Quantum speed limit

Let ${\displaystyle ds^{2}}$ be the Bures metric, defined by

$${\displaystyle ds(\rho ,\rho +d\rho )^{2}={\frac {1}{2}}\sum _{jk}{\frac {|<j|d\rho |k>|^{2}}{p_{j}+p_{k}}}}$$

If a quantum system is evolving under a time-dependent Hamiltonian ${\displaystyle H}$, then its velocity according to Bures metric is upper bounded by

$${\displaystyle ds\leq {\frac {\sigma _{H}(t)}{\hbar }}dt}$$

where ${\displaystyle \sigma _{H}(t)}$ is the uncertainty in energy at time ${\displaystyle t}$.

Two corollaries:

• The time taken to evolve from ${\displaystyle \rho }$ to ${\displaystyle \rho '}$ is ${\displaystyle \tau \geq {\frac {\hbar }{{\bar {\sigma }}_{H}}}dist(\rho ,\rho ')}$, where is ${\displaystyle {\bar {\sigma }}_{H}:={\frac {1}{\tau }}\int _{0}^{\tau }\sigma _{H}(t)dt}$ the time-averaged uncertainty in energy.
• The time taken to evolve from one pure state to another pure state orthogonal to it is ${\displaystyle \tau \geq {\frac {\pi }{2}}{\frac {\hbar }{{\bar {\sigma }}_{H}}}}$.

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Computation machinery is constructed out of physical matter that follows quantum mechanics, and each operation, if it is to be unambiguous, must be a transition of the system from one state to an orthogonal state. Suppose the computing machinery is a physical system evolving under Hamiltonian that does not change with time. Then according to the Margolus–Levitin theorem, its operations per time per energy obey

$${\displaystyle {\frac {1/\delta t_{\perp }}{E}}\leq {\frac {1}{\hbar {\frac {\pi }{2}}}}=6\times 10^{33}\mathrm {s} ^{-1}\cdot \mathrm {J} ^{-1}.}$$

That is, the processing rate of all forms of computation cannot be higher than about 6 × 10³³ operations per second per joule of energy. This includes "classical" computers since even classical computers are still made of matter that follows quantum mechanics.