Quantum Mechanics - Time evolution operator , bra ket states.

[binbagsss]

The question is to calculate the time evoution of S$_{x}$ wrt <$\Psi$(t)$\pm$ l where <$\Psi$$\pm$ (t) l= ( $\frac{1}{\sqrt{2}}$(exp($^{+iwt})$< $\uparrow$ l , $\pm$ exp($^{-iwt}$)<
$\downarrow$ l ) [1]

Sx=$\frac{}{2}$($^{0}_{1}$$^{1}_{0}$ )

Here is my attempt:

- First of all from [1] I see that l $\Psi$$\pm$ (t) > = ( $\frac{1}{\sqrt{2}}$(exp($^{-iwt})$ l $\uparrow$ > , $\pm$ exp($^{+iwt}$) l
$\downarrow$ > )*

where * denotes transposing the matrix so it's now a column matrix

So <$\Psi$(t)$\pm$ l Sx l $\Psi$$\pm$ (t) > = $\frac{ħ}{4}$( $\frac{1}{\sqrt{2}}$(exp($^{+iwt})$< $\uparrow$ l , $\pm$ exp($^{-iwt}$)< $\downarrow$ ) S$_{x}$( $\frac{1}{\sqrt{2}}$(exp($^{-iwt})$ l $\uparrow$ > , $\pm$ exp($^{+iwt}$) l
$\downarrow$ > )* = $\frac{ħ}{4}$ < $\uparrow$l S$_{x}$l $\uparrow$> $\pm$ exp $^{+iwt}$<$\uparrow$ lS$_{x}$ l $\downarrow$> $\pm$ exp $^{-iwt}$< $\downarrow$l S$_{x}$l $\uparrow$> $\pm$$^{2}$<$\downarrow$ lS$_{x}$ l $\downarrow$>

= $\frac{ħ}{4}$ ( 1 $\pm$ exp $^{+iwt}$<$\uparrow$ lS$_{x}$ l $\downarrow$> $\pm$ exp $^{-iwt}$< $\downarrow$l S$_{x}$l $\uparrow$> $\pm$$^{2}$1)Okay, so my solution goes straight from line 2 to the answer:

= $\frac{ħ}{4}$ ( $\pm$ exp $^{+iwt}$<$\uparrow$ lS$_{x}$ l $\downarrow$> $\pm$ exp $^{-iwt}$< $\downarrow$l S$_{x}$l $\uparrow$>)


So my questions are:

- what happends to < $\uparrow$l S$_{x}$l $\uparrow$> and < $\downarrow$l S$_{x}$l $\downarrow$> terms? I multiply the bra and ket matrix explicitly , and attain 1 in both cases, so what has happened to these in the answer?
- Also, the $\pm$$^{2}$ looks messy. should have i got this? can it be simplified to $\pm$

Many thanks for any help, greatly appreciated.

 

[binbagsss]

anyone?