Wave Function Question: Understanding |0,f\rangle & \hat{c}_{\vec{k},\sigma}^+

[LagrangeEuler]

http://books.google.rs/books?id=vrc...=0CB0Q6AEwAA#v=onepage&q=nolting RKKY&f=false

Here in the page 203 is defined $|\vec{k}_i^{(i)},m_{s_i}^{(i)}\rangle$ and also $|0,f\rangle=|0\rangle|f\rangle$
what that notation means?
What is $|0,f\rangle=|0\rangle|f\rangle$?
If operator $\hat{c}_{\vec{k},\sigma}^+$ creates electron with wave vector $\vec{k}$ and spin $\sigma$.
What is
\hat{c}_{\vec{k},\sigma}^+|0,f\rangle?

 

[ZapperZ]

This is not clear. Are you asking what a bra-ket notation mean, or are you specifically asking what a state |a>|b> mean?

Zz.

 

[LagrangeEuler]

Only in this case. How operator $\hat{c}$ attacks $|0,f\rangle$?

 

[LagrangeEuler]

For example
\langle 0;f|\hat{C}_{i\sigma}^+|0;f\rangle=?

 

[Matterwave]

You have to define the state the the operator...these are defined things...

 

[andrien]

I think it is just for two separate things so it can be written as product.Any operator associated with electron will act on second one not on first one

 

[LagrangeEuler]

Well I can give more information. But I gave in first post book and the page. But ok...
\hat{s}_i^z=\frac{\hbar}{2}(\hat{c}_{i\uparrow}^+ \hat{c}_{i\uparrow}-\hat{c}_{i\downarrow}^+\hat{c}_{i\downarrow})
\hat{s}_i^+=\hbar \hat{c}_{i\uparrow}^+ \hat{c}_{i\downarrow}
\hat{s}_i^-=\hbar \hat{c}_{i\downarrow}^+ \hat{c}_{i\uparrow}

In text.
''Without perturbation electron exist in their unpolarised ground state. In addition since they don't interact with each other, the unperturbed electron ground state can be written as the antisymetrised product of single electron states
|\vec{k}_i^{(i)},m_{s_i}^{(i)}\rangle=|\vec{k}_i^{(i)}\rangle |m_{s_i}^{(i)}\rangle
where the spin magnetic quantum number $m_{s_i}^{(i)}$ takes the values $\pm \frac{1}{2}$. $|\vec{k}_i^{(i)}\rangle$ is wave vector where superscript refers to the particle number. Furter since we want to treat conduction electrons as s electrons, which excludes spin orbit interraction, we can separate the spin and the space parts. Let
|0;f\rangle=|0\rangle|f\rangle
''
Can you now give me explanation. What is $|0\rangle$? What is $|f\rangle$? And how defined operators act on this state? Tnx.