User User198 | chat.stackexchange.com

User198

14:28

@ACuriousMind Oks thank you

User198

13:45

@Slereah Ok thank you

User198

13:43

2.) Momentum in CM: $\{\cdot,p\}=\frac{\partial \cdot}{\partial q}$

in QM $\hat p=-i\hbar\nabla$ its similar.

What about $\{\cdot,q\}=-\frac{\partial \cdot}{\partial p}$ in QM, is there expression akin to the momentum one in QM?

User198

13:42

1.) Momentum is the generator of translation $\{f,p\}=\frac{\partial f}{\partial q}$

$q$ is the generator of momentum, but from the Poisson brackets, I get it with the $-$ sign: $\{f,q\}=-\frac{\partial f}{\partial p}$ by definition. Is it correct than to say that "$q$ is the generator of momentum" or should I now say that the "$q$ is the negative generator of momentum" or sth like that? It bothers me that both expressions are not the same, but we talk like they are. Or am I missing something?

User198

09:50

@SirCumference Sir Cumference really cool name xD

User198

09:49

@ACuriousMind True

User198

09:40

@naturallyInconsistent Yeah, thats the whole formalism. It would be cool if it was like the Poisson-Hamilton equation or Hamilton-Poisson.

User198

yst 21:11

Ok thanks

User198

yst 21:11

@ACuriousMind It doesn't exist, yeah

User198

yst 21:03

If asking such question makes sense xD

User198

yst 21:02

In theory at the exact same time*

User198

yst 21:02

What do I get when I measure spin along 2 different axis, but at the same time?

User198

yst 19:31

I want to learn more about density matrix, mixed states, phase space formalism; but I am looking for a gentle intro to those topics. Does such a book exist?

User198

yst 16:56

@RyderRude Hah ok thanks.

User198

yst 16:55

@Feynmate Ok thanks

User198

yst 16:08

What is this equation called in the literature $\dot f = \{f,H\}$ ? Classical Heisenberg equation of motion? I couldn't find much sources that name it explicitly.

User198

Wed 17:12

@Feynmate xD

User198

Tue 21:16

@TobiasFünke Maybe I overcomplicated, disregard the equations. All in all, saying that "Schrodinger equation is a statement of conservation of energy" is not a correct statement (even meaningless), right?

User198

Tue 21:14

@TobiasFünke Ok thanks.

User198

Tue 19:36

I saw on one page today (not about physics, but a page connected with education in general; some university information posts and some administration): "Schrodinger equation can be regarded as a statement of conservation of energy in quantum mechanics." Idk why they are writing about the Schrodinger equation at all. xD

But how wrong is that?

I think it is wrong, because (I guess) the person writing that thinks that $i\hbar\frac{\partial \psi}{\partial t}\equiv \hat E= \hat H \psi$ so $\hat E= \hat H \psi$ and that is wrong. Should I notify them that that post is wrong?

User198

Tue 19:05

Good evening

User198

Tue 15:04

Alright, thanks.

User198

Tue 14:56

We can integrate the action of a system over arbitrary small time intervals, and hence get arbitrary small values. But does it physically have any meaning when the action of a system is smaller than $h$?

User198

Tue 14:14

@SillyGoose But that is from maths perspective; differential geometry. I was thinking more about mechanics but described via geometry.

User198

Tue 14:12

@Slereah Hm. Thanks. That occured to me, but I wanted to be more specific, otherwise I could just say its "mechanics".

User198

Tue 13:55

When talking about geometric mechanics (1) , symplectic geometry (2), contact geometry (3), Poisson geometry (4) , usage of Lie groups in mech,...

What would be the "roof" term that accompanies all of that? Is it geometric mechanics?

User198

Tue 13:37

@DIRAC1930 Boltzmann would agree

User198

Tue 13:36

@RyderRude Doing too much philosophy without physical exercise is not a good thing. Mens sana in corpore sano.

User198

Feb 15 21:52

@ACuriousMind Ok thanks.

User198

Feb 15 21:52

@ACuriousMind Its not. I just found about it today. Like Schrodinger equation for open systems.

User198

Feb 15 21:51

@ACuriousMind So for most systems considered in QM the ehrnfest theorem is useless?

User198

Feb 15 21:50

@ACuriousMind Have you heard of Lindbladian ?en.wikipedia.org/wiki/Lindbladian

User198

Feb 15 21:49

It is a coincidence that $V(\langle x\rangle) = \langle V(x)\rangle$ when $U$ is quadratic.

User198

Feb 15 21:49

@ACuriousMind True

User198

Feb 15 21:47

But only for "nice" systems. That have nice quadratic at max potentials.

User198

Feb 15 21:46

@RyderRude I was reading about Ehrnfest theorem, the expecectation values of the operators correspond to the classical behaviour of the system.

User198

Feb 15 21:39

Here is a really nice answer by Terrence Tao that talks about different types of viewing mechanics:
https://mathoverflow.net/questions/225814/does-quantum-mechanics-ever-really-quantize-classical-mechanics
Might be helpfull for someone

User198

Feb 15 12:38

@TobiasFünke Ah ok thanks.

User198

Feb 15 12:37

@ACuriousMind Understandable. I will restrain for asking such questions furthermore. Although some of the answers I got in this chat were helpful, so I am thankful for that.

User198

Feb 15 12:32

True true

User198

Feb 15 12:30

But I agree with you 100%

User198

Feb 15 12:30

Yes yes I understand that. But it is a super fast resource

User198

Feb 15 12:29

xD

User198

Feb 15 12:28

Often, however, the Schrödinger equation is difficult to solve (even with a computer). "Therefore, physicists have developed mathematical techniques to simplify these problems and clarify what is happening physically. One such technique is to apply a unitary transformation to the Hamiltonian. Doing so can result in a simplified version of the Schrödinger equation which nonetheless has the same solution as the original."

User198

Feb 15 12:27

I read this on the wiki en.wikipedia.org/wiki/…

User198

Feb 15 12:26

2.) Do some people do unitary transformations on the Hamiltonian, and calculate the evolution on a system that way, instead of solve the Schrodinger equation?

User198

Feb 15 12:25

No until now. I think I understand it now. We use unitary transformations because they preserve the inner product, so measurements will always be correct.

And it is U dagger on the left because that is how a bra transforms under a unitary transformation and U on the right because that is how a ket transforms under a UT.

User198

Feb 15 11:45

It reminds me of kind of when you act on a vector with a rotation matrix from one side and its transponse on the left to make a rotation.

User198

Feb 15 11:45

Hm. My question is than this:

When considering a unitary transformation of a state $\psi$ if we want $\psi '$ we get it by $\psi '= \hat U(t) \psi$. Ok.

For operators we do:

$\hat A'=\hat U \hat A U^\dagger$

Why do we do it in that way for transforming operators?

User198

Feb 15 11:36

So just because $U(t) is time dependant it turns $\hat A_S$ into a time dependant quantity?

User198

The h Bar