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 26143

In introductory mechanics, the momentum of a particle is its mass times its velocity. In electrodynamics, the momentum of a field is proportional to the cross-product of the electric field with the magnetic field. In special relativity, momentum is generalized to four-momentum.

21 votes

How to get the position operator in the momentum representation from knowing the momentum op...

Assume one dimensional and $\hbar=1$ By \begin{array} \hat \hat{p} |p \rangle &= p | p \rangle \\ \langle x | \hat{p} |p \rangle &= p \langle x | p \rangle \\ -i \frac{\partial}{\partial x} \lan …
user26143's user avatar
  • 6,451
10 votes

Expectation of momentum in the bound state

Is it logically correct to assert that the expectation of the momentum $\langle p \rangle=0$ for any bound state because it is bound to some finite region? …
user26143's user avatar
  • 6,451
7 votes
Accepted

QED: basic process which violates energy-momentum conservation

Put the initial electron in rest, the four momentum is $(m,0,0,0)$. The emission photon is $(E,0,0,E)$. The electron after emitting photon is $(m−E,0,0,−E)$. …
user26143's user avatar
  • 6,451
8 votes
2 answers
2k views

What is the most general expression for the coordinate representation of momentum operator?

I have a question about deriving the coordinate representation of momentum operator from the canonical commutation relation, $$[x,p]= i.$$ One derivation (ref W. …
user26143's user avatar
  • 6,451