13.7: No Cloning Theorem

One of the most natural operations in classical information is to copy bits, it happens all the time in our computers. Quantum logic diverges from classical logic already at this level. Qubits cannot be copied, or as it is usually stated: qubits cannot be cloned.

There are several intuitive arguments that can help us understand why it is so. Remember that in Chapter 10 we emphasized that the act of measuring changes the system being measured; if the system is in a superposition, the result of the measurement will be one of the states of the superposition. And the superposition is destroyed. Intuitively, if measuring is at all required to do the cloning, then it will be impossible to have two clones, since we cannot learn anything about the initial superposition. Furthermore, the superposition itself is destroyed by the act of measuring. This implies that a viable cloning device cannot use measurement.

Assume we have such a device and it operates without requiring measurement. One of the foundations of quantum mechanics is the uncertainty principle introduced by Heisenberg. The principle says that certain physical variables cannot be measured at the same time to an arbitrary precision. The example is position and momentum; if the position of a particle is measured with a given precision \(\Delta x\), the precision with which its momentum is measured is limited: \(\Delta p > \hbar /2\Delta x\). With the presumed cloning machine at hand it should be possible to clone the particle and measure momentum to arbitrary precision in one clone and position to arbitrary precision in the other, possibly violating Heisenberg’s principle.

These arguments by themselves do not prove the impossibility of cloning, but suggest that the matter is by no means trivial.

Figure 13.1: Suggested cloning device

To show that cloning is not possible, let us assume that it were possible to clone, and that we could set up a “machine” like the one in Figure 13.1. The cloning device takes the information of one qubit \(|\; \phi_1\rangle\) and copies it into another “blank” qubit, the result is a qubit \(|\; \psi_1\rangle\) identical to \(|\; \phi_1\rangle\), and the original \(|\; \phi_1\rangle\) is unmodified. According to what we saw in our overview of bracket notation, such a machine is an operator (we will call it \(\widehat{C}\)) because it transforms two qubits into two other qubits; and as an operator, \(\widehat{C}\) must be unitary. Thus we define \(\widehat{C}\)

\[\mid \;Original\rangle\; \otimes \mid Blank \rangle \quad \stackrel{\widehat{C}}{\longrightarrow} \quad \mid Original\rangle \;\otimes \mid clone \rangle \tag{13.41} \]

We are now ready to clone two arbitrary qubits \(|\; \phi_1\rangle\) and \(|\; \phi_2\rangle\) separately.

\begin{align*}
\widehat{C}\;|\;\phi_{1}\rangle \mid blank \rangle &=\left|\;\phi_{1}\right\rangle\left|\;\psi_{1}\right\rangle \tag{13.42}\\
\widehat{C}\;|\;\phi_{2}\rangle \mid blank \rangle &=\left|\;\phi_{2}\right\rangle\left|\;\psi_{2}\right\rangle \tag{13.43} \\
& \tag{13.44}
\end{align*}

where it is understood that \(|\; \phi_1\rangle =|\; \psi_1\rangle\) and \(|\; \phi_2\rangle =|\; \psi_2\rangle\), and we have given them different names to distinguish original from copy.

Since the cloning machine is unitary, it preserves the dot products, so we can compare the dot product before and after cloning

\[\left.\left\langle\phi_{2}\;\right|\left\langle blank \;||\; \phi_{1}\right\rangle \mid blank \right\rangle=\left\langle\phi_{2}\;\left|\;\left\langle\psi_{2}\;|| \;\phi_{1}\right\rangle\;\right|\; \psi_{1}\right\rangle \tag{13.45} \]

Recall the rules for taking the dot product of tensor products, each element in the tensor product of kets is multiplied by the bra in the same position in the tensor product of bras, therefore

\[\left.\left\langle\phi_{2} \;\mid \; \phi_{1}\right\rangle\langle blank \;|\; blank \right\rangle=\left\langle\phi_{2} \mid \phi_{1}\right\rangle\left\langle\psi_{2} \mid \psi_{1}\right\rangle \tag{13.46} \]

The requirements that kets be normalized imposes that \(\langle blank \;|\; blank\rangle\) = 1. The above equation can only be true in two cases:

• \(\langle \phi_2 \;|\; \phi_1\rangle\) = 0, which means that \(|\; \phi_1\rangle\) and \(|\; \phi_2\rangle\) are orthogonal. This means that we can clone states chosen at random from a set of orthogonal states. And is equivalent to say that we can clone \(|\; 0\rangle\) and \(|\; 1\rangle\), which we already knew since we do that classically all the time.
• \(\langle \psi_2 \;|\; \psi_1\rangle\) = 1, which means that \(\psi_2 = \psi_1\), that is, that clones obtained in each operation are identical. If the two originals were different, as we had assumed, what this result says is that the clone is independent from the original, which is quite a bizarre property for a clone!.

This proof shows that perfect cloning of qubits cannot be achieved. We can certainly store the result of a measurement (this is another way of phrasing the first case), but we cannot clone the superpositions.