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After taking some measure, how can a qunit be "unmeasured"? Is unmeasurement (ie reverse quantum computing) possible?

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I am not really sure about what you mean by "unmeasuring" a qubit, but if you mean to recover the qubit that was measured by manipulating the post-measurement state then I am afraid that the answer is no. When a quantum state is measured, the supoerposition state of such is collpased to one of the possible outcomes of the measurement, and so the qubit is lost.

The third postulate of quantum mechanics explains measurments in the quantum world, and such postulate says the following:

Quantum measurements are described by a collection $\{M_m\}$ of measurement operators. These are operators acting on the state space of the system being measured. The index $m$ referes to the measurement outcomes that may occur in the experiment. If the state of the quantum system is $|\psi\rangle$ immediately before the measurement, then the probability that result $m$ occurs is given by \begin{equation} p(m)=\langle\psi|M_m^\dagger M_m|\psi\rangle, \end{equation} and the state of the sytem after the measurement is \begin{equation} \frac{M_m|\psi\rangle}{\sqrt{\langle\psi|M_m^\dagger M_m|\psi\rangle}}. \end{equation}

So the post-measurement state collapses into another state defined by the postulate 3, and the previous quantum state is lost irreversibly. See also this wikipedia entry for wave function collapse, where it explains the collapse of quantum states after measurement.

Consequently, if the same measurment wants to be done, the quantum state must be prepared again before the measurement and so the xperiment can be repeated.

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  • $\begingroup$ It seems to me what is occuring is as follows: a pure state (eg white light) is undergoing a fourier transform (eg splitting in a crystal to produce rainbow) & then we are asking "which color of the rainbow is white?" $\endgroup$ – meowzz Jun 29 '18 at 16:19
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    $\begingroup$ @meowzz Neither a QFT nor an inverse QFT necessarily involve measurement. The circuit diagram on Wikipedia certainly doesn't involve any. So if you were to perform a QFT, then an inverse QFT, on a state, without measurement in between, then (as long as your gates were free of noise and error) you should be able to recover the initial state. $\endgroup$ – probably_someone Jun 29 '18 at 21:44
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    $\begingroup$ @meowzz You can certainly predict the Wigner function of a state at each step. Actually extracting the Wigner function at each step requires lots of measurements (each measurement contributes one data point to the sample, which, together, gives you a "shadow" probability distribution). So to determine the intermediate Wigner function, you have to measure in between the QFT and the inverse QFT, which collapses the superposition and makes the output of the inverse QFT not equivalent to the initial state. $\endgroup$ – probably_someone Jun 30 '18 at 20:33
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    $\begingroup$ I think that @probably_someone is on the point in his insight of your questions. If a unitary operation $U$ is applied to the quantum state, then such operation is reversible by applying the complex conjugate operation $U^\dagger$. However, when involving measurements, the wave function collapses and the inverse unitary operation will not give back the initial quantum state. $\endgroup$ – Josu Etxezarreta Martinez Jul 2 '18 at 7:50
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    $\begingroup$ However, sometimes indirect measuremets can be done in order to get information about the quantum state without actually destroying the state. For example, when doing quantum error correction, ancilla qubits are used in order to measure the syndromes associated with the errors that have happened to the quantum state. This way, recovery can be done in the corrupted state in order to get the original state without having collpased its wave function. $\endgroup$ – Josu Etxezarreta Martinez Jul 2 '18 at 8:00
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You can compute by measuring - see cluster-based quantum computation - but the whole thing that makes measurement different in quantum mechanics is that it destroys the superposition. It can't be undone. Once you measure, the qudit isn't in a state $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle + ... +\gamma|n\rangle$ but in a state $|\psi\rangle = |0\rangle$ or $|\psi\rangle = |1\rangle$ or what have you based upon probability. When you measure the qubit again soon after, it stays as either $|0\rangle$ or $|1\rangle$. The superposition is gone. We can't get it back (except by doing the same operations that lead our qubit to that point, in which case it'll be very similar) because we can't clone a qubit, so we can't figure out what $\alpha$ and $\beta$ are.

Tl;dr: No.

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  • $\begingroup$ Part of my confusion is around the 'destruction' of the superposition. Doesn't this violate the 1st law of thermodynamics (ie conservation of energy)? $\endgroup$ – meowzz Jun 29 '18 at 16:01
  • $\begingroup$ Re: exception (performing initialization operations). How could this be done? Given some initial state, could you determine the computation that led to it & iterate through that process until you returned to the initial state? $\endgroup$ – meowzz Jun 29 '18 at 16:08
  • $\begingroup$ @meowzz I was assuming that you knew the gates you had applied to get to a certain point. $\endgroup$ – heather Jun 29 '18 at 16:26
  • $\begingroup$ @meowzz also, wrt the 1st law of thermodynamics, the inbetween states now have 0 for a coefficient, but the probabilities still add up to 1. $\endgroup$ – heather Jun 29 '18 at 16:27
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    $\begingroup$ @meowzz (cc heather) Talking about the 1st law of thermo & energy conservation (and trying relating that to the conservation of probability) doesn't make much sense. When you're performing a measurement, the decoherence occurs due to the interaction of your qunit with the surrounding environment. During that process of measurement, energy can be transferred to and fro between the qunit and the environment. If you have to verify "energy conservation", you'd have to calculate the initial and final total energy of the "entire system" (including the environment), which is practically undefined. $\endgroup$ – Sanchayan Dutta Jun 29 '18 at 17:39
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Unitary operation is revesible, but measurement is a projection operation, which is not reveaible. Think about matrix inverse, projection matrix has lower rank and does not have inverse

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  • $\begingroup$ Thoughts on section 4? $\endgroup$ – meowzz Jul 16 '18 at 21:04

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