Coherent state and squeezed states can be swapped why cloning coherent states can't be perfect to clone we obtain $$|\cos(|z|)b\rangle \otimes |\sin(|z|)b\rangle,$$

If we set $|z| = \pi/4$ then

$$|b\rangle \otimes |0\rangle = |b/\sqrt 2\rangle \otimes |b/\sqrt 2\rangle$$

which is imperfect cloning is there a way to perfectly clone?


1 Answer 1


The no-cloning theorem states that an unknown quantum state cannot be copied exactly --- so this rules out any algorithm that attempts to produce perfect copies of an arbitrary quantum state (including squeezed and coherent states).

As you note, however, the no-cloning theorem does not rule out the production of approximate quantum state clones. Andersen et al. introduced the most-optimal scheme for cloning of Gaussian states --- their algorithm produces a clone with fidelity of 65% to the original state (the proven optimum possible is $f=2/3$).

The circuit they use in their algorithm is here:

enter image description here

and can be implemented using Strawberry Fields like so:

# state to be cloned
Coherent(0.7+1.2j) | q[0]

# 50-50 beamsplitter
BS = BSgate(pi/4, 0)

# symmetric Gaussian cloning scheme
BS | (q[0], q[1])
BS | (q[1], q[2])
MeasureX | q[1]
MeasureP | q[2]
Xgate(scale(q[1], sqrt(2))) | q[0]
Zgate(scale(q[2], sqrt(2))) | q[0]

# after the final beamsplitter, modes q[0] and q[3]
# will contain identical approximate clones of the
# initial state Coherent(0.1+0j)
BS | (q[0], q[3])

You can check out the Gaussian cloning tutorial in the Strawberry Fields documentation for more information.

  • $\begingroup$ Thank you It's really helpful. $\endgroup$ Commented Sep 20, 2019 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.