Is the Cross Product of Orbital Angular Momentum Always Zero?

[ognik]

Hi - from orbital angular momentum components, $[L_x, L_y] = iL_z$

My book claims 'Hence, $ \vec{L} \times \vec{L} = i\vec{L} $' I'm keen to know how they get that, an also why that cross products isn't = 0, like $A \times A$ would be ?

 

[topsquark]

Hi - from orbital angular momentum components, $[L_x, L_y] = iL_z$

My book claims 'Hence, $ \vec{L} \times \vec{L} = i\vec{L} $' I'm keen to know how they get that, an also why that cross products isn't = 0, like $A \times A$ would be ?

This is a Quantum Mechanics problem and the reason that L x L isn't 0 is due to the fact that the components of L are not commutative: \(\displaystyle L_i ~ L_j \neq L_j ~ L_i \).

You need two more relations to do this problem. In total we have
\(\displaystyle [ L_x, ~ L_y ] = i L_z\)

\(\displaystyle [ L_y, ~ L_z ] = i L_x\)

\(\displaystyle [ L_z, ~ L_x ] = i L_y\)

(This is more compactly stated as \(\displaystyle [ L_i, ~ L_j ] = i \epsilon _{ijk} L_k\). (The "i" in the coefficient is the usual complex number, whereas \(\displaystyle i,~j,~k = x,~y~,z \) respectively otherwise.)

So.
\(\displaystyle L \times L = \left | \begin{matrix} \hat{x} & \hat{y} & \hat{z} \\ L_x & L_y & L_z \\ L_x & L_y & L_z \end{matrix} \right | \)

\(\displaystyle = \hat{x} \left ( L_y ~ L_z - L_z ~ L_y \right ) + \hat{y} \left ( L_z ~ L_x - L_x ~ L_z \right ) + \hat{z} \left ( L_x ~ L_y - L_y ~ L_x \right )\)

\(\displaystyle = \hat{x} ~ [ L_y, ~ L_z ] + \hat{y} ~ [ L_z, ~ L_x ] + \hat{z} ~ [ L_x, ~ L_y ] \)

\(\displaystyle = \hat{x} i L_x + \hat{y} i L_y + \hat{z} i L_z = i L\)

-Dan

Addendum: I should mention that if you look at L x L = i L you will notice that the units don't match up. This is because many Physicists have an annoying habit of setting \(\displaystyle \hbar = 1\). (I don't mind the other convention: c = 1.) The rule of thumb is that any quantity that contains an \(\displaystyle \hbar\) is a purely Quantum property/operator and thus vanishes in the Classical limit. In this case we should have \(\displaystyle L \times L = i \hbar L\). This equation contains an \(\displaystyle \hbar\) so in the Classical limit \(\displaystyle \lim_{\hbar \to 0} L \times L = 0\), which it should. By setting \(\displaystyle \hbar = 1\) we would not be able to see this.

 

[ognik]

\(\displaystyle L \times L = \left | \begin{matrix} \hat{x} & \hat{y} & \hat{z} \\ L_x & L_y & L_z \\ L_x & L_y & L_z \end{matrix} \right | \)

\(\displaystyle = \hat{x} \left ( L_y ~ L_z - L_z ~ L_x \right ) + \)...

- assume the $L_x$ is a typo.I get $ \hat{x} \left ( L_y ~ L_z - L_y ~ L_z \right ) +... $ what makes the order change around please?

Addendum: I should mention that if you look at L x L = i L you will notice that the units don't match up. This is because many Physicists have an annoying habit of setting \(\displaystyle \hbar = 1\). (I don't mind the other convention: c = 1.) The rule of thumb is that any quantity that contains an \(\displaystyle \hbar\) is a purely Quantum property/operator and thus vanishes in the Classical limit. In this case we should have \(\displaystyle L \times L = i \hbar L\). This equation contains an \(\displaystyle \hbar\) so in the Classical limit \(\displaystyle \lim_{\hbar \to 0} L \times L = 0\), which it should. By setting \(\displaystyle \hbar = 1\) we would not be able to see this.

Immensely valuable point, thanks.

 

[topsquark]

- assume the $L_x$ is a typo.

Yes, it's a typo. I'll fix it up in a moment. Thanks for the catch!

I get $ \hat{x} \left ( L_y ~ L_z - L_y ~ L_z \right ) +... $ what makes the order change around please?

The x component of a x b is \(\displaystyle a_y b_z - a_z b_y\). It's a definition.

-Dan

 

[ognik]

Aaaargh, I've been doing it wrong for years, outside of operators I suppose it didn't matter, will remember this now - but 'by definition' never sits well in my weird ol' head...