Quantum speed limit theorems are quantum mechanics theorems concerning the orthogonalization interval, the minimum time for a quantum system to evolve between two orthogonal states, also known as the quantum speed limit.

Consider an initial pure quantum state expressed as a superposition of its energy eigenstates

${\displaystyle \left|\psi (0)\right\rangle =\sum _{n}c_{n}\left|E_{n}\right\rangle }$.

If the state ${\displaystyle \left|\psi (0)\right\rangle }$ is let to evolve for an interval ${\displaystyle \delta t}$ by the Schrödinger equation it becomes

${\displaystyle \left|\psi (\delta t)\right\rangle =\sum _{n}c_{n}e^{-i{\frac {E_{n}\delta t}{\hbar ))}\left|E_{n}\right\rangle }$,

where ${\displaystyle \hbar ={\frac {h}{2\pi ))}$ is the reduced Planck constant. If the initial state ${\displaystyle \left|\psi (0)\right\rangle }$ is orthogonal to the evolved state ${\displaystyle \left|\psi (\delta t)\right\rangle }$ then ${\displaystyle \left\langle \psi (0)|\psi (\delta t)\right\rangle =0}$ and the minimum interval ${\displaystyle \delta t_{\perp ))$ required to achieve this condition is called the orthogonalization interval[1] or time.[2]

## Mandelstam-Tamm theorem

The Mandelstam-Tamm theorem[1] states that

${\displaystyle \delta E\delta t_{\perp }\geq \hbar {\frac {\pi }{2))}$,

where

${\displaystyle (\delta E)^{2}=\left\langle \psi |H^{2}|\psi \right\rangle -(\left\langle \psi |H|\psi \right\rangle )^{2}={\frac {1}{2))\sum _{n,m}|c_{n}|^{2}|c_{m}|^{2}(E_{n}-E_{m})^{2))$,

is the variance of the system's energy and ${\displaystyle H}$ is the Hamiltonian operator. The theorem is named after Leonid Mandelstam and Igor Tamm. In this case, quantum evolution is independent of the particular Hamiltonian used to transport the quantum system along a given curve in the projective Hilbert space; it is the distance along this curve measured by the Fubini-Study metric.[3]

### 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]

## Margolus–Levitin theorem

The Margolus–Levitin theorem[4] states that

${\displaystyle E_{avg}\delta t_{\perp }\geq \hbar {\frac {\pi }{2))}$,

where

${\displaystyle E_{avg}=\left\langle \psi |H|\psi \right\rangle =\sum _{n}|c_{n}|^{2}E_{n))$,

is the system's average energy and ${\displaystyle H}$ is the Hamiltonian operator, such that

• ${\displaystyle H}$ does not depend on time;
• ${\displaystyle H}$ has zero ground state energy.

The theorem is named after Norman Margolus and Lev B. Levitin.

### Proof

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]

## Other relevant theorems

Relevant theorems concerning the Margolus–Levitin and the Mandelstam-Tamm theorems were proved[2] in 2009 by Lev B. Levitin and Tommaso Toffoli.

### 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 ${\displaystyle E_{1}=E_{max))$.

#### Proof

Let

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

Assume a contrario that ${\displaystyle E_{max}>{\frac {2\pi \hbar }{\delta t_{\perp ))))$. We can define ${\displaystyle E_{l}\doteq E_{max}-{\frac {2\pi \hbar }{\delta t_{\perp ))}>0}$. But then

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

Thus, replacing ${\displaystyle E_{max))$ with ${\displaystyle E_{l}>E_{max))$ does not change ${\displaystyle S(\delta t_{\perp })}$ and therefore the set of energy eigenvalues is bounded from above.[2] To prove the existence of the lower bound on ${\displaystyle E_{max))$, let the average energy be ${\displaystyle E_{avg}^{(1)))$. We note that replacing energy levels ${\displaystyle E_{n))$ in ${\displaystyle S(\delta t_{\perp })}$ with ${\displaystyle E_{max}-E_{n))$ will not affect its validity. But after such a replacement, the average energy is ${\displaystyle E_{avg}^{(2)}=E_{max}-E_{avg}^{(1)))$, and we can choose ${\displaystyle E_{avg}=\min \left(E_{avg}^{(1)},E_{avg}^{(2)}\right)}$. Thus ${\displaystyle E_{avg}\leq {\frac {E_{max)){2))}$. Using the bound on ${\displaystyle E_{avg))$ from the Margolus–Levitin theorem completes the proof.[2]

Furthermore, if ${\displaystyle \delta t_{\perp }={\frac {\pi \hbar }{E_{max))))$ then

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

which is satisfied[2] iff ${\displaystyle E_{0}=0}$, ${\displaystyle E_{1}=E_{max}={\frac {\pi \hbar }{\delta t_{\perp ))))$, and ${\displaystyle |c_{n}|^{2}={\frac {1}{2))}$.