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Value of a qubit and its state - is there any difference between these two terms in sense of terminology?

For example, can we name this state of a qubit also a value of a qubit: $$ |\psi\rangle = \frac{1}{\sqrt{2}} |0\rangle + \frac{1}{\sqrt{2}} |1\rangle $$

(By value, I don't mean eigenvalue)

UPD: By the way, I always use a word "state" for states like $|\psi\rangle$ and never call it a "value". The question is more about if someone use "value" for $|\psi\rangle$, is it a complete mistake? For example, if you saw it on Wikipedia what would you say?

UPD2: And what about states like $|0\rangle$ and $|1\rangle$? They are without a doubt states but also frequently called value which doesn't look like a big mistake.

UPD3: As for me, "value" more of a service word and by itself sounds not so clear as "result of a measurement", which is more informative. For example, like in a statement: "the result of the measurement has a value 1". What do you think?

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    $\begingroup$ I think both mean the same, and the difference is in the area of usage. The term state is a general physical term, and can be used anywhere, the term value is specific for Quantum Information Science. $\endgroup$
    – kludg
    Commented Mar 22, 2021 at 4:50
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    $\begingroup$ Can you show us any "reputable source" which calls that the "value" of the qubit? $\endgroup$
    – user1271772 No more free time
    Commented Mar 25, 2021 at 19:00
  • $\begingroup$ I think the only example you've given so far is: stackoverflow.com/q/43241775/1271772 where they put the word "value" in quotes. It's a question by a non-expert who asked a non-expert question, on StackOverflow of all places (not a quantum computing community). They used "value" because they're used to classical computing and are a newbie at quantum. Every expert in quantum computing uses "state". Can you tell us more about why you're asking this question? The bounty has not attracted any new answers except mine, and while my answer is correct, it doesn't seem to be what you're seeking $\endgroup$
    – user1271772 No more free time
    Commented Mar 27, 2021 at 13:27

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For the $|\psi\rangle$ in your question, please call it a "state", not "value". The word "value" is not "mainstream" or popular in this context.

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I'm not aware of any widespread technical distinction between the "value" and "state" of qubits. I'd expect any paper or textbook or presentation using such a distinction to define its terms before expecting them to be understood in different ways.

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Whenever you mention 'state' of a qubit, you are referring to the "Quantum State" defined in Quantum Mechanics as:

In quantum physics, a quantum state is a mathematical entity that provides a probability distribution for the outcomes of each possible measurement on a system.

So $|\psi\rangle$ is a quantum 'state' via which we can access the probable 'values' that it can take, so for your example, the 'values' it can take is $0$ and $1$ and it's 'state' is $\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle$. Using which we can infer that the two values are equally probable.

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    $\begingroup$ But we also can say that $|\psi\rangle = \frac{1}{\sqrt{2}} |0\rangle + \frac{1}{\sqrt{2}} |1\rangle$ is a state before measurement and $|0\rangle$ (or $|1\rangle$) is a state after the measurement (result of the measurement) $\endgroup$
    – doktr
    Commented Mar 17, 2021 at 11:11
  • $\begingroup$ And you'd be completely correct in saying that, but think about the implication that your statement has, say, for example, say your qubits are physically existing with the help of electronic spin - you saying that you have $|0\rangle$ state implies that you have the value of $0$(physically electron-spin is upwards) with $100$% probability. Kinda redundant way to say that your electron is in the upward spin. But it won't be silly to talk about it before, since you can perform mathematical analysis on it to see how would they evolve with different unitary operations. $\endgroup$
    – MetaInformation
    Commented Mar 17, 2021 at 11:26
  • $\begingroup$ Also, state vectors are not the only way you can make sense of it, check out density operators, an alternative way of formulation of states. $\endgroup$
    – MetaInformation
    Commented Mar 17, 2021 at 11:29
  • $\begingroup$ Thanks for the link, it is really helpful even though I've already known about the density matrix while ago. My question was more about what words can be used for description of a qubit state. It is clear for me that the word 'state' can be used after the measurement as well as before. At least because if we get $|0\rangle$ after the first measurement we then can measure the qubit again in the Hadamard basis and get $|+\rangle$ or $|-\rangle$ with equal probability. I'm more interested in knowing is it completely prohibited to use a word 'value' before the measurement of a qubit. $\endgroup$
    – doktr
    Commented Mar 17, 2021 at 11:58
  • $\begingroup$ Because in some discussions 'value' is not used as the synonym of the result of a measurement. For example, like in one of the answers to this topic in the phrase "to copy the qubit-being-Hadamarded's value onto a second qubit": stackoverflow.com/questions/43241775/… $\endgroup$
    – doktr
    Commented Mar 17, 2021 at 12:19

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