7

It is written in some places that the unitarity of time evolution is what prevents quantum cloning. However, consider the typical definition of a cloning operator $A$. For all $\left|\psi\right>$ and a standard state $\left|0\right>$, $$ A[\left|\psi\right>\otimes \left|0\right>] = \left|\psi\right>\otimes \left|\psi\right> $$ Without using the unitarity of $A$, I can follow the proof in these notes to demonstrate no-cloning. With a superposition state $\left|\chi\right> = a\left|\psi\right>+b\left|\phi\right>$, $A$ can be appled to find, $$ A[\left|\chi\right>\otimes \left|0\right>] = a(\left|\psi\right>\otimes \left|\psi\right>)+b(\left|\phi\right>\otimes\left|\phi\right>) $$ $A$ could also be applied to find, $$ A[\left|\chi\right>\otimes \left|0\right>] =\left|\chi\right>\otimes\left|\chi\right> = (a\left|\psi\right>+b\left|\phi\right>)\otimes(a\left|\psi\right>+b\left|\phi\right>) $$

These expressions are not equal, so there is a contradiction. This appears to be a proof that no-cloning is impossible for any linear time evolution, even in an alternate universe where quantum time evolution does not have to be unitary. Is this reasoning correct?

Community
  • 1
Display Name
  • 745
  • 5
  • 13
  • 1
    This looks right on a first reading, but I would suggest the following as a check. No-cloning can be stated in words as follows: you can't make a separable copy of a pure state unless you have a preferred basis and you have been assured in advance that the state is in one of the basis states, not a superposition; and even if you do have this assurance, you can't copy the phase. I would go through the list of assumptions in this verbal statement and check that they're all either explicitly or implicitly used in your proof. Then I would go through the conclusions and [...] –  Feb 20 '19 at 00:08
  • 1
    [...] check that they're all things you've proved. If there are no differences, then it seems like you've proved no-cloning without assuming unitarity. I think the assumptions in this statement are all necessary, and it's easy to find counterexamples if you leave them out. Check that your proof doesn't omit these assumptions in ways that allow it to be disproved by these counterexamples. –  Feb 20 '19 at 00:11
  • 1
    Checking back over my notes on this topic, yes, the theorem only depends on linearity, although proofs using unitarity may be simpler sometimes. See https://physics.stackexchange.com/a/73623/4552 –  Feb 21 '19 at 01:37
  • 2
    can't one make the even simpler argument that if $A$ is linear and $A(\boldsymbol v\otimes\boldsymbol e_1)=\boldsymbol v\otimes\boldsymbol v$ for all $\boldsymbol v$, then $A((\lambda \boldsymbol v)\otimes\boldsymbol e_1)=\lambda (\boldsymbol v\otimes\boldsymbol v)=\lambda^2(\boldsymbol v\otimes\boldsymbol v)$ (because we can "duplicate" before and after using the linearity of the tensor product), and thus the cloning property implies nonlinearity of the operator? @BenCrowell – glS Feb 22 '19 at 09:11

1 Answers1

3

In fact, this is one of the two popular kinds of proofs by contradiction of the no-cloning theorem which claims the nonexistence of quantum operation that can duplicate arbitrary unknown quantum state. It may be helpful to clearly give two kinds of proofs here:

(i) proof by contradiction based on the linearity: just as what you have done, assume that there is a quantum operation $U_{\mathrm{clone}}$ which can duplicate arbitrary unknown quantum state. Then for arbitrary state $|\Psi\rangle=\alpha|\phi_1\rangle+\beta|\phi_2\rangle$, $$U_{\mathrm{clone}}|\Psi\rangle|0\rangle=|\Psi\rangle|\Psi\rangle.$$ However, by the assumption of linearity of this quantum cloning operation, we have $$U_{\mathrm{clone}}|\Psi\rangle|0\rangle=\alpha U_{\mathrm{clone}}|\phi_1\rangle|0\rangle+\beta U_{\mathrm{clone}}|\phi_2\rangle|0\rangle =\alpha |\phi_1\rangle|\phi_1\rangle+\beta |\phi_2\rangle|\phi_2\rangle.$$ We thus arrive at a contradiction. Here I would like to point out that this proof is actually the initial proof of no-cloning theorem used by Wotters and Zurek in their paper in 1982 and also by Dieks in his 1982 paper where he also indicated that the linearity of quantum mechanics can be used to prove the impossibility of superluminal communication.

(ii) proof by contradiction based on the unitarity of the quantum operation: we first assume that the quantum clone operation $U_{\mathrm{clone}}$ is unitary, then for arbitrary states $|\Psi\rangle$ and $|\Phi\rangle$, $U_{\mathrm{clone}}|\Psi\rangle|0\rangle=|\Psi\rangle|\Psi\rangle$ and $U_{\mathrm{clone}}|\Phi\rangle|0\rangle=|\Phi\rangle|\Phi\rangle$. Taking the inner product of two sides of the above two equations and using the unitarity assumption of $U_{\mathrm{clone}}$, we arrive at $\langle \Psi|\Phi\rangle=(\langle\Psi|\Phi\rangle)^2$ which is the case only when $|\Psi\rangle$ and $|\Phi\rangle$ are orthogonal. This proof is first proposed by Yuen in his paper in 1986. Now this proof is more popular in quantum information books, e.g., Peres' book Quantum Theory: Concepts and Methods, Nielsen and Chuang's book Quantum Computation and Quantum Information.

Zhian Jia
  • 124
  • Any idea why the second proof is more popular? The first proof seems to be more general, since it proves no-cloning for linear operators which might still be non-unitary, – dnaik Jun 27 '23 at 06:16