Normalization of the Angular Momentum Ladder Operator

[PatsyTy]

Homework Statement

Obtain the matrix representation of the ladder operators $J_{\pm}$.

Homework Equations

Remark that $J_{\pm} | jm \rangle = N_{\pm}| jm \pm 1 \rangle$

The Attempt at a Solution

[/B]
The textbook states $|N_{\pm}|^2=\langle jm | J_{\pm}^\dagger J_{\pm} | jm \rangle$, from here it is straightforward:

$\langle jm | J_{\mp} J_{\pm} | jm \rangle = \langle jm | (J^2 - J_z^2 \pm J_z | jm \rangle = \langle jm | \big( j(j+1)|jm\rangle-m^2\hbar^2 |jm\rangle \pm m \hbar^2 | jm \rangle\big)$

Giving us $|N_\pm |^2=j(j+1)-\hbar^2 m (m\pm 1)$

What I do not understand is the very first step where we write out $|N_{\pm}|^2=\langle jm | J_{\pm}^\dagger J_{\pm} | jm \rangle$ from the given equation $J_{\pm} | jm \rangle = N_{\pm}| jm \pm 1 \rangle$.

I have tried multiplying the first equation by the hermitian conjugate of $J_\pm|jm\rangle$ giving

$\langle jm | J_\mp J_\pm | jm \rangle = \langle jm | J_\mp N_{\pm}| jm \pm 1 \rangle = N_{\pm} \langle jm | J_\mp | jm \pm 1 \rangle$ however I don't think this is correct as I don't see how I can get a second $N_\pm$ from $\langle jm | J_\mp | jm \pm 1 \rangle$.

The text gives the explanation "Since both $|jm\rangle$ and $|jm+1\rangle$ are normalized to unity..." and that's the justification for this step. I don't fully trust this qualitative description so I am trying to write it out mathematically. Any help would be appreciated!

 

[stevendaryl]

I have tried multiplying the first equation by the hermitian conjugate of $J_\pm|jm\rangle$ giving

$\langle jm | J_\mp J_\pm | jm \rangle = \langle jm | J_\mp N_{\pm}| jm \pm 1 \rangle = N_{\pm} \langle jm | J_\mp | jm \pm 1 \rangle$ however I don't think this is correct as I don't see how I can get a second $N_\pm$ from $\langle jm | J_\mp | jm \pm 1 \rangle$.

Here's the part that you are missing: For any matrix element of the form $\langle A |O|B \rangle$, we have:

$\langle A |O|B \rangle = \langle B |O^\dagger|A \rangle^*$

So in particular,
$\langle jm | J_\mp | jm \pm 1 \rangle = \langle jm\pm 1 | J_\pm | jm \rangle^*$

See if that helps.

 

[PatsyTy]

Here's the part that you are missing: For any matrix element of the form $\langle A |O|B \rangle$, we have:

$\langle A |O|B \rangle = \langle B |O^\dagger|A \rangle^*$

So in particular,
$\langle jm | J_\mp | jm \pm 1 \rangle = \langle jm\pm 1 | J_\pm | jm \rangle^*$

See if that helps.

Thanks! I got it right away with that.