de.wikipedia.org/wiki/Heisenbergsche_Unsch%C3%A4rferelation
1 correction found
\[ H , p \] \= − i ℏ ⋅ V ′ .
The sign of this commutator is reversed. With the page’s own convention [A,B] = AB − BA, one gets [H,p] = [V(x),p] = +iℏ V′(x), not −iℏ V′(x).
Full reasoning
Earlier on the page, the commutator is defined as [A,B] = AB − BA. For a 1D Hamiltonian (H = p^2/(2m) + V(x)), the kinetic term commutes with (p), so
[
[H,p] = [V(x),p].
]
Using the standard momentum operator (p = -i\hbar,\partial_x), this evaluates to
[
[V(x),p] = V(-i\hbar\partial_x) - (-i\hbar\partial_x)V = +i\hbar,V'(x).
]
A Durham University quantum-mechanics lecture note states this directly: "[ H , p ] = [ V(x) , p ] = i ħ ∂V(x)/∂x". So the sign in the article’s displayed commutator is opposite to the standard result.
(The uncertainty relation written immediately after this is unaffected, because it uses an absolute value.)
1 source
- Commutators and Uncertainty Principle
A similar computation shows that [ H , p ] = [ V(x) , p ] = i ħ ∂V(x)/∂x . So momentum p and energy H are compatible only if V(x) is constant.