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Suppose we have the quantum circuit below with a quantum register of 2 qubits and a classical register of 2 bits. The Hadamard gates and CNOT gate are not important for the question. When we measure a qubit we collapse it into one of the computational basis states (let's suppose we are working with the basis $ \{|0\rangle,|1\rangle\}$). But on Nielsen and Chuang's book, it says

...this operation [measurement] converts a single qubit state $|\psi \rangle=\alpha |0\rangle +\beta |1\rangle$ into a probabilistic bit $M$ (...) which is $0$ with a probability of $|\alpha|^2$, or 1 with a probability of $|\beta|^2$.

Now here's what I don't understand:

  1. If the measurement is an irreversible process then the probability of the resultant state is towards the initial quantum state it was in, that is, if the resultant state is $1$ with a probability of, say, $63\%$, that means that is has a $63\%$ probability that the initial state was $|\psi \rangle= |1\rangle$. However, after it has collapsed to a bit, then, there is no probability associated with it, it certainly is $1$ for all future measurements, it shouldn't be a probabilistic bit.

  2. In the circuit below, what is the need of registering the state of the qubit in a classical bit if it has collapsed irreversibly into that same state? Does it only make sense to do so when it is the end of the circuit?

enter image description here

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If the measurement is an irreversible process then the probability of the resultant state is towards the initial quantum state it was in, that is, if the resultant state is $1$ with a probability of, say, $63\%$, that means that is has a $63\%$ probability that the initial state was $|\psi \rangle= |1\rangle$.

This is a fundamental misinterpretation. If the resultant state after measurement is $|1\rangle$ with $63\%$ probability, then it simply means that the initial quantum state was of the form $\sqrt{0.27}e^{i\theta}|0\rangle + \sqrt{0.63}e^{i\phi}|1\rangle$ where $\theta,\phi\in \Bbb R$ (of course, assuming we measured in the $\{|0\rangle, |1\rangle\}$ basis). This initial quantum state collapses to the state $|1\rangle$ with $63\%$ probability! Don't think of it as "if the resultant state is $|1\rangle$ with $63\%$ probability it means that the initial quantum state was $|1\rangle$ with $63\%$ probability." That's wrong. Qubits can totally exist in a linear superposition of $|0\rangle$ and $|1\rangle$ and do not necessarily need to be either one of them i.e. either $|0\rangle$ or $|1\rangle$. A superposition is a perfectly valid state of existence for a qubit! It also does not make sense to say: it's both $|0\rangle$ and $|1\rangle$ at the same time. Quantum mechanics is a mathematical model of reality and it does not need to match with your classical intuitions and need not be describable with the limited English vocabulary we apply for classical objects. We need to extend our vocabulary to describe qubits and that's where mathematics helps us. The word "superposition" is a mathematical term in this context.

However, after it has collapsed to a bit, then, there is no probability associated with it, it certainly is $1$ for all future measurements, it shouldn't be a probabilistic bit.

I believe the word in the term that's confusing you is "converts". I'd say don't overthink this. They simply mean that on measurement of a qubit in the state $a|0\rangle + b|1\rangle$ there $|a|^2$ probability of getting $|0\rangle$ and $|b|^2$ probability of getting $|1\rangle$. The term "probabilitic bit" here is simply an abstraction. They mean that upon measurement you can imagine your qubit to be a "bit" with a value of either 0 or 1. That is, it has the value 0 with probability $|a|^2$ and value 1 with probability $|b|^2$. But this is assuming they don't know the measurement result yet. They're simply mathematically modelling the measured qubit as a probabilitic bit without the knowledge of the result of the measurement. Of course, if you know the result of the measurement it's going to be either 0 or 1 with $100\%$ probability.

In the circuit below, what is the need of registering the state of the qubit in a classical bit if it has collapsed irreversibly into that same state? Does it only make sense to do so when it is the end of the circuit?

This is already answered in Why do we need a Classical Register for carrying out Quantum Computations?

The results of the measurements are basically classical information i.e. 0 or 1 if you're measuring in the $\{|0\rangle, |1\rangle\}$ basis. Remember that in reality you're measuring things like electron spin. You need to store that classical information somewhere for future use, and that is what copying the results of the measurement to the classical bits achieves. And yes, in the gate model it's generally done after the measurements (which is usually the final step of a quantum algorithm).

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  • $\begingroup$ Thank you for the answer, but you say Qubits can totally exist in a linear superposition of |0⟩ and |1⟩ and do not necessarily need to be either one of them, doesn't this contradict the discreteness that quantum mechanics describes? That is, without measurement, reality evolves/ lives in continuum and not in discrete levels of energy and so any experiments we ever do they never depict the rawest form of reality, the one that isn't biased by our human experience of the world $\endgroup$ – Bidon Jan 28 at 14:51
  • $\begingroup$ Also, this answer and the comment inspired me to ask this question: physics.stackexchange.com/questions/457287/… $\endgroup$ – Bidon Jan 28 at 15:27
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    $\begingroup$ @Bidon "Discreteness" in quantum mechanics does not mean what you think it means. Please go through Reason for the discreteness arising in quantum mechanics? It's unfortunate that popular science sources throw around buzzwords like "discreteness" casually, which in turn causes so many misconceptions among beginners. If you're having trouble understanding the answers on Physics SE thread I linked, it might be worth asking a fresh question about this, on our site. $\endgroup$ – Sanchayan Dutta Jan 28 at 15:35
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    $\begingroup$ [cont.] Wikipedia has a quick and dirty explanation of what discreteness means: "Quantum mechanics differs from classical physics in that energy, momentum, angular momentum and other quantities of a bound system are restricted to discrete values (quantization)". $\endgroup$ – Sanchayan Dutta Jan 28 at 15:44

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