For a qubit to be in "actual state" $\psi=\alpha_0|0\rangle+\alpha_1|1\rangle$ cannot be viewed as the qubit is either in $|0\rangle$ or $|1\rangle$ with probability of $|\alpha_0|^2$ or $|\alpha_1|^2$ respectively prior to measurement. But its in superposition form. However on measurement, the qubit collapse to $|0\rangle$ or $|1\rangle$. Post this measurement why do we say the qubit is no longer in superposition state? Shouldn't it be still in superposition state of $|0\rangle$ and $|1\rangle$, but on every sub-sequent measurement (without any other linear transformation) the qubit always collapse to same value as previous one.
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1$\begingroup$ Does this answer your question? What does "measuring a state" mean? $\endgroup$– MonteNeroJan 9 at 5:34
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$\begingroup$ @MonteNero That question gives the explanation on what "measuring" means. I understand that on measurement state collapse into |0⟩ or |1⟩. But in some papers/Q&A/text books, its mentioned that the qubit is no longer in superposition. Shouldn't we say after measurement, its in superposition with $|\alpha_0|^2=1 and |\alpha_1|^2=0$ (if collapsed to |0⟩ ) and on measurement we will always get |0⟩. Isn't it technically wrong to say "not in superposition". In reality we do not what is the state of qubit until next measurement, but we only know that it will be always |0⟩ on next measurement. $\endgroup$– AjaiSJan 9 at 5:48
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1$\begingroup$ Once the state you mentioned collapses into a definite state like 0 or 1 it becomes classical information. There is no point to talk about superposition anymore. Moreover, we know exactly what the post measurement state is. $\endgroup$– MonteNeroJan 9 at 14:06
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1$\begingroup$ Besides, by your logic everything in our macro world is in a superposition then. I guess you could view it that way, but it is just strange. $\endgroup$– MonteNeroJan 9 at 14:08
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1$\begingroup$ So I guess we could (rather should) say the qubit is no longer in superposition state post measurement from mathematical and logical point of view, but from quantum reality perspective its not wrong to consider them still in superposition (with knowledge that always collapse to a know state with 100% probability when ever its measured). $\endgroup$– AjaiSJan 9 at 18:09
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3 Answers
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Any pure state can be written as a superposition of other states in infinitely many ways, exactly like any vector can be written as a linear combination of other vectors in infinitely many ways.
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Shouldn't it be still in superposition state of |0⟩ and |1⟩, but on every sub-sequent measurement (without any other linear transformation) the qubit always collapse to same value as previous one.
Good question! One thing I'd ask about operating under this idea is how do things work under a change of basis?
For example, let us say we start with the initial $X$-eigenstate $|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$. If we measure it in the $Z$-basis, we measure $|0\rangle$ with probability $1/2$ and $|1\rangle$ with probability $1/2$. Let's say we measured $|0\rangle$. Now the reason we say the state collapses into $|0\rangle$ is because if we measure it again in the $Z$-basis we always get $|0\rangle$, but if we instead use the $X$-basis we measure $|+\rangle$ half the time and $|-\rangle$ the other half because $|0\rangle = \frac{1}{\sqrt{2}}(|+\rangle+|-\rangle)$. So even though we started off in $|+\rangle$, because of the intermediate measurement, the odds we measure $|+\rangle$ again at the end is only half.
It's change of basis experiments in this same vein like Stern-Gerlach that let us figure out that electrons have a fundamentally quantum up/down property we now call 'spin'
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Studying the comments in the original post, it seems you are interpreting the term "superposition state" to mean what I would call "element of the Hilbert space spanned by your classical basis states".
With this understanding, you are absolutely correct: even after measurement, qubits remain intrinsically quantum objects, belonging to the same Hilbert space as before. They've just been projected onto a particular "axis" of that Hilbert space.
The dissenting opinions in the comments are merely suggesting that we reserve the term "superposition state" to discriminate states off an axis from those on an axis. But either way, a qubit is in a quantum state, not a classical one.
The other recent answers are essentially agreeing with you, by pointing out that whether or not a qubit is on or off an axis depends entirely on how we define our axes, ie. which vectors we use as our basis elements.
Personally I find words like "superposition", "interference" and "entanglement" are crutches for people trying to understand quantum mechanics from a classical perspective, and that we should really just abandon them entirely, but that's a radical opinion. ^_^
