13
votes
Accepted
What's the Difference between T2 and T2*?
The naming started in NMR and it has become the difference between the following two experiments.
Experiment one: Prepare the qubit in a superposition state (apply a H gate) and vary the wait time ...
- 561
9
votes
Is Quantum Biocomputing ahead of us?
"Is Quantum Biocomputing ahead of us?"
There has been some work done on biocomputing, quantum computing, spin chemistry, and magnetochemical reactions.
Correlated radical pairs — pairs of ...
- 2,287
8
votes
Accepted
If a Hamiltonian is quadratic in the ladder operator, why is its time evolution linear in the ladder operator?
Hint: Instead of using the BCH formula in the form usually presented, for example at the top of this Wikipedia page, use this consequence of Hadamard's Lemma:
$$\tag{1}
e^{iHt}\hat{a}e^{-iHt} = \hat{a}...
- 13.5k
6
votes
If a Hamiltonian is quadratic in the ladder operator, why is its time evolution linear in the ladder operator?
Use the differential form of the time evolution,
$$dO/dt=i[H, O]\ .$$
- 5,682
5
votes
Passive improving of nanodiamond surfaces for NV centers?
I have worked with NVs in nanodiamonds a little bit, and you are totally right, surface characteristics have a huge influence on how far we can push them. There are definitely multiple groups working ...
- 824
5
votes
Accepted
Why does joint ground state not change under action of beam splitting unitary operator?
Calculate
$$
\begin{align}
\hat{U}|00\rangle &= \exp\left(-igt(\hat{a}^\dagger_2\hat{a}_1+\hat{a}^\dagger_1\hat{a}_2)\right)|00\rangle \\
&= \sum_{k=0}^\infty \frac{(-igt)^k}{k!}(\hat{a}^\...
- 20.8k
4
votes
Why does joint ground state not change under action of beam splitting unitary operator?
There's more than one way, and I'll suggest two of them here:
Expand $\hat{U}$ using the formula for the Taylor series of an exponential ($e^\hat{A}$) centered around $\hat{A}=\hat{0}$, and then you ...
- 13.5k
4
votes
Accepted
Do we know anything about the computational complexity of the exchange-correlation functional?
Computing the exchange-correlation functional to sufficiently high accuracy is QMA-hard, where QMA is the quantum version of NP. In particular, this means that in all likelihood, it will be hard even ...
- 5,682
4
votes
Accepted
Difference between coherence transfer, polarization transfer and population transfer?
Given a quantum system in a state defined by a density matrix $\rho$, it is an accepted terminology to use the term population for the diagonal matrix elements (not necessarily in the computational ...
- 2,495
3
votes
What's the Difference between T2 and T2*?
From Chapter 15 of NII's quantum information lecture series on "Fundamentals of Noise processes" (link here):
An applied DC field $H_0$ is not completely uniform in all space points. If many spin ...
- 1,616
3
votes
Is Quantum Biocomputing ahead of us?
There has been a great deal of scientific debate over evidence of quantum effects in biology due to the difficulties of reproducing scientific evidence. Some have found evidence of quantum coherence ...
- 31
3
votes
Is Quantum Biocomputing ahead of us?
Much has been written about Quantum Biology. A somewhat old -and yet, solid- take is that of Phillip Ball, The dawn of Quantum Biology (Nature 2011, 474, 271-274). For now, let's not review that and ...
- 3,777
3
votes
Accepted
Fermionic occupation operator and nearest neighbor Fermionic hopping interaction as a qubit operator
The oldest and most commonly known way is the Jordan-Wigner transformation. The qubit operators will be $\mathcal{O}(N)$-local for $N$ occupiable orbitals.
A significantly more complicated way is the ...
- 13.5k
2
votes
Accepted
Fermionic commutation relation using Jordan-Wigner transformation
Based on my answer to this: Fermionic occupation operator and nearest neighbor Fermionic hopping interaction as a qubit operator, you can see that we have:
\begin{align}
\hat{a}_i &= \frac{1}{2} ...
- 13.5k
2
votes
Why does joint ground state not change under action of beam splitting unitary operator?
Let $|\psi\rangle$ be an eigenstate of an operator $A$, $A|\psi\rangle=\lambda|\psi\rangle$.
Then
$$e^A |\psi\rangle = \sum_{k=0}^\infty \frac{A^k}{k!}|\psi\rangle = \sum_{k=0}^\infty \frac{\lambda^k}{...
glS♦
- 23.4k
2
votes
If a Hamiltonian is quadratic in the ladder operator, why is its time evolution linear in the ladder operator?
Note that
$$[(a^\dagger)^n,a] = -n(a^{\dagger})^{n-1},
\qquad [(a^\dagger)^n a^m,a] = -n (a^\dagger)^{n-1}a^m,
\qquad [a^n,a]=0.$$
Consider an arbitrary function of the mode operators, that we assume ...
glS♦
- 23.4k
2
votes
Fermionic occupation operator and nearest neighbor Fermionic hopping interaction as a qubit operator
Use the Jordan-Wigner transformation. For a 1D chain with NN interaction it will yield a spin Hamiltonian with NN interaction (specifically, the hopping will map to a XX term and the on-site term to a ...
- 5,682
2
votes
Accepted
What is the most optimistic perspective of room-temperature solid-state QC?
Articles on technology from 10 years ago are often outdated, to some extent the same can be said of last year's information. Occasionally something will stand for decades, or fall into decline only to ...
- 2,287
2
votes
Translation of color/toric code to a small network of solid-state spins
I don't know the translation into physics, but the circuit you want for the most basic demonstration is the following:
Here, $|+\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$, and the gates are controlled-...
- 55.7k
1
vote
Do avoided crossings / CTs /ZEFOZs optimize quantum fidelity in practice?
The best I have it's this generic answer, which I put here for clarity, hoping for improvements/corrections or even to be superseded by something better:
If the limiting factor for fidelity in a ...
- 3,777
1
vote
Which temperature has been the highest to achieve a quantum logic operation?
I think your reference has the answer: nitrogen vacancy centers in diamond, where you can do one qubit gates at room temperature. In fact, even higher temperatures are possible, but you will have to ...
- 508
1
vote
Decoherence of spin-entangled triplet-pair states in the solid state: local vs delocalized vibrations
Let me go for a self-learner experience. After some reading, my short answer to my own question
Would the calculation of the loss of entanglement be necessarily related to delocalized vibrational ...
- 3,777
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