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Math Help - Quantum Mechanics - Angular Momentum (Eigen Functions) part 5

  1. #1
    Senior Member bugatti79's Avatar
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    Quantum Mechanics - Angular Momentum (Eigen Functions) part 5

    Hi folks,

    I have 2 queries

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

     <br />
\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}<br />

    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)

     <br />
\mathbf r = x \mathbf i +y \mathbf j + z \mathbf k <br />
\implies
     <br />
\mathbf r = r sin (\theta) cos (\phi) \mathbf i + r sin (\theta) sin (\phi) \mathbf j + r cos (\theta) \mathbf k<br />

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

    Any ideas?
    Attached Thumbnails Attached Thumbnails Quantum Mechanics - Angular Momentum (Eigen Functions) part 5-dsc_0822.jpg  
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  2. #2
    A Plied Mathematician
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    For (1), I would just look at your picture and use the right-hand rule for cross products.

    For (2), check this out.
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  3. #3
    Senior Member bugatti79's Avatar
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    Thats really useful thanks,

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

     <br />
\displaystyle \frac{d }{d \theta}(r cos (\theta) sin (\phi)) = - sin (\theta)<br />
    I thought it would have been - r sin (\theta) sin (\phi)??
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    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?
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    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Ackbeet View Post
    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
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    Quote Originally Posted by bugatti79 View Post
    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.
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  7. #7
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Ackbeet View Post
    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!
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    You're welcome!
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