Question about commutation and operator

[KFC]

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

[tex]\langle \alpha', jm|J_z A |\alpha, jm\rangle[/tex]

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

[tex]m\hbar\langle \alpha', jm|A|\alpha, jm\rangle[/tex]

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

 

[olgranpappy]

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

[tex]\langle \alpha', jm|J_z A |\alpha, jm\rangle[/tex]

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

[tex]m\hbar\langle \alpha', jm|A|\alpha, jm\rangle[/tex]

As you see, this just like have [tex]J_z[/tex] acting on the bar and returns the [tex]m\hbar\langle \alpha', jm|[/tex]. My question is: how can [tex]J_z[/tex] 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
[tex]
\langle \alpha |\beta\rangle=(\langle \beta | \alpha\rangle)^*
[/tex]

You can use the above relation to rewrite the quantity of your interest with [itex]J_z^\dagger=J_z[/itex] 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
[tex]
\langle \alpha |\beta\rangle=(\langle \beta | \alpha\rangle)^*
[/tex]

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

Thank you so much. I forget the relation [tex]J_z^\dagger=J_z[/tex], 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 [itex]|\alpha\rangle\in H[/itex], there's a bra [itex]\langle\alpha|\in H^*[/itex] that takes an arbitrary ket [itex]|\beta\rangle[/itex] to the scalar product [itex](|\alpha\rangle,|\beta\rangle[/itex]).

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.

[tex](|\alpha\rangle,|\beta\rangle)=\langle\alpha|(|\beta\rangle)=\langle\alpha||\beta\rangle=\langle\alpha|\beta\rangle[/itex]

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

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

but in bra-ket notation, this equation is just

[tex]\langle\alpha|(X|\beta\rangle)=(\langle\alpha|X)|\beta\rangle[/tex]