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1

If you want to hard code a specific set of qubits you could measure the spin (pass it through an inhomogeneous magnetic field) of a spin 1/2 system and then if you get the wrong spin, flip it (tune a photon to a wavelength based on a given uniform magnetic field strength). If you want to read an output you could do the same thing, measure the spin of each ...


-1

It is assumed and proven that before measurement there is no distinct spin up or spin down state unless of course it is somehow prepared. General state of a particle is described as a linear combination of spin up and spin down state. If you measure the z projection you get 50 % upz and 50 % downz. If you then take upz particles and measure the x projection ...


0

If you prepare many silver atoms in the exact same state each time you might get 1000 out of 1000 to come out spin up. Or if you prepare many silver atoms in the exact same state (but different than the one above) each time you might get 0 out of 1000 to come out spin up. And when I say might it depends on if you pick the right state to prepare them and ...


1

Since $|H\rangle= \frac{1}{\sqrt2}(|R\rangle + |L\rangle)$ and $|V\rangle= \frac{1}{i\sqrt2}(|R\rangle - |L\rangle)$ the state $|\psi\rangle = \frac{1}{\sqrt2}(|H\rangle_1|H\rangle_2|H\rangle_3 + |V\rangle_1|V\rangle_2|V\rangle_3)$ and the state $\frac{1}{\sqrt2}(|H\rangle_1 \frac{1}{\sqrt2}(|R\rangle_2 + |L\rangle_2) \frac{1}{\sqrt2}(|R\rangle_3 + ...


0

The movie "Prince of Darkness" used the idea of tachion transmissions as a means by which 'information' (in this case a radio transmission) could be streamed into the past.


0

k and c can be set to 1 in a more real sense than the other constants because they do not have any fundamental units like length/time, mass/energy, or charge. By relativity, c's length over time cancels, but you should always conserve the sqrt(-1) that goes with time so its unit is 1/i. In other words, whenever you see "seconds" in units, like G and h, ...


0

To answer your last 2 questions: "Are there bounds on optimal encoding, whose purpose would be to delay a release of a finite fraction of the information?" and "Can there be encoding which releases a finite fraction of the information only at the very last stages of the decoding?" It's not a potential purpose of optimal encoding, but a required ...


0

Here $\rho$ is the density matrix for a pure state, but it's nonetheless a density matrix (simply one with von Neumann entropy of nought, but that doesn't matter). Recall that for any Hermitian operator $\hat{M}$, the $n^{th}$ moment of the probability distribution of the measurement made by $\hat{M}$ is $\mathrm{tr}(\rho\,\hat{M}^n)$. See, for example, the ...


0

Wave function collapse (or a measurement on the QM system) is not unitary since it is a projection of the state vector.


2

The answer depends on the implementation, as is often the case when asking practical questions about quantum computing. To give you an example of the state of the art in trapped ions, the Lucas group in Oxford can achieve less than one error in 1 million single-qubit gates, which they claim is less than the fault-tolerance threshold. Their error rate for ...


0

Short answer: This is impossible in principle for all states. A bit more elaborate: Given any state, by the postulate of quantum mechanics, it will be projected into the eigenspace of the eigenvalue measured when measuring an observable (this can be extended to quantum instruments). This means that the state will not be altered if and only if it is in the ...


1

The reason a measurement outcome on the second qubit forces a particular state for the first qubit is that the two qubits are entangled. You are correct about the probability of measuring state $\psi_b$, but the issue is that $\psi_b$ is under-determined, since $j$ can take any value in $\{0,1\}^n$. So when you perform the measurement on the second qubit, ...


3

Consider any purification of $\rho$: $|\psi\rangle \in \mathbb{H}_A \otimes \mathbb{H}_B$. This purification necessarily admits a Schmidt decomposition of the form: $$|\psi \rangle=\sum_i \sqrt{\lambda_i} |\alpha_i\rangle_A |\beta_i \rangle_B $$ with the state $\rho$ being of the form: $$\rho=\sum_i \lambda_i |\alpha_i\rangle \langle \alpha_i |$$ This ...


1

The NOT gate can be written as $$ \sigma_x \propto \exp[i\pi/2\,\sigma_x] $$ (up to an irrelevant global phase). Its roots are therefore of the form $$ \exp[i\phi\,\sigma_x] = \cos(\phi)\,I + i\sin(\phi)\,\sigma_x $$ with $n\phi = \pi/2 + 2\pi k$ with $k\in\mathbb N$.


2

This topic seems to me as functions of operators, so I'll explain this issue and will use "n-th root NOT gate" as example. Any function you apply on the operator - is applied on it's eigenvalues. If the operator is diagonal - all the eigenvalues are on the diagonal and applying the function is simply apply it to any element on the diagonal. $$A=D=\Sigma_i ...


1

I'm guessing that by $n^{\text{th}}$ root you mean some gate which if you pass a signal through $n$ of them the output will be identical to the output of the gate you want to "root". For odd $n$ it's pretty obvious that the classical NOT gate is it's own $n^{\text{th}}$ root. In that an odd number of NOT gates one after the other are identical to a single ...


0

Suppose you have a standard spin qubit: spin "up" is $|1\rangle$ and spin "down" is $|0\rangle$ . Your qubit is in some storage location which does not couple the two spin directions to each other or have any different energy, you store $|\psi\rangle = a |0\rangle + b |1\rangle$ in this qubit. ` Now some proton flies by, or a minor fluctuation in the Earth's ...



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