# Does this commutation relation hold?

I was wondering whether it is true that $[L_x^2,x^2+y^2+z^2]=0$. I could not find it in the internet and therefore I wanted to ask here whether anybody here knows that this is true or false.

• You could work it out yourself, using the basic properties of the commutator and the basic commutation relation $[X_i,P_j]=i\hbar\delta_{ij}$ Furthermore, does $L^2_x$ mean $L_xL_x$? – Federico Carta Jan 10 '14 at 13:46

It is certainly true. The standard way to prove it is computing the LHS. However it is true due to simple theoretical reasons: If $U_t = e^{-it L_x}$ is the unitary representation of a rotation, of an angle $t$, around the $x$ axis, you have:

$$U_t |\hat{\bf x}|^2 U^\dagger_t = |\hat{\bf x}|^2$$

since $|\hat{\bf x}|^2= \hat{x}^2+ \hat{y}^2+ \hat{z}^2$ transforms as a scalar. Taking the $t$ derivative for $t=0$ you get:

$$[L_x,|\hat{\bf x}|^2] = 0$$

Finally:

$$[A^2,B]= A[A,B] + [A,B]A$$

ends the computation.

• Why does $\hat{x}$ transform as a scalar? It is a vector! I believe there is a typo in your answer, and you meant $|\hat{x}|^2$ – Federico Carta Jan 10 '14 at 13:54
• Sure, the question concerns $|\hat{x}|^2$, which is a scalar, and your answer is right. I am referring to the fourth line of your answer, in which you write "since $\hat{x}$ transforms as a scalar. Taking the...". There is a typo in there, since you write $\hat{x}$ is a scalar but instead it is a vector, and its modulus square is a scalar. – Federico Carta Jan 10 '14 at 14:03
• Sorry, I did not see the typo you pointed out. I have corrected. Thanks – Valter Moretti Jan 10 '14 at 14:04
• many thans to all of you – Xin Wang Jan 10 '14 at 14:28