Twierdzenia o prędkości ewolucji stanu kwantowego

${\displaystyle \delta t_{\perp }}$

${\displaystyle |S(\delta t_{\perp })|^{2}=|\left\langle \psi (0)|\psi (\delta t_{\perp })\right\rangle |^{2}=0}$.

Korzystając z wzoru Eulera i faktu, że sinus jest funkcją nieparzystą mamy

${\displaystyle {\begin{aligned}|S(\delta t)|^{2}&=|\left\langle \psi (0)|\psi (\delta t)\right\rangle |^{2}=\sum _{n,m}|c_{n}|^{2}|c_{m}|^{2}e^{-i{\frac {\delta t}{\hbar }}\left(E_{n}-E_{m}\right)}=\\&=\sum _{n,m}|c_{n}|^{2}|c_{m}|^{2}\cos \left({\frac {\delta t}{\hbar }}\left(E_{n}-E_{m}\right)\right)\end{aligned}}}$.

Zauważamy^[2], że

${\displaystyle {\begin{aligned}|S(\delta t)|^{2}&\geqslant 1-{\frac {4}{\pi ^{2}}}\sum _{n,m}|c_{n}|^{2}|c_{m}|^{2}{\frac {\delta t}{\hbar }}\left(E_{n}-E_{m}\right)\sin \left({\frac {\delta t}{\hbar }}\left(E_{n}-E_{m}\right)\right)\\&-{\frac {2}{\pi ^{2}}}\sum _{n,m}|c_{n}|^{2}|c_{m}|^{2}\left({\frac {\delta t}{\hbar }}\left(E_{n}-E_{m}\right)\right)^{2}\end{aligned}}}$,

ponieważ ${\displaystyle \cos(x)\geqslant 1-{\frac {4}{\pi ^{2}}}x\sin(x)-{\frac {2}{\pi ^{2}}}x^{2},}$ ${\displaystyle \forall x\in \mathbb {R} }$. Zauważamy, że

${\displaystyle {\frac {d|S(\delta t)|^{2}}{d\delta t}}=-\sum _{n,m}|c_{n}|^{2}|c_{m}|^{2}\sin \left({\frac {\delta t}{\hbar }}\left(E_{n}-E_{m}\right)\right){\frac {E_{n}-E_{m}}{\hbar }}}$.

Tym samym

${\displaystyle |S(\delta t)|^{2}\geqslant 1+{\frac {4}{\pi ^{2}}}\delta t{\frac {d|S(\delta t)|^{2}}{d\delta t}}-{\frac {1}{\pi ^{2}}}\left({\frac {2\delta t}{\hbar }}\delta E\right)^{2}}$.

Ponieważ ${\displaystyle |S(\delta t)|^{2}\geqslant 0}$ więc ${\displaystyle {\frac {d|S(\delta t)|^{2}}{d\delta t}}=0}$ jeżeli ${\displaystyle S(\delta t)=0.}$ zatem drugi wyraz zanika dla ${\displaystyle \delta t=\delta t_{\perp }}$ oraz

${\displaystyle 0\geqslant 1-{\frac {1}{\pi ^{2}}}{\frac {4\delta t_{\perp }^{2}}{\hbar ^{2}}}\left(\delta E\right)^{2}}$.

${\displaystyle \cos(x)=1-{\frac {4}{\pi ^{2}}}x\sin(x)-{\frac {2}{\pi ^{2}}}x^{2}}$

${\displaystyle x=0}$

${\displaystyle x=\pm \pi }$

${\displaystyle {\frac {\delta t_{\perp }}{\hbar }}\left(E_{n}-E_{m}\right)=0\quad {\text{lub}}\quad {\frac {\delta t_{\perp }}{\hbar }}\left(E_{n}-E_{m}\right)=\pm \pi \quad \forall n,m,c_{n}\neq 0,c_{m}\neq 0}$,

co zachodzi jedynie dla dwóch energetycznych stanów własnych ${\displaystyle E_{0}=0}$ oraz ${\displaystyle E_{1}=\pm {\frac {\pi \hbar }{\delta t_{\perp }}}}$, czyli dla stanu

${\displaystyle \left|\psi _{q}\right\rangle ={\frac {1}{\sqrt {2}}}\left(e^{i\varphi _{0}}\left|0\right\rangle +e^{i\varphi _{1}}\left|\pm {\frac {\pi \hbar }{\delta t_{\perp }}}\right\rangle \right)}$

unikalnego z wyłączeniem degeneracji energii ${\displaystyle E_{1}}$ oraz dowolnych współczynników fazowych ${\displaystyle \varphi _{0}}$ i ${\displaystyle \varphi _{1}}$ stanów własnych^[2].

${\displaystyle \delta t_{\perp }}$

${\displaystyle S(\delta t_{\perp })=\left\langle \psi (0)|\psi (\delta t_{\perp })\right\rangle =\sum _{n}|c_{n}|^{2}e^{-i{\frac {E_{n}\delta t_{\perp }}{\hbar }}}=0}$.

Zauważamy^[2], że

${\displaystyle {\begin{aligned}{\text{Re}}(S(\delta t))&=\sum _{n}|c_{n}|^{2}\cos \left({\frac {E_{n}\delta t}{\hbar }}\right)\geqslant \\&\geqslant \sum _{n}|c_{n}|^{2}\left(1-{\frac {2}{\pi }}{\frac {E_{n}\delta t}{\hbar }}-{\frac {2}{\pi }}\sin \left({\frac {E_{n}\delta t}{\hbar }}\right)\right)=\\&=\sum _{n}|c_{n}|^{2}-{\frac {2\delta t}{\pi \hbar }}\sum _{n}|c_{n}|^{2}E_{n}-{\frac {2}{\pi }}\sum _{n}|c_{n}|^{2}\sin \left({\frac {E_{n}\delta t}{\hbar }}\right)=\\&=1-{\frac {2\delta t}{\pi \hbar }}E_{avg}+{\frac {2}{\pi }}{\text{Im}}(S(\delta t))\end{aligned}}}$,

gdyż ${\displaystyle \cos(x)\geqslant 1-{\frac {2}{\pi }}x-{\frac {2}{\pi }}\sin(x),\forall x\geqslant 0}$. Ponieważ ${\displaystyle S(\delta t_{\perp })=0}$ wymaga ${\displaystyle {\text{Re}}(S(\delta t_{\perp }))={\text{Im}}(S(\delta t_{\perp }))=0}$ zatem

${\displaystyle 0\geqslant 1-{\frac {2}{\pi }}{\frac {E_{avg}\delta t_{\perp }}{\hbar }}}$.

${\displaystyle \cos(x)=1-{\frac {2}{\pi }}(x+\sin(x))}$

${\displaystyle x=0}$

${\displaystyle x=\pi }$

${\displaystyle {\frac {E_{n}\delta t_{\perp }}{\hbar }}=0\quad {\text{lub}}\quad {\frac {E_{n}\delta t_{\perp }}{\hbar }}=\pi \quad \forall n,c_{n}\neq 0}$.

Zatem ${\displaystyle E_{0}=0}$ oraz ${\displaystyle E_{1}={\frac {\pi \hbar }{\delta t_{\perp }}}}$, a jedynym stanem, który to zapewnia jest kubit

${\displaystyle \left|\psi _{q}\right\rangle ={\frac {1}{\sqrt {2}}}\left(e^{i\varphi _{0}}\left|0\right\rangle +e^{i\varphi _{1}}\left|{\frac {\pi \hbar }{\delta t_{\perp }}}\right\rangle \right)}$

o zrównoważonej superpozycji energetycznych stanów własnych ${\displaystyle \left|E_{0}\right\rangle }$ i ${\displaystyle \left|E_{1}\right\rangle }$ unikalny z wyłączeniem degeneracji energii ${\displaystyle E_{1}}$ oraz dowolnych współczynników fazowych ${\displaystyle \varphi _{0}}$ i ${\displaystyle \varphi _{1}}$ stanów własnych^[2].

${\displaystyle S(\delta t_{\perp })=\left\langle \psi (0)|\psi (\delta t_{\perp })\right\rangle =\sum _{n}|c_{n}|^{2}e^{-i{\frac {\delta t_{\perp }}{\hbar }}E_{n}}=0}$.

Załóżmy a contrario, że ${\displaystyle E_{max}>{\frac {2\pi \hbar }{\delta t_{\perp }}}}$. Możemy zdefiniować ${\displaystyle E_{l}\doteq E_{max}-{\frac {2\pi \hbar }{\delta t_{\perp }}}>0}$. Ale wówczas

${\displaystyle e^{-i{\frac {\delta t_{\perp }}{\hbar }}E_{l}}=e^{-i{\frac {\delta t_{\perp }}{\hbar }}E_{max}}e^{2\pi i}=e^{-i{\frac {\delta t_{\perp }}{\hbar }}E_{max}}}$.

Zatem zamiana ${\displaystyle E_{max}}$ na ${\displaystyle E_{l}>E_{max}}$ nie zmienia ${\displaystyle S(\delta t_{\perp })}$, a tym samym zbiór wartości własnych energii jest ograniczony od góry^[2]. Aby udowodnić istnienie kresu dolnego na ${\displaystyle E_{max}}$, niech średnią energią będzie ${\displaystyle E_{avg}^{(1)}}$. Zauważamy, że zamiana poziomów energii ${\displaystyle E_{n}}$ w ${\displaystyle S(\delta t_{\perp })}$ na ${\displaystyle E_{max}-E_{n}}$ nie ma wpływu na ich ważność. Ale po takiej zamianie, średnia energia to ${\displaystyle E_{avg}^{(2)}=E_{max}-E_{avg}^{(1)}}$ i możemy wybrać ${\displaystyle E_{avg}=\min \left(E_{avg}^{(1)},E_{avg}^{(2)}\right)}$. Zatem ${\displaystyle E_{avg}\leqslant {\frac {E_{max}}{2}}}$. Wykorzystanie kresu na ${\displaystyle E_{avg}}$ z twierdzenia Margolusa–Levitina kończy dowód^[2].

${\displaystyle \delta t_{\perp }={\frac {\pi \hbar }{E_{max}}}}$

${\displaystyle S(\delta t_{\perp })=\sum _{n}|c_{n}|^{2}e^{-i\pi {\frac {E_{n}}{E_{max}}}}=\sum _{n}|c_{n}|^{2}\left(\cos \left(\pi {\frac {E_{n}}{E_{max}}}\right)-i\sin \left(\pi {\frac {E_{n}}{E_{max}}}\right)\right)=0}$,

co zachodzi^[2] wtedy i tylko wtedy gdy ${\displaystyle E_{0}=0}$, ${\displaystyle E_{1}=E_{max}={\frac {\pi \hbar }{\delta t_{\perp }}},}$ oraz ${\displaystyle |c_{0}|^{2}=|c_{1}|^{2}={\frac {1}{2}}}$.