Search Results
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 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
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 |
In physics, an operator is almost always either a square matrix or a linear mapping between two function spaces (defined on, say, $\mathbb R^n$). Operators serve as observables and as time evolution operators in Quantum Mechanics. This tag will most often find valid use in quantum mechanics; don't use this tag just because your equations contain "everyday operations" like $\times$, $+$!
3
votes
1
answer
140
views
Can Hadamard's formula be used for fermionic operators?
[B, [B,A]] + \dots$$
for fermionic operators?
Suppose I have fermionic operators that obey anticommutation relations
$\{a,a^{\dagger}\}=1$ and $\{a,a\}=\{a^{\dagger},a^{\dagger}\}=0$. … [a, (1-2a^{\dagger}a)]+\dots$
Is this formula universal for fermionic and bosonic operators? …