Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why must quantum logic gates be linear operators? I mean, is it just a consequence of quantum mechanics postulates?

share|cite|improve this question
up vote 1 down vote accepted

Suppose you pick a state $|\psi_i\rangle$ at random with probability $p_i$ and send it through a logic gate denoted by $G$. This random state is written as a density matrix $\rho = \sum_i p_i |\psi_i\rangle\!\langle\psi_i|$. Denote $G(|\psi_i\rangle\!\langle\psi_i|)$ as the result of applying $G$ to a particular state.

Now if the input is $|\psi_i\rangle\!\langle\psi_i|$ with probability $p_i$, then output is $G(|\psi_i\rangle\!\langle\psi_i|)$ also with probability $p_i$. Thus, the output state must be $\sum_i p_i G(|\psi_i\rangle\!\langle\psi_i|)$ and therefore $$G(\rho) \;=\; G\left(\sum_i p_i |\psi_i\rangle\!\langle\psi_i|\right) \;=\; \sum_i p_i G\Bigl(|\psi_i\rangle\!\langle\psi_i|\Bigr)\,.$$ This can be extended to random mixed input $\rho_i$ in place of $|\psi_i\rangle$, leading us to conclude that $G(\sum_i p_i \rho_i) = \sum_i p_i G(\rho_i)$, which is precisely the definition of linearity.

share|cite|improve this answer

Linear operators that are unitary conserve probability. Losing such a conservation makes computation with quantum mechanics meaningless.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.