# Thread: Quantum Mechanics - Angular Momentum (Eigen Functions) part 5

1. ## Quantum Mechanics - Angular Momentum (Eigen Functions) part 5

Hi folks,

I have 2 queries

1) Finding Lx, Ly and Lz in spherical co ordinates

$\displaystyle \displaystyle \nabla = \hat r \frac{\partial}{\partial r}+ \hat \theta \frac{1}{r} \frac{\partial}{\partial \theta}+ \hat \phi \frac{1}{ rsin \theta} \frac{\partial}{\partial \phi}$

Based on attached schematic its given that $\displaystyle \mathbf r=r \hat r$, $\displaystyle (\hat r \times \hat r)=0$, $\displaystyle (\hat r \times \hat \theta)=\hat \phi$,$\displaystyle (\hat r \times \hat \phi)=-\hat \theta$

How are these determined?

2) The unit vectors $\displaystyle \hat \theta$ and $\displaystyle \hat \phi$ are to be resolved into their cartesian components. I believe I can get $\displaystyle \mathbf r$(based on schematic and that $\displaystyle s= r sin \theta$)

$\displaystyle \mathbf r = x \mathbf i +y \mathbf j + z \mathbf k \implies$
$\displaystyle \mathbf r = r sin (\theta) cos (\phi) \mathbf i + r sin (\theta) sin (\phi) \mathbf j + r cos (\theta) \mathbf k$

but I dont know how to get the units vectors in terms of their cartesian components $\displaystyle \hat \theta$ and $\displaystyle \hat \phi$

Any ideas?

2. For (1), I would just look at your picture and use the right-hand rule for cross products.

For (2), check this out.

3. Thats really useful thanks,

I am just wondering how they derived at equation 21, ie

$\displaystyle \displaystyle \frac{d }{d \theta}(r cos (\theta) sin (\phi)) = - sin (\theta)$
I thought it would have been $\displaystyle - r sin (\theta) sin (\phi)$??

4. You're correct. However, that's not quite what they're doing. For one thing, they're dividing by the length. So, if you first compute

$\displaystyle \displaystyle\frac{d\mathbf{r}}{d\theta}=\begin{bm atrix}-r\sin(\theta)\sin(\phi)\\ r\cos(\theta)\sin(\phi)\\ 0\end{bmatrix},$ and then find its length, which is

$\displaystyle \displaystyle\left|\frac{d\mathbf{r}}{d\theta}\rig ht|=r\sqrt{\sin^{2}(\theta)\sin^{2}(\phi)+\cos^{2} (\theta)\sin^{2}(\phi)}=r|\sin(\phi)|=r\sin(\phi),$

on account of the range of $\displaystyle \phi$ (the polar angle on this website), then you'll find that

$\displaystyle \hat{\mathbf{\theta}}=\dfrac{\dfrac{d\mathbf{r}}{d \theta}}{\left|\dfrac{d\mathbf{r}}{d\theta}\right| }=\dfrac{1}{r\sin(\phi)}\begin{bmatrix}-r\sin(\theta)\sin(\phi)\\ r\cos(\theta)\sin(\phi)\\ 0\end{bmatrix}=\begin{bmatrix}-\sin(\theta)\\ \cos(\theta)\\ 0\end{bmatrix},$

as they have. Does that make it clearer, perhaps?

5. Originally Posted by Ackbeet
You're correct. However, that's not quite what they're doing. For one thing, they're dividing by the length. So, if you first compute

$\displaystyle \displaystyle\frac{d\mathbf{r}}{d\theta}=\begin{bm atrix}-r\sin(\theta)\sin(\phi)\\ r\cos(\theta)\sin(\phi)\\ 0\end{bmatrix},$ and then find its length, which is

$\displaystyle \displaystyle\left|\frac{d\mathbf{r}}{d\theta}\rig ht|=r\sqrt{\sin^{2}(\theta)\sin^{2}(\phi)+\cos^{2} (\theta)\sin^{2}(\phi)}=r|\sin(\phi)|=r\sin(\phi),$

on account of the range of $\displaystyle \phi$ (the polar angle on this website), then you'll find that

$\displaystyle \hat{\mathbf{\theta}}=\dfrac{\dfrac{d\mathbf{r}}{d \theta}}{\left|\dfrac{d\mathbf{r}}{d\theta}\right| }=\dfrac{1}{r\sin(\phi)}\begin{bmatrix}-r\sin(\theta)\sin(\phi)\\ r\cos(\theta)\sin(\phi)\\ 0\end{bmatrix}=\begin{bmatrix}-\sin(\theta)\\ \cos(\theta)\\ 0\end{bmatrix},$

as they have. Does that make it clearer, perhaps?
Yes, thats almost making sense. Just couldnt find any reference in my books regarding the magnitude of a 3*1 matrix, ie your second line. Im guessing its an extended form of pythagoras, ie SRSS's...

Thanks

6. Originally Posted by bugatti79
Yes, thats almost making sense. Just couldnt find any reference in my books regarding the magnitude of a 3*1 matrix, ie your second line. Im guessing its an extended form of pythagoras, ie SRSS's...

Thanks
That is the Euclidean norm, or Euclidean length, or Euclidean magnitude of any vector in $\displaystyle \mathbb{R}^{n};$ that is, a vector of any dimension. So, you always have the formula

$\displaystyle \displaystyle\|\mathbf{r}\|_{2}=\sqrt{\sum_{j}r_{j }^{2}},$

or the SRSS's, as you abbreviated it. The subscript $\displaystyle 2$ there on the norm symbol $\displaystyle \|\cdot\|$ denotes the Euclidean norm.

In a more concrete fashion, think about the equation for a sphere of radius $\displaystyle r$ centered at the origin:

$\displaystyle r^{2}=x^{2}+y^{2}+z^{2}.$

$\displaystyle r$ is also the magnitude of any vector going from the center of the sphere to a point on the sphere. The Euclidean length comes from that.

7. Originally Posted by Ackbeet
That is the Euclidean norm, or Euclidean length, or Euclidean magnitude of any vector in $\displaystyle \mathbb{R}^{n};$ that is, a vector of any dimension. So, you always have the formula

$\displaystyle \displaystyle\|\mathbf{r}\|_{2}=\sqrt{\sum_{j}r_{j }^{2}},$

or the SRSS's, as you abbreviated it. The subscript $\displaystyle 2$ there on the norm symbol $\displaystyle \|\cdot\|$ denotes the Euclidean norm.

In a more concrete fashion, think about the equation for a sphere of radius $\displaystyle r$ centered at the origin:

$\displaystyle r^{2}=x^{2}+y^{2}+z^{2}.$

$\displaystyle r$ is also the magnitude of any vector going from the center of the sphere to a point on the sphere. The Euclidean length comes from that.
Perfect. Thanks!

8. You're welcome!