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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 …
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? …
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 $(m−E,0,0,−E)$. …
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. …