Not getting why applying an one qubit operation on the first qubit updates the rest of them if the rest of them are not entangled in some way to the first qubit. They must be entangled to be updated, otherwise they will stay unchanged. Right?
It's not the qubits that are getting updated. The quantum state is a superposition of a number of different states each with its own amplitude, and applying a single qubit operation to the first qubit does change the amplitudes of the other states too. Try this: take a 2 qubit system a|00> + b|01> + c|10> + d|11> and apply the Hadamard transform to only the first qubit. When you regroup all the terms you'll see that the amplitudes of all the states have been updated.
I don't think so. Let's think of 2 qubits sitting next to each other. If they are NOT entangled than e.g. I can use the X-gate to bit-flip the first qubit. The 2nd will not be changed by this. Since they are not entangled, we can factorize their superposition state and treat each one as if it were alone. Things will be different if they are entangled, because than we can not factorize their combined state. Did I get this wrong?
@@ernestrutherford5854 What changed here is the amplitude, which represents the probability. Assume 2 independent Qbits, even the second one is never touched, applying the gate to the 1st one will change the probability distribution of the combined quantum states of the two. It's like the Monty Hall problem in probability theory, the conditional probability changes, joint probability distribution also changes, which is irrelevant to entanglement
Not getting why applying an one qubit operation on the first qubit updates the rest of them if the rest of them are not entangled in some way to the first qubit. They must be entangled to be updated, otherwise they will stay unchanged. Right?
It's not the qubits that are getting updated. The quantum state is a superposition of a number of different states each with its own amplitude, and applying a single qubit operation to the first qubit does change the amplitudes of the other states too.
Try this: take a 2 qubit system a|00> + b|01> + c|10> + d|11> and apply the Hadamard transform to only the first qubit. When you regroup all the terms you'll see that the amplitudes of all the states have been updated.
I don't think so. Let's think of 2 qubits sitting next to each other. If they are NOT entangled than e.g. I can use the X-gate to bit-flip the first qubit. The 2nd will not be changed by this. Since they are not entangled, we can factorize their superposition state and treat each one as if it were alone.
Things will be different if they are entangled, because than we can not factorize their combined state.
Did I get this wrong?
@@ernestrutherford5854 What changed here is the amplitude, which represents the probability. Assume 2 independent Qbits, even the second one is never touched, applying the gate to the 1st one will change the probability distribution of the combined quantum states of the two. It's like the Monty Hall problem in probability theory, the conditional probability changes, joint probability distribution also changes, which is irrelevant to entanglement
Shouldn't it be identity of dimension 2n-4 in the tensor multiplication at 6:40 ?
que? How are you getting 2n-4 squire?