Commutation relationsl, angular momentum

[rayman123]

Homework Statement

calculate the following commutation relations
[L_{x}L_{y}]=
[L_{y}L_{z}]=
][L_{z}L_{x}]=

Homework Equations

[L_{x},L_{y}]= -\hbar^2[y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}]
where the expression in the parentheses describes L_{z}
but i still have \hbar^2 can someone explain to me how we get i \hbar L_{z} where did the \hbar^2 go? And how did we get i back again?

here is my calculation
[L_{x},L_{y}]= L_{x}L_{y}-L_{y}L_{x}=-\hbar^2[(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y})(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z})-(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z})(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y})]


here is my second commutation
[L_{z},L_{x}]= in this one i get plus instead of minus...do not know where i make mistake...
[L_{z},L_{x}]= }=-\hbar^2[(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x})(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y})-(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y})(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x})]= -\hbar^2(x\frac{\partial}{\partial z}+z\frac{\partial}{\partial x})

Homework Statement

 

[Mindscrape]

Ummm, yeah, you're missing the ihbar because you haven't equated the
<br /> [L_{x},L_{y}]= -\hbar^2[y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}]<br />
part to how Lz is supposed to look. You'll hit your head as you notice that to get to Lz you need a factor of ihbar. Same for the others. :)

 

[rayman123]

hm i still do not get it...first it dissapears and then suddenly comes upp again...
and why in the last commutation the sign is wrong?

 

[lanedance]

[L_{x},L_{y}]<br /> = -\hbar^2[y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}]<br /> = (i^2)\hbar^2[y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}]<br /> = (i\hbar) \left(i\hbar[y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}]<br /> \right)<br /> <br />

 

[vela]

[L_{x},L_{y}]= -\hbar^2[y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}]
where the expression in the parentheses describes L_{z}

That's your mistake. You're forgetting the factor of i\hbar.

L_{z} = i\hbar \left(y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}\right) \ne y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}

here is my second commutation
[L_{z},L_{x}]= in this one i get plus instead of minus...do not know where i make mistake...
[L_{z},L_{x}]= }=-\hbar^2[(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x})(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y})-(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y})(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x})]= -\hbar^2(x\frac{\partial}{\partial z}+z\frac{\partial}{\partial x})

We can't see where you made the mistake either since you didn't show your work. The set-up looks fine, though.

 

[thebigstar25]

if you don't want things to get messy try to start from the fact that L= rxp
then, Lx= yp(z) - zp(y) .. Ly= zp(x) - xp(z) .. Lz = xp(y) - yp(x)

also make use of : [x,y]=0, [x,z]=0, [y,z]=0, [px,y]=0, [px,z]=0, [py,x]=0, [py,z]=0, [pz,x]=0, [pz,y]=0, [px,x]=-ih-par, [py,y]=-ih-par, [pz,z]=-ih-par ..