The h Bar

12:01 PM

@ACuriousMind Yeah it is. I hate this binomial coefficients and identity stuff. It's like high school trig where you just trying stuff until it works.

Phase

Hi, sorry for intruding but I have a really basic, fundamental confusion over something, and put it in a question but still haven't had any responses, and wondered if putting this in chat could help, if not, no worries :^)

skill patrol

@YashasSamaga try this: when you add 2 feet and 2 inches you get 2 feet 2 inches.

2' + 2" = 2'2"

Yashas Samaga

HAHAH! I love that.

skill patrol

:-)

ACuriousMind

@Phase Asking questions in chat is fine, although we prefer you don't just come here to advertise your main site questions.

skill patrol

12:05 PM

Askaway

Phase

Sorry, I can try asking it in new words here:

Yashas Samaga

@skillpatrol it is a contradiction, it is both 4 and 14 so math is wrong

LOL

Phase

I'm just confused because in Lectures they never really went into detail on the Unitary operator, and I feel like I'm misunderstanding it. In my head I can only imagine it as a constant 'rotation', so if it's an orthonormal basis of two basis states, the classical analogue of it in my head would be a constant rate of rotation between two orthogonal basis vectors. And I guess the only real evidence i have of this is that if H is constant, then $\hat U$ is just a function of t

and that if H|psi> is constant then so too would ihbar d/dt |psi>

skill patrol

yesterday, by skill patrol

@YashasSamaga you got it pal. Don't let them drag you down to their level.

ACuriousMind

@Phase 1. It's not "the" unitary operator. The time evolution operator $U$ you get from the Hamiltonian is one particular family of unitary operators, but there are many others.

Yashas Samaga

12:10 PM

It is really surprising that super religious people have become anti-science/math to an extent that they have problems accepting that 2 + 2 = 4.

This is absolutely insane.

ACuriousMind

2. Indeed, a unitary operator is the complex analogue of a rotation in a real vector space

Phase

re: 2) So the 'rate of change of state' [I don't know if that's actually something that makes sense] is constant if H is constant?

ACuriousMind

However, in two complex dimensions you have more than just a simple rotation - the two-dimensional unitary group $\mathrm{SU}(2)$ is almost the same as the three-dimensional group of real rotations $\mathrm{SO}(3)$.

skill patrol

Ingnorance is infinite @YashasSamaga

Yashas Samaga

I wonder how much $ would he pay to the retailer when he buys 2 packs of apples each consisting of 2 apples,

14$ :D

ACuriousMind

12:13 PM

@Phase That...depends on your state. For eigenstates of $H$ that is true, for non-eigenstates the "rate of change" does not really make much sense - the "rate of change" would usually be the time derivative, which is just given by the Hamiltonian applied to the state in that instant (that's what the Schrödinger equation tells us!), and if the state is not an eigenstate that doesn't give you a number but a (non-constant) vector.

Yashas Samaga

"US Senate passes new bill, makes it legal to kill hibernating bear families in den"

I thought Republicans were religious people.

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