The h Bar

9:03 PM

Ah ok, since $(a-b*)\langle a\lvert b \rangle = 0$ holds for all eigenvectors $\lvert a\rangle,\lvert b\rangle$ thus it must hold for the subset of eignevectors $\lvert b\rangle = \lvert a\rangle$ that give the equation $(a-a*)\langle a\lvert a \rangle = 0$ and from this any eigenvalues will satisfy $a=a^*$ thus the case $a=b^*$ where a,b not necessary real will be ruled out

NeuroFuzzy

Zee's section on Feynman diagrams is so confused. There are errata and corrections to the errata

at least one subsection.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Corrections to the corrections?

NeuroFuzzy

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ kitp.ucsb.edu/members/PM/zee/nuts.html p. 43

"the denominator should contain (4!)^2 instead of (4!)^3 (Thanks to Jean Orloff) INCORRECT: The numerical factor stated in the book is in fact correct. Thanks to the many readers who sent in this anti-erratum. " etc.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

:-/

NeuroFuzzy

And this is for the first edition but the second edition has similar things that still apply, apparently

maybe not

ACuriousMind

9:10 PM

@NeuroFuzzy Everything on Feynman diagrams tends to be rather confused :P

Noone has the patience to actually determine all the symmetry factors properly, so most people just wing it, I suspect

NeuroFuzzy

@ACuriousMind :D Sounds like I should write a program to do it for me

@ACuriousMind :(

ACuriousMind

In tenuously related news, the literature on spinors in arbitrary dimensions is also rather confused, and not two authors seem to agree on their sign conventions. Several introduce signs that "may be determined" but never determine them.

And the mathematicians insist on calling some spinors "pinors" which does not help the terminology at all :P

NeuroFuzzy

"pnors" is the next logical step

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

No wonder @yuggib calls it "sloppy."

ACuriousMind

@NeuroFuzzy Well, not everything. The stuff without explicit examples is safe against making sign and prefactor errors in the examples ;P

@NeuroFuzzy lol, I pnorted.

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ I have never heard yuggib talk about spinors.

The issue is not actually a lack of rigor in this case, it's a lack of clear terminology and conventions

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

9:17 PM

I just meant in general :-)

Secret

Feb 1 at 11:21, by Secret

I see

Combine your discussion with the many PSE, in that case I guess our picture is complete for that FAQ:
Experiment->Found observables that can be directly measured are non commutative and real (most natural to choose reals)->algebra of observables there are linear functionals (states) that associated a numerical value to each observable->A theory of non commutative probability requires the states to be vectors in some hilbert space-> non commutativity lead to commutation relation (CCR), the real-ality of directly measured observables means the operators must be hermitian-> hermitian op…

@0celo7 @Acuriousmind
This summary of a discussion in the distant past with you and yuggib have explained why quantum mechanics need to be modelled by a complex hilbert space. But the recent discussion on hermitian operators have sparked a new question when mixed with this summary:

What rules out the possibility of having a degenerate inner product operating in a hilbert space as the model to describe quantum mechanics. Specifically, what (experimental or mathematical or both) disallow null state vectors (i.e. $\langle a\lvert a\rangle=0$ but $\lvert a\rangle\neq \mathbf{0}$ to appear in n…

ACuriousMind

@Secret The non-degeneracy of the inner product is necessary to have a Hilbert space. For one, you cannot talk about convergence w.r.t. the inner product properly if it is degenerate, and important results such as the spectral theorem do not hold.

Far more physically, $\langle \psi \vert \psi \rangle =0$ just doesn't make any sense in light of the Born rule: The state $\lvert \psi \rangle$ has zero probability to be found to be the state $\lvert\psi\rangle$???

Secret

Hmm, make sense

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