(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

calculate the following commutation relations

[tex][L_{x}L_{y}]=[/tex]

[tex][L_{y}L_{z}]=[/tex]

[tex]][L_{z}L_{x}]=[/tex]

2. Relevant 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]

1. The problem statement, all variables and given/known data

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# Homework Help: Commutation relationsl, angular momentum

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