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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.
@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?
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
1:36 AM
@Sanjana we have gone block crazy in this chat
@Relativisticcucumber Oops. Is it that common, nowadays here?
@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)
2:27 AM
🎩
🧐
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?
2:43 AM
@Sanjana i don't mind if you ask it :D
h o n k
h o n k
@naturallyInconsistent bolbteps & ryder cant hear u ;)
@Sanjana i think the block density is higher than other chats i have participated in XD
@Relativisticcucumber actually, that's also false, because RR doesnt block ppl and bolb somehow keeps seeing what I said.
H O N K ~
@naturallyInconsistent i swear rr blocked me
3:12 AM
lol good?
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
miao miao iz naat taiwanese either
@naturallyInconsistent 0.0
what is "modern prof"?
ok but my prof is certainly in shanghai rn
3:13 AM
modern physics?
it is nice to be compared to a prof~
he was the best prof ever
yay~
sing, sing a song~
SING SING
3:21 AM
🎵🎶
 
3 hours later…
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?

SanjanaAnomalies 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...

6:28 AM
woop woop
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
Is it possible for you to not be this level of abstract?
Conceptually, I am looking at an ensemble of spin-1/2s but all definitions are in terms of bosonic creation operators
7:05 AM
Which is confusing because spin half are fermions
I mean, are you trying to use Schwinger's trick or not?
@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
Then, what are your $\hat a^\dagger$ and $\hat b^\dagger$ operators?
7:33 AM
hi
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.
@Sanjana r u also looking for situations for situations where $[H,A]=0$, but $\{H,A\}\neq 0$?
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"
 
5 hours later…
@SillyGoose i think a friend of mine is doing that in the lab in bologna
@Sanjana i wondered about this too before
Do you really want to be known as working for the balloney lab
12:55 PM
The bologney lab has a long history of studying sandwiched research :P
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 :-)
 
1 hour later…
2:09 PM
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"
 
2 hours later…
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
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?
 
3 hours later…
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
 
1 hour later…
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?
9:42 PM
@ClaudioMenchinelli Ed Witten said this about Lubos?
@Sanjana ...do you really think Witten ever used "bro" :P
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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