The electronic structure of CNT is discussed on the basis of band structure of
graphene. Graphene has a linear dispersion relation:
E = h_cut vF |k|
where k is the 2D wavevector and vF is the Fermi velocity. CNTs are
macroscopic along the axis but have a circumference of atomic dimensions, which...
Okay, so I read the following:
"Single Photons as Qubits
Single photons are largely free of the noise or
decoherence that plagues other systems..."
reference: http://arxiv.org/abs/0803.1554
I am trying to come up with reasons to justify the decoherence point.
I completely understand that coherence is experienced by a collective state only. I misunderstood what mfb had written.
Please just confirm if the following claim of mine is correct or not: "Photons do not interact much with the matter because they do not have color, flavor and charge. The...
secondly, why doesn't the interaction of photons with vacuum fluctuations "... lead to decoherence..."?
Also, in the presence of charge, there is vacuum polarization. Do these vacuum fluctuations happen only in the presence of charged particles?
You don't have to scream at me. I was considering photonS.
I was confused about "Decoherence is not a property of particles". Electrons are elementary particles. They interact with each other and lose information. Photons are particles too, no?
I read it some where that there is very small decoherence for photons. The reason being that photons do not interact with each other (Is that because photons are chargeless, colorless and flavourless particles?) and hence the information that they contain tends to stay with them. They have a...
Simplification -- complicated summation involving delta functions
Homework Statement
\frac{1}{\sqrt{(2^3)}}\sum[δ(k+1)+δ(k-1)]|k> for k=0 to 7
Homework Equations
The Attempt at a Solution
I am trying to simplify the above expression. I get \frac{1}{∏*\sqrt{(2^3)}} |1>, which is...
Homework Statement
define a function f:H--> gHg^{-1}
Homework Equations
prove if f is 1-1 and onto.The Attempt at a Solution
1-1:
f(h1)=f(h2)
gh1g^{-1}=gh2g^{-1}
h1=h2 (left and right cancellations)
onto:
f(g^{-1}hg)=gg^{-1}hgg^{-1}=h
so every h belonging to H has an image of g^{-1}hg...
The string S cannot be called a cycle. Its not 1-1, hence it can't have an inverse. Although I understand what you have stated but I think giving the example i.e the string S, would have been enough :smile:
Thank you!
You answered that in your last to last reply. The proof is constructed this way that i pick first element.
What I was thinking was what if some other element, lying in the middle somewhere, got repeated, but then we can write a representation such that this element becomes the first element...
But why should x_k be the first element? when there can be different representations for the same cycle eg {1,5,7}={7,1,5}, we can't really fix the first element, or can we?
When you say "... your sequence has to have a repeated element just because the number of choices is finite", could you please give an example of it? Isn't the sequence {1,2,3} finite despite there being no repetition?