# I Understanding the no-clone theorem

1. Dec 7, 2017

### fog37

Hello,

The well known no-clone theorem states that it is impossible to exactly copy the quantum state of a system. The explanation seems to reside in the uncertainty principle...

But, for example, if we measure a photon with known polarization (say vertical) with a vertical linear polarizer we obtain a vertically polarized photon as output. Wouldn't that be an example of copying the original photon since the measurement result isa photon with equal polarization as the incoming photon (i.e. a copy of the incoming one)? Where is the flaw in my thinking?

Thanks!
Fog37

2. Dec 7, 2017

### Staff: Mentor

It has nothing to do with the HUP since it doesn't concern non-commuting observables.

A definite polarization is what can be considered classical information. The problem comes about when trying to copy a system that is in a superposition of states in a particular basis.

What the no cloning theorem says is not that particular states can't be reproduced. (QM theory is based on the idea of identically prepared systems, so there must be a way to prepare systems in a given state reproducibly.) The no cloning theorem shows that it is impossible to build an operator that will clone any arbitrary quantum state (in other words, it is impossible to make a universal copying machine).

3. Dec 7, 2017

### Staff: Mentor

Let me add that if you have one photon coming in and one coming out in the same state, you have not violated the no-cloning theorem. Your scenario is more like quantum teleportation. What you couldn't have is the original photon still in its state an additional photon in the same state.

4. Dec 7, 2017

### fog37

Thanks.

I follow your last comment: copying, in general, implies creating something new that is an exact replica of the original. So two entities are involved (the original and the replica). In my example, the incoming photon is measured and we only get one outcoming photon (with the same polarization). That is not exactly copying...

But I am still not fully grasping the difference between classical copying (possible) and quantum copying (impossible). For example, what if we had a photon with arbitrary and unknown polarization (say diagonal at 45 degrees) and we wanted to measure that polarization and know it. How would the no-cloning theorem fit in this situation? We would try to measure the photon's polarization with a polarizer but we would not know which orientation to choose. We may choose the right orientation, or a different orientation...

5. Dec 7, 2017

### Staff: Mentor

As soon as you measure, the original state is lost. You have to proceed by entanglement without losing the original state.

I prefer working with the spin of massive particles than photons, since it is almost impossible to interact with a photon without destroying it. If you have a spin-1/2 particle that is prepared in the spin-up or spin-down state only, then it is possible to measure the spin without modifying the state, and creating another particle in the same state. This is classical copying. But if you take the particle to be in any superposition of spin-up and spin-down, then it is impossible to create an operation that will copy this exact superposition (since a measurement will destroy the superposition). At best, one can use entanglement to teleport the state to another particle, bu then the original particle has lost its original state.

6. Dec 7, 2017

### Demystifier

No, it resides on the superposition principle.

The non-cloning principle states that it is impossible to clone an arbitrary unknown state.

7. Dec 7, 2017

### fog37

Thanks Dr. Claude.

So we can create entanglement in such way that a spin-1/2 particle is prepared in the spin-up and the other particle, entangled with the first, is also in the same spin-up. The second particle would then be a clone of the first entangled particle. But, you mention, this is not possible if the entangled particles have a spin state that is a superposition. The two entangled particles will have states that are still entangled but cannot be identical and therefore cannot be a copy of each other. Is that correct? What is the problem with the superposition of states?

Also, when we measure a photon we always destroy it. What happen to a massive particle when we measure it instead instead? I guess it simply remains that particle but with a different quantum state...

8. Dec 8, 2017

### Staff: Mentor

In the usual quantum teleportation protocol, you have particle A and B entangled, and you want to transfer the state of particle C to B (or A, the label doesn't matter). You do that by manipulating A and C together and make some measurement that tells you what to do with B such that it recovers the original state of C. But doing this changes the state of particle C.

There is no problem a priori with superposition of states, since all states are superpositions (it is a question of choice of basis). But as Demystifier pointed out, I didn't clarify that it is impossible to construct an operation that will copy an unknown state. For a spin-1/2 particle, you could come up with a machine that will copy the particle when it is in any state $| \psi \rangle$, and that machine will work also when it is in state $| \phi \rangle$ where $\langle \phi | \psi \rangle = 0$, but that machine will not work when the particle is in state $| \chi \rangle = ( | \psi \rangle + | \phi \rangle) / \sqrt{2}$. Then you could decide to make instead a machine that will copy state $| \chi \rangle$, which will also work with state $| \xi \rangle$ where $\langle \chi | \xi \rangle = 0$, but then this new machine will not copy states $| \psi \rangle$ and $| \phi \rangle$ correctly.

Not always: there are some methods for non-destructive measurements of photons in cavities. But most of the time, measurement of a photon means destruction. I also do not like using photons to think about basic QM because they are very special particles that introduce lots of difficulties (such that there is no wave function for a photon) that make the discussion of QM too complicated.

Ans yes, for a massive particle, you are simply manipulating its quantum state, be it the orientation of its spin, its position, momentum, internal state (for composite particles), etc.

9. Dec 8, 2017

### jerromyjon

I believe that is where the HUP comes in, if you could copy an unknown state then you could measure position of one and momentum of the other which could give you precise results, and the uncertainty principle excludes that from being possible.

10. Dec 8, 2017

### Staff: Mentor

I don't see why this would violate the HUP, since you would have two particles, one in a well defined momentum state, the other in a well defined position state. It would tell you nothing about the position/momentum state of the other.