# Deriving the matrix for the rising ladder operator


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).

• 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. Jun 21 '21 at 13:54
• 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. Jun 22 '21 at 6:21

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