- #1
binbagsss
- 1,291
- 12
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} < \uparrowl S_{x}l \uparrow> \pm exp ^{+iwt}<\uparrow lS_{x} l \downarrow> \pm exp ^{-iwt}< \downarrowl 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}< \downarrowl 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}< \downarrowl S_{x}l \uparrow>)
So my questions are:
- what happends to < \uparrowl S_{x}l \uparrow> and < \downarrowl 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.
\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} < \uparrowl S_{x}l \uparrow> \pm exp ^{+iwt}<\uparrow lS_{x} l \downarrow> \pm exp ^{-iwt}< \downarrowl 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}< \downarrowl 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}< \downarrowl S_{x}l \uparrow>)
So my questions are:
- what happends to < \uparrowl S_{x}l \uparrow> and < \downarrowl 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.