Why can't a fanout be made with a CNOT gate?

Question

I know this question has been answered here, but the answers leave some things confusing to me. When broken down (and used to calculate), the idea that the states are entangled doesn't seem to be right. I'm sure there is a mistake in my reasoning, but I'm not sure where. Here goes:

Say we take qubits in states $a|0\rangle + b|1\rangle$ and send them through a CNOT gate, which looks like this, the gate on the bottom being an XOR (that outputs 1 if the inputs are not the same binary value).

First, we use a fanout, which, according to other answers, leaves us with two qubits in state $a|00\rangle + b|11\rangle$. The set of three qubits, before two are passed through the XOR gate, can be expressed as $a|00\rangle\otimes|0\rangle + b|11\rangle\otimes|0\rangle$.

When we put this through the XOR gate, there is probability $a^2$ that the qubit will be $0$ which would result in an output of $0$. This means that exiting the gate, the probability amplitude of $|0\rangle$ is $a$. There is probability $b^2$ that the qubit will be $1$ so the probability amplitude of getting $|1\rangle$ as the second ouput is $b$. So our qubit is in state $a|0\rangle + b|1\rangle$, a clone of the original state. We now have two qubits in this state, as the first one comes out of the first output unchanged. As far as I can tell, they are not entangled.

Obviously, there is something wrong in this logic, I just don't know where. Could someone help me out?

Answer 1

I think you agree that if you start with the state $(a|0\rangle+b|1\rangle)|0\rangle$, the cnot produces $a|00\rangle+b|11\rangle$. The issue is why is the state of the first qubit not the same as $a|0\rangle+b|1\rangle$.

The answer is if you only look at that one qubit and you only look in the standard, $Z$ basis, then they do look the same. But those are two very restrictive assumptions.

For example, if I look at the qubit in the $|\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt{2}$ basis, then the original state is $$ \frac{a+b}{\sqrt{2}}|+\rangle+\frac{a-b}{\sqrt{2}}|-\rangle, $$ so we get the answer $|+\rangle$ with probability $\frac12|a+b|^2$. Meanwhile, our entangled state can be written as $$ \frac{1}{\sqrt{2}}|+\rangle(a|0\rangle+b|1\rangle)+\frac{1}{\sqrt{2}}|-\rangle(a|0\rangle-b|1\rangle), $$ so we get the answer $|+\rangle$ on the first qubit with probability 1/2. They are different.

Similarly, if you look at two qubits, let's return to the standard, $Z$ basis. For two copies of the original state, the probability of finding them both in $|0\rangle$ should be $|a|^4$, while for the entangled state it's just $|a|^2$. Again, they're different.

Answer 2

I'm a little confused about which gates operate on which qubits and how, but following the linked question, I think I understand that you are wondering why, given a single qubit in the state in $a|0\rangle+b|1\rangle$ and preparing two qubits in a state $a|00\rangle+b|11\rangle$ does not qualify as cloning the first bit, especially because the probabilities of measuring $0$ and $1$ on the second bit is exactly the same as that of the first bit.

The important answer which must be stressed is that the second qubit is not so much a copy of the first qubit, as it is entangled with the first qubit - they are married together (in a non-classical way.)

It might be an oversimplification, but cloning is classical, entangling is quantum.

The best example of the difference between cloning and entangling that I can think of would be that Eve the eavesdropper can follow your scheme to entangle qubits sent in a BB84 channel between Alice and Bob, but she would not be able to deduce the secret key based on such entanglement, without getting noticed by Alice/Bob.