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.


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.

  • 1
    $\begingroup$ I do not quite understand this. Unitary operators take a Hilbert space to itself. But your $|\psi>$ and your $|\psi>|\psi>$ live in two different Hilbert spaces (the second being a tensor product of two copies of the first), so I don't see how it makes any sense to even ask whether a unitary operator can take $|\psi>$ to $|\psi>|\psi>$. $\endgroup$ – WillO Nov 16 '15 at 17:26

"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 $(\alpha \lvert\Psi\rangle +\beta \lvert\Phi\rangle)(\alpha \lvert\Psi\rangle +\beta \lvert\Phi\rangle)$.

  • $\begingroup$ Merely being additive is not enough for linearity. You need an extra condition, or if one wishes to be brief, $\hat{A}(\alpha|\Psi\rangle + \beta|\Phi\rangle) = \alpha\hat{A}|\Psi\rangle + \beta\hat{A}|\Phi\rangle$ instead. $\endgroup$ – Stan Liou Aug 3 '13 at 21:54
  • $\begingroup$ @stan Thanks for the clarification. I wasn't saying additivity implied linearity, I was saying that being additive followed from the definition of linearity. $\endgroup$ – Malabarba Aug 3 '13 at 22:53
  • $\begingroup$ Reading it again, I see how the text was misleading. Thanks. $\endgroup$ – Malabarba Aug 3 '13 at 22:59
  • $\begingroup$ @BruceConnor Yes, that makes sense, it was along the lines of what I was thinking about. Now, at this point, my question is more like - what is a proof-writer actually referring to when (s)he says "from the linearity of QM, ..." in the no-cloning theorem? $\endgroup$ – Ahaan S. Rungta Aug 4 '13 at 2:04
  • $\begingroup$ @Ahaan Could you quote the rest of that sentence? Like I did, he is probably saying that his next statement follows easily from the fact that operators used in quantum mechanics are linear operators. $\endgroup$ – Malabarba Aug 4 '13 at 9:17

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