# What is the nature of the correspondence between unitary operators and reversible change?

Why does the formalism of QM represent reversible changes (eg the time evolution operator, quantum gates, etc) with unitary operators?

To put it another way, can it be shown that unitary transformations preserve entropy?

The fact that evolutions of quantum mechanics are unitary after finite periods of time can be proven from the Schrödinger equation, and hinges on the characterization of unitary operators as those linear operators which are norm-preserving.

Recall the Schrödinger equation: $$\frac{\mathrm d}{\mathrm d t} |\psi\rangle \;=\; -i H |\psi\rangle \;,$$ where $H$ may or may not be time-dependent, but in any case has real eigenvalues, so that $H = H^\dagger$. As a result, the way in which $|\psi\rangle$ changes instantaneously with time is in such a way that its magnitude, as a vector in Hilbert space, does not increase. We can see this by simply computing: $$\frac{\mathrm d}{\mathrm dt} \langle\psi|\psi\rangle \;=\; \left[\frac{\mathrm d}{\mathrm dt} \langle\psi|\right]|\psi\rangle + \langle\psi|\left[\frac{\mathrm d}{\mathrm dt}|\psi\rangle\right] \;=\; i\langle\psi|H^\dagger|\psi\rangle - i\langle\psi|H|\psi\rangle \;=\; 0.$$ because $H = H^\dagger$. Then at all times the state vector remains the same length. That is to say, the norm is preserved under finite-time evolution.

Because unitary operators are exactly those ones which preserve the norm, it follows that the finite-time evolution will be unitary.

• That's helpful, thank you. Please could you explain in a little more detail why it is that finite-time evolutions of this sort should be reversible? – Benjamin Hodgson Jan 16 '13 at 23:33
• Reversibility is a consequence of unitarity: precisely because $|\psi\rangle \mapsto U|\psi\rangle$ preserves norms, we have $1 = \langle \psi | \psi \rangle = \langle \psi | U^\dagger U | \psi \rangle$ for all unit vectors $|\psi\rangle$. It follows by linearity that $U^\dagger U = I$, which in particular means that $U$ is invertible. – Niel de Beaudrap Jan 16 '13 at 23:50
• @NieldeBeaudrap Thanks for your previous comment. I would like to know if the converse is true (and if so, how it can be seen), i.e., is unitarity of the propagator necessary for the evolution to be invertible? I can only see that the propagator should admit an inverse for the evolution to be reversible--not why the inverse should necessarily be the complex conjugate. Sorry if I am missing very something basic. – Dvij Mankad Oct 10 '18 at 10:40
• ^something very basic* – Dvij Mankad Oct 10 '18 at 11:07

Like all 20th century physics, the formalism is invariant with respect to time reversal. This was true in classical mechanics and it remains true in QM because canonical quantization does not alter the meaning of energy - it just becomes an evolution operator. Unitary operators satisfying $A A^{\dagger} = I$ are associated logarithmically to Hermitian ones with real eigenvalues. Since measured quantities must be real, this was historically the inevitable route for QM.

Entropy should not be confused with unitary evolution. It is a thermodynamic quantity requiring a system of many states and it always increases for one direction in time, even when the underlying laws are time reversal invariant. This was Boltzmann's original insight (see kinetic theory for gases). With black holes we can talk about other forms of entropy, but the general view is that unitary evolution applies to all the usual quantum states so that information is conserved even for black hole processes. This is the so called Information Paradox.

The story is much more complicated in QFT but time reversal symmetry is still maintained at a fundamental level, even though there is a time operator with symmetry broken for specific interactions by CP violation.