Solving Projector Question on Mixed States

[raisin_raisin]

A few months ago I wrote this line down, but it does not seem to follow any more. Am I mistaking a mistake now or when I first wrote it down? thanks
$$( | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 |) \rho ( | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 |)$$


$$= | 0 \rangle \langle 0 | \rho | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 | \rho | 1 \rangle \langle 1 |$$

where $$\rho$$ is an arbitrary mixed state.

(It is not letting me preview my latex so fingers crossed this works as expected)

 

[Edgardo]

You have to use the distributive law
1. a*(b+c) = ab + ac
2. (a+b)*(c+d) = (ac + ad + bc + bd)$$( | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 |) \rho ( | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 |)$$

$$=( | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 |) (\rho | 0 \rangle \langle 0 | +\rho | 1 \rangle \langle 1 |)$$

$$= \langle 0 | \rho | 0 \rangle \cdot | 0 \rangle \langle 0 | + \langle 0 | \rho | 1 \rangle \cdot | 0 \rangle \langle 1 | + \langle 1 | \rho | 0 \rangle \cdot | 1 \rangle \langle 0 | + \langle 1 | \rho | 1 \rangle \cdot | 1 \rangle \langle 1 |$$A fast online latex equation editor is here:
http://www.codecogs.com/latex/eqneditor.php

 

[Fredrik]

To workaround the preview problem, hit your browser's refresh button when the incorrect image comes up, and then confirm that you want to send the information one more time. You can also edit your post up to 24 hours after you posted it.

 

[Edgardo]

Thanks for this trick, didn't know about it.