Quantum Circuits Must Be Reversible

A closed quantum circuit cannot throw information away. Its gates are unitary operators, and unitary operators have inverses.

This is the basic reason quantum circuits are reversible.

Unitary means invertible

A closed-system quantum gate is represented by a unitary operator $U$. By definition,

$$U^{†}U=I.$$

Therefore

$$U^{-1}=U^{†}.$$

So every valid closed-system quantum gate can be undone. If a state evolves as

$$|\psi ⟩\mapsto U|\psi ⟩,$$

then applying $U^{†}$ recovers the original state:

$$U^{†}U|\psi ⟩=|\psi ⟩.$$

Irreversible classical functions lose information

Many classical functions are not reversible. For example,

$$f(a,b)=a+b$$

loses information if only the sum is kept.

The output $5$ could have come from many inputs:

$$(0,5),(1,4),(2,3),\dots$$

So the map

$$(a,b)\mapsto a+b$$

cannot be inverted in general. The output does not contain enough information to recover the input.

That kind of collapse is not allowed for a unitary gate, because a unitary gate must have a unique way back.

Reversible embeddings keep the input

To use a classical function inside a quantum circuit, one usually embeds it into a reversible transformation.

For a classical function $f$, the standard pattern is

$$|x⟩\otimes |y⟩\mapsto |x⟩\otimes |y\oplus f(x)⟩,$$

where $\oplus$ is bitwise XOR.

The important point is that $x$ is preserved. The function value is written into a separate target register by modifying $y$.

This transformation is reversible because applying it twice recovers the original target:

$$\begin{array}{rl}y& \mapsto y\oplus f(x),\\ y\oplus f(x)& \mapsto y\oplus f(x)\oplus f(x)=y.\end{array}$$

The input $x$ was kept all along, so the second application uses the same $f(x)$ again.