Linearity of Quantum Mechanics?

Question

The proof of the No-Cloning Theorem states "By the linearity of quantum mechanics, ..." -- Could someone please give me a rough sketch/outline of what this means? Does it have to do with the Hilbert Space that wave functions live in?

I apologize if this question isn't specific enough, I just wanted to fully understand this concept.

Answer 1

The no-cloning theorem states that it is not possible to have a quantum state $|\psi\rangle$ evolve into two separable (non-entangled) copies described by the tensor product state $|\psi\rangle|\psi\rangle$.

The proof boils down to the simple observation that when expressing $|\psi\rangle$ in some basis ${|0\rangle, |1\rangle, |2\rangle, ...}$:

$$|\psi\rangle = \alpha_0 |0\rangle + \alpha_1 |1\rangle + \alpha_2 |2\rangle + ... $$

the cloning operation would be a unitary evolution of the form:

$$U(\alpha_0 |0\rangle + \alpha_1 |1\rangle + ... ) = \alpha_0^2 |0\rangle|0\rangle + \alpha_0 \alpha_1 |0\rangle|1\rangle + \alpha_1 \alpha_0 |1\rangle|0\rangle + \alpha_1^2 |1\rangle|1\rangle... $$

This leads to a contradiction, as the unitary operator $U(..)$ is linear and can never create amplitudes like $\alpha_0^2$ and $\alpha_0\alpha_1$ that are quadratic functions of the $\alpha_i$.

So the linearity the author is referring to is the linearity of the unitary evolution. In quantum physics evolution is described by unitary operators which transform incoming states into outgoing states that are a linear combination of the ingoing states.

Answer 2

"By the linearity of quantum mechanics" is actually a reference to the linearity of the operators used it quantum mechanics. It means that, for a linear operator $A$ (by the very definition of linearity), $$A\bigl(\alpha\lvert\Psi\rangle +\beta\lvert\Phi\rangle\bigr)=\alpha A\lvert\Psi\rangle + \beta A\lvert\Phi\rangle,$$ where $\alpha$ and $\beta$ are complex numbers.

The way this applies to the no cloning theorem, is quite simple. A cloning operator would have to satisfy the following: there is a vector $\lvert \Xi \rangle$ such that, for any $\lvert \Psi \rangle$, we have $$A\lvert\Psi\rangle \lvert\Xi\rangle= \lvert \Psi\rangle \lvert\Psi\rangle.$$

However, $\alpha\lvert\Psi\rangle +\beta\lvert\Phi\rangle$ is just as much of a valid state as $\Psi$. Thus, combined with the equation above for linear operators (which is like saying "due to the linearity of quantum mechanics"), this implies that $$A (\alpha\lvert\Psi\rangle +\beta\lvert\Phi\rangle\bigr) \lvert\Xi\rangle= \alpha A\lvert\Psi\rangle \lvert\Xi\rangle +\beta A\lvert\Phi\rangle\bigr \lvert\Xi\rangle =\alpha \lvert\Psi\rangle \lvert\Psi\rangle +\beta \lvert\Phi\rangle\bigr \lvert\Phi\rangle.$$ Which is not what we wanted, as an actual a copy of the initial state would have been $(\alpha \lvert\Psi\rangle +\beta \lvert\Phi\rangle)(\alpha \lvert\Psi\rangle +\beta \lvert\Phi\rangle)$.