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

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

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
Homework Helper
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.

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..

nrqed
Homework Helper
Gold Member
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

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

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

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

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

nrqed
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 $i \hbar L_z$.