I can't see how get matrix representation of ladder operators of angular momentum from their Kronecker-delta matrix elements,

$$\langle l,m'|L_{\pm}|l,m\rangle = \hbar \sqrt{l(l+1)-m(m\pm1)} \delta _{m',m\pm1},$$

(9-33, Gasiorowicz 3rd ed.) which defines matrix components from Kronecker delta ($l$ fixed, $l=1$). As I understand, first index are rows, second index are columns (in fact, from 9-18: $A_{mn} = \langle u_m | A | u_n \rangle)$. Then, lets get the matrix form of $L_{+}$. These are all pretty obvious and slow steps, but I am doing it this way to see clearly where I am wrong. For the first row:

  • $m'=-1$ (1st row), $m=-1$ (1st column), so $\delta_{m',m+1} = \delta_{-1,0}=0$
  • $m'=-1$ (1st row), $m=0$ (2nd col), so $\delta_{m',m+1} = \delta_{-1,1}=0$
  • $m'=-1$ (1st row), $m=1$ (3rd col), so $\delta_{m',m+1} = \delta_{-1,2}=0$

The second row:

  • $m'=0$ (2nd row), $m=-1$ (1col), so $\delta_{m',m+1} = \delta_{0,0}=1$
  • $m'=0$ (2nd row), $m=0$ (2col), so $\delta_{m',m+1} = \delta_{0,1}=0$
  • $m'=0$ (2nd row), $m=1$ (3col), so $\delta_{m',m+1} = \delta_{0,2}=0$

And the third row:

  • $m'=1$ (row), $m=-1$ (1col), so $\delta_{m',m+1} = \delta_{1,0}=0$
  • $m'=1$ (row), $m=0$ (2col), so $\delta_{m',m+1} = \delta_{1,1}=1$
  • $m'=1$ (row), $m=1$ (3col), so $\delta_{m',m+1} = \delta_{1,2}=0$

So the only values that do not vanish are (2,1) and (3,2), as we could get directly before. Then with the normalization, $\sqrt2$ on both cases, I get:

$L_{+,obtained}= \hbar \begin{pmatrix} 0 & 0 & 0 \\ \sqrt{2} & 0 & 0 \\ 0 & \sqrt{2} & 0 \end{pmatrix}$

And $L_+$ should be (9-35):

$L_{+}= \hbar \begin{pmatrix} 0 & \sqrt{2} & 0 \\ 0 & 0 & \sqrt{2} \\ 0 & 0 & 0 \end{pmatrix}$

In fact, what I am obtaining is $L_-$, and the opposite ($L_+$ when I am looking for $L_-$). I do not know where I am messing the rows/columns, the signs... but fixing this should be pretty straightforward.

  • 1
    $\begingroup$ Yes you need to make sure that your ordering of basis states $-1,0,1$ is same as text, which could use $1,0,-1$. The acid test are the commutation relations, which must be satisfied for both orderings. $\endgroup$ Jun 20, 2017 at 11:44
  • $\begingroup$ Just checked this and definitely you are right, in that order just works fine, and I obtain the result of the text. Thanks! $\endgroup$
    – DavidOrti
    Jun 20, 2017 at 12:00

1 Answer 1


It's not a problem, you get the right answer (just transposed one). It's just a question of matrix indexation, not quantum physics.

P.S. Answer in book is given in form $L_{mn} = <l,n|L_{+}|l,m>$.

Also this equation may help you: $L_+|l,m> = \sqrt{(l-m)(l+m+1)}|l,m+1>$.


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