No cloing

The no-cloning theorem is a fundamental principle in quantum mechanics that states it is impossible to create an exact copy of an arbitrary unknown quantum state. This is a cornerstone of the security in quantum key distribution protocols like BB84.

Another interesring property of quantum systems is the no-cloning theorm. Check No-cloning for proof. In short, for an example, if we would like to have a two-qubit quantum gate $U$ that will be able to copy the first qubit into the second. we would need,

$U(\lvert\psi\rangle \otimes \lvert0\rangle) = \lvert\psi\rangle \otimes \lvert\psi\rangle$

from above, we know that $U\lvert00\rangle = \lvert00\rangle$ and $U\lvert10\rangle = \lvert11\rangle$ . Therefore,

$U\bigg(\frac{1}{\sqrt{2}}(\lvert00\rangle + \lvert10\rangle)\bigg) = \frac{1}{\sqrt{2}}(U\lvert00\rangle + U\lvert10\rangle) = \frac{1}{\sqrt{2}}(\lvert00\rangle + \lvert11\rangle).$

Then, from previous example, we also know that $\frac{1}{\sqrt{2}}(\lvert00\rangle + \lvert10\rangle$ can be written in a product form,

$\frac{1}{\sqrt{2}}(\lvert00\rangle + \lvert10\rangle = \frac{1}{\sqrt{2}}\bigg(\lvert0\rangle + \lvert1\rangle\bigg)\lvert0\rangle$

Now, let's apply gate $U$ , we should have,

$U\frac{1}{\sqrt{2}}(\lvert00\rangle + \lvert10\rangle = U\bigg(\frac{1}{\sqrt{2}}\bigg(\lvert0\rangle + \lvert1\rangle\bigg)\lvert0\rangle\bigg) = \frac{\lvert0\rangle + \lvert1\rangle}{\sqrt{2}}\frac{\lvert0\rangle + \lvert1\rangle}{\sqrt{2}}$

which is a product state, and we also know have $U\bigg(\frac{1}{\sqrt{2}}(\lvert00\rangle + \lvert10\rangle)\bigg) = \frac{1}{\sqrt{2}}(\lvert00\rangle + \lvert11\rangle)$ ,

$\frac{\lvert0\rangle + \lvert1\rangle}{\sqrt{2}}\frac{\lvert0\rangle + \lvert1\rangle}{\sqrt{2}}\neq \frac{1}{\sqrt{2}}(\lvert00\rangle + \lvert11\rangle)$

therefore, no such gate $U$ exist.

What is the No-Cloning Theorem?

The no-cloning theorem ensures that:

Given a qubit in an unknown quantum state $\lvert\psi\rangle$ = $\alpha\lvert0\rangle$ + $\beta\lvert1\rangle$ , it is impossible to produce another qubit in the exact same state $\lvert\psi\rangle$ without disturbing the original.

Why?

The proof is rooted in the linearity of quantum mechanics:

To clone a quantum state, we would need a cloning operation U that works like this: $$ U(\lvert\psi\rangle \otimes \lvert0\rangle) = \lvert\psi\rangle \otimes \lvert\psi\rangle $$ where $\lvert0\rangle$ is a blank state (e.g., an auxiliary system). See Non-cloing for tensor.
For two arbitrary quantum states $\lvert\psi\rangle$ and $\lvert\phi\rangle$ , linearity requires: $$ U(\lvert\psi\rangle \otimes \lvert0\rangle) = \lvert\psi\rangle \otimes \lvert\psi\rangle $$ $$ U(\lvert\phi\rangle \otimes \lvert0\rangle) = \lvert\phi\rangle \otimes \lvert\phi\rangle $$
If $\lvert\psi\rangle \neq \lvert\phi\rangle$ , linearity leads to a contradiction because the superposition of the states cannot preserve the cloning operation: $$ U(a\lvert\psi\rangle + b\lvert\phi\rangle) \neq a\lvert\psi\rangle \otimes \lvert\psi\rangle + b\lvert\phi\rangle \otimes \lvert\phi\rangle $$ Thus, no such universal cloning operation $U$ can exist.

1. The No-Cloning Theorem

The no-cloning theorem states that it is impossible to create an exact copy of an arbitrary quantum state. Given a quantum state : $$ \lvert \psi \rangle = \alpha \lvert 0 \rangle + \beta \lvert 1 \rangle, $$ there is no operation that can produce: $$ U(\lvert \psi \rangle \otimes \lvert 0 \rangle) = \lvert \psi \rangle \otimes \lvert \psi \rangle, $$ where $ \lvert 0 \rangle $ is an auxiliary "blank" state.

2. Why Cloning is Impossible

The proof relies on the linearity of quantum mechanics.

Suppose we have two distinct quantum states $\lvert \psi \rangle$ and $\lvert \phi \rangle$ . For a universal cloning operation $U$ , we require: $$ U(\lvert \psi \rangle \otimes \lvert 0 \rangle) = \lvert \psi \rangle \otimes \lvert \psi \rangle, $$ $$ U(\lvert \phi \rangle \otimes \lvert 0 \rangle) = \lvert \phi \rangle \otimes \lvert \phi \rangle. $$

Now, consider a superposition of these states: $$ \lvert \chi \rangle = a \lvert \psi \rangle + b \lvert \phi \rangle, $$ where $a$ and $b$ are complex coefficients.

By the linearity of $U$ , applying it to $ \lvert \chi \rangle $ gives: $$ U(\lvert \chi \rangle \otimes \lvert 0 \rangle) = U((a \lvert \psi \rangle + b \lvert \phi \rangle) \otimes \lvert 0 \rangle). $$ Expanding this: $$ U(\lvert \chi \rangle \otimes \lvert 0 \rangle) = a U(\lvert \psi \rangle \otimes \lvert 0 \rangle) + b U(\lvert \phi \rangle \otimes \lvert 0 \rangle). $$

Substituting the cloning results: $$ U(\lvert \chi \rangle \otimes \lvert 0 \rangle) = a (\lvert \psi \rangle \otimes \lvert \psi \rangle) + b (\lvert \phi \rangle \otimes \lvert \phi \rangle). $$

However, the desired cloning result would be: $$ \lvert \chi \rangle \otimes \lvert \chi \rangle = (a \lvert \psi \rangle + b \lvert \phi \rangle) \otimes (a \lvert \psi \rangle + b \lvert \phi \rangle). $$

Expanding this: $$ \lvert \chi \rangle \otimes \lvert \chi \rangle = a^2 (\lvert \psi \rangle \otimes \lvert \psi \rangle) + ab (\lvert \psi \rangle \otimes \lvert \phi \rangle) + ab (\lvert \phi \rangle \otimes \lvert \psi \rangle) + b^2 (\lvert \phi \rangle \otimes \lvert \phi \rangle). $$

3. The Contradiction

The two results are not equal: 1. The linearity of $U$ only gives: $$ a (\lvert \psi \rangle \otimes \lvert \psi \rangle) + b (\lvert \phi \rangle \otimes \lvert \phi \rangle). $$ This result lacks the cross terms $ab(\lvert \psi \rangle \otimes \lvert \phi \rangle)$ and $ab (\lvert \phi \rangle \otimes \lvert \psi \rangle)$ .

Thus, the desired cloning result requires the cross terms because it represents the tensor product of the superposition with itself.

This mismatch proves that a universal cloning operation $U$ cannot exist.

4. Key Insights

The no-cloning theorem arises because quantum mechanics is linear, and linear operations cannot replicate the behavior of classical copying.
Quantum states cannot be duplicated without violating the mathematical structure of quantum mechanics.

This makes cloning of arbitrary quantum states impossible, providing a foundation for the security of quantum cryptography protocols like BB84. Let me know if you'd like further clarification!