Prove [L^x,L^y] = ihL^z using angular momentum operators

[foranlogan2]

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 can't seem to get my head around this question to find the answe of ihL^z . any help would be very much appreciated

 

[Sojourner01]

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

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

Solve this for the angular momentum operators $$\hat{L}_x$$ and $$\hat{L}_y$$ - which you know. The result should cancel down to the form of the angular momentum operator $$\hat{L}_z$$.

 

[dextercioby]

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.

 

[foranlogan2]

yeah I am aware commutator rules but i just can't 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..

 

[nrqed]

yeah I am aware commutator rules but i just can't 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

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

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

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

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

 

[emotion55]

Lx=yPz-zPy
Ly=zPx-XPz
Pk=ih d/dk use these relations, careful derivatives and then eliminate.
and result is [Lx,Ly]=ihLz .

 

[foranlogan2]

thanks patrick , do i use [AB,CD] relation for each 4 commutators?,like [YPz,ZPx] and also [ZPy,ZPx] etc

 

[nrqed]

thanks patrick , do i use [AB,CD] relation for each 4 commutators?,like [YPz,ZPx] and also [ZPy,ZPx] etc

Yes. And you use $[X_i, X_j]= [P_i,P_j] =0$ and $[X_i,P_j] = i \hbar \delta_{ij}$. you'll see, you will get $i \hbar L_z$.

You are welcome

Patrick