- #1
rayman123
- 152
- 0
Homework Statement
calculate the following commutation relations
[tex][L_{x}L_{y}]=[/tex]
[tex][L_{y}L_{z}]=[/tex]
[tex]][L_{z}L_{x}]=[/tex]
Homework Equations
[tex] [L_{x},L_{y}]= -\hbar^2[y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}][/tex]
where the expression in the parentheses describes [tex] L_{z}[/tex]
but i still have [tex] \hbar^2[/tex] can someone explain to me how we get [tex] i \hbar L_{z}[/tex] where did the [tex] \hbar^2[/tex] go? And how did we get i back again?
here is my calculation
[tex][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})][/tex]
here is my second commutation
[tex] [L_{z},L_{x}]= [/tex] in this one i get plus instead of minus...do not know where i make mistake...
[tex][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})[/tex]