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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?

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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 $-$.

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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 $\lambda\in\mathbb C$.

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  • $\begingroup$ What do you mean when you say take multiple copies?. Can i make i duplicate of that state? But that would violate the no cloning theorem or do I attach an ancilla with my state $\alpha|0\rangle+\beta|1\rangle$ and do a CNOT with that. Do this for say 100 times each time adding an ancilla. After the completion of those 100 rounds the measurement outcome that is in majority is the one my quantum state was in most probably $\endgroup$ – Upstart Nov 13 at 19:27
  • $\begingroup$ @Upstart as per no cloning, you cannot implement a reversible operation duplicating an arbitrary input state. What I'm saying here is simply that in practice you can usually prepare the same state multiple times. Eg shine your laser through the same quantum optical apparatus to alway get the same output. No-cloning has no bearing here because you are not using a state as input to make a copy of the same state $\endgroup$ – glS Nov 13 at 19:31
  • $\begingroup$ okay. But can still do what I proposed that I can do a CNOT between the state and my ancilla state $|0\rangle$. And repeat these steps and take the majority measurements. $\endgroup$ – Upstart Nov 13 at 19:34
  • $\begingroup$ @Upstart I'm not totally clear about the protocol you are proposing but it seems like a probabilistic protocol so no-cloning does not apply in that scenario. Anyway, you might be better off asking that as a separate question as it does not seem to be related to this one. $\endgroup$ – glS Nov 13 at 19:37

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