"Quantum mechanics is reversible" this statement is everywhere, some even said it's just an observed fact about the universe. I can't find a layman explanation or example why is it reversible?

  • $\begingroup$ What does you mean by "reversible"? $\endgroup$
    – Name YYY
    Sep 28, 2017 at 15:32
  • $\begingroup$ "Quantum gates have to be reversible because quantum mechanics is reversible (and even more specifically it is unitary)" what this really means $\endgroup$
    – Consy
    Sep 28, 2017 at 15:44

1 Answer 1


So before college even, we students learn that there is this wonderful alternate way to look at a lot of physical problems in terms of energies. But, they seem to lose some information about the problem. Like, the kinetic energy loses directional information. When we get to college, we learn that two people actually have formalisms named after them, the Lagrangian formalism of Joseph-Louis Lagrange, and the Hamiltonian formalism of William Rowan Hamilton, which use these energies to describe the world. So you can use energies to understand the world, if only you track the information some other way! So for example in Hamiltonian mechanics the system occupies a point in “phase space” where all of the particles’ position and momentum components are specified quite precisely, but the total energy function, the Hamiltonian, tells the point how to move across phase space over time. All of that information is re-added elsewhere by this invention of phase space.

It seems to be a property of these formalisms that they do not do well with irreversible processes like energy loss. They are “reversible” theories of classical mechanics, meaning that if you describe classical mechanics with them, if you reversed all of the momentums in the problem everything would appear to move backwards in time. When I say “do not do well” please do not understand me as saying “impossible,” I just mean “not a first-class citizen.” To model something in contact with a system held at constant temperature, for example, you need a phase-space model of that other system—say a bunch of harmonic oscillators with random phases and diverse frequencies, each having an average amount of energy given by that temperature. You then need to make approximations and average over the states of the oscillators, to get this time irreversibility and information loss out of a description where information by its nature can never be lost. (I don't mean that this is a physical detail of information but a mathematical detail of its encoding in the approach. Information is a positional coordinate in the phase space, how do you lose a coordinate? Does the space change dimensionality? So I’ve gotta throw away the information myself because the theory is not going to.)

We have not managed to build quantum mechanics with the old approach of forces, where information was implicit so irreversibility was easy to model. Let me give you a quick overview of the Hamiltonian construction, classical quantum mechanics: a way of looking at the world where we represent the world with matrices[1] instead of numbers. Matrix multiplication has the quirk that $AB\ne BA$ in general, so that the order of multiplication usually matters: we use this to understand things like the Heisenberg uncertainty principle and the like. The general principle is that

  • the state of the world is some complex column vector[2] that we call $|\psi\rangle$
  • it corresponds to some complex-conjugate row vector that we call $\langle \psi|$
  • we predict only averages of measurements, and we predict them by associating to that measurable quantity some matrix $M$
  • what we actually observe in our experiments, the “quanta” or mysterious whole numbers, are some discrete numbers which characterize the matrix called its eigenvalues
  • and the average that we predict for this measurement $\langle M \rangle$ in the state $|\psi\rangle$ is then given by the matrix multiplication,$$\langle M \rangle = \langle \psi|M|\psi \rangle.$$

This last operation leaves a 1x1 matrix on the right hand side which we just interpret as a number. In order to guarantee that it is a real number, we must restrict $M$ to be a special sort of matrix: it must be equal to its "adjoint," which is its matrix-transpose with all of the elements complex-conjugated. We call these matrices "Hermitian" matrices, and the key point is that they make all of the bad imaginary numbers disappear for all of our predictions about averages, which is good because most of the numbers that we deal with in this world are not complex numbers.

So one really nice thing about this formalism is that these eigenvalues come with corresponding eigenstates, which is a state where there is absolutely no uncertainty, the system has exactly such-and-so measured value. I therefore might be able to understand a given vector both by its components according to the momentum operator and the position operator, and either set of eigenstate components could be used to rebuild up the same vector.

The key question you have asked is, what happens over time? The answer is that we are chasing the Einstein-Planck equation, that the energy of a photon is given proportional to its frequency $f$ by some constant $\hbar$ as: $$ E = 2\pi~\hbar~f. $$ And we do this in the way the formalism prescribes, the total energy is given by a Hermitian matrix that we call $H$, for the "Hamiltonian" of the system. This has a bunch of eigenvalues $E_n$ with eigenstates $|n\rangle$. And we build your physical state out of these eigenstates, $$ |\psi\rangle=\sum_{n=0}^{\infty} \psi_n |n\rangle $$ and finally we send them rotating around the space of complex numbers with the given frequency, $$|n\rangle\mapsto e^{-iE_nt/\hbar} |n\rangle.$$Once we describe the matrix exponential function $\exp$ we can see that this amounts to saying that $$ |\psi(t)\rangle= \exp\left({-i~H~t\over\hbar}\right) |\psi_0\rangle $$ which can also be described via the famous Schrödinger equation, $$ i\hbar \frac{\mathrm d\vphantom{t}}{\mathrm d t}|\psi(t)\rangle=H|\psi(t)\rangle.$$ What we mean when we say that the physics is entirely reversible is that the above expression for $|\psi(t + \delta t)\rangle$ does not lose any information: it cannot produce the same output given two different input $|\psi(t)\rangle$ terms. There is always a different Hamiltonian, usually $H\mapsto -H$, which will undo any time-evolution that you originally did. The term “unitary” in particular refers to this operator $\exp\left({-i~H~t\over\hbar}\right)$ which is not just invertible, but its inverse is also its conjugate transpose.

There is one aspect that is not "unitary" like this, and it is the not-well-understood process by which we do not actually see the averages but the eigenvalues, what is sometimes called “wavefunction collapse” or just the “measurement problem,” and e.g. the many-worlds interpretation for quantum mechanics is very rigorous about trying to make everything unitary including this measurement process, by inventing these other “worlds” to hold the lost information.

There is also a slight elaboration of the above formalism which does better at representing information loss, and this is called the state matrix formalism or “density matrices” or so, instead of $\langle \psi|M|\psi\rangle$ you have $\operatorname{Tr}(\rho~M)$ and instead of the Schrödinger equation you have the Lindblad equation, $$ i\hbar \frac{\partial \rho}{\partial t} = H\rho-\rho H +\sum_{k=1}^N \left( A_k \rho A_k^\dagger -\frac12 A_k^\dagger A_k \rho - \frac12 \rho A_k^\dagger A_k \right) $$ which reduces to the Schrödinger equation if these off-terms $A_k=0$, and which can be interpreted as a continuous measurement (the above nonunitary operation) applied to a larger system which then gets ignored. So this is a quantum mechanics that no longer has that unitary reversible dynamics.

  1. Technically, the "matrices" that I'm talking about are more often differential operators with similar properties to matrices; sometimes they are infinite-dimensional.
  2. Again, if the "matrix" is a differential operator then the "vector" is a function.

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