What does "measuring a state" mean?

Question

I have been reading about secret sharing schemes, and they regularly come up with a line that says 'that a person upon receiving a state measures the state in one of the basis say the computational basis $|0\rangle,|1\rangle$ or the Hadamard basis $\dfrac{|0\rangle+|1\rangle}{\sqrt{2}}$ and $\dfrac{|0\rangle-|1\rangle}{\sqrt{2}}$. Now what I don't understand is suppose I have state say $$\alpha|0\rangle+\beta|1\rangle$$ that I get from someone and that someone will give me the secret when I 'measure' the state in the right Basis.

1. Now If I chose the standard basis $|0\rangle,|1\rangle$ what will be the result I will get? If I measure with respect of the projection operator $\langle 0|$ I get $\alpha$ and If I measure with respect of the projection operator $\langle 1|$ I get $\beta$.

2. If I measure with respect to the Hadamard basis $\dfrac{|0\rangle+|1\rangle}{\sqrt{2}}$ I get $\dfrac{\alpha+\beta}{\sqrt{2}}$ and if i measure with respect $\dfrac{|0\rangle-|1\rangle}{\sqrt{2}}$ I get $\dfrac{\alpha-\beta}{\sqrt{2}}$.

3. My first question basically means just selecting a Basis wouldn't serve the cause, because the basis themselves involve different projection operators and hence different operators. can somebody explain this concept?

4. Suppose I have a state $\omega|0\rangle$ and I measure in the standard basis then I either get a $0$ or $1$ depending on the projection operator chosen. So does that mean that $\omega$ has nothing to do with the result?

Answer 1

Look like a lot of misunderstanding.

If you measure a state $\alpha|0\rangle+\beta|1\rangle$ in computational basis, the state collapses either to $|0\rangle$ or $|1\rangle$. In Quantum Information we say that we measured state $|0\rangle$ or $|1\rangle$, or simply we measured $0$ or $1$. There is no chance to know $\alpha$ and $\beta$ from the measurement.

If you measure a state in Hadamard basis, the state collapses either to $|+\rangle =\dfrac{|0\rangle+|1\rangle}{\sqrt{2}}$ or to $|-\rangle = \dfrac{|0\rangle-|1\rangle}{\sqrt{2}}$, and we say we measured either $+$ or $-$.

Answer 2

Now If I chose the standard basis $|0\rangle,|1\rangle$ what will be the result I will get? If I measure with respect of the projection operator $\langle 0|$ I get $\alpha$ and If I measure with respect of the projection operator $\langle 1|$ I get $\beta$.

This is wrong. First of all, arguably the most natural kind of measurement in QM consists in choosing a basis and having the state collapse in that basis. For example, if you have a single qubit, you can "measure in the computational basis", which makes the state collapse in either $|0\rangle$ or $|1\rangle$ with some probabilities. You can think of measurements as the types of "questions" you can ask to a state. If you ask the system whether it's in the state $|0\rangle$ or $|1\rangle$, the state will collapse into one of these possibilities. You could instead ask whether the state is $|0\rangle+|1\rangle$ or $|0\rangle-|1\rangle$, and that will make the state collapse into of the elements of this other basis.

When you write $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$, what you are saying is that $|\psi\rangle$ is a state which, if you measure it in the computational basis (that is, the basis $\{|0\rangle,|1\rangle\}$), will be found in the state $|0\rangle$ with probability $|\alpha|^2$ and in the state $|1\rangle$ with probability $|\beta|^2$. If you take many copies of this state and measure it in the computational basis every time, you will get sometimes one result and sometimes the other (unless of course $\alpha=0$ or $\alpha=1$).

It is also worth noting that when I say "you get $|0\rangle$ or $|1\rangle$", what I mean is that you can have two different experimental outcomes, and you associate these two experimental outcomes with the labels "$|0\rangle$" and "$|1\rangle$". How this association is made depends on the context and is at least partially a matter of convention. For example, if you are talking about a photon's polarisation, you can decide that $|0\rangle$ means "horizontal polarisation" and $|1\rangle$ means "vertical polarisation".

Suppose I have a state ω|0⟩ and I measure in the standard basis then I either get a 0 or 1 depending on the projection operator chosen. So does that mean that ω has nothing to do with the result?

In the bra-ket formalism, states are defined up to multiplication by complex scalars. This means that $|\psi\rangle$ and $\lambda|\psi\rangle$ represent the same physical state for any $\lambda\in\mathbb C$.