Skip to main content
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
Results tagged with
Search options not deleted user 251599

A mathematical construct quantifying the difference in effect of applying two operators in two alternate successions. It is the defining product of a Lie algebra, the efficient underlying description of Lie groups, of use in several areas of physics, most notably quantum field theory.

1 vote
0 answers
138 views

Do the number operator and general Hamiltonians commute?

In a proof of the equivalence of the canonical and grand canonical ensembles (in the thermodynamic limit) in Jochen Rau's Statistical Physics and Thermodynamics (highly recommended!), the author evalu …
EE18's user avatar
  • 1,261
3 votes
3 answers
581 views

If an operator commutes with the Hamiltonian then do all of its projectors commute also?

Suppose I have an (self-adjoint) operator (representing an observable) $O$ which commutes with the system Hamiltonian $H$. Does it follow that $[P_O(\Delta),H] = 0$ where $P_O(\Delta)$ is an arbitrary …
EE18's user avatar
  • 1,261
1 vote
1 answer
131 views

Commuting normal operators have common eigenbasis?

One often finds in quantum mechanics textbooks (for physicists) a "proof" that self-adjoint and commuting operators $A,B$ have a common eigenbasis. However, in the standard proof of Bloch's theorem (s …
EE18's user avatar
  • 1,261
0 votes
0 answers
51 views

Showing that time ordering does not matter for the measurement of commuting observables

Suppose I have two observables $R$ and $S$ who are represented by operator $R$ and $S$ which commute (I will hereafter ignore the distinction between observables and the operators representing them), …
EE18's user avatar
  • 1,261
2 votes
1 answer
130 views

Why can't commuting operators allow for full state determination?

For the purposes of this question, suppose that the operator $R$ representing the observable $\mathsf{R}$ has nondegenerate eigenspaces. In discussing state determination (for some given situation, ho …
EE18's user avatar
  • 1,261