# Commutation relations of P and H

1. Aug 14, 2010

### orienst

Can we always calculate the commutation relations of two observables? If so, what’s the commutator of P (momentum) and H (Hamiltonian) in infinite square well, considering that the momentum is not a conserved quantity?

2. Aug 15, 2010

### dextercioby

For a quantum observable, call it A, it's not important that it doesn;t commute with the Hamiltonian. It only matters that

D(AH)∩D(HA)≠$\emptyset$

and the domain of the product operator is the subset of D(H), such as

I(H)⊂D(A)

3. Aug 15, 2010

### genneth

Remember that [P,.] works like a derivative for x. So, generically, for H = P^2/2m + V(x) where V is any potential, [P,H] = -dV/dx. In classical mechanics, this is the force. For infinite square potential, it gives two delta spikes at the edges of the box.