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: \(\displaystyle L_i ~ L_j \neq L_j ~ L_i \).ognik said: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 ?
- 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?topsquark said:\(\displaystyle L \times L = \left | \begin{matrix} \hat{x} & \hat{y} & \hat{z} \\ L_x & L_y & L_z \\ L_x & L_y & L_z \end{matrix} \right | \)
\(\displaystyle = \hat{x} \left ( L_y ~ L_z - L_z ~ L_x \right ) + \)...
topsquark said: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.
Yes, it's a typo. I'll fix it up in a moment. Thanks for the catch!ognik said:- assume the $L_x$ is a typo.
The x component of a x b is \(\displaystyle a_y b_z - a_z b_y\). It's a definition.ognik said:I get $ \hat{x} \left ( L_y ~ L_z - L_y ~ L_z \right ) +... $ what makes the order change around please?
The cross product, also known as the vector product, is a mathematical operation that takes two vectors as input and produces a third vector that is perpendicular to both input vectors. It is denoted by the symbol "×" or "x" and is defined as the product of the magnitudes of the two vectors and the sine of the angle between them.
The cross product should be zero when the two input vectors are parallel or one of the vectors has a magnitude of 0. This is because the sine of the angle between two parallel vectors is 0, resulting in a cross product of 0. Additionally, if one of the vectors has a magnitude of 0, the resulting cross product will also be 0.
The cross product being zero indicates that the two input vectors are either parallel or one of the vectors has a magnitude of 0. This can be useful in various mathematical and scientific applications, such as determining if two lines are parallel or finding the area of a parallelogram.
No, the cross product can never be negative or a complex number. This is because the magnitude of the sine function is always between 0 and 1, and the cross product is the product of the magnitudes of the two input vectors and the sine of the angle between them. Therefore, the resulting cross product will always be a positive real number.
The cross product and the dot product are both types of vector products, but they have different properties and results. The dot product results in a scalar quantity, while the cross product results in a vector quantity. Additionally, the dot product measures the projection of one vector onto another, while the cross product measures the perpendicularity of two vectors.