How to define a 'clone' of a mixed state?

Question

State clone of a pure state is clear. But how to define a clone of a mixed state?

For example, for a proper mixed state A, $\tfrac12(|0\rangle\langle 0|+|1\rangle\langle 1|)$, if there is a clone of A as A', then the joint system AA' should be in
$$\frac{|0\rangle\langle 0|+|1\rangle\langle 1|}{2} \otimes \frac{|0\rangle\langle 0|+|1\rangle\langle 1|}{2} \quad\text{or}\quad \frac{|00\rangle\langle 00|+|11\rangle\langle 11|}{2},$$ as somebody calls it a 'copy' instead of 'cloning'?

What's more, for improper mixture, for example an EPR pair AB as $(|00\rangle+|11\rangle)/\sqrt{2}$, how to define a 'clone' of A? Should the clone A' just be a normal mixture given by $(|0\rangle\langle 0|+|1\rangle\langle 1|)/2$ or the entanglement with B should be considered, i.e., A' should also be entangled with B? Of course if A' is entangled with B, we will violate the monogamy of entanglement. But is there a possibility that since the reduced density matrix of A is not really a 'state' of A, so that the clone of A is meaningless?

Answer 1

A 'clone' of a general mixed state $\rho$ is typically understood to mean the tensor product $$\rho\otimes\rho.$$ This is because you want to be able to do experiments independently on both copies, which is what the separable tensor product means. On the other hand, if you produce a state of the form $$\frac{|11⟩⟨11|+|00⟩⟨00|}{2}$$ then any experiment you do on the first copy will be replicated by the second one (i.e. detecting $|1⟩⟨1|$ on the original precludes $|0⟩⟨0|$ coming up on the copy). This can be useful, but it is not what we mean by cloning.