# Question about commutation and operator

1. Nov 3, 2008

### KFC

I am reading the book by J.J.Sakurai, in chaper 3, there is a relation given as

$$\langle \alpha', jm|J_z A |\alpha, jm\rangle$$

Here, j is the quantum number of total angular momentum, m the component along z direction, $$\alpha$$ is the third quantum number. $$J_z$$ is angular momentum operator, A is arbritary operator. Generally, $$J_z$$ is not commutate with A, but Sakurai just give the result directly as following

$$m\hbar\langle \alpha', jm|A|\alpha, jm\rangle$$

As you see, this just like have $$J_z$$ acting on the bar and returns the $$m\hbar\langle \alpha', jm|$$. My question is: how can $$J_z$$ acting on the bar vector?

2. Nov 3, 2008

### olgranpappy

It is correct to let J_z act on the bra in the way it does. If you are worried about this you can go back to chapter one and review the properties of bras and kets. For example
$$\langle \alpha |\beta\rangle=(\langle \beta | \alpha\rangle)^*$$

You can use the above relation to rewrite the quantity of your interest with $J_z^\dagger=J_z$ acting on a ket if you want.

3. Nov 3, 2008

### KFC

Thank you so much. I forget the relation $$J_z^\dagger=J_z$$, thanks again :)

4. Nov 3, 2008

### olgranpappy

you're welcome.

5. Nov 3, 2008

### Fredrik

Staff Emeritus
You already got the answer, but you may find this interesting too: