# What do we mean by Unitary Dynamics in Quantum Computing?

In the afterword to the Tenth Anniversary Edition of the book Quantum Computation and Quantum Information the authors say:

For many years, the conventional wisdom was that coherent superposition-preserving unitary dynamics was an essential part of the power of Quantum Computers.

Since I am just starting reading this subject, so I am unable to understand what exactly is being said here.

Exactly, what do we mean by Unitary Dynamics in general and with reference to Quantum Computing? If possible , kindly explain me in simple English, instead of rigorous maths.

• Unitary dynamics is required in QM in order to preserve the scalar products, i.e. the transition amplitudes $\langle\phi(t),\psi(t)\rangle$, and thus also normalization $\langle\psi(t),\psi(t)\rangle$ with time evolution. That means that those quantities would have the same value at time zero and any time $t$ afterwards. This is important to correctly interpret quantum states as suitable "probability distributions" on the space of observables. – yuggib Jun 14 '15 at 10:16

Unitary dynamics means that the evolution of quantum states are describen by unitary operators. Physically this means that there is no dissipation in the system. This is important, because inherently quantum phenomena needed for quantum computation eg. entanglement and quantum superposition are only observable for a long period if no dissipation or other kind of noise is present. Otherwise noise "washes" away these quantum effects. E.g. if you have thermal noise it will make your system go into random states, and you can't have it controallably in one certain state, which is the key for quantum computing. Basically the reason you don't see quantum effetcs in the macroscopic world is because thermal and other kinds of noises wash them away and also because a large number of particles show an average behavior, and averaging also gets rid of them. This lecture might help you: online lecture for unitary evolution.

You didn't ask for math, but here is just a little, you will bump into it really soon, if you keep on studying quantum computing especially the physical principles beind it. The Schrödinger equation governing the time evolution of the state of a system (eg. a qubit, or system of qubits) tells you that $$i\hbar\frac{\partial \left|\psi(t)\right>}{\partial t} = \hat{H}\left|\psi(t)\right>$$ $$i\hbar\frac{\partial \hat{U}(t) \left|\psi_0\right>}{\partial t} = \hat{H}\hat{U}(t) \left|\psi_0\right>$$ $$i\hbar\frac{\partial \hat{U}(t)}{\partial t}\left|\psi_0\right> = \hat{H}\hat{U}(t) \left|\psi_0\right>,$$ where $\left|\psi_0\right>$ is the initial state of the system (which is arbitrary) and $\hat{U}(t)$ is a time dependent unitary operator describing its time evolution. At the end of the answer I will tell you why it is unitary. You can solve the equation for the operator, which will be $$\hat{U}(t)=e^{-\frac{i}{\hbar}\hat{H}t}.$$

The Schrödinger equation is only good for closed system however, and can't handle open systems. One of the usual equations capable for handling open quantum systems and frequently used in quantum computing (at least in circuit qed, one of its possible realizations) is the Lindblad-Kossakowski equation: $$\frac{\textrm{d}}{\textrm{dt}}\hat{\rho}^R(t) = -\frac{i}{\hbar}\left[\hat{H}(t), \hat{\rho}^R(t)\right] + \sum_n \frac{1}{2} \left[2 \hat{C}_n \hat{\rho}^R(t) \hat{C}_n^+ - \left\{\hat{C}_n^+ \hat{C}_n, \hat{\rho}^R(t)\right\}\right],$$ where $\hat{\rho}^R$ is the density matrix of the system, and $\hat{C}_n$s are the operators through which the system interacts with its environment. If the $\hat{C}_n$s are zero, that means a case of no interaction with the environment, that is a closed system, and the Lindblad equation reduces to the von Neumann equation $$\frac{\textrm{d}}{\textrm{dt}}\hat{\rho}^R(t) = -\frac{i}{\hbar}\left[\hat{H}(t), \hat{\rho}^R(t)\right],$$ which is equivalent with the Schrödinger equation, see this wiki article. And since the Schrödinger equation gives you a solution with unitary evolution, you get that if no noise present, you have unitary evolution.

Unfortunately unitary evolution is a very mathematical thing, here is just one more hint to help you understand why a system without dissipation evolves by a unitary operator, and why a system under dissipative effects doesn't.

Unitary operators have the property that their inverse is their adjoint. It is easy to write the adjoint for the Schrödinger equation for the time evolution operator, and it can be shown that this adjoint operator is really the inverse of the former (see 3rd slide of this lecture). From this you get a very important property: if an operator describing the time evolution of a non-dissipative system is a solution, also its inverse is a solution. This means that if a certain process can physically occur, an other process can also physically occur, and this process is like former process with time going backwards. To illustrate this, here is an example from simple classical mechanics: if a brick moves on a frictionless table, it is natural that that an other process can occur, namely the same brick going backwards on its trajectory. However if there is friction, and the brick moves, and the brick and the table heat up due to dissipation, the inverse process is not physically possible: a heated brick won't go back on the trajectory gaining kinetic energy while the brick and the table cool down.

• Thank you for this detailed explanation. I understood most of this, but i think i should study more before i cross-question. – gpuguy Jun 15 '15 at 15:27
• by the way when you say 'dissipation' you are referring to dissipation of what? – gpuguy Jun 15 '15 at 15:30
• @gpuguy Dissipation of energy. But actually not only dissipation matters, but all forms of energy change with the environment. As I have mentioned, the environment can not only dissipate energy but is also able to take the system into a statistical mixture of excited states (i.e. is you won't know in what state your system is). Btw if you think I have answered your question, an "accept answer" would be appreciated :) Otherwise pls ask. – user3237992 Jun 16 '15 at 8:19