Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Favorites infavorites:mine
infavorites:1234
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with Search options user 56599

Fundamental characteristic property of particles which together with orbital angular momentum acts as the generator of rotations and which doesn't have a classical equivalent but is sometimes compared to and contrasted with classical intrinsic angular momentum.

2
votes
Yes, the factorizability is an axiom. Formally, it is states as The Hilbert space of a composite system is the Hilbert space tensor product of the state spaces associated with the component syst …
answered Mar 19 by InertialObserver
0
votes
If I understand your question correctly, there seems to be a misunderstanding in your statement As far as I understand, in quantum mechanics, you can identify any spin half particle to be fermion …
answered Feb 24 by InertialObserver
2
votes
The addition rule has an absolute value sign. That is $$ s_{tot}= |s_1 + s_2|,|s_1 + s_2 - 1|, \ldots, |s_1 - s_2| $$ And so the smallest value you can have is $1/2 - 1/2 = 0$. $s_{tot} $ is the th …
answered Feb 27 by InertialObserver
1
vote
The components of every field must satisfy the KG equation, regardless of its spin. This makes sense, since it is nothing more than just the Einstein energy momentum relation. The idea is that the KG …
answered Sep 9 '18 by InertialObserver
0
votes
Your equation for $| \psi \rangle $ is \begin{equation} |\psi\rangle = \frac{1}{\sqrt{6}}\left(2\big\uparrow\big\uparrow\big\downarrow - \big\downarrow\big\uparrow\big\uparrow - \big\uparrow\big\ …
answered Mar 9 by InertialObserver
1
vote
Since it appears you are actively working on the problem I will offer a series of hints. Please comment if you need further guidance. Hint 1: Muonium is a bound state of $\mu^+ e^-$. Note that they …
answered Jan 5 by InertialObserver
2
votes
No, the intrinsic spin of a particle is just a consequence of a "particle" being represented by an irreducible representation of the the Poincaré group. The spin of the particle is determined by the …
answered Feb 11 by InertialObserver
3
votes
For completeness I'll summarize the answer here. After a fun conversation in the comments, we saw that it will be more illuminating to write $$H=-\sum_ {i=1}^{N-1} \sigma_i^x \sigma_ {i+1} ^x + h \ …
answered Dec 29 '18 by InertialObserver
2
votes
You're reading Griffiths, so I will try to stay within his vocabulary---but to answer your question I have to introduce perhaps some formalism that Griffiths doesnt. In general, this is the story. M …
answered Dec 4 '16 by InertialObserver