- #1
binbagsss
- 1,259
- 11
The question is to calculate the time evoution of S[itex]_{x}[/itex] wrt <[itex]\Psi[/itex](t)[itex]\pm[/itex] l where <[itex]\Psi[/itex][itex]\pm[/itex] (t) l= ( [itex]\frac{1}{\sqrt{2}}[/itex](exp([itex]^{+iwt})[/itex]< [itex]\uparrow[/itex] l , [itex]\pm[/itex] exp([itex]^{-iwt}[/itex])<
[itex]\downarrow[/itex] l ) [1]
Sx=[itex]\frac{}{2}[/itex]([itex]^{0}_{1}[/itex][itex]^{1}_{0}[/itex] )
Here is my attempt:
- First of all from [1] I see that l [itex]\Psi[/itex][itex]\pm[/itex] (t) > = ( [itex]\frac{1}{\sqrt{2}}[/itex](exp([itex]^{-iwt})[/itex] l [itex]\uparrow[/itex] > , [itex]\pm[/itex] exp([itex]^{+iwt}[/itex]) l
[itex]\downarrow[/itex] > )*
where * denotes transposing the matrix so it's now a column matrix
So <[itex]\Psi[/itex](t)[itex]\pm[/itex] l Sx l [itex]\Psi[/itex][itex]\pm[/itex] (t) > = [itex]\frac{ħ}{4}[/itex]( [itex]\frac{1}{\sqrt{2}}[/itex](exp([itex]^{+iwt})[/itex]< [itex]\uparrow[/itex] l , [itex]\pm[/itex] exp([itex]^{-iwt}[/itex])< [itex]\downarrow[/itex] ) S[itex]_{x}[/itex]( [itex]\frac{1}{\sqrt{2}}[/itex](exp([itex]^{-iwt})[/itex] l [itex]\uparrow[/itex] > , [itex]\pm[/itex] exp([itex]^{+iwt}[/itex]) l
[itex]\downarrow[/itex] > )* = [itex]\frac{ħ}{4}[/itex] < [itex]\uparrow[/itex]l S[itex]_{x}[/itex]l [itex]\uparrow[/itex]> [itex]\pm[/itex] exp [itex]^{+iwt}[/itex]<[itex]\uparrow[/itex] lS[itex]_{x}[/itex] l [itex]\downarrow[/itex]> [itex]\pm[/itex] exp [itex]^{-iwt}[/itex]< [itex]\downarrow[/itex]l S[itex]_{x}[/itex]l [itex]\uparrow[/itex]> [itex]\pm[/itex][itex]^{2}[/itex]<[itex]\downarrow[/itex] lS[itex]_{x}[/itex] l [itex]\downarrow[/itex]>
= [itex]\frac{ħ}{4}[/itex] ( 1 [itex]\pm[/itex] exp [itex]^{+iwt}[/itex]<[itex]\uparrow[/itex] lS[itex]_{x}[/itex] l [itex]\downarrow[/itex]> [itex]\pm[/itex] exp [itex]^{-iwt}[/itex]< [itex]\downarrow[/itex]l S[itex]_{x}[/itex]l [itex]\uparrow[/itex]> [itex]\pm[/itex][itex]^{2}[/itex]1)Okay, so my solution goes straight from line 2 to the answer:
= [itex]\frac{ħ}{4}[/itex] ( [itex]\pm[/itex] exp [itex]^{+iwt}[/itex]<[itex]\uparrow[/itex] lS[itex]_{x}[/itex] l [itex]\downarrow[/itex]> [itex]\pm[/itex] exp [itex]^{-iwt}[/itex]< [itex]\downarrow[/itex]l S[itex]_{x}[/itex]l [itex]\uparrow[/itex]>)
So my questions are:
- what happends to < [itex]\uparrow[/itex]l S[itex]_{x}[/itex]l [itex]\uparrow[/itex]> and < [itex]\downarrow[/itex]l S[itex]_{x}[/itex]l [itex]\downarrow[/itex]> 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 [itex]\pm[/itex][itex]^{2}[/itex] looks messy. should have i got this? can it be simplified to [itex]\pm[/itex]
Many thanks for any help, greatly appreciated.
[itex]\downarrow[/itex] l ) [1]
Sx=[itex]\frac{}{2}[/itex]([itex]^{0}_{1}[/itex][itex]^{1}_{0}[/itex] )
Here is my attempt:
- First of all from [1] I see that l [itex]\Psi[/itex][itex]\pm[/itex] (t) > = ( [itex]\frac{1}{\sqrt{2}}[/itex](exp([itex]^{-iwt})[/itex] l [itex]\uparrow[/itex] > , [itex]\pm[/itex] exp([itex]^{+iwt}[/itex]) l
[itex]\downarrow[/itex] > )*
where * denotes transposing the matrix so it's now a column matrix
So <[itex]\Psi[/itex](t)[itex]\pm[/itex] l Sx l [itex]\Psi[/itex][itex]\pm[/itex] (t) > = [itex]\frac{ħ}{4}[/itex]( [itex]\frac{1}{\sqrt{2}}[/itex](exp([itex]^{+iwt})[/itex]< [itex]\uparrow[/itex] l , [itex]\pm[/itex] exp([itex]^{-iwt}[/itex])< [itex]\downarrow[/itex] ) S[itex]_{x}[/itex]( [itex]\frac{1}{\sqrt{2}}[/itex](exp([itex]^{-iwt})[/itex] l [itex]\uparrow[/itex] > , [itex]\pm[/itex] exp([itex]^{+iwt}[/itex]) l
[itex]\downarrow[/itex] > )* = [itex]\frac{ħ}{4}[/itex] < [itex]\uparrow[/itex]l S[itex]_{x}[/itex]l [itex]\uparrow[/itex]> [itex]\pm[/itex] exp [itex]^{+iwt}[/itex]<[itex]\uparrow[/itex] lS[itex]_{x}[/itex] l [itex]\downarrow[/itex]> [itex]\pm[/itex] exp [itex]^{-iwt}[/itex]< [itex]\downarrow[/itex]l S[itex]_{x}[/itex]l [itex]\uparrow[/itex]> [itex]\pm[/itex][itex]^{2}[/itex]<[itex]\downarrow[/itex] lS[itex]_{x}[/itex] l [itex]\downarrow[/itex]>
= [itex]\frac{ħ}{4}[/itex] ( 1 [itex]\pm[/itex] exp [itex]^{+iwt}[/itex]<[itex]\uparrow[/itex] lS[itex]_{x}[/itex] l [itex]\downarrow[/itex]> [itex]\pm[/itex] exp [itex]^{-iwt}[/itex]< [itex]\downarrow[/itex]l S[itex]_{x}[/itex]l [itex]\uparrow[/itex]> [itex]\pm[/itex][itex]^{2}[/itex]1)Okay, so my solution goes straight from line 2 to the answer:
= [itex]\frac{ħ}{4}[/itex] ( [itex]\pm[/itex] exp [itex]^{+iwt}[/itex]<[itex]\uparrow[/itex] lS[itex]_{x}[/itex] l [itex]\downarrow[/itex]> [itex]\pm[/itex] exp [itex]^{-iwt}[/itex]< [itex]\downarrow[/itex]l S[itex]_{x}[/itex]l [itex]\uparrow[/itex]>)
So my questions are:
- what happends to < [itex]\uparrow[/itex]l S[itex]_{x}[/itex]l [itex]\uparrow[/itex]> and < [itex]\downarrow[/itex]l S[itex]_{x}[/itex]l [itex]\downarrow[/itex]> 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 [itex]\pm[/itex][itex]^{2}[/itex] looks messy. should have i got this? can it be simplified to [itex]\pm[/itex]
Many thanks for any help, greatly appreciated.