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$\newcommand{\Ket}[1]{\left|#1\right>}$ $\newcommand{\Bra}[1]{\left<#1\right|}$ $\newcommand{\Braket}[2]{\left<#1|#2\right>}$

Notation:

$\vec{l}$ is the orbital angular momentum

$\vec{s}$ is the electron spin angular momentum

$\vec{j}$ is the total angular momentum

We consider a one-electron-system for a p-electron i.e. $l=1, s=1/2$.

Since the operators $\hat{\vec{l}}^2$ and $\hat{l}_z$ resp. $\hat{\vec{s}}^2$ and $\hat{s}_z$ commute we can find a common basis for them. We find the vector spaces

$ V_l = span\bigg\{ \Ket{1,1}, \Ket{1,0}, \Ket{1,-1} \bigg\} \tag{1} $

$ V_s = span\bigg\{ \Ket{\frac{1}{2}, \frac{1}{2}}, \Ket{\frac{1}{2}, -\frac{1}{2}} \bigg\} \tag{2} $

Using these two vector spaces we can get the common vector space $V_{ls}$ by "multiplying" the basis for $V_l$ and $V_s$.

In the decoupled display (this might be a bad translation, but should become clear) we assume that $\hat{\vec{l}}$ and $\hat{\vec{s}}$ don't interact with each other. Which means that $l, m_l, s, m_s$ are "good" quantum numbers to describe the system. So we can choose the following basis functions:

$\Ket{l, m_l, s, m_s} = \Ket{l, m_l}\Ket{s, m_s}$

Note: One might use the shorter notation $\Ket{m_l, m_s}$ (although I never do here).

We get the following six basis functions in the uncoupled display (for the basis $B_{l,s}$

\begin{align*} \Ket{l, m_l, s, m_s} &= \Ket{l, m_l} \Ket{s, m_s}\\ \Ket{1,\ 1,½,\ ½} &= \Ket{1,\ 1}\Ket{½,\ ½} \\ \Ket{1,\ 1,½,\text{-}½} &= \Ket{1,\ 1}\Ket{½,\text{-}½} \\ \Ket{1,\ 0,½,\ ½} &= \Ket{1,\ 0}\Ket{½,\ ½} \tag{3} \\ \Ket{1,\ 0,½,\text{-}½} &= \Ket{1,\ 0}\Ket{½,\text{-}½} \\ \Ket{1,\text{-}1,½,\ ½} &= \Ket{1,\text{-}1}\Ket{½,\ ½} \\ \Ket{1,\text{-}1,½,\text{-}½} &= \Ket{1,\text{-}1}\Ket{½,\text{-}½} \end{align*}

Now we want to find the matrix for $\hat{l}_+$ (which describes the rising ladder operator) in the basis $B_{l,s}$.

We know

\begin{align*} \hat{l}_+ \Ket{1, m_l} &= \hbar [1(1+1) - m_l(m_l + 1)]^{1/2}\\ &= \hbar [2 - m_l(m_l + 1)]^{1/2} \tag{4} \end{align*}

so we get

$m_l = 1 \quad \Rightarrow \quad \hat{l}_+\Ket{1,1} = 0\Ket{1,1} = 0 \tag{4}$

$m_l = 0 \quad \Rightarrow \quad \hat{l}_+\Ket{1,0} = \sqrt{2}\hbar\Ket{1,1} \tag{5}$

$m_l = -1 \quad \Rightarrow \quad \hat{l}_+\Ket{1,-1} = \sqrt{2}\hbar\Ket{1,0} \tag{6}$

Now I learned that if we want to write an operator as a matrix we do

$ A_{nm} = \Bra{n}\hat{A}\Ket{m} \tag{7} $

so in our case that'd be

$ (l_+)_{l, m_l, s, m_s;}= \Bra{l, m_l, s, m_s} \hat{l}_+ \Ket{l, m_l, s, m_s} \tag{8} $

Sorry for the above notation abomination.

Since $\hat{l}_+$ does not act on the spin part of the wavefunction, we get:

$ \hat{l}_+ = \sqrt{2}\hbar \begin{pmatrix} 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \tag{9} $

Now to my question: I can't see how we got (9). Is it just experience? I just don't get their line of thought.

What I'd do is:

using (4), (5), and (6) we get

$ m_l = 1 \quad \Rightarrow \quad \Bra{1,1}\hat{l}_+\Ket{1,1} = \Bra{1,1}0\Ket{1,1} = 0 \Braket{1,1}{1,1}\tag{10} $

$ m_l = 0 \quad \Rightarrow \quad \Bra{1,0}\hat{l}_+\Ket{1,0} = \Bra{1,0}\sqrt{2}\hbar\Ket{1,1} = \sqrt{2}\hbar\Braket{1,0}{1,1} \tag{11} $

$ m_l = -1 \quad \Rightarrow \quad \Bra{1,-1}\hat{l}_+\Ket{1,-1} = \Bra{1,-1}\sqrt{2}\hbar\Ket{1,0} = \sqrt{2}\hbar\Braket{1,-1}{1,0} \tag{12} $

from which I get the following matrix for $\hat{l}_+$ in the basis of $V_l$

$ \hat{l}_+^{V_l} = \sqrt{2}\hbar \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{pmatrix} \tag{13} $

whereas the

  • first row denotes $\Bra{l, m_l} = \Bra{1,1}$

  • second row denotes $\Bra{l, m_l} = \Bra{1,0}$

  • third row denotes $\Bra{l, m_l} = \Bra{1,-1}$

  • first column denotes $\Bra{l, m_l} = \Ket{1,1}$

  • second column denotes $\Bra{l, m_l} = \Ket{1,0}$

  • third column denotes $\Bra{l, m_l} = \Ket{1,-1}$

since we are in the decoupled display and thus the spin doesn't interact with the angular momentum, we get the identity matrix for the spin part i.e. for $V_s$.

We could then use the kroenecker product to get something "similar" to (9).

So I basically presented two approaches. The first approach being we derive the $6\times 6$ matrix directly and the second approach is we derive the $2\times 2$ matrix for the spin part and the $3\times 3$ matrix for the angular momentum part and take the kroenecker product.

I now have three questions:

  1. How exactly did they map the basis (3) to the rows and columns of (9)? I can't figure it out.
  2. Is (13) correct for the described "mapping" of rows and columns to the basis functions?
  3. How do I know which way I should take the kroenecker product? I could do $A \otimes B$ or $B \otimes A$.

I know that probably is doesn't matter as long as I respect the order of the basis but the issue here is that I don't think that happened in (9). I think what I got in (13) respects the ordering they used in (3).

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  • 1
    $\begingroup$ Are you sure that the text leading up to your questions is correct? Something about all this seems a bit sketchy to me.. But I could be wrong. $\endgroup$
    – Noumeno
    Jun 21 '21 at 13:54
  • $\begingroup$ I mean sure it could be a typo. If it is, then several hundred students didn't notice/complain. Anyway, I'm gonna ask the one who created it then. $\endgroup$
    – xotix
    Jun 22 '21 at 6:21
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1: It appears (9) is up-down inverted w.r.t. (3). Preserving the ordering in (3), it would be:

$ \hat{l}_+ = \sqrt{2}\hbar \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \end{pmatrix} \tag{9} $

2: Yes. The top entry is deleted, the middle on is sent to the top, and the bottom on is sent to the middle. (Along with a factor of $\sqrt 2\hbar$.)

3: The order of the Kronecker product is up to you. The second multiplicand is the "fast" index, which is the spin index according to (3).

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  • $\begingroup$ Thank you for the confirmation. $\endgroup$
    – xotix
    Jun 22 '21 at 6:21

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