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I found this website which shows how to derive the matrices for $L_{+}, L_{-}$ and while I understand the derivation of the equation for $<lm|L_{+}|lm'>$ and $<lm|L_{-}|lm'>$ I do not understand how to use this to find the values, and positions of the values, that are given in the matrix.

Taking $L_{+}$ as an example, I know the delta function will only be non zero when $m'=m+1$ so for $l=1$ the values would be $\sqrt{2}$ for $m'=0,-1$ (corresponding to $m=-1,0$) and $0$ for $m'=1, m=0$. I got the values by plugging the value of $m'$ into the equation given ie $$<lm|L_{+}|lm'>=\sqrt{l(l+1)-m(m+1)}\hbar\delta_{m'(m+1)}.$$

After arriving at these values, I don't know how to place these values into the matrix, and though it may sound like a stupid question, I would really appreciate some help. Also if you know of a general form for this matrix, that would great as well.

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  • $\begingroup$ I don't have time to write a thorough answer demonstrating this, but perhaps you should try writing the bra, matrix, ket multiplication out for a simple system (say two basis vectors). The fact is that $ \langle i | M | j \rangle$ is the matrix element $M_{ij}$, and this may help you see that. $\endgroup$ – zeldredge Mar 8 '16 at 14:48
  • $\begingroup$ @zeldredge Actually that fact is all I really needed. I just wanted to know how to correlate the bra-ket notation to the actual matrix values $\endgroup$ – Mecury-197 Mar 9 '16 at 14:23
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The $L_+$ and $L_-$ operators have block matrix representation for every $l$. Sorting the basis vectors from $m=-l$ to $m=l$ in each block - all the nonzero elements are exactly above or below the main diagonal, i.e the equation couple elements only to closest basis vectors (nearest up for $L_-$ or down for $L_+$ in vector basis order as presented later. The elements above the diagonal couple to $m'=m-1$ and below the diagonal to $m'=m+1$)

Notice that my 'tag' ($m'$, mtag) notation is opposite to yours.

Try the following mathematica code (example for $l=3$)-

l = 3;
Lminus = Table[
  Sqrt[l (l + 1) - m (m - 1)] KroneckerDelta[mtag, m - 1], {mtag, -l, 
   l}, {m, -l, l}]
Lplus = Table[
  Sqrt[l (l + 1) - m (m + 1)] KroneckerDelta[mtag, m + 1], {mtag, -l, 
   l}, {m, -l, l}]

The order of basis vector is as following ($|l,m\rangle$ format) - $$\begin{pmatrix} |3,-3\rangle \\ |3,-2\rangle \\ |3,-1\rangle \\ |3,0\rangle \\ |3,1\rangle \\ |3,2\rangle \\ |3,3\rangle\end{pmatrix}$$ The results look like - $$L_-=\left( \begin{array}{ccccccc} 0 & \sqrt{6} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & \sqrt{10} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 \sqrt{3} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 2 \sqrt{3} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \sqrt{10} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \sqrt{6} \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} \right)$$

and
$$L_+=\left( \begin{array}{ccccccc} 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \sqrt{6} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \sqrt{10} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 \sqrt{3} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 \sqrt{3} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \sqrt{10} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \sqrt{6} & 0 \\ \end{array} \right)$$

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