4 clarified edited Jun 30 at 17:51 SpaceChicken 4844 bronze badges 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). More simply First, we will 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 $$0$$ (whichwhich would result in an output of zero as our ancilla bit is $$0$$), so. 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 $$1$$$$|1\rangle$$ 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? 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. More simply, we will 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 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 zero as our ancilla bit is $$0$$), so exiting the gate, the probability amplitude is $$a$$. There is probability $$b^2$$ that the qubit will be $$1$$ so the probability amplitude of $$1$$ 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. Obviously, there is something wrong in this logic, I just don't know where. Could someone help me out? 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? 3 deleted 74 characters in body edited Jun 30 at 17:37 SpaceChicken 4844 bronze badges 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. Also, pardon the janky notation, I don't know how to type what I need to. Here goes: Say we take qubits in states $$a|0\rangle + b|1\rangle$$ and send them through a CNOT gate. More simply, we will 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 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 zero as our ancilla bit is $$0$$), so exiting the gate, the probability amplitude is $$a$$. There is probability $$b^2$$ that the qubit will be $$1$$ so the probability amplitude of $$1$$ 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. Obviously, there is something wrong in this logic, I just don't know where. Could someone help me out? 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. Also, pardon the janky notation, I don't know how to type what I need to. Here goes: Say we take qubits in states $$a|0\rangle + b|1\rangle$$ and send them through a CNOT gate. More simply, we will 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 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 zero as our ancilla bit is $$0$$), so exiting the gate, the probability amplitude is $$a$$. There is probability $$b^2$$ that the qubit will be $$1$$ so the probability amplitude of $$1$$ 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. Obviously, there is something wrong in this logic, I just don't know where. Could someone help me out? 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. More simply, we will 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 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 zero as our ancilla bit is $$0$$), so exiting the gate, the probability amplitude is $$a$$. There is probability $$b^2$$ that the qubit will be $$1$$ so the probability amplitude of $$1$$ 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. Obviously, there is something wrong in this logic, I just don't know where. Could someone help me out? 2 latexify the ket states edit approved Jun 30 at 16:57 Mark S 2,02644 silver badges2727 bronze badges 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. Also, pardon the janky notation, I don't know how to type what I need to. Here goes: Say we take qubits in states a|0> + b|1>$$a|0\rangle + b|1\rangle$$ and send them through a CNOT gate. More simply, we will use a fanout, which, according to other answers leaves us with two qubits in state a|00> + b|11>$$a|00\rangle + b|11\rangle$$. The set of three qubits can be expressed as a|00>⊗|0> + b|11>⊗|0>$$a|00\rangle\otimes|0\rangle + b|11\rangle\otimes|0\rangle$$. When we put this through the XOR gate, there is probability a squared$$a^2$$ that the qubit will be 0 $$0$$ (which would result in an output of zero as our ancilla bit is 0$$0$$), so exiting the gate, the probability amplitude is a$$a$$. There is probability b squared$$b^2$$ that the qubit will be 1$$1$$ so the probability amplitude of 1$$1$$ is b$$b$$. So our qubit is in state a|0> + b|1>$$a|0\rangle + b|1\rangle$$, 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. Obviously, there is something wrong in this logic, I just don't know where. Could someone help me out? 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. Also, pardon the janky notation, I don't know how to type what I need to. Here goes: Say we take qubits in states a|0> + b|1> and send them through a CNOT gate. More simply, we will use a fanout, which, according to other answers leaves us with two qubits in state a|00> + b|11>. The set of three qubits can be expressed as a|00>⊗|0> + b|11>⊗|0>. When we put this through the XOR gate, there is probability a squared that the qubit will be 0 (which would result in an output of zero as our ancilla bit is 0), so exiting the gate, the probability amplitude is a. There is probability b squared that the qubit will be 1 so the probability amplitude of 1 is b. So our qubit is in state a|0> + b|1>, 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. Obviously, there is something wrong in this logic, I just don't know where. Could someone help me out? 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. Also, pardon the janky notation, I don't know how to type what I need to. Here goes: Say we take qubits in states $$a|0\rangle + b|1\rangle$$ and send them through a CNOT gate. More simply, we will 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 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 zero as our ancilla bit is $$0$$), so exiting the gate, the probability amplitude is $$a$$. There is probability $$b^2$$ that the qubit will be $$1$$ so the probability amplitude of $$1$$ 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. Obviously, there is something wrong in this logic, I just don't know where. Could someone help me out? 1 asked Jun 30 at 5:30 SpaceChicken 4844 bronze badges