What do they mean by "qubit can't be copied"?

Question

What does it mean by ''qubit can't be copied'' ?

In a note, it is saying that:

Copying a qubit means $$U|\psi\rangle_A|0\rangle_B=|\psi\rangle_A|\psi\rangle_B$$ i.e; applying a unitary transformation on the qubit state. It is explained as, if the copy operation is possible then there will be a unique unitary matrix $U$ which will work on all qubit state, and then shown that existence of such $U$ is not possible.

I am not getting how it can be written in such a way, unitary matrix $U$ will operate on $|\psi\rangle_A$ only I think, how it can copy it to second $|0\rangle$ state?

Secondly, why we are making the assumption that, "if such unitary matrix exists then that will be a unique unitary matrix which will work on all qubit states" why we can't use different unitary matrix to copy different qubit state(if possible, as $|+\rangle$ can't be copied)?

E.g, we can copy $|0\rangle_A$ to another state $|0\rangle_B$ $$U|0\rangle_A=|0\rangle_B\\U|1\rangle_A=|1\rangle_B$$ as classical bit can be copied, it is possible to find such $U$.

Answer 1

All operations on quantum states are unitary operations. We don't make the rules, this is just how nature seems to work. So if you want to define an operation that copies a qbit, it has to be a unitary operation. That unitary operation would look like this:

$U|\psi\rangle_A|0\rangle_B=|\psi\rangle_A|\psi\rangle_B$

So you have the qbit you want to copy, $|\psi\rangle_A$, and the qbit into which you want to copy it, $|0\rangle_B$. This is the most general way to write the copy operation, although any other way of writing it reaches the same conclusion: it cannot be done.

The reason for this is as follows. Consider your starting state:

$|\psi\rangle|0\rangle = \begin{bmatrix} \alpha \\ \beta \end{bmatrix} ⊗ \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} \alpha \\ 0 \\ \beta \\ 0 \end{bmatrix}$

And now consider your desired ending state:

$|\psi\rangle|\psi\rangle = \begin{bmatrix} \alpha \\ \beta \end{bmatrix} ⊗ \begin{bmatrix} \alpha \\ \beta \end{bmatrix} = \begin{bmatrix} \alpha^2 \\ \alpha\beta \\ \beta\alpha \\ \beta^2 \end{bmatrix}$

So you want to go from here to here:

$\begin{bmatrix} \alpha \\ 0 \\ \beta \\ 0 \end{bmatrix} \rightarrow \begin{bmatrix} \alpha^2 \\ \alpha\beta \\ \beta\alpha \\ \beta^2 \end{bmatrix}$

But see those exponents? They mean this is not a linear operation! And since we can only perform linear operations on quantum states, no operation exists which can take us from the first state to the second (other than an operation which itself uses the values of $\alpha$ and $\beta$). Thus, copying (cloning) is impossible when you don't know $\alpha$ or $\beta$.

As for why we don't just use a different unitary transformation for each copy, that would require us knowing the exact quantum state we want to copy. If we know the exact quantum state, we can just take a blank qbit and reconstruct the same quantum state on the qbit. Which is fine, but pretty useless considering the reason we want to be able to copy a quantum state is so we can find the value of the quantum state in the first place.

Classical bits can always be copied, as you discovered. Of course, we copy classical bits all the time in the real world (you're reading copied classical bits right now!).

Answer 2

As already mentioned in the other answers, the crucial point is that copying means implicitly that the state of the original qubit is unknown, i.e. given a qubit in an unknown state, you want to prepare a second qubit to be in exactly the same state.

To make it more intelligible, there is a less mathematical argument that this should not be possible: By the uncertainty relation you cannot determine the values of two complementary observables (e.g. orthogonal spin directions) on the qubit with arbitrary precision at the same time. If you could copy the qubit, you could make copies and measure each of the observables on the copies with arbitrary precision, which contradicts the uncertainty relation.

Answer 3

To answer the first part of the question (whether unitary matrix $U$ operates on $|\psi_A \rangle$ only):

A unitary matrix can operate on an arbitrary number of qubits. Single-qubit gates, like Pauli X, Y and Z gates, operate on one qubit and are represented by 2x2 matrices; CNOT gate operates on two qubits and is represented by a 4x4 matrix, etc.

In this case $U$ denotes a unitary transformation operating on two qubits, represented by a 4x4 matrix.

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To answer the second part of the question (why should there be only one unitary to copy all possible states):

It is possible to find unitaries which copy some qubit states. The easiest example is CNOT gate which copies the states $|0 \rangle$ and $|1 \rangle$:

$$CNOT|0\rangle_A |0 \rangle_B=|0\rangle_A|0\rangle_B\\ CNOT|1\rangle_A |0 \rangle_B=|1\rangle_A|1\rangle_B$$

But this unitary will not work to copy an unknown superposition of the states $|0 \rangle$ and $|1 \rangle$:

$$CNOT(\alpha |0\rangle + \beta |1\rangle)_A |0 \rangle_B=\alpha |0\rangle_A|0\rangle_B + \beta |1\rangle_A |1\rangle_B \neq (\alpha |0\rangle + \beta |1\rangle)_A (\alpha |0\rangle + \beta |1\rangle)_B$$

You can follow the proof given in the Wikipedia article to see that any one unitary can copy at best two orthogonal states.

We need to find one unitary that would work for all states because the no-cloning theorem addresses only copying of an unknown quantum state. If we know exactly what state we need to create, we can just create it from scratch without using the prototype qubit at all.