Going to post a few pictures of a derivation, my apologies. I fear that full context needed. At any rate, I'm unable to follow how equation (3.64) obtains from (3.22) by inserting (3.63). We seem to be switching the vectors which are dotted using the BAC-CAB rule, but then it's not clear to me why the cross product $\mathbf k \times (\mathbf k \times \nabla_r E_{F_n})$ vanishes if that's true

@Arjun then the answer is yes

And in case 3.22, the current equation under the parabolic band approximation, is needed:

@ACuriousMind They'll span the whole of L^2[0,1] you mean?

@Arjun Yes. Note that the domain of definition is dense, meaning all vectors not in it are limits of sequences of vectors in it, and that the usual definition of a Hilbert basis is also just that you can obtain every vector as a "limit" of the basis vectors (because the sum is allowed to be infinite), so this is consistent.

but at the point where you're worrying about details like dense domains of definitions at all, I would suggest looking into the rigorous version of the spectral theorem in terms of spectral measures or stop worrying, really :P

@ACuriousMind I guess I'll stop worrying soon lol,I definitely lack the mathematical background :D

@ACuriousMind Okay two last questions: 1)So the above thing holds for any arbitrary conditions given that the domain remains dense in L^2? 2)So the "regular" eigenfucntions would lie in the domain of definition right?Will these themselves span whole of L^2 or we need to also consider the generalized eigenfuctions,I remember reading in ballentine we also need to consider the generalized ones.

@Arjun if generalized eigenfunctions exist, you need them to span the whole space

@ACuriousMind why you removed it?

because it was wrong :P

ookie : )

I have a question about irreducible tensor operators. In the space of some n-rank tensor operator, do these irreducible tensor operators, play the same role that basis kets in a vector space play?

@ACuriousMind I have a moral question for you. If you noticed after hours that a message you'd written was wrong, would you delete it?

@Mr.Feynman not if anyone had reacted to it already - then the damage is already done, after all, and I'd just ping the people I'd told that with the correction

The question is about using your powers to do something I can't: delete old messages

You can right the wrong

You have the power to make this virtual world perfect

as I said - removing the message after people have already seen it doesn't really undo the damage, so I don't see the point

what if the chat was unusually quiet and no one had replied? I assume you would

@Mr.Feynman what if someone read it and silently accepted it? :p

@Arjun I don't know what "arbitrary conditions" you're talking about here

The (generalized) eigenfunctions of a self-adjoint operator span the entire Hilbert space, period.

@Mr.Feynman probably, yes

Oh this damn formatting ruined my moment

I mean I would also delete your (or any other users) last message no one had reacted to if they asked me to

Ah

I thought that was a big no no, like "you can't change the past, young one"

@ACuriousMind Can you also see,who starred a given statement? :p

@Arjun no

more starkiller than astronomer

You are a GoD, basically

@ACuriousMind lol

You can't create, you just destroy

And since I suppose Dragon Ball is not your thing (at least not Super), GoD=God of destruction

I was just about to remind you that I won't get anime references :P

Is there a geometrical interpretation of what it means when a rotation is applied to a 2nd order tensor/ matrix, the same way there is for a vector ?

Bold of you to assume that I expected you to catch it :P

There must be something I can flex :P

@Mr.Feynman Bertrand Russell disapproves of this definition

@EE18 That is not what is happening. Ignore the $\times$ in Equation (3.64) because it is actually a scalar (dot) multiplication, not a cross product at all.

you are, however, absolutely correct that the full context was needed. Definitely needed all 3 pages just to decipher what is happening

Is the space of 2nd order tensors, 9 dimensional ?

@naturallyInconsistent Agreed that the times there is not cross product related. But what I'm saying is that we essentially go from $(\mathbf k \cdot \nabla E_F) \mathbf k$ to $(\mathbf k \cdot \mathbf k) \nabla E_F)$

That sort of transfer is why I was thinking BAC-CAB rule (and the corresponding cross product). How did they do it? This is from page 71 of Nelson's Physics of Solar Cells in case it helps

@EE18 you are correct. That's pretty bad, and not justified

But in the absence of a magnetic field, the only thing that destroys spherical symmetry in favour of cylindrical symmetry is $\vec\nabla E_F$, and that would thus also be the direction of $\vec k$

so that both expressions turn out to be the same

@ACuriousMind Only references I've seen u make/get are monty python ones :P

and star trek

recently been rewatching some of them. Very much enjoy the cheese shop sketch and the argument clinic.

"Yes, I'd like to have an argument"

Must be very relatable as a SE mod

@naturallyInconsistent I don't follow this argument. Are you saying $\V k$ and $\nabla E_F$ are in the same direction? (I must be misunderstanding since that surely isn't true)