Help with commutator question please!

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consider a general one dimensional potential v(x) drive an expression for the commutator [H,P] where h is hamiltonian operator and momentum operator. i keep getting zero and i dont think i should. since next part of hw question sais what condition must v(x) satisfy so that momentum will be constant of the motion recall that this is true if the momentum communtes with the hamiltonian. so i dont think it should zero. any help would be great trtying to do all hw problems to prepare for final!
 

samalkhaiat

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[tex] [P^{2}/2m + V(x) , P] = [ V , P] = -i \frac{dV}{dx}[/tex]
 
thats how i started but didnt arrive at that can u show a little bit more detail if u dont mind.
thanks alot though
 
Consider the action of the commutator on a wavefunction

[H, P] psi(x)

(so, we work in the position representation). Write out the commutator explicitely:

H P psi - P H psi
 
yes thats what im doing but im still getting zero
 
Expand the potential as a Taylor series and use the commutation relation [tex][x,p]=i \hbar[/tex].
 

samalkhaiat

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[tex] [P^{2}/2m + V(x) , P] = [ V , P] = -i \frac{dV}{dx}[/tex]
[tex][P^{2}/2m, P] + [V , P] = [V,P][/tex]

because [P,P] = 0.

then

[tex]
[H,P] \phi(x) = [V(x) , id/dx] \phi(x) = iV \frac{d \phi}{dx} -i\frac{d}{dx} \{V(x)\phi(x)\} = \{-i dV/dx\} \phi(x)
[/tex]
 
um is there an easier way? like the first respnse seems right but idk how to get there its an elementary problem that i have im sure
 
samal where is the h bar term? is that a property of commutator i have never seen that before
 
i think the problem is i dont know what to do when we have an additon in the commutator i dont know how to treat it
 
The commutator is linear [A+B,C] = [A,C] + [B,C]. Just write it out and you'll see that.
 

samalkhaiat

Science Advisor
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samal where is the h bar term? is that a property of commutator i have never seen that before
I always set my h bar equal to 1. you can put it back in the equation. The first point you need to know is

[A + B , C] = [A,C] + [B,C]

the second point, an (differential) operator in the commutator acts on all the stuff on the right, so

[tex][A , i\hbar d/dx] f(x) = A(i\hbar df(x)/dx) - i\hbar (d/dx)(Af(x))[/tex]
 
oh got it thanks alot guys!
 

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