Recent content by phyky

I want to clear numeric content(without clear others) in certain range with a button. example: $E$10:$H$26 and $K$11:$Z$15. What macro code should I have? Appreciate your help

|A>=a|0>+b|1> so <A|A>=a^2<0|0>+b^2<1|1>+a*b<0|1>+b*a<1|0> where <i|j>=0 <i|i>=1. where the interference term?

given two system is entangled, |A>=(|0_{A}>+|1_{A}>)/√2, |B>=(|0_B{}>+|1_B{}>)/√2. entangle state |AB>what is the probability to find |0_A{}0_B{}> and |1_A{}1_B{}>. are there still 1/2 just like normal inner product formulation?

so if transition from A to B it need |<B|A>|^2 to find interference term? and <A|A> show only probability but nothing about interference? but its outer product |A><A| does show interference?

|A>=a|0>+b|1> <A|A>=probability amp |<A|A>|^2=probability If the state is same, so <A|A>= a^2+b^2=1. then i find the probability and no interference term in it! so can i say <A|A>=|<A|A>|^2 and no square is needed? since if i square it become a^4+b^4=?? what wrong with it! and where...

Entanglement is inseparable state |ψ>=\Sigma\Sigmac(n1,n2)|n1n2> but for decoherence where system and environment coupling |ψ>(s,e)=\Sigmac|s>\otimes|e> which is normal tensor product, can we still say entanglement cause decoherece or just the tensor product between system and environment?

some inconsistency where since A and B is non commute, why we can compute <|AB|> or <|BA|>? so i think it is impossible from beginning because Heisenberg uncertainty forbid us to compute both A and B accurately.

conclusion is if AB is non commute. we can only compute 1 of the expectation value, either <|AB|> or <|BA|>?

is projection kinda same as the ψ "collapse" to 1 of the eigenstate \phi?

so we can only compute 1 of it? but how about <[A,B]>, we can compute it?

1.) an inner product of a state vector represent by <\psi|\psi>. sometimes the notation is like <\phi|\psi> is mean transfer from state |\psi> to <\phi|.it mean the former 1 do not transfer the state? what is the difference between both? 2.) what is mean by <x|\psi>? is it mean x(position)...

if 2 hermitian operator A, B is commute, then AB=BA, the expectation value <.|AB|.>=<.|BA|.>. how about if A and B is non commute operator? so we can not calculate the exp value <.|AB|.> or <.|BA|.>?

wave function always refer to the position of particle with probability amplitude. since wave function is not a wave but field, frequency seems meaningless to it. can i imagine it is a vibration field with certain frequency f in complex plane or something?

1. v=∇S/m is still discrete? i mean the motion of particle still quanta or continuous? 2. so v is the velocity of the particle(group) or wave(phase velocity)? 3. ∇S can be find by measure the velocity of wave or particle? 4. So the wave is not longer associated with particle (wave particle...

try see this answer, and i think it is reasonable. http://www.sciforums.com/showthread.php?73210-Single-slit-single-electron/page2