# Quantum Mechanics - Angular Momentum

• Feb 23rd 2011, 12:41 PM
bugatti79
Quantum Mechanics - Angular Momentum
Determine $[\hat L_x, \hat L_y]f(x,y,z)$

where $\displaystyle \hat L_x=\frac{\hbar}{i}(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y})$ and

$\displaystyle \hat L_y=\frac{\hbar}{i}(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z})$

$
\displaystyle [\hat L_x, \hat L_y]f(x,y,z)=\left(\frac{\hbar}{i}\right)^2 \left[ y\frac{\partial}{\partial z}\left(z \frac{\partial f}{\partial x}\right) -y \frac{\partial}{\partial z} \left( x \frac{\partial f}{\partial z} \right)
$

$
\displaystyle -z \frac{\partial}{\partial y} \left( z \frac{\partial f}{\partial x} \right)+z \frac{\partial}{\partial y} \left( x \frac{\partial f}{\partial z} \right)-z \frac{\partial}{\partial x} \left( y \frac{\partial f}{\partial z} \right) \right)
$

$
\displaystyle +z \frac{\partial}{\partial x} \left( z\frac{\partial f}{\partial y} \right)+x \frac{\partial}{\partial z} \left( y \frac{\partial f}{\partial z} \right)-x \frac{\partial}{\partial z} \left( z \frac{\partial f}{\partial y} \right) \right]$

What I dont understand is what's done next... ie if you look at the first term

$
\displaystyle y\frac{\partial}{\partial z}\left(z \frac{\partial f}{\partial x}\right)
$
. This expands out to $
\displaystyle y\frac{\partial f}{\partial x} +yz\frac{\partial^{2} f}{\partial z \partial x}
$

What is happening here?..
• Feb 23rd 2011, 12:55 PM
Ackbeet
That's just the product rule, isn't it?
• Feb 23rd 2011, 01:00 PM
TheEmptySet
Quote:

Originally Posted by bugatti79
Determine $[\hat L_x, \hat L_y]f(x,y,z)$

where $\displaystyle \hat L_x=\frac{\hbar}{i}(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y})$ and

$\displaystyle \hat L_y=\frac{\hbar}{i}(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z})$

$
\displaystyle [\hat L_x, \hat L_y]f(x,y,z)=\left(\frac{\hbar}{i}\right)^2 \left[ y\frac{\partial}{\partial z}\left(z \frac{\partial f}{\partial x}\right) -y \frac{\partial}{\partial z} \left( x \frac{\partial f}{\partial z} \right)
$

$
\displaystyle -z \frac{\partial}{\partial y} \left( z \frac{\partial f}{\partial x} \right)+z \frac{\partial}{\partial y} \left( x \frac{\partial f}{\partial z} \right)-z \frac{\partial}{\partial x} \left( y \frac{\partial f}{\partial z} \right) \right)
$

$
\displaystyle +z \frac{\partial}{\partial x} \left( z\frac{\partial f}{\partial y} \right)+x \frac{\partial}{\partial z} \left( y \frac{\partial f}{\partial z} \right)-x \frac{\partial}{\partial z} \left( z \frac{\partial f}{\partial y} \right) \right]$

What I dont understand is what's done next... ie if you look at the first term

$
\displaystyle y\frac{\partial}{\partial z}\left(z \frac{\partial f}{\partial x}\right)
$
. This expands out to $
\displaystyle y\frac{\partial f}{\partial x} +yz\frac{\partial^{2} f}{\partial z \partial x}
$

What is happening here?..

This is just the product rule

$\displaystyle y\frac{\partial}{\partial z}\left(z \frac{\partial f}{\partial x}\right)=y\left[\frac{\partial}{\partial z}\left( z \frac{\partial f}{\partial x}\right) \right]=y\left[ \frac{\partial f}{\partial x}\frac{\partial }{\partial z}z+z\frac{\partial }{\partial z} \frac{\partial f}{\partial x}\right]=y\left[ \frac{\partial f}{\partial x}(1)+\frac{\partial^2 f}{\partial z\partial x}\right]$
• Feb 23rd 2011, 01:39 PM
bugatti79
Quote:

Originally Posted by TheEmptySet
This is just the product rule

$\displaystyle y\frac{\partial}{\partial z}\left(z \frac{\partial f}{\partial x}\right)=y\left[\frac{\partial}{\partial z}\left( z \frac{\partial f}{\partial x}\right) \right]=y\left[ \frac{\partial f}{\partial x}\frac{\partial }{\partial z}z+z\frac{\partial }{\partial z} \frac{\partial f}{\partial x}\right]=y\left[ \frac{\partial f}{\partial x}(1)+\frac{\partial^2 f}{\partial z\partial x}\right]$

ok, what if for arguments sake that the function f only depend on x and y then

$\displaystyle y\frac{\partial}{\partial z}\left(z \frac{\partial f}{\partial x}\right)= y \frac{\partial f}{\partial x}$?

Any how, the original reduces down to

$\displaystyle \left(\frac{\hbar}{i}\right)^2 \left(y \frac{\partial}{\partial x} - x \frac{\partial}{\partial y}\right) f(x,y,z) = \left( \frac{\hbar}{i} \right) \left( \frac{\hbar}{i} y \frac{\partial}{\partial x} - x \frac{\partial}{\partial y} \right) f (x,y,z)= i \hbar \hat L_z f (x,y,z)$

What is happening in the middle terms with the hbar over i moved inside the brackets? It may be a typo as I was writing furiously trying to keep up with the lecturer! (Giggle)

Thanks
• Feb 23rd 2011, 02:19 PM
Ackbeet
Quote:

Originally Posted by bugatti79
ok, what if for arguments sake that the function f only depend on x and y then

$\displaystyle y\frac{\partial}{\partial z}\left(z \frac{\partial f}{\partial x}\right)= y \frac{\partial f}{\partial x}$?

Yes.

Quote:

Any how, the original reduces down to

$\displaystyle \left(\frac{\hbar}{i}\right)^2 \left(y \frac{\partial}{\partial x} - x \frac{\partial}{\partial y}\right) f(x,y,z) = \left( \frac{\hbar}{i} \right) \left( \frac{\hbar}{i} y \frac{\partial}{\partial x} - x \frac{\partial}{\partial y} \right) f (x,y,z)= i \hbar \hat L_z f (x,y,z)$

What is happening in the middle terms with the hbar over i moved inside the brackets? It may be a typo as I was writing furiously trying to keep up with the lecturer! (Giggle)

Thanks
The hbar over i should distribute and multiply both terms. Your

$\hat{L}_{z}=-i\hbar\left[x\,\dfrac{\partial}{\partial y}-y\,\dfrac{\partial}{\partial x}\right].$

Computing $i\hbar\hat{L}_{z}$ should get you the far left term (minus the test function f).