Quantum Mechanics - Angular Momentum (Eigen Functions) part 5

**bugatti79** · March 3rd 2011, 11:06 AM

Hi folks,

I have 2 queries

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

$ \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 $\mathbf r=r \hat r$ , $(\hat r \times \hat r)=0$ , $(\hat r \times \hat \theta)=\hat \phi$ , $(\hat r \times \hat \phi)=-\hat \theta$

How are these determined?

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

$ \mathbf r = x \mathbf i +y \mathbf j + z \mathbf k \implies$
$ \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 $\hat \theta$ and $\hat \phi$

Any ideas?

**Ackbeet** · March 3rd 2011, 11:40 AM

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

For (2), check this out.

**bugatti79** · March 3rd 2011, 01:19 PM

Thats really useful thanks,

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

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

**Ackbeet** · March 4th 2011, 05:06 AM

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

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

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

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

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

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

**bugatti79** · March 9th 2011, 10:44 AM

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

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

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

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

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

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

**Ackbeet** · March 9th 2011, 10:59 AM

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 $\mathbb{R}^{n};$ that is, a vector of any dimension. So, you always have the formula

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

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

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

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

$r$ 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.

**bugatti79** · March 9th 2011, 11:05 AM

Originally Posted by Ackbeet

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

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

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

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

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

$r$ 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!

**Ackbeet** · March 9th 2011, 11:08 AM

You're welcome!

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