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Jun
26
comment What operation does a beam splitter apply?
@Strilanc You can't do CNOTs with linear optics.
Jun
26
comment What operation does a beam splitter apply?
@Strilanc I think you should first make clear to yourself and to us what these matrices mean. Otherwise, there is no way we can agree on the difference between those two matrices. -- My interpretation is that you are talking about building a linear optical quantum computer with a dual rail encoding of a qubit (i.e., one photon in one of two paths), you insert a beam splitter, and you want to know the gate action in the quantum computer. -- But unless you clearly explain what these matrices describe these are just wild guesses.
Jun
25
comment What operation does a beam splitter apply?
"unitary matrix equivalent to the operation of a beam splitter": On what? On the creation operators? -- Other than that, both matrices are equivalent up to a multiplication from left and right with $\mathrm{diag}(1,-i)$, which depends on how you define your modes (or whatever else your matrices act on).
Jun
24
comment Entangled Photon (laser pointer)
Of course you can entangle more than two particles: en.wikipedia.org/wiki/Multipartite_entanglement
Jun
24
comment Why do Transformations on Qubits for Quantum Computation have to be Unitary?
@ThomasElliot Also note that irreversibility means that information is lost to the environment, i.e., the system becomes less quantum, which is not what we want for a quantum computer.
Jun
24
comment Why do Transformations on Qubits for Quantum Computation have to be Unitary?
@ThomasElliot The other model is not commonly used, because the unitary model is easier. But as I said: You don't need to assume unitarity, but any circuit model defined with physical gates is equivalent to it, as it can be simulated using unitaries (known as the Stinespring dilation).
Jun
24
answered Why do Transformations on Qubits for Quantum Computation have to be Unitary?
Jun
23
comment What is the condition for local operations on bipartite entangled state?
@sasha Exactly, $M_A=m_A\otimes I$ and $M_B=I\otimes m_B$. It is however customary (though sloppy) to use $M_A$ both for the operator on $\mathcal H_A$ and $\mathcal H_A\otimes \mathcal H_B$.
Jun
23
reviewed No Action Needed Question about kinetic energy
Jun
23
reviewed No Action Needed What will happen if 0°C water is slowly added into a vacuum container adiabatically/isothermally?
Jun
23
answered What is the condition for local operations on bipartite entangled state?
Jun
23
comment What is the condition for local operations on bipartite entangled state?
Ok, I'll make a short answer out of it.
Jun
23
comment What is the condition for local operations on bipartite entangled state?
Locality of $M_A$ and $N_B$ certainly implies that $[M_A,N_B]=0$, so the latter is a valid property which can be used to derive properties of local measurements. Does this answer your question?
Jun
23
reviewed No Action Needed Wick's Theorem: Why is the vacuum expectation value of uncontracted operators zero?
Jun
23
comment What is the condition for local operations on bipartite entangled state?
Where exactly? You are not being very specific.
Jun
23
comment What is the condition for local operations on bipartite entangled state?
The condition that $E_A$ and $E_B$ commute is a natural generalization of locality, but in general not the same. However, one can indeed do resource theory based on such a commutation relation. What paper are you reading?
Jun
22
comment Equation 2.27 from Pachos's introduction to topological quantum computing
Unitaries can always be taken out of exponentials, i.e. $\exp(UAU^\dagger)=U\exp(A)U^\dagger$. Is this what you are asking about?
Jun
21
reviewed No Action Needed When to use Kelvin over Rankine and vice versa
Jun
21
reviewed Reviewed Is it possible to have infinite combinations in reality?
Jun
21
reviewed No Action Needed Why can we set mass to zero in Yukawa RGE derivation?