If I understand correctly, for two orthonormal states $\left|\psi_1\right\rangle$ and $\left|\psi_2\right\rangle$ in the Hilbert space H, there must exist a unitary transformation $U$, such that:

$$U\left|\psi_1\right\rangle\left|\alpha\right\rangle = \left|\psi_1\right\rangle\left|\psi_1\right\rangle$$

$$U\left|\psi_2\right\rangle\left|\alpha\right\rangle = \left|\psi_2\right\rangle\left|\psi_2\right\rangle$$

where $\left|\alpha\right\rangle$ is the initial state of the second subsystem, on which orthonormal states are cloned.

My question is: where can I find the reference for the formal proof of this folklore lemma?


2 Answers 2


I'm not sure about a reference for this lemma, but maybe this will help.

$\left|\psi_1\right\rangle\left|\alpha\right\rangle, \left|\psi_2\right\rangle\left|\alpha\right\rangle$ are an orthonormal basis of a two-dimensional subspace of your initial Hilbert space. The $U$ in your equations (it shouldn't be on the right side, by the way) explicitly maps this to orthonormal basis vectors of a subspace of your final Hilbert space. This restricted $U$ is already unitary.

Now choose any unitary mapping between the orthogonal complements of these two-dimensional subspaces and the whole map together will be unitary.

  • $\begingroup$ Thanks for pointing out the editing issue - yes, $U$ should not be on the right side. I have edited that. $\endgroup$
    – user36125
    Dec 22, 2014 at 8:27

Though the question was asked almost 2 years ago, yet since it hasn't been closed yet, maybe the following answer will be helpful.

Given your 2 equations

  1. $$U|\psi_1\rangle|\alpha\rangle = |\psi_1\rangle|\psi_1\rangle$$
  2. $$U|\psi_2\rangle|\alpha\rangle = |\psi_2\rangle|\psi_2\rangle$$

take the inner product of (1) with (2):

left-hand side: $$\langle \psi_1|\langle \alpha|U^\dagger U |\psi_2\rangle|\alpha\rangle = \langle \psi_1|\langle \alpha |\psi_2\rangle|\alpha\rangle = \langle \psi_1|\psi_2\rangle \langle \alpha |\alpha\rangle = \langle \psi_1|\psi_2\rangle $$

right-hand side: $$ \langle \psi_1| \langle \psi_1| \psi_2\rangle |\psi_2 \rangle = \langle \psi_1|\psi_2\rangle ^2 $$

Notice that the results on the left-hand side and on the right-hand side differ while they should be equal if cloning were possible. In fact, they are equal only in 2 cases:

  1. If the states $|\psi_1\rangle$ and $|\psi_2\rangle$ are orthogonal, i.e. $\langle \psi_1|\psi_2\rangle = 0$

  2. If the states $|\psi_1\rangle$ and $|\psi_2\rangle$ are identical, i.e. $\langle \psi_1|\psi_2\rangle = 1$

In other words, only orthogonal states can be cloned by the same unitary transformation.

You may also want to refer to "Introduction to Quantum Physics and Information Processing" by Radhika Vathsan where this proof is presented in slightly different form.


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