The h Bar

2:25 AM

from what I can gather, a microwave (appliance) applies waves in the range where water molecules in food absorb the most heat, why does the plate, then, get hot? is it simply from the transfer of heat from the food?

I could just put a plate in the microwave to test this but I have a feeling it would still get hot.

John Rennie

6:01 AM

In general crockery doesn't get hot if you put them in the microwave. When I make coffee I always put milk in the mug then put the mug in the microwave to heat the milk, and the handle of the mug doesn't get hot. The body gets hot, but that's conduction from the hot milk because you can feel the temperature difference above and below the top of the milk.

Relativisticcucumber

the handle can definitely get hot. ive burned myself many times - depends on the material i think

John Rennie

I guess it's possible for some types of ceramics to absorb microwaves.

Pure silica, i.e. glass, does not absorb microwaves and does not get hot in a microwave.

rob

6:56 AM

I bought some new stoneware dishes a year ago and actually (gasp) read the paper that came with them.

Mine have an unglazed circle at the bottom, which I guess is where the dishes sit in the kiln.

If you let them soak in water, they can absorb water through this unglazed section into the bulk of the dish, which makes them stop being microwave-safe.

The way to test is to put an empty dish in the microwave next to a full teacup or measuring cup of water or the like, and heat the water.

If the empty dish gets hot, whether it says “microwave safe” on it or not, it’s not microwave safe any more.

John Rennie

I need some more coffee. If I never post again you may assume the microwaved mug has immolated me.

rob

[waits with bated breath]

[fixes autocorrect-induced pun about “baited breath” — @JohnRennie and I do not have that kind of relationship]

John Rennie

7:17 AM

Phew, survived again!

Silly Goose

9:26 AM

what is the big picture importance of unitary operators in quantum mechanics? I feel like I have bits and pieces like:
1) generators of unitary operators are hermitian
2) they commute with the hamiltonian if the hamiltonian is invariant under said unitary transformation (and so lead to their generators being conserved quantities)

Slereah

9:36 AM

Unitary operators do not change the probability of a state

ACuriousMind

@SillyGoose Wigner's theorem says symmetries/transformations must be (anti-)unitary operators

Silly Goose

9:53 AM

what is a non copenhagen justification of sticking to rays instead of vectors in QM?

Slereah

You're gonna have to be more specific

there are many interpretations and many of them don't use Hilbert spaces in a natural way at all

Silly Goose

hm i guess if we ignore any interpretation relating to probabilities, and if we are sticking to the Hilbert space formulation of QM, is there still the postulate that vectors up to phase are the same states?

Slereah

I mean do you have one in mind?

Silly Goose

also what does it physically mean to perform a unitary transformation on your system (Hamiltonian?). In any physical theory where reality is mapped in coordinates i feel like it is intuitive what a rotation transformation is, but in QM i mean you are working just in abstract Hilbert space?

i do not, but I do not understand why you would want to make the claim states are same up to phase unless you wanted to be able to arbitrarily normalize your states (which has a probability tinge to it)

Slereah

It is an experimental fact that QM does have probabilistic results, so it is something that you will have to deal with at some point of your theory

whether that is something fundamental or not

but Bohmian mechanics doesn't use Hilbert spaces as a basic component, no

At least I don't think so?

Who's our local Bohmian expert

I forget his name

ACuriousMind

10:03 AM

@SillyGoose that's not interpretation-dependent, it's just the Born rule

@SillyGoose I'm not exactly sure what you mean by "physically". You can, as usual, interpret transformations as active - "I'm rotating my system!" - or passive - "I'm rotating my coordinates/description of the system".

I don't see why the "abstract" nature of the Hilbert space would matter here: After all, classical mechanics also works with generalized coordinates that don't have to have anything directly to do with real-space position

Silly Goose

but i guess i don't see how you can identify coordinates with the usual x-y-z of Euclidean space that we experience

Slereah

which coordinates?

Silly Goose

well that is the thing also

i mean also, perhaps a related thing, you cannot visualize in space as we experience it first of all a complex wave function and so definitely not a "spatially" translated wave function, or at least I think it would be incorrect to do so

Slereah

I mean you have a position operator

and you do have coordinate invariant measurements you can perform on such things

For instance describe a system as your measuring apparatus and another as your studied system

The distance between the two will be independent of coordinates

you can also set coordinates and then your unitary operators may act on them, ie a rotation on a coordinate system

Silly Goose

but i mean the position operator for each system is a different operator right

so what tells you you can treat them as the same quantity

Slereah

10:17 AM

what do you mean

Silly Goose

the position operator $\hat{x}_1$ defined over $H_1$ representing system 1 is a different operator than the position operator $\hat{x}_2$ defined over $H_1$, no?

because one is a map from vectors in H_1 to H_1 and the other is a map from vectors in H_2 to H_2

ACuriousMind

the coordinate $x : \mathbb{R}\to M_1$ for the configuration space $M_1$ of one system is a different coordinate then $x: \mathbb{R}\to M_2$ for the space $M_2$ of another

this is not a quantum concern :P

Silly Goose

what do you mean by that?

Slereah

Although the fact that they are about the particles being in the same space is represented by how your spatial transformations act on those

ACuriousMind

@SillyGoose I mean if you're worried that two systems have different state spaces and hence their positions are "different operators" that's no different in classical mechanics where technically each system also has a different configuration space

Slereah

10:20 AM

You have some representation of your Galilean/Lorentz group acting on those

ACuriousMind

if you don't worry about the "x" in classical mechanics technically being a different coordinate for each system I don't see why you'd worry about it in quantum mechanics

Mr. Feynman

@ACuriousMind didn't you swap $\mathbb{R}$ and $M$?

Slereah

In the case of two particles it will be some reducible representation that acts on those two

ACuriousMind

@Mr.Feynman which way coordinates go is an annoying convention

you can swap them if you like

Slereah

like a "one particle system" in QM will have its relation to the underlying space by some irrep of the kinematic group you're using

Mr. Feynman

10:23 AM

I'll do :P

Slereah

And if you add additional particles, that is some reducible rep which is some product of the irreps

[up to braiding]

ACuriousMind

@SillyGoose also, I don't see why the position operator being different matters at all: By definition of what we mean by a rotation, any eigenstate $\lvert \vec x\rangle$ of the 3d position operators should be mapped to $\lvert R\vec x\rangle$ by a rotation $R$, where $R\vec x$ is the classical rotation of the vector $\vec x$.

Slereah

That's the difference between say, a particle in $\mathbb{R}^2$ and two particles in $\mathbb{R}^1$

Even though they have the same Hilbert space technically

Since you have $$L^2(X, \mu) \otimes L^2(Y, \lambda) \cong L^2(X \times Y, \mu \times \lambda)$$

ACuriousMind

this is exactly the same logic as figuring out how the momentum operator acts as $\mathrm{i}\partial_x$ by saying it has to act as $\lvert x\rangle \mapsto \lvert x+a\rangle$ (see e.g. this answer of mine)

@Slereah I mean, all separable Hilbert spaces are isomorphic

I know what you want to say but saying these spaces are isomorphic as Hilbert spaces is kinda vacuous

Slereah

Some Hilbert spaces are isomorphiers than others

Very wise pigs

Silly Goose

10:33 AM

XD

Slereah

It's one of those QM thing that I think isn't as well talked about rly

A quantum theory is as much about the operators acting on the Hilbert space as the Hilbert space itself

Silly Goose

i mean how can you classically rotate a system when the system itself does not have a definite position necessarily

ACuriousMind

@SillyGoose easy: I have a probability distribution $\rho(x,p)$ on phase space and the rotation is $\rho(Rx,Rp)$

Slereah

There are no definite positions in the real world either and you can still rotate things

ACuriousMind

likewise, you have the quantum wavefunction $\psi(x)$ and the rotation is $\psi(Rx)$

(possibly with $R^{-1}$ instead of $R$ depending on how you think about rotations vs. coordinates)

Slereah

10:39 AM

But yes rotation is kind of a thing that requires you to set some coordinates first

but then again if you don't have coordinates, you don't have much need for rotation

ACuriousMind

@Slereah I think the non-vacuous statement here would be a statement that that isomorphism is given by the canonical embedding of functions $F(X)\otimes F(Y)\to F(X\times Y)$

Slereah

the two systems also share the same number of canonical operators and commutation relations

But they are acted on differently by the Galilean group

Mr. Feynman

12:44 PM

How would you make sense of the definition of $T$-matrix in ordinary QM given in $(6.9)$?

6.11 seems more reasonable as a definition. I mean, I can't get over the definition of $T$ depending on a small parameter $\varepsilon$ here (which of course is there to prevent a divergence)

Should I accept $(6.9)$ as a definition of $T$ up to $\mathcal{O}(\varepsilon)$?

ACuriousMind

@Mr.Feynman the text explicitly says you're supposed to take $\epsilon\to 0$ and then $t\to\infty$

Mr. Feynman

@ACuriousMind Sure, but I mean $T$ appears before doing so

ACuriousMind

it's just that since 6.10 isn't written as $T= \dots$ there's no nice way of explicitly writing those limits into that equation

Mr. Feynman

(Oh sorry I wrote 6.9 above :P)

So the "real" definition is 6.11

ACuriousMind

but they do certainly intend to say that the definition of $T$ is via that limit, i.e. you get some $T(\epsilon,t)$ from 6.10 and then the actual $T$ is $T = \lim_{t\to\infty}\lim_{\epsilon\to 0} T(\epsilon,t)$

Mr. Feynman

12:54 PM

I ask this also because then the text does some manipulations only considering $t_0=-\infty$ and the upper extremum as $t$

But it might just be it is easier to work this way since in those calculations they eventually disappear

One more silly question, by using the integral representation of the delta function in $(6.11)$ they are assuming the spectrum is continuous (?), so wouldn't be better to replace $\delta_{ni}$ with a Dirac delta too?

So I can write $S=1-2\pi i T$ (?)

ACuriousMind

@Mr.Feynman no, there's a difference between $\delta_{ni}$ and $\delta(E_n- E_i)$ - the former is saying the indices have to be the same, the latter is saying the energies have to be the same

you can have $n\neq i$ but $E_n = E_i$

Mr. Feynman

1:14 PM

Oh gosh right, that was a stupid question indeed. I got tricked not thinking of non degenerate states but even then thing wouldn't change

Moreover, that delta is a statement of conservation of energy which is a real variable so no wonder it's inside a delta

So isn't there an easy relation between the operators $S$ and $T$?

I've occasionally seen $S=1-2\pi i T$ but now I'm not sure that makes sense comparing to 6.11

ACuriousMind

@Mr.Feynman That's correct if you enforce energy conservation by something else

Mr. Feynman

@ACuriousMind Hm, I'm kinda uncomfortable with the delta disappearing in that case

Now I have a good reason to hate this notation for distributions :P

But I guess it's like Fermi's golden rule and it's really integrated over a range of energy and what the last remark means is that this range contains the support of the delta distribution

bolbteppa

2:56 PM

@SillyGoose Do you get it now or do you want another explanation

More Anonymous

4:24 PM

Random question which genre is this?? I seem to dig it

Listen to The mixtape by Anant Saxena on #SoundCloud
https://on.soundcloud.com/a2yP8

Silly Goose

8:01 PM

@bolbteppa id like to hear another definition for sure

bolbteppa

8:25 PM

Given the Schrodinger equation with a time-independent Hamiltonian, you know the wave function reduces to a sum in terms of orthonormal stationary states $\Psi(x,t) = \sum_n c_n e^{-(i/\hbar) E_n t} \psi_n(q) = \sum_n c_n(t) \psi_n(q)$ where $\int dq \psi_m^*(q) \psi_n(q) = \delta_{mn}$. Interpreting the $\psi_n(q)$ as basis vectors, we can ask whether transformations of the basis $\psi_n'(q) = \hat{S} \psi_n(q)$ exist which preserve orthonormality.

If $\Psi(q) = \sum_n c_n \psi_n$ then $c_n = \int \psi_n^*(q) \Psi dq$. If we set $\hat{S} \psi_n = \sum_m c_m \psi_m$ then $c_m = \int \psi_m^*(q) \hat{S} \psi_n(q) dq := S_{mn}$ so that $\psi_n' = \hat{S} \psi_n = \sum_m S_{mn} \psi_m(q)$. So we want
$$\delta_{mn} = \int \psi_m'^*(q) \psi_n'(q) dq = \int \int S_{am}^* \psi_a^*(q) S_{bn} \psi_b(q) dq = S_{am}^* S_{bn} \int \psi_a^*(q)\psi_b(q) dq = S_{am}^* S_{bn} \delta_{ab} = S_{cm}^* S_{cn} $$
i.e. $\hat{S}^{\dagger} \hat{S} = I$. Thus $\hat{S}^{-1} = \hat{S}^{\dagger}$. Studying $\psi_n = \hat{S}^{-1} \psi_n'$ gives you $\hat{S} \hat…

Thus from $\hat{H} \psi_n = E_n \psi_n$ we have $\hat{H} \hat{S}^{-1} \psi_n' = E_n \hat{S}^{-1} \psi_n'$ so that $\hat{S} \hat{H} \hat{S}^{-1} \psi_n' = E_n \psi_n'$. If we also have that $[\hat{S},\hat{H}] = 0$ then $ \hat{H} \psi_n' = E_n \psi_n'$, meaning that the $\psi_n'$ obtained from a unitary operator $\hat{S}$ acting on $\psi_n$ of $\hat{H}$ will have the same energy eigenvalues $E_n$ if $[\hat{S},\hat{H}] = 0$ holds.

Setting $\hat{S} = e^{i \hat{T}}$ for $\hat{T}$ hermitian means this reduces to $[\hat{H},\hat{T}] = 0$. Since $\hat{T}$ is hermitian, it always has an eigenvector, so we can find simultaneous eigenvectors of $\hat{H}$ and $\hat{T}$, this can be used to simplify solving a given Schrodinger equation.

For example, since the Hermitian generators of the rotation group $\hat{J}_i$ commute with a central potential Hamiltonian (e.g. hydrogen atom), it means we can rotate the eigenfunctions and they will still have the same energy. Since the generators are Hermitian, the eigenfunctions could be simultaneous eigenstates of $\hat{H}$ and all the $\hat{J}_i$,

however since the $\hat{J}_i$ don't commute amongst themselves, $\hat{H}$ and only one of them (usually $\hat{J}_z$, with eigenvalue $m$) admit simultaneous eigenstates. Since the casimir $\sum_i \hat{J}_i^2$ (with eigenvalue $l(l+1)$) does commute with the $\hat{J}_i$ this gives is a further operator having the same eigenstates, so the simultaneous eigenfunctions are denoted $\psi_{n,m,l}$.

If we chose $\hat{J}_z$, then rotations about the z-axis would not disturb the eigenfunctions $\psi_{n,l,m}$. However rotating around some other axis does disturb the eigenfunctions: it wont change the energy eigenvalue because $[\hat{J}_i,\hat{H}] = 0$, nor the value of $l$ because $[\hat{J}^2,\hat{J}_i] = 0$, but the value of $m$ will be affected.

Thus we need to study all irreducible representations of the rotation group, where we find that $m$ can take $2l+1$ values $-m,..,m$ increasing in steps of $+1$, with $l$ an integer or half integer (how spin arises), and then in a given irreducible representation, rotations about a general axis will send $\psi_{..,m}$ into a linear combination of all the other vectors living in this representation.

A second way they arise is by noting we can write
$$\Psi(x,t) = \sum_n c_n e^{-(i/\hbar) \hat{H}t} \psi_n(q) = e^{-(i/\hbar) \hat{H}t} \Psi(x,0) = \hat{S}(t) \Psi(x,0),$$
where $\hat{S}(t)$ is unitary. It can be used to set up the Heisenberg picture, again coming from the same starting point (this is the same thing you get by formally solving Schrodinger $i \hbar \partial_t \Psi = \hat{H} \psi$.

(Should be: $m$ can take the $2l+1$ values $m \in \{-l,...,+l \}$...)

bolbteppa

8:54 PM

Then by looking at an operator $\hat{f}$ in the basis $\psi_n'$ we see from it's matrix elements in this basis that they are the same matrix elements of the operator $\hat{S}^{-1} \hat{f} \hat{S}$ in the original basis, since $\int dq \psi_m'^* \hat{f} \psi_n'(q) dq = \int \psi_n \hat{S}^{-1} \hat{f} \hat{S} \psi_n dq$.

Thus the commutators like $[\hat{x}',\hat{p}'] = \hat{S}^{-1}[\hat{x},\hat{p}]\hat{S} = i \hbar$ etc... are preserved, so unitary transformations are the quantum analog of canonical transformations in classical mechanics, since canonical transformations preserve the poisson brackets of these variables (and commutators are the quantum analog of poisson brackets).

Him

10:33 PM

Is entropy something that is studied on cosmological scales? Like, in a lab, we can measure the change in entropy for a mole of water when it transitions from a liquid to a gas - we can assign this a number, and improve our processes to measure it more accurately or whatever, it's a thing that is studied. On a cosmological scale, we might ask something like "what is the change in entropy of our sun, Sol, over the period of this year?"

Is this the sort of question that physicists ask? Are there processes by which we could measure such a thing?

bolbteppa

11:14 PM

12

Q: Entropy of the Sun

Is it possible to measure or calculate the total entropy of the Sun? Assuming it changes over time, what are its current first and second derivatives w.r.t. time? What is our prediction on its asymptotic behavior (barring possible collisions with other bodies)?

Him

11:33 PM

Sorry to keep asking random questions here. I am trying to figure out why a question on physics.stackexchange is not mainstream, and the commenters told me to figure out why not here.

A comment on the question above suggests that the Sun is generally increasing entropy in it's vicinity. I have phrased this as something like "The change in entropy for the Sun is large", but I don't think that this is quite right.

What is the correct way to phrase this property that the Sun is generally dumping heat like crazy on its surroundings, and that it is somehow therefore "generating entropy", whatever that means?

Is there some way to make sense of this concept in an English phrase that won't make hardcore physicists gag? :)

here is my original question for reference. I thought initially that the "not mainstream" was in reference to entropy on cosmological scales. I edited based on the feedback here, and a new commenter suggested that maybe the "not mainstream" involves the fact that the Sun does not, in fact, induce changes in entropy, or that the way in which I'm phrasing this concept is not entirely sensical.

Transcript for