The h Bar

Sanjana

12:40 AM

Anybody knows any time dependent but time translation invariant systems?

@Sanjana There is this go-to Hamiltonian system that is the example for "Hamiltonian $\neq$ Energy", and it has that behaviour.

Sanjana

@naturallyInconsistent which is???

I mean. I know some systems with $H \ne E$, but are you saying all of them has this behaviour or do u have some particular example in mind?

naturallyInconsistent

I have that particular example in mind. It is the extremely standard one. Bead on a rigidly rotated hoop.

You can choose for its Hamiltonian to be time-independent or dependent

Relativisticcucumber

1:36 AM

@Sanjana we have gone block crazy in this chat

Sanjana

@Relativisticcucumber Oops. Is it that common, nowadays here?

naturallyInconsistent

no

Sanjana

@SillyGoose I have myself become interested in this question so much that I want it to be posted on the main site. Would you consider doing it or can I do it (and give a hat tip to you :p)

user 85795

2:27 AM

🎩

🧐

SnoopyKid

Hello, not sure if this is physics related. But what do we call the phenomenon where if we look at the LED computer monitor the screen appears brighter and then from another point of view like from the left the screen appears to be darker and the colors seem to be inverted?

Silly Goose

2:43 AM

@Sanjana i don't mind if you ask it :D

Sanjana

3:00 AM

@SnoopyKid Twisted nematic field effect

user 85795

h o n k

Silly Goose

h o n k

Relativisticcucumber

@naturallyInconsistent bolbteps & ryder cant hear u ;)

@Sanjana i think the block density is higher than other chats i have participated in XD

naturallyInconsistent

@Relativisticcucumber actually, that's also false, because RR doesnt block ppl and bolb somehow keeps seeing what I said.

H O N K ~

Relativisticcucumber

@naturallyInconsistent i swear rr blocked me

naturallyInconsistent

3:12 AM

lol good?

Relativisticcucumber

yesh

schweg

@naturallyInconsistent you remind me so much of my modern prof

but alas he is not taiwanese

he did live in sing for awhile tho

naturallyInconsistent

miao miao iz naat taiwanese either

Relativisticcucumber

@naturallyInconsistent 0.0

naturallyInconsistent

what is "modern prof"?

Relativisticcucumber

ok but my prof is certainly in shanghai rn

naturallyInconsistent

3:13 AM

modern physics?

Relativisticcucumber

yes

naturallyInconsistent

it is nice to be compared to a prof~

Relativisticcucumber

he was the best prof ever

naturallyInconsistent

yay~

sing, sing a song~

Silly Goose

SING SING

user 85795

3:21 AM

🎵🎶

Sanjana

6:10 AM

@SillyGoose I posted it.

0

Q: Is there a conceptual inverse of anomalies i.e. a notion of quantum enhancement of symmetries?

Anomalies usually occur when a classical symmetry ceases to be a symmetry of the theory when quantized. Are there quantum systems with certain symmetries which cease to exist when you take classical limits? In other words, are there systems which when quantized gets a symmetry enhancement? I woul...

Silly Goose

6:28 AM

woop woop

Silly Goose

6:44 AM

Say I have two bosonic operators $a^\dagger$ and $b^\dagger$. Define a spin coherent state as $\lvert \alpha, \beta \rangle \rangle = (\alpha a^\dagger + \beta b^\dagger)^N\lvert \text{vac} \rangle$.

For simplicity, let $N = 2$. So, we have $\lvert \alpha, \beta \rangle \rangle = (\alpha a^\dagger + \beta b^\dagger)^2\lvert \text{vac} \rangle$.

This is supposed to model two two-level systems. $a$ and $b$ corresponding to the two levels.

How do I write this state in a basis that reflects that each system occupies one of two levels (or a superposition thereof)?

It is apparently equivalent to writing $\lvert \alpha, \beta \rangle \rangle = (\alpha a^\dagger + \beta b^\dagger)\lvert \text{vac} \rangle \otimes (\alpha a^\dagger + \beta b^\dagger)\lvert \text{vac} \rangle $ in the $\lvert a \rangle \otimes \lvert b \rangle,$ etc. basis

but saying that these two states are the same seems quite wrong

one thought is that the vacuum state is not the same in the first representation of the state i wrote versus in the second

naturallyInconsistent

Is it possible for you to not be this level of abstract?

Silly Goose

Conceptually, I am looking at an ensemble of spin-1/2s but all definitions are in terms of bosonic creation operators

naturallyInconsistent

7:05 AM

Which is confusing because spin half are fermions

I mean, are you trying to use Schwinger's trick or not?

Silly Goose

@naturallyInconsistent well I am confused about this separately as well.

It seems in an actual BEC one can use, for example, $^{87}$Rb atoms restricted to occupy one of two states of definite angular momentum

and treat this situation as an actual spin-1/2 system

naturallyInconsistent

Then, what are your $\hat a^\dagger$ and $\hat b^\dagger$ operators?

Ryder Rude

7:33 AM

hi

naturallyInconsistent

I mean, it is clearly suggestive of $\hat a^\dagger$ creating spin up and $\hat b^\dagger$ creating spin down. There are many potential confusions; This seems to be a two-level system as in having two operators $\hat a$ and $\hat b$, so you ought not to pick $N=2$ because that is another two-ness that is potentially confused with this.

Ryder Rude

@Sanjana r u also looking for situations for situations where $[H,A]=0$, but $\{H,A\}\neq 0$?

naturallyInconsistent

In any case, let $\hat c^\dagger=\alpha\hat a^\dagger+\beta\hat b^\dagger$ so that your state looks like $(\hat c^\dagger)^N\left|\text{vac}\right>$, and in this form it is manifestly "excited the same state twice"

Ryder Rude

12:14 PM

Why plague doctors wore beaked masks

►

lucabtz

@SillyGoose i think a friend of mine is doing that in the lab in bologna

@Sanjana i wondered about this too before

Slereah

Do you really want to be known as working for the balloney lab

user 85795

12:55 PM

The bologney lab has a long history of studying sandwiched research :P

Bologna sandwich

The bologna sandwich is a sandwich common in the United States and Canada. Also known as a baloney sandwich, it is traditionally made from sliced bologna sausage between slices of white bread, along with various condiments, such as mayonnaise, mustard, and ketchup. The bologna sandwich is a regional specialty in the East, Midwest, Appalachia, and South. It is a sandwich served at lunch counters of small, family-run markets that surround the Great Smoky Mountains, and fried bologna sandwiches can be found on restaurant menus in many places in the South. The fried version is likewise sometimes sold...

no offense intended :-)

Slereah

2:09 PM

Claudio Menchinelli

2:21 PM

lol what kind of discussion did I join in lol

I've read the wiki page about Lubos and now I feel miserable :P

a paper on matrix theory as still an undergraduate is insane, or in the words of Ed Witten: "bro was actually built different"

Claudio Menchinelli

4:07 PM

If I'm dealing with two particles of bosonic nature (spin 1) and the system state is represented by: $$\psi(x) = \psi_1(x)(A|1,-1\rangle + B|-1,1\rangle) + C\psi_2(x)|0,0\rangle, A,B,C \in \mathbb{C}$$ Where $\psi_i(x) \equiv \text{ eigenfunction of the 1D QHO }$, and $|s_1,s_2\rangle$ refers to the eigenvalues of $\{S_{1z},S_{2z}\}$

what condition must I impose upon the $A,B$ coeffiecients in order for $|\psi\rangle$ to be symmetric under the action of $\hat{P}_{12}$. I know that $\psi_1(x)$ is antisymmetric, so I need the term in brackets to be symmetric but how do I do it? I feel like it's trivial but my mind is not working properly hahaha

I meant to say antisymmetric in my last sentence. Anyway, I've just realized I need to apply the anti-symmetrizer to both terms :P

no wait, I'm not getting any condition on $A,B$ by doing so

It should be a stupid question, what is happening to me. I'll come back to this later my brain isn't properly functioning rn lol

ACuriousMind

5:02 PM

@ClaudioMenchinelli To be the answer seems to be obviously $A=B$ since $A\lvert 1,-1\rangle$ becomes $A\lvert -1,1\rangle$ and $B\lvert -1,1\rangle$ becomes $B\lvert 1,-1\rangle$ under exchange, no?

Claudio Menchinelli

7:55 PM

Yeah I checked the solutions a bit after because I felt bad and realized I was overthinking again, thus forgetting to simply swap the indices...Anyways, the sol says $A = -B$, and it's correct because it needs to be antisymmetric so that the minus sign from the HO wavefunction evens all out

@ACuriousMind but yeah that's the way it had to be done hahaha

Sanjana

9:24 PM

@RyderRude I was looking for situations only where $[H,A]=0$ . No condition on the anticommutator.

But now I am convinced that this won't be possible because of the precise reason u stated: boosts are time dependent transformations

@lucabtz Did you come up with an answer?

Sanjana

9:42 PM

@ClaudioMenchinelli Ed Witten said this about Lubos?

ACuriousMind

@Sanjana ...do you really think Witten ever used "bro" :P

Claudio Menchinelli

10:38 PM

@Sanjana that was a good one

I'm pretty sure that's what he thought :)

guys I have a question

I think there's a mistake in a solution to an exam exercise which I've solved

We have two bosons of spin 1 as before, and their initial state is: $$\psi = f(\mathbf{r}_1, \mathbf{r}_2}) [\frac{1}{2}(|1,-1\rangle +|-1,1\rangle)+ \frac{1}{\sqrt{2}}|1,1\rangle]$$ with the same notation as before, namely $|s_{1z},s_{2z}\rangle$

$$ \psi = f(\mathbf{r}_1, \mathbf{r}_2 ) \left[\frac{1}{2} (|1,-1\rangle +|-1,1\rangle)+ \frac{1}{\sqrt{2}}|1,1\rangle\right] $$

I shall retreat lol, I've realized that there may be no mistake at all :P

there aren't indeed any mistakes, I'm just dumb

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