[benj]

uhmm...we have this take home exam on quantum chem and i dunno if im on the right track..can anyone tell me if i'm doing this correctly??pls...

the question: evaluate the commutators (a) [H,p_x], and (b) [H,x] where H= p²_x/2m + V(x). choose (i)V(x) = V, a constant, (ii) V(x) = (1/2)kx², (iii) V(x) :rarrow:V(r) = e²/4(pi)(epsilon)r.

i'll just show my solutions in (a.i),(a.ii) and (a.iii)
(a.i)letting [H,p_x] operate on an arbitrary function A_x,
[H,p_x]A_x = Hp_xA_x - p_xHA_x
 :rarrow:p_x= -ihd/dx  ; H= (-h²/2m)d²/dx² + V

           Hp_xA_x = (ih³/2m)d³A_x/dx³ - ihVdA_x/dx
           p_xHA_x = (ih³/2m)d³A_x/dx³ - ihVdA_x/dx
           thus [H,p_x] = 0.
(a.ii)when V(x) = (1/2)kx²; H = (-h²/2m)d²/dx² + (1/2)kx²
           Hp_xA_x = (ih³/2m)d³A_x/dx³ - (ihk/2)(x²)dA_x/dx
           p_xHA_x = (ih³/2m)d³A_x/dx³ - (ihk/2)(x²)dA_x/dx - ihkxA_x
           thus  [H,p_x] = ihkxA_x
(a.iii)(i dunno...T_T) i just let x=r and likewise dx= dr since (i think) its only considering 1 dimension and (i think x and r are just identical,hahaha..i neglected the cosine part..)..but i dunno..i really think its off..here it is..waaahhh!!!T_T
p_r = -ihd/dr ; H = -(h²/2m)d²/dr² + e²/4(pi)(epsilon)r.

           Hp_rA_r = (ih³/2m)d³A_x/dx³ - (ihe²/4(pi)(epsilon))(1/r)dA_r/dr
           p_rHA_r = (ih³/2m)d³A_x/dx³ - (ihe²/4(pi)(epsilon))(1/r)dA_r/dr + (ihe²/4(pi)(epsilon))A_r/r²
           and thus..taddaahhh...[H,p_x] = [H,p_r] = -(ihe²/4(pi)(epsilon))A_r/r²
huhuhu..i really dunno..pls help me...T_T

[benj]

 
help me pls...

[Hunt]

Hi benj,

It's a bit difficult to follow your work. Try to attach images or use Latex next time. Anyway your answers seem to be correct but u need to remove the "A" function at the end. That is , [H,p] = C A must be [H,p] = C ( no "A" ). You should review that in your book.

For the last part I would agree with you that the potential must be in terms of x i.e. r =x , otherwise, the hamiltonian would be in spherical polar coordinates and so must be the linear momentum operator , which doesnt seem to be the case ( according to the .pdf u attached a while back where the subscript x was shown below p ).

[benj]

oh right!!!hahaha..i missed that one..gee..thanks a lot!!!really really appreciate it..thanks!!!