4.2.1 No cloning

When we have a classical bit, it is easy to copy or clone it – simply look at it and copy what you see:

[0]\displaystyle[0] ↦[00],\displaystyle\mapsto[00],
[1]\displaystyle[1] ↦[11].\displaystyle\mapsto[11].

Can we also clone quantum bits?

Let us assume for a moment that this is possible. This would mean that there exists a quantum operation CC that, given any qubit in state |ψ⟩\left|\psi\right\rangle and a fresh qubit in state |0⟩\left|0\right\rangle, acts as

C⁢(|ψ⟩⊗|0⟩)=|ψ⟩⊗|ψ⟩C(\left|\psi\right\rangle\otimes\left|0\right\rangle)=\left|\psi\right\rangle% \otimes\left|\psi\right\rangle (4.16)

to produce two copies of |ψ⟩\left|\psi\right\rangle out of a single copy. (Why do we supply the fresh qubit? This is so that CC has the same number of input qubits as output qubits.)

For example, the cloner would operate as follows on the basis states:

C⁢|00⟩\displaystyle C\left|00\right\rangle =|00⟩,\displaystyle=\left|00\right\rangle, (4.17)
C⁢|10⟩\displaystyle C\left|10\right\rangle =|11⟩.\displaystyle=\left|11\right\rangle.

Just like we can easily clone a classical bit, it is easy to find a quantum operation that clones the basis states. For example, the controlled-NOT operation CNOT1→2\mathrm{CNOT}_{1\to 2} from Eq. 3.63 does this job.

But is there a quantum operation that can clone an arbitrary unknown qubit state, not just a basis state? In the following homework problem, you will show that this is not possible.

Homework 4.1 (No cloning).

In this homework problem we want to prove that there exists no quantum operation CC that satisfies Eq. 4.16. We will use a trick called proof by contradiction. This means that we will show that if such a cloning operation CC existed then this would imply something that we know to be wrong (e.g., “0=10=1”). From this we can then conclude that no such CC can exist.
Thus let us start by assuming that there is a quantum operation CC that satisfies Eq. 4.16. Now you can calculate C⁢(|+⟩⊗|0⟩)C(\left|+\right\rangle\otimes\left|0\right\rangle) in two different ways:

  1. 1.
    ​

    First use Eq. 4.16 and then write the result in the form of Eq. 3.31.

  2. 2.
    ​

    First expand |+⟩⊗|0⟩\left|+\right\rangle\otimes\left|0\right\rangle in the form of Eq. 3.31, then use that CC is linear, and finally apply Eq. 4.16.

Do you get the same answer in both cases? If not, what can you conclude?

Hack.

Assuming such an operation is possible, let us see how it would act on |ψ⟩=|+⟩\left|\psi\right\rangle=\left|+\right\rangle. On one hand, since CC is supposed to clone any single-qubit states, including |+⟩\left|+\right\rangle, we should get

C⁢(|+⟩⊗|0⟩)\displaystyle C(\left|+\right\rangle\otimes\left|0\right\rangle) =|+⟩⊗|+⟩=12⁢(|0⟩+|1⟩)⊗12⁢(|0⟩+|1⟩)\displaystyle=\left|+\right\rangle\otimes\left|+\right\rangle=\frac{1}{\sqrt{2% }}\left(\left|0\right\rangle+\left|1\right\rangle\right)\otimes\frac{1}{\sqrt{% 2}}\left(\left|0\right\rangle+\left|1\right\rangle\right)
=12⁢(|0⟩⊗|0⟩+|0⟩⊗|1⟩+|1⟩⊗|0⟩+|1⟩⊗|1⟩)\displaystyle=\frac{1}{2}\left(\left|0\right\rangle\otimes\left|0\right\rangle% +\left|0\right\rangle\otimes\left|1\right\rangle+\left|1\right\rangle\otimes% \left|0\right\rangle+\left|1\right\rangle\otimes\left|1\right\rangle\right)
=12⁢(|00⟩+|01⟩+|10⟩+|11⟩).\displaystyle=\frac{1}{2}\left\lparen\left|00\right\rangle+\left|01\right% \rangle+\left|10\right\rangle+\left|11\right\rangle\right\rparen.

On the other hand, since all quantum operations must be linear,

C⁢(|+⟩⊗|0⟩)=C⁢(12⁢|00⟩+12⁢|10⟩)=12⁢C⁢|00⟩+12⁢C⁢|10⟩=12⁢(|00⟩+|11⟩),\displaystyle C(\left|+\right\rangle\otimes\left|0\right\rangle)=C\left\lparen% \frac{1}{\sqrt{2}}\left|00\right\rangle+\frac{1}{\sqrt{2}}\left|10\right% \rangle\right\rparen=\frac{1}{\sqrt{2}}C\left|00\right\rangle+\frac{1}{\sqrt{2% }}C\left|10\right\rangle=\frac{1}{\sqrt{2}}\left\lparen\left|00\right\rangle+% \left|11\right\rangle\right\rparen,

where the last equality follows from Eq. 4.17. Since these two equations give different results, we get a contradiction. Hence we conclude that there is no quantum operation that can clone an unknown quantum state.

This famous result is known as the no cloning theorem. The same conclusion (and likely also the argument that you gave in 4.1) applies also to probabilistic bits! Here is an intuitive explanation for why we can copy neither probabilistic nor quantum information. If this were possible then, given a probabilistic bit in an unknown state pp or a qubit in an unknown state |ψ⟩\left|\psi\right\rangle, we could first produce as many copies of pp and |ψ⟩\left|\psi\right\rangle as we like. Given these copies, we could then measure them in various ways and use the obtained data to estimate the probabilities of pp or the amplitudes of |ψ⟩\left|\psi\right\rangle to arbitrary precision (just like we did in in Section 2.5.1 to figure out the inner workings of the yellow mystery box). Thus, we could from a single probabilistic or quantum bit learn an arbitrary amount of information. This should clearly not be possible!

Indeed, if this were possible then we would live in a very strange world (much stranger than the one described by quantum mechanics)! For example, imagine a coin whose probability of heads is p=0.1011010010⁢…p=0.1011010010\ldots where the binary digits encode the whole content of Wikipedia as well as all YouTube videos and all pictures of cats you can find on the internet. If cloning probabilistic bits were possible, I could flip this coin once and write down which outcome I got. This is a probabilistic bit of information that is equal to 0 with probability pp. If I send this probabilistic bit to you and you have the ability to clone it, you could produce as many copies of it as you want and then measure them all. By looking at the measurement outcomes and counting how many zeroes you got, you could estimate the probability pp. In fact, by producing sufficiently many copies of the original bit, you could estimate this probability arbitrarily well! In particular, you would be able to extract out any binary digit of pp and hence also all the information encoded in pp, including the cat picture number 65535!

This should clearly be impossible, since otherwise we would not need USB drives, data centers, or to pay for our mobile phone data connection – we could just store all our information in a single probabilistic bit and transmit it all by sending this bit to somebody else! This is certainly too good to be true…