he did

Semiclassical

7 hours ago, by Semiclassical

(my own sense is that, while Bell might have not been doing justice to von Neumann's own thinking, he was probably responding to how other people were using von Neumann's proof. so the 'not just false but foolish' is really more a commentary on how von Neumann's impossibility theorem was received by the mainstream community. iirc the Bub paper really doesn't address this)

and, again, I never claimed that Bell was right on it. what I was saying was emphatically "the conventional wisdom", and conventional wisdom is often wrong

bolbteppa

I don't read that first paragraph as an insult haha, dBB genuinely looks that way to me which is a real disappointment

user1271772

00:59

Would there be anyone here interested in this SE proposal? area51.stackexchange.com/proposals/119810/scientific-software/…

Semiclassical

@bolbteppa yeah, well, I think it's full of **** and I have no interest in eating a meal prepared with **** on top.

bolbteppa

That was just for context to your saying "Bell comes along and points out that the assumptions von Neumann made are silly" as in, it's debatable whether what von Neumann did was silly the way Bell implied it was

I know you said that too, was merely complementing your point

Semiclassical

And had you said that, I would have agreed.

But you didn't. You just plopped a huge whole steaming pile of LM

Captain Giraffe

I had a fly fly an extremely odd flight path straight into my frying pan. I've never seen something so odd. My stove-top is an induction stove. My google searches gives me zilch. Is this a reasonable question to post?

bolbteppa

I don't know why his emotive comments have any bearing on the details he brings up in his posts

Semiclassical

01:05

You thought they were important enough to include.

bolbteppa

I can disagree with him on all the emotional stuff, especially the political stuff etc... but on the facts/details of physics you have to be very very careful because he knows his sh*t... and that first paragraph is context for his second paragraph on hidden variables being the positions ('undermining the very foundations of QM' as mentioned in the first paragraph)

I still don't understand the whole spin thing with dBB but he makes a really interesting point about dBB forcing us to at least theoretically be able to know the spin in some sense which immediately breaks rotational invariance, things like that really need to be addressed, and I have not even seen a dBB article which indicates the author even knows what spin is tbh

Semiclassical

at the level of non-relativistic QM, spin is just a property of the wavefunction in deBB

if your point is that the use of spinor wavefunctions is not well-motivated in dBB, that's fine

I don't know a great response to that off the top of my head

bolbteppa

Spin arises because the rotation group is not simply connected so that representations may be projective, and probability distributions representable as e.g. wave functions are allowed to be projective representations of the rotation groups, so you get this extra indeterminacy very naturally... It's intrinsic to normal QM, I literally cannot find a single Bohm article that even hints at group theory let alone this stuff

user1271772

Does anyone here use any software such as QuTip (for open quantum systems) or Spinach (for spin dynamics) ?

Semiclassical

more to the point, it is not the case that one knows the spin prior to measurement in deBB

there is no 'spin property' that a quantum measurement reveals in the deBB setup

bolbteppa

01:17

"Try to add the spin to a particle. If the logic of Bohmian mechanics – the wave function "is" a classical field and we should also add some classical values of a maximum set of commuting observables – were universally valid, it's clear that aside from the spinor-valued wave function (up,down), we should also assume that Nature "objectively knows" about the classical bit of information that tells you whether the spin is "actually" up or down.

Because they sort of realize that the rotational symmetry holds exactly and the hypothesis that the classical value exists with respect to one axis would break the symmetry kind of maximally, they decide that the Bohmian rules must be "skipped" in the case of the spin – they just manually omit some degrees of freedom that should be there according to the general prescription of Bohmian mechanics

and hope that the spin measurements are ultimately reduced to position measurements so that it doesn't hurt if some degrees of freedom are not doubled in the usual Bohmian way."

Semiclassical

@bolbteppa Except, no, that's not how the dBB story works.

bolbteppa

The dBB story I've seen is far worse than this, randomly introducing spinor-valued wave functions out of thin air

Semiclassical

Suppose I have two SG devices, both set up to measure the spin z-component, with the same initial wavefunction and even the same initial particle position. It does not follow that both would give the same output.

bolbteppa

SG is already invoking magnetism which is a relativistic phenomenon, it's banned in non-rel QM for all intents and purposes

Semiclassical

riiiight

bolbteppa

01:22

Non-relativistic hydrogen atom with a coulomb potential leads to spin naturally, obviously it's also an approximation but that's irrelevant to why it arises, irrelevant to the origin of the potential, all that matters is spherical symmetry

Semiclassical

again, if you're objecting to the use of spinor wavefunctions as being unmotivated on principled grounds, that's an interesting enough objection

I don't know a good response to it off the top of my head.

But you quote LM to the effect that, if you assume a spinor-valued wavefunction, then "we should also assume that Nature "objectively knows" about the classical bit of information that tells you whether the spin is "actually" up or down". and the moment you do that, I lose all confidence in what you're talking about

bolbteppa

Well the point LM makes is that there should be a motivation, the normal motivation, but if you try to use that and ALSO take the dBB logic of a wave function as being real seriously, it leads to a contradiction

Semiclassical

because I know for a fact, having actually worked through that story myself recently, that that's simply false.

bolbteppa

Nature "objectively knows" where those hidden variables (the position) are, even if we can never see/measure them, otherwise hidden variables don't exist at all obviously right?

Semiclassical

---false.---- read that too fast

that's not the dBB story.

bolbteppa

01:26

If the hidden variables do not exist in nature, then they do not exist

Semiclassical

what you're saying is true for position.

bolbteppa

Just because we can't see them, they still clearly exist

Semiclassical

it is not true for spin in the dBB story

bolbteppa

Yeah that's what I'm talking about, position

Semiclassical

yes, now you are. but before that you were talking about spin

bolbteppa

01:28

So if you take the idea of a wave function seriously in any sense, then spin is an immediate consequence of wave functions being projective representations of the rotation group, so spin variables are undeniable, they absolutely arise

Semiclassical

You keep telling me that, and yet I know for a damn fact that there's no spin-vector that an SG device reveals.

that's not how the dBB story works.

a spinful particle in the dBB story has a spinor wavefunction and a position. that is -all-

there is no extra classical bit.

back later.

Semiclassical

01:53

back

bolbteppa

If you take the Schrodinger equation for the Hydrogen atom, and assume the solutions can be plugged into $\int |\Psi|^2 dx$ as probabilities, then the $\Psi$ wave functions are projective representations of the rotation group immediately, even if you had no idea what "spin" was you could discover it by these facts alone, you immediately have spinor wave functions...

and dBB actually has to immediately invoke this stuff by this alone... I don't think any dBB authors even know QM well enough to understand this tbh which is shocking... but this means our wave function not only depends on the position $(x,y,z)$, it also depends on a new spin variable $s$, $\Psi = \Psi(s,x,y,z)$,

so already we expect $s$ should have a "classical" analogue the way $(x,y,z)$ have classical analogues (i.e. the normal classical position) even though we can't measure the right $(x,y,z)$ we can only set up probabilities for the right $(x,y,z)$, and you can justify this from the fact the spin operator in some direction is part of a maximal set of commuting observables and so is measurable,

even worse if $\Psi$ is a "real" field and it depends on this $s$ the $s$ has to have a "real" meaning, so 'nature' should objectively know what value the spin is going to take when you measure, just as 'nature' knows what position you'll find the particle at when you measure (even if we can't know where it will be), but this would break rotational invariance and invalidate the very idea of spin in the first place, a contradiction...

Semiclassical

You keep saying that dBB "has to have s"

bolbteppa

However dBB just has to, for no reason, deny that spin should have this role, we just pretend that $s$ in $\Psi(s,x,y,z)$ is not a degree of freedom with a classical analogue like $x,y,z$, calling it a 'property of the wave function' is wrong since it has to arise in path integrals and density matrices but mainly because of rotational invariance, note this same logic applies to relativity

Semiclassical

And I'm telling you that this is just blatantly false. There is no $s$ in $\Psi$.

bolbteppa

The $\Psi$ in the non-relativistic hydrogen atom problem is a spinor wave function, it has to have an $s$

Semiclassical

01:58

Do you mean in the sense of whether it's $\binom{1}{0}$ vs. $\binom{0}{1}$?

bolbteppa

Yeah e.g. $s = 1/2$ or $s = -1/2$

Semiclassical

Okay, though to be clear that's really $m_z=\pm 1/2$

bolbteppa

What do you mean?

Semiclassical

I'm just saying that that's the more standard notation

$s=1/2$, $m_z=\pm 1/2$ (or is it $m_s=\pm 1/2$? now I forget)

vs. there being some $\vec{s}$

bolbteppa

I've usually seen $m_z$ or just $m$ as the notation for the eigenvalue of the $\hat{l}_z$ angular momentum generator, and $s$ is the spin

Semiclassical

02:00

I've usually seen $s$ for the spin quantum number

bolbteppa

If you had no idea about anything, and just set up a hydrogen atom Schrodinger equation for $\Psi(x,y,z)$, you could, by group theory alone, realize one should include an extra $s$ variable to $\Psi$, $\Psi(s,x,y,z)$ - two ways to see it:

one can do it locally from Lie algebra representation theory, noting one needed to fix $\hat{l}_z$ but noting the direction it was fixed in doesn't matter (leading to spinors), or globally from the representation theory of non-simply connected Lie groups (again leading to spinors), one can go more primitive than this but lets ignore that for now, this logic also applies to Bohmian mechanics

Semiclassical

What you're saying is not quite right, though there's a fix to it

Suppose I have a spinor wavefunction which factorizes as $\Psi(x,y,z)=\phi(x,y,z)\frac{\uparrow_z+\downarrow_z}{\sqrt{2}}$

I'm not going to be able to claim that such a state will have a definite label of $m_z$

So to say that that's really $\Psi(x,y,z,m_z)$ seems basically incorrect to me

bolbteppa

Not sure what that means, just work with eigenfunctions for now, what does it mean for an eigenfunction to have an extra degree of freedom (the $s$), how does it even arise in the first place

Semiclassical

One could imagine a story where the particle carries a label $m_z$ in addition to its position $(x,y,z)$

but that's not true in the dBB story.

The fix I'd note (though this isn't a perfect fix either) is that if you've got a factorizable state $\Psi(x,y,z)=\phi(x,y,z)\chi$, then $\chi$ will be an eigenstate of some spin operator $\hat{n}\cdot \vec{S}$

and without loss of generality I can take it to be a spin-up eigenstate, i.e. $\chi=\uparrow_n$

in that case I can credibly claim that my wavefunction is of the form $\Psi(x,y,z,m_n=+1/2)$

But, not all $\Psi(x,y,z)$ are factorizable as such

So that's not a generic scenario

Overall I'm not really getting what you're after, though. Sure, you can write $\Psi(x,y,z;m_z)$ to denote what the spinor wavefunction would look like in the Sz basis

But what does not follow is that dBB must treat $m_z$ the same way as $x,y,z$. Whether or not you think dBB should treat them all the same way---well, the simple fact is that it doesn't.

dBB only insists on position values and the wavefunction having an objective meaning, independent of experimental conditions. it makes no such claim about any $m_s$.

"In short, for the (deterministic, hidden variable) pilot-wave theory, the outcome of “measuring the z-component of the spin of the particle” is not simply a function of the initial state of the “particle” (i.e., the particle+wave complex). The exact same initial state can yield either of the two possible measurement outcomes, depending on which of two possible experimental devices is chosen for performing the experiment."

bolbteppa

02:24

So $\Psi$ has an objective meaning, $x,y,z$ have objective meanings, but one of the variables of the $\Psi$ is not objective?

Semiclassical

Sure. I mean, there are cases when you can attribute an objective meaning to it: If I send an electron through an SG device and have it come out along one direction, then I'm guaranteed to have that same electron come out along the same direction if I send it through again.

bolbteppa

Man come on, that's the end of dBB right there haha

Semiclassical

Why?

Spin isn't a classical property.

bolbteppa

$f(x,y,?????) = :)$

Semiclassical

Why should it be objective in the same way that position is?

I do so hate it when one source quotes X as having said Y without actually citing them. it makes tracking down sources so much harder.

I guess this is the closest instance at hand:

Bell: "We have here a picture in which although the wave has two components, the particle has only position.... The particle does not ‘spin’, although the experimental phenomena associated with spin are reproduced. Thus the picture resulting from a hidden-variable account of quantum mechanics need not very much resemble the traditional classical picture that the researcher may, secretly, have been keeping in mind."

(source: books.google.com/…)

bolbteppa

02:42

Well, if paths do exist in a dBB world (but are just hidden from measurement), then things like positions, velocities, and angular momentum also really exist and are just hidden from measurement also, and so spin also really exists (hand-wavingly due to it's relationship to angular momentum, more seriously because it's a representation of angular momentum) and is also just hidden from measurement,

so it's theoretically just another well-defined but hidden variable, and immediately contradicts the rotational invariance and representation theory it arose from, so you just have to artificially, and completely unjustifiably, pretend spin is different for no reason to try make things work

Semiclassical

Why, exactly, must all variables in a theory be given the same status?

The way one makes a spin measurement is by observing a particle to move along some direction in space.

bolbteppa

If paths exist (but are just hidden from measurement), then in terms of the hidden variables, positions, velocities, momenta, angular momenta and, by therefore spin, must also exist for the hidden variables and all be treated equally

Semiclassical

Which is to say, one infers angular momentum by making a spin measurement.

@bolbteppa Whether or not you think they should exist, it is simply not the case in any dBB story I know that they all have the same status.

bolbteppa

The idea that spin must be treated differently is pretty much a cop out tbh

Semiclassical

Well, tell me then: How do you measure a spin component without making a position measurement?

bolbteppa

02:46

The problem is why a spin component even arises in the first place

Why not flubber components also

Lets throw in the old skinniness variable

Semiclassical

If flubber components were required in order to describe experiments, then we'd use them

They're not, so we don't

What's lurking around the corner here, I should note, is the Kochen-Specker theorem

"A second line [of development] is the Kochen–Specker one. The essential difference from Bell's approach is that the possibility of underpinning quantum mechanics by a hidden variable theory is dealt with independently of any reference to locality or nonlocality, but instead a stronger restriction than locality is made, namely that hidden variables are exclusively associated with the quantum system being measured; none are associated with the measurement apparatus. This is called the assumption of non-contextuality. Contextuality is related here with incompatibility of quantum mechanical ob…

The point of relevance here is the point about hidden variables being "exclusively associated with the quantum system being measured; none are associated with the measurement apparatus." the K-S theorem says that no hidden variables theory like that can reproduce the predictions of QM

the point is, though, that dBB doesn't deal exclusively with such noncontextual properties

it takes position to be primitive and noncontextual, but it basically doesn't treat anything else in that way

quoting from a different source: "Bell has said that (for Bohmian mechanics) spin is not real. Perhaps he should better have said: “Even spin is not real,” not merely because of all observables, it is spin which is generally regarded as quantum mechanically most paradigmatic, but also because spin is treated in orthodox quantum theory very much like position, as a “degree of freedom”—a discrete index which supplements the continuous degrees of freedom corresponding to position—in the wave function.

ugh, now I'm just being spammy

bolbteppa

03:02

I can't believe it so egregiously ignores the group theory underlying spin, to pretend we can treat spin as different to other variables, I mean spin wouldn't exist if we weren't dealing with a non-simply connected group and the possibility of projective representations, because of this it's kind of just angular momentum, there is really no reason to treat it differently in terms of being 'real'

This is another shocker with Bohm tbh

Semiclassical

I strongly suspect part of the issue here is that the people who have been most interested in dBB aren't people who do much group theory

bolbteppa

Yeah the group perspective on QM is the hardest part of it all, it's really a question of picking up tiny things here and there and hoping for the best :p

Semiclassical

Which to me points to the real embarrassment of pilot-wave theory: I'll defend it as being internally consistent and compatible with experiment. But it is remarkably detached from the actual practice of physics.

i mean, I think there's a self-fulfilling prophecy there: physicists don't like what they see, so they don't work on it and it remains obscure

bolbteppa

It sounds like you're just defending normal QM and calling it pilot wave theory by doing a few extra random calculations (e.g. characteristics)

Semiclassical

eh. I don't think they're as random as you say they are.

but there is a fact: if at the end of the day you get the same predictions as QM, just with more work, then is there really a point to it?

I think the main response by people in the dBB community would be that orthodox QM is more than just a wave equation or unitary time evolution; it's also the measurement postulates. And a big reason philosophers have tended to like dBB is because it has the measurements postulates arise from the theory rather than needing to be included as axioms.

But this again points to the distance to actual physics practice: How many physicists stay up at night worrying about the measurement problem?

bolbteppa

03:10

I wonder if 'orthodox' qm, i.e. assuming random axioms, is responsible for this thinking that we can just pick and choose different axioms and get different QM theories

Semiclassical

my own feeling is that, whatever appeal dBB may have to philosophers when it comes to the measurement problem, this is simply not a sufficient argument

bolbteppa

We all know they really like dBB because of determinism, that's the whole draw of this theory to like 99.9% of people interested haha

Semiclassical

feh. I don't think 'determinism' is the chief thing, if only because a lot of the same people who like dBB are also willing to put up with stuff like GRW with its intrinsic stochasticity

I think the motivation is more along the lines of wanting to be able to talk about quantum processes as taking place in spacetime, in a concrete way

I think this is probably a bit too pat, but to a first approximation I'd say the point is not "determinism" but "visualizability"

in that regard, GRW and dBB aren't so far off. with dBB, you insist that there's a trajectory description at all times; this comes at the cost of having to invoke an equilibrium hypothesis and to accept nonlocality.

with GRW, you insist that there's a trajectory description except at those random points where there's a discontinuous quantum jump. that gets you out of some issues, but it leaves the quantum jumps as inexplicable.

in both cases, though, you still have a story where there's some tidy visualization of the system as a particle undergoing a (possibly non-smooth and therefore indeterministic) trajectory

bolbteppa

03:29

"So what do the Bohmists do? They simply deny that the spin exists in the same way as the positions or velocities do. While the wave functions - now real waves - still depend on the spin (because the wave functions must clearly be spinorial for an electron), there is no additional degree of freedom associated with them.

That's a tragic tumor, a seed of self-destruction, inserted into the very heart of the pilot wave theory but the Bohmists must obviously try to transform this tumor into a virtue."

"Imagine a world where not all observables (which are linked to Hermitean operators in quantum mechanics) are "fundamental" at a hypothetically deeper (Bohmian) level. Only a subset of the "primitive" ones can be associated with real "properties" (additional classical degrees of freedom that accompany the wave function).

The position and the velocity are "real properties" and so is the orbital angular momentum - which is just the cross product of the position and the momentum. However, the spin can't be a "real property", as the Bohmists (and I) explained: it must be just "contextual", oth…

Semiclassical

more

oops

I think that dBB people might take issue with even claiming that velocity is at the same status as position

but that requires a citation

bolbteppa

If a path exists in any sense of the word, even a hidden path, a velocity exists

Hidden path implies hidden positions and hidden velocities, otherwise paths don't exist

Semiclassical

right, this article: mdpi.com/1099-4300/20/6/440/htm

is the one I have in mind

so the claim that 'hidden velocities must be at the same status as hidden positions' isn't one which all dBB people would assent

though, as I look at it, they're talking more about momentum than velocity per se

not sure what to make of that

bolbteppa

It's breaking math to say that the positions at each point in time are hidden variables and that they generate a hidden path but also the velocity doesn't exist

Semiclassical

it also doesn't seem to really match the very point of the guidance equation

from that you get that the Bohmian velocity is, via the wavefunction, a function of position and time

so the velocity is not independent of position, like it would be in newtonian physics

but it still exists in the dBB story, so I'm puzzled as to what that author could possibly mean

and once you grant that, you seemingly do grant at least some meaning to the orbital angular momentum $\vec{L}=m\vec{v}\times \vec{r}$

bolbteppa

03:47

Hidden position at each time generates a hidden path, this hidden path has a hidden tangent vector which is it's velocity at each point along the hidden path, the hidden positions and hidden velocities are different from the measured positions and measured velocities or no

Semiclassical

Eh, that last line is a bit tricky

Semiclassical

04:06

the reason i'm hesitant to sign onto it is because I've seen statements like this (from a paper I can't access atm):

Aug 4 at 3:49, by Semiclassical

"Thus, while a measurement of [the Bohmian] position is always a measurement of the position operator, a measurement of the position operator is not necessarily a genuine measurement of [the Bohmian] position!"

oh, wait. i was thinking a different paper. that particular remark is from here: books.google.com/…

Semiclassical

04:27

(this paper has more discussion of such, c.f. page 5: arxiv.org/pdf/1509.00767.pdf)

Cows

04:52

:D

Ok done reading for the night :)

vzn

in theory salon, 3 mins ago, by vzn

reddit on Terence Tao vs QM (wrt NYT profile) https://www.reddit.com/r/math/comments/9az4th/til_terry_tao_thought_he_didnt_hav‌e_to_study_for/

user157860

06:11

@JohnRennie, Hi John, concerning the formula we discussed:$$E = \sqrt{p^2c^2 + m^2c^4}$$, I realized that $p$ is equivalent to $m\gamma v$ in the case of a particle and $E/c$ if $m=0$, if it is so, is it possible for a formula to have 2 different readings (in math or physics) ?

John Rennie

← 1 message moved from Problem Solving Strategies

@user157860 it depends what you mean

The momentum is also equal to $h/\lambda$ where $\lambda$ is the de Broglie wavelength. For a photon the de Broglie wavelength is just the light wavelength.

user157860

@JohnRennie, I mean one term of any formula should not have only one referent?

John Rennie

So all the quantities E, p, $\lambda$ are all inter-related.

user157860

yes , but is it possible that a formula must be examined in order to decide what substitute to P?

John Rennie

For a photon you can write $E=pc$, $E=h\nu$, $E=hc/\lambda$ and all three equations are really the same thing written in different ways.

Semiclassical

06:18

more generally, one has $E/pc = v/c$

and that limit remains valid in the limit $v\to c$

user157860

yes, but in each equation each term means only one thing. In our formula we must first decide if it regards a particle or a photon, and then interpret p consequently, we choose if it is $m\gamma v$ or E/c. Isnt't that rather odd or at least unusual?

Semiclassical

blah, should've said $E/pc = c/v$

user157860

@Semiclassical, do you get my point?

Semiclassical

point stands, though: regardless of velocity, one has $p=\gamma m v = Ev/c^2$

and that last result still makes sense even as $v\to c$

John Rennie

Moment as in $mv$ is a classical concept, but photons require quantum mechanics to describe them.

Semiclassical

06:22

so if one has a particle going at the speed of light, one had better have $p=E/c$

John Rennie

Once you move to QM momentum has a precise meaning as the eigenvalue of the momentum operator.

$$ \hat{p} = -i\hbar\frac{d}{dx} $$

Note that the mass does not appear in the operator.

For both massive and massless particles we get the momentum by applying this operator to the wavefunction, so the distinction between massive and massless particles disappears.

user157860

yes, it's like that, but it's a SR formula, you can't let a term of a formula be ambiguous, right? You can't have a formula a-la-carte, :"..insert here the value for the case may be". Do you think the formula is fully legit?

If you have a master-key term,you should use that, and not p

Semiclassical

i mean, you're necessarily going to have the photon be a weird case: the only speed it's allowed to have is $c$. you can't have a photon at rest

John Rennie

Yes, the formula is fully legit, and indeed is routinely experimentally confirmed every day at particle colliders like the LHC.

Semiclassical

as such, one shouldn't really be shocked to find that the case of a photon's momentum must be dealt with differently than that of, say, an electron

user157860

06:28

of course it works, if you pick the right choice, what I am objecting is the choice itself , p should be replaced by only one term

John Rennie

You are worrying about something that is not a problem. In QM the problem you're worrying about disappears, and in classical mechanics we know the problem isn't there in the fundamental theory (QM) so we aren't fussed about treating massive and massless particles differently.

Semiclassical

anyways, to me the most fundamental thing is the invariant $E^2-(pc)^2=(mc^2)^2$

to the extent that we take that seriously, then when $m=0$ one must have $E=pc$ as the correct formula

it's the only limiting case possible

(that's even true when you start formulating the quantum mechanics of a relativistic electron: you start from the desire for $E^2-(pc)^2=(mc^2)^2$ to hold true in some appropriate sense. you have to go through a lot of work to the Dirac equation from that, but it's still the starting point)

user157860

@Semiclassical, :"then when m=0 one must have E=pc as the correct formula " that is what I am objecting, the fact that you must choose how to interpret p,

That is not allowed in math

It surely unortodox, is it?

Semiclassical

Suppose I've got a particle of mass $m$ with energy $E$. Then that formula gives its momentum as $pc=\sqrt{E^2-(mc^2)^2}$.

If I suppose $E\gg mc^2$, then I can express that as $pc=E\sqrt{1-(mc^2/E)^2}=E(1-mc^2/E)^2/2+\cdots)=E-\frac{mc^2}{2E}$

So if $m$ is small but nonzero, then $pc$ is approximately $E$. in the limit $m\to 0$, you get $pc=E$

the only thing I'm assuming is that I take the mass to be small while holding the total energy fixed.

once I do that, I find that the only possible relation between the momentum and the energy is $p=E/c$

user157860

@Semiclassical, you have 2 formulae embedded in one $$E = \sqrt{m\gamma^2^2c^2 + m^2c^4}$$ and $$E = \sqrt{(E/c)^2c^2 + m^2c^4} $$

Semiclassical

06:36

That's not a strange alternative case. That's the only appropriate limit.

What?

The first equation, latex issues aside, seems fine. But I have no idea what you mean in the second one

$p=E/c$ is only valid when $m=0$, in which case what you've written amounts to $E=\sqrt{(E/c)^2 c^2}$ which is trivial

@user157860 (can't have squared squared in latex)

user157860

@Semiclassical, if it's a particle the equation in its full form is $$E = \sqrt{m\gamma^2v^2c^2 + m^2c^4$$ and if m=0 it becomes $$E = \sqrt{(E/c)^2 + m^2c^4$$

Semiclassical

no, it doesn't. That's not a valid equation

for one, the algebra looks wrong. moreover, you're neglecting that as $v\to c$ one has $\gamma\to\infty$

So one can't naively take $m\to 0$ in $p=\gamma m v$

Nor can you do so in $E=\gamma mc^2$.

the energy-momentum relation in full is $E=\sqrt{p^2 c^2+m^2 c^4}=\sqrt{\gamma^2 m^2 c^2 v^2+m^2 c^4}$

user157860

@Semiclassical, this formula is correct for particles, I discussed it with John:$$E = \sqrt{m^2\gamma^2v^2c^2 + m^2c^4}$$

Semiclassical

you're misquoting it. $m$ has to enter as $m^2$ for units to work

okay, sure. point still stands: a photon is both massless and travels at the speed of light

therefore you'd need both $m=0$ and $v=c\implies \gamma=1/\sqrt{1-v^2/c^2}=\infty$

So one can't take $m=0$ in the first term and expect to get a meaningful result

user157860

@Semiclassical, please develop the original formula into a full blown one which is always valid, if you can

Semiclassical

06:47

on the other hand, for nonzero mass you can write the above as $E=\sqrt{p^2 c^2+m^2 c^4}=pc\sqrt{1+m^2 c^2/p^2}$

and that last equation is valid regardless of velocity, so long as $p$ is finite

with gamma removed, we now can validly take the limit as $m\to 0$.

The only way to 'develop the original formula into a one which is always valid' is to write it in a way that doesn't involve $\gamma$, since $\gamma\to \infty $ as $v\to c$

once you do that, then it's clear that when $m=0$ the only possible relation between $E$ and $p$ is $E=pc$

user157860

@Semiclassical, can you do that, I think it's impossible

Semiclassical

I just did. $E=\sqrt{p^2 c^2+m^2 c^4}$

That relation between $E$ and $p$ is valid for all velocities and all masses.

alternatively, if you want a result that doesn't involve mass, you can do as I did earlier and rewrite that to $E=pc^2/v$

in both cases, it's clear that the photon---a massless particle moving at the speed of light---will satisfy $E=pc$.

John Rennie

07:09

@user157860 there is a sense in which $p$ is the fundamental property and mass and wavelength are derived from it. In relativity the rest mass is the norm of the four-momentum. So if you regard the four-momentum as fundamental then mass is a derived quantity.

Semiclassical

mass is what ensures $E^2>p^2 c^2$

John Rennie

It is as a result of this that we get situations where the mass of a system is different to the mass of its components. For example the mass of a hydrogen atom is less than the mass of a proton plus the mass of an electron.

Semiclassical

when $m=0$, then you have $E=pc$

John Rennie

In that case the expression $E^2 = p^2c^2 + m^2c^4$ is the fundamental equation that applies to anything. We may choose to replace $p$ by one of the two derived quantities $m$ or $\lambda$.

But all we're doing is taking the master equation and writing it in two different ways.

Anurag Baundwal

07:57

Help ... I've opened this page on Chrome (mobile) ... how do I render mathjax?

Bernardo Meurer

@JohnRennie Heya!

John Rennie

@BernardoMeurer morning stranger! :-)

Bernardo Meurer

Long time no see! :D

How have you been?

John Rennie

Life goes on ...

Things tend to move along pretty steadily when you're 57 :-)

Bernardo Meurer

Hehehe, the dream!

John Rennie

08:04

It's a lot more ... erm ... eventful when you're 18 :-)

How did the summer job go? Good fun?

Bernardo Meurer

Glad to hear everything is smooth, is your mother doing okay? Chester not too cold?

Summer job was awesome! They are hiring me part-time while i am in school :D

John Rennie

Cool :-)

Bernardo Meurer

Which is great timing since I need to start paying for my own education

John Rennie

Ah, your Dad isn't playing ball?

Bernardo Meurer

He, much like the economy in Brazil, keeps getting more and more unstable

So I don't want to depend on either anymore

John Rennie

08:07

Of course the IT industry is full of millionaires who enjoyed working in IT so much they dropped out of college ...

Bernardo Meurer

Yes, they offered me to drop out and join...

But I don't feel comfortable doing that

John Rennie

It's worth considering.

Bernardo Meurer

Also, I left CS as a major

Well, had I taken it I'd have a lot of money right now

John Rennie

You did?

Bernardo Meurer

I did, not because I wanted to though

13 hours ago, by Bernardo Meurer

@ACuriousMind I'm sick and tired of being in a CC, I can't stand it anymore. This means I need to finish IGETC (Basically a list of classes you have to take to transfer) and my major requirements ASAP. CS has a lot of requirements, meaning I'd have to be here another two years (totalling 3), so I'm changing majors. My major is not Math (obviously focusing on CS) and I'll be in some UC (Berkeley or LA) hopefully this time next year. Once there I'll declare a CS double major everything going well.

John Rennie

08:09

What's a CC?

Bernardo Meurer

Community College

John Rennie

Don't rule out dropping out of college.

Bernardo Meurer

Basically a cheap, not-so-good, uni you go to do your lower-level classes and then transfer to a proper uni to do your higher-level classes

John Rennie

If you're interested in a career in IT, like that startup you worked at, then I doubt a CS degree is that important.

Bernardo Meurer

Agreed, but I want to conquer a degree

I am a math major now

John Rennie

08:12

Really? Degree level math is an odd discipline. I could never do it because it has no motiviation other than studying maths for its own sake. I'm surprised you find that attractive.

Bernardo Meurer

I don't, but it's less bad than the rest

and I can get away with only going up to multivariable calculus and diffeq and then taking only discrete math classes

Which are the ones I do like

John Rennie

Well, OK, but you're paying an arm and leg for this as opposed to being paid to do something that's fun.

Bernardo Meurer

Sure, but with it I can come back and do a PhD if I want to

I don't want to kill that chance, it's something I think I want to do

John Rennie

Fair enough. The best reason for doing something is that it's fun.

Though I have to say that if the IT revolution had come round 20 years earlier I probably wouldn't have done a PhD.

Bernardo Meurer

I don't know, I really want to get a degree from a good school

And as a Math major I have a lot of chances of getting into Berkeley

John Rennie

08:18

My advice is worth what you pay for it, but I'd ask yourself why you want to go to Berkeley. Is it something you'll really enjoy? Or is it just a target that you've always had without ever really questioning why you have it?

If you're really enjoying coding for a startup then I'd code for a startup.

You can always go back to college after a couple of years when you have some money behind you.

Bernardo Meurer

I want to go to Berkeley so I have a shiny star on my CV that I can use to work where I want to work afterwards, and make it easier for me to get into a good PhD program

I like coding for a startup, but I also have little place to fall down if I make a mistake, or if the startup goes under. I think a degree will give me more security

John Rennie

Will it though?

Do many IT companies specifically want a degree?

Or PhD?

Bernardo Meurer

For what I want to do, yeah

At least in my experience

PhD not really, I want that for the fun

John Rennie

Fair enough then.

I certainly found doing a PhD enormous fun.

Bernardo Meurer

Yeah, I think doing a PhD is a shaping experience

For better or for worse

John Rennie

08:23

It's the first time in your life that you're not just learning from someone else. You're deciding what to do not just doing what you're told.

Bernardo Meurer

That freedom is a bit scary to be honest

John Rennie

It won't be when you get there. The reality is that you'll find there are shedloads of things you want to study. The only difficulty is choosing what to not study.

Jasper Loy

I heard that ocelot got suspended from chat for a year, sorry to hear that.

John Rennie

It used to annoy me that the lab didn't open till 9 a.m. so I couldn't go in at the crack of awn to work :-)

Jasper Loy

Hi @BernardoMeurer I see you are still handsome!

Bernardo Meurer

08:26

@JohnRennie Yeah, that's exactly why I find it a bit scary

@JasperLoy Eh, thanks!

John Rennie

Hey, I'm handsome too (in my dreams :-)

Jasper Loy

@JohnRennie You look handsome too. You remind me of John Lennon.

John Rennie

How is life otherwise? Are you living somewhere nice? HiFi good? Computers all working?

@JasperLoy the glasses I suppose :-)

Jasper Loy

@JohnRennie Well, you also have this cool musician type look.

By the way, I recently got a very nice basic physics textbook, A Course in Classical Physics 1--4 by Alessandro Bettini. I guess it's not well-known, so I thought I would share here.

It's a bit expensive to get the four paperbacks, but still better than the colourful first year college textbooks.

John Rennie

7

That's the whole picture. Not really the sort of gear John Lennon would have worn :-)

Jasper Loy

08:31

@JohnRennie What occasion was that that you were dressed like that?

John Rennie

It was a production of Die Lustige Witwe by Lehar. I used to be in an amateur opera group.

That was me dressed up as a nobleman.

Jasper Loy

I love opera! I am an amateur singer, lol.

John Rennie

Doing the amateur opera was really good fun. I only sung in the chorus, but that was fine by me. Less scary than having a major role. I recommend it if you have the time.

Bernardo Meurer

@JohnRennie Life is good, I think :)

HiFi is working well, although the remote function on the volume knob of my Marantz has stopped working

which is no biggie since I've never used it much

Jasper Loy

Is it true that macOS is vastly superior to Windows?

John Rennie

08:35

Damn, you mean you have to get up off the couch to change the volume!!!

Bernardo Meurer

The laptop you gave me has been retired and know lives a happy life as a Linux server in my closet

@JohnRennie I don't own a couch :P

All I have is a desk, so it's always nearby

John Rennie

@BernardoMeurer I guess it makes a good Linux server, though the disk space would be a bit limited given you can only fit two disks in it.

Bernardo Meurer

Yeah, but I only use it as a distcc node since my macbook is so underpowered

Soon it will also host some more fun stuff, like a public copy of all my music

John Rennie

My latest project has been squeezing ridiculously overpowered CPUs into tiny PCs.

Bernardo Meurer

"public", you'll need user/pw

Are you one of these people who like to make CPUs throttle? :P

John Rennie

08:38

I'm currently using an i7-4790 squeezed into a Dell microPC 3020m.

Jasper Loy

I am thinking of using SSD instead of HDD cos I heard it is much better.

Bernardo Meurer

Ooooh, nice!!

I wish I had one of those as a home server!

Although Intel only gives me headaches nowadays

John Rennie

@BernardoMeurer it seems OK. I thought it would overheat and throttle, but in fact it stabilisies around 80C.

Dell obviously did a good job on the cooling.

Bernardo Meurer

That's not bad at all!

How much did it cost total?

John Rennie

The 3020m was about £100 from eBay and the CPU was about £150.

Jasper Loy

08:40

Dell laptops seem to be cheaper than HP ones.

Bernardo Meurer

Nice price!

John Rennie

@BernardoMeurer This sort of thing

Bernardo Meurer

I still want myself a Thinkpad

John Rennie

Though you can get them with a Celeron CPU even cheaper - you'll be junking the CPU anyway.

Bernardo Meurer

fascinating that that can cool a CPU

Jasper Loy

08:41

My Lenovo Ideapad is not working well.

John Rennie

@JasperLoy both OSX and Windows have their strengths and weaknesses, use whichever you find nicer. SSDs are much faster than regular hard disks. Dell laptops are good value.

2

@BernardoMeurer well it's only a 45W CPU

and not overclocked.

Bernardo Meurer

Interesting

What's the core count? Per/core freq? Cache sizes?

John Rennie

Details here

It's pretty powerful for a four core. You need to jump to six cores to get a significant power increase.

Bernardo Meurer

Bah, darned shared L3 cache

Gave me issues recently

Yashas

10:50

$W = \int\vec{F}.d\vec{r} or W = -\int\vec{F}.d\vec{r}$

it doesn't matter, right?

conventions?

After starting uni, I feel like I know less physics than I knew in high school :|

2

Anonymous

11:17

@Yashas Depends on what $F$ is and by whom the work is being done. In general, I'd just find the magnitude and logically deduce the sign later.

Anonymous

@BernardoMeurer Hey! How're you doing? :D I saw you worked at an AI startup during the summer. How was it? Writing new code for them? Or coming up with new algos, etc?

Anonymous

Also, what are the advantages of Rust? I've absolutely no idea about it :/

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