Can anyone help? in quantum mechanics commutator prove [L^x,L^y] = ihL^z

  • #1
can anyone help?? in quantum mechanics commutator prove [L^x,L^y] = ihL^z

given

:L^x =(y^(pz)^-z^(py)^)
:L^y =(z^(px)^-x^(pz)^)
:L^z =(x^(py)^-y^(px)^) where ^ is just showing its operator


prove comutator [L^x,L^y] = ihL^z

I am swamped at every hurdle and cant seem to get my head around this question to find the answe of ihL^z . any help would be very much appreciated:surprised
 

Answers and Replies

  • #2
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You're aware of the form of the commutation relation, correct?

[tex][\hat{L}_x,\hat{L}_y]\psi = \hat{L}_x (\hat{L}_y\psi) - \hat{L}_y (\hat{L}_x\psi) [/tex]

Solve this for the angular momentum operators [tex]\hat{L}_x[/tex] and [tex]\hat{L}_y[/tex] - which you know. The result should cancel down to the form of the angular momentum operator [tex]\hat{L}_z[/tex].
 
  • #3
dextercioby
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I like the nice, geometric proof of the commutation relations for angular momentum that Sakurai, for example, gives in his delightful book. Unfortunately not mathematically rigorous, but very well looking.
 
  • #4
yeah im aware commutator rules but i just cant prove the relation ,i am confused about how to rearange and put in that form. i.e my maths might not be up to scratch but if u can help me with this little proof i might be able to crack the rest... here it goes ...
z(py)x(pz) - x(pz)z(py) ,how do i break this down mathematically without breaking any rules to get ih{x(py)} if [z,pz] = ih..
 
  • #5
nrqed
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yeah im aware commutator rules but i just cant prove the relation ,i am confused about how to rearange and put in that form. i.e my maths might not be up to scratch but if u can help me with this little proof i might be able to crack the rest... here it goes ...
z(py)x(pz) - x(pz)z(py) ,how do i break this down mathematically without breaking any rules to get ih{x(py)} if [z,pz] = ih..
Write

[tex] [ L_x, L_y] = [Y P_z - Z P_y, Z P_x - X P_z ] [/tex]

and then use the fact that the commutator of sums is equal to the sum of commutators

[tex] = [Y P_z, Z P_x] - [Z P_y, Z P_x] - [Y P_z , X P_z] + [ Z P_y , X P_z] [/tex]

Now do each of those commutators.

(My favorite trick is to use directly [AB,CD] = A[B,C]D + B[A,D]C + AC[B,D] +[A,C] DB which can be proven by simply expanding or starting from the simpler and obvious [AB,C] = A[B,C] + [A,C] B )

Patrick
 
  • #6
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Lx=yPz-zPy
Ly=zPx-XPz
Pk=ih d/dk use these relations, careful derivatives and then eliminate.
and result is [Lx,Ly]=ihLz .
 
  • #7
thanks patrick , do i use [AB,CD] relation for each 4 commutators?,like [YPz,ZPx] and also [ZPy,ZPx] etc
 
  • #8
nrqed
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thanks patrick , do i use [AB,CD] relation for each 4 commutators?,like [YPz,ZPx] and also [ZPy,ZPx] etc
Yes. And you use [itex] [X_i, X_j]= [P_i,P_j] =0 [/itex] and [itex] [X_i,P_j] = i \hbar \delta_{ij} [/itex]. you'll see, you will get [itex] i \hbar L_z [/itex].

You are welcome

Patrick
 

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