Proof

We want to find the smallest interval ${\displaystyle \delta t_{\perp }}$ such that

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

We note^[2] that

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

using Euler's formula and noting that the sine function is odd. Then

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

since ${\displaystyle \cos(x)\geq 1-{\frac {4}{\pi ^{2}}}x\sin(x)-{\frac {2}{\pi ^{2}}}x^{2}}$, ${\displaystyle \forall x\in \mathbb {R} }$. We note that

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

Thus

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

Since ${\displaystyle |S(\delta t)|^{2}\geq 0}$ then ${\displaystyle {\frac {d|S(\delta t)|^{2}}{d\delta t}}=0}$ if ${\displaystyle S(\delta t)=0}$. So the second term vanishes for ${\displaystyle \delta t=\delta t_{\perp }}$ and

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

For this bound to become an equality we demand ${\displaystyle \cos(x)=1-{\frac {4}{\pi ^{2}}}x\sin(x)-{\frac {2}{\pi ^{2}}}x^{2}}$, that is ${\displaystyle x=0}$ or ${\displaystyle x=\pm \pi }$. Thus

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

which holds for only two energy eigenstates ${\displaystyle E_{0}=0}$ and ${\displaystyle E_{1}=\pm {\frac {\pi \hbar }{\delta t_{\perp }}}}$. Thus, the only state that attains this bound is a two-level pure quantum state (qubit) in an equal superposition

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

of energy eigenstates ${\displaystyle \left|E_{0}\right\rangle }$ and ${\displaystyle \left|E_{1}\right\rangle }$, unique up to degeneracy of the energy level ${\displaystyle E_{1}}$ and arbitrary phase factors ${\displaystyle \varphi _{0}}$, ${\displaystyle \varphi _{1}}$ of the eigenstates.^[2]

Proof

Graphs of trigonometric functions used in inequalities of Mandelstam-Tamm and Margolus–Levitin theorems.

We want to find the smallest interval ${\displaystyle \delta t_{\perp }}$ such that

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

We note that^[2]

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

as ${\displaystyle \cos(x)\geq 1-{\frac {2}{\pi }}x-{\frac {2}{\pi }}\sin(x),\forall x\geq 0}$. Since ${\displaystyle S(\delta t_{\perp })=0}$ requires ${\displaystyle {\text{Re}}(S(\delta t_{\perp }))={\text{Im}}(S(\delta t_{\perp }))=0}$ then

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

For this bound to become an equality we demand ${\displaystyle \cos(x)=1-{\frac {2}{\pi }}(x+\sin(x))}$, that is ${\displaystyle x=0}$ or ${\displaystyle x=\pi }$. Thus

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

which holds for only two energy eigenstates ${\displaystyle E_{0}=0}$ and ${\displaystyle E_{1}={\frac {\pi \hbar }{\delta t_{\perp }}}}$. Thus, the only state that attains this bound is a two-level pure quantum state (qubit) in an equal superposition

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

of energy eigenstates ${\displaystyle \left|E_{0}\right\rangle }$ and ${\displaystyle \left|E_{1}\right\rangle }$, unique up to degeneracy of the energy level ${\displaystyle E_{1}}$ and arbitrary phase factors ${\displaystyle \varphi _{0}}$, ${\displaystyle \varphi _{1}}$ of the eigenstates.^[2]

Time-varying Hamiltonian

The Margolus-Levitin theorem generalizes to the case with time-varying Hamiltonian and mixed states.^[5]

Let ${\displaystyle H_{\delta t}}$ be the Hamiltonian at time interval ${\displaystyle \delta t}$, such that ${\displaystyle H_{\delta t}}$ still has zero energy in the ground state. Let the system start at some mixed state with density operator ${\displaystyle \rho _{0}}$ and evolve by the Schrödinger equation over time. Then

${\displaystyle \int _{0}^{\delta t}|tr(\rho _{0}H_{\delta t})|dt\geq \hbar D_{B}(\rho _{0},\rho _{\delta t})}$,

where ${\displaystyle D_{B}}$ is the Bures distance between the starting state and the ending state.

To obtain the original theorem, set ${\displaystyle H_{\delta t}}$ to be independent of time, and ${\displaystyle \rho _{0}=\left|\psi (0)\right\rangle \left\langle \psi (0)\right|}$, then since pure states evolve to pure states, ${\displaystyle \rho _{\delta t}=\left|\psi (\delta t)\right\rangle \left\langle \psi (\delta t)\right|}$, and so by the formula for the Bures distance between pure states,

${\displaystyle E_{avg}\delta t\geq \hbar \arccos |\left\langle \psi (0)|\psi (\delta t)\right\rangle |}$,

and when the starting and ending states are orthogonal, we obtain ${\displaystyle E_{avg}\delta t_{\perp }\geq \hbar {\frac {\pi }{2}}}$. However, the Margolus–Levitin theorem has not yet been established in time-dependent quantum systems, whose Hamiltonians ${\displaystyle H_{\delta t}}$ are driven by arbitrary time-dependent parameters, except for the adiabatic case.^[6]

Theorem

In the case ${\displaystyle E_{avg}\neq \delta E}$ the orthogonalization interval satisfies

${\displaystyle \delta t_{\perp }\leq {\frac {\pi \hbar \left(1+e^{\ln {\left|{\frac {\delta E}{E_{avg}}}\right|}}\right)}{2E_{avg}\left(1+{\frac {\delta E}{E_{avg}}}\right)}}\left(1+\epsilon \right)={\frac {\pi \hbar }{2E_{avg}}}\left(1+\epsilon \right),\quad \forall \epsilon >0}$

Theorem

For any state ${\displaystyle \left|\psi \right\rangle }$

${\displaystyle {\frac {E_{max}}{4}}\leq E_{avg}\leq {\frac {E_{max}}{2}}}$,

where ${\displaystyle E_{max}}$ is the maximum energy eigenvalue of ${\displaystyle \left|\psi \right\rangle }$ and

${\displaystyle \pi \hbar \leq E_{max}\delta t_{\perp }\leq 2\pi \hbar }$,

wherein ${\displaystyle E_{max}\delta t_{\perp }=\pi \hbar }$ for the qubit state ${\displaystyle \left|\psi _{q}\right\rangle }$ with ${\displaystyle E_{1}=E_{max}}$.