question about commutation and operator

[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?

[olgranpappy]

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?

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.

[KFC]

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.

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

[olgranpappy]

you're welcome.

[Fredrik]

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

A bra is a linear function that takes kets to complex numbers, i.e. it's a member of the dual space H*. There's a theorem that guarantees that for each ket |\alpha\rangle\in H, there's a bra \langle\alpha|\in H^* that takes an arbitrary ket |\beta\rangle to the scalar product (|\alpha\rangle,|\beta\rangle).

Recall that when T is a linear function acting on x, it's conventional to write Tx instead of T(x). This convention is used with bras. Also, whenever two | symbols should appear next to each other, only one is written out. So we have e.g.

(|\alpha\rangle,|\beta\rangle)=\langle\alpha|(|\be ta\rangle)=\langle\alpha||\beta\rangle=\langle\alp ha|\beta\rangle[/itex]

This takes some getting used to. This is one place where it gets confusing: If X is an operator, X^\dagger is defined by

[tex](|\alpha\rangle,X|\beta\rangle)=(X^\dagger|\alpha\ rangle,|\beta\rangle)

but in bra-ket notation, this equation is just

\langle\alpha|(X|\beta\rangle)=(\langle\alpha|X)|\ beta\rangle