About the electromagnetic 4-vector ##A^\mu=(\Phi/c,\mathbf{A})##. If it is indeed a four-vector, then its squared length ##A_\mu A^\mu=\Phi^2/c^2-\mathbf{A}^2## should be a Lorentz invariant. What is the physical significance of ##\Phi^2/c^2-\mathbf{A}^2##?
Thanks for any help.
OK, I have ##|\psi\rangle=\displaystyle\sum_{j=0}^{2^n-1}a_j|j\rangle## with a peak at ##j=r##, i.e., ##|a_r|^2\approx 1## and ##|a_{j\neq r}|^2\approx 0##. I take the observable ##\hat{O}=\displaystyle\sum_{k=0}^{2^n-1}k|k\rangle\langle k|## and I make the measurement...
In quantum computing, there are algorithms such as the quantum Fourier transform or Grover's algorithm that ends up with a superposition ##\sum_j a_j|j\rangle## with one amplitude ##a_x\sim 1## and the other amplitudes ##\sim 0##. I am looking for a way to recover the ##x##...
Let me rephrase the question another way:
I have a wave function resulting from a quantum Fourier transform. This wave function is a superposition with one of the amplitudes very high (because it corresponds to the period). How do I recover the period?
Let be a superposition ##|\psi\rangle=\sum_j a_j|j\rangle## with one amplitude ##a_x## much greater than the others, where ##x## is not known. For example, ##|\psi\rangle## may result from the quantum Fourier transform of a periodic wave function with an unknown period. I expect a measurement of...
##P_{AB}\equiv P(\theta_A,\theta_B)=\frac{1}{2}\cos^2(\theta_A-\theta_B)##, right from Born's rule, is the probability that both Alice and Bob observe a photon. The probability that Alice's photon is absorbed and not Bob's is ##P_{\bar{A}B}=P(\theta_A+90°,\theta_B)##. One can check that...
The experiments to test the Bell inequalities such as the Aspect's one are explained using the XIX century Malus law of classical optics. I think, with entangled photons, using Born's rule is much better, even if it happens to give the same result.
"Then you just have to do the algebra and calculate ##P=<\psi|\hat{P}_A\otimes\hat{P}_B|\psi>##"
Let me do it:
##P=(\cos\psi<HH|+\sin\psi<VV|)|\theta_A><\theta_A|\otimes|\theta_B><\theta_B|(\cos\psi|HH>+\sin\psi|VV>)##
##\hphantom{P}=(\cos\psi<H|\theta_A><H|+\sin\psi<V|...
Thank you for the answers.
I note ##|HH>\equiv|H>\otimes|H>##, the same with V. With this notation, the wave function of the entangled pair is ##|\psi>=\cos\psi|HH>+\sin\psi|VV>##. The projectors of Alice and Bob are, respectively ##\hat{P}_A=|\theta_A><\theta_A|## and...
Let be 2 non-entangled photons ##|\phi_i>=\cos\phi_i|H>+\sin\psi_i|V>## with i=1 or 2. The wave function of the 2 photons is then
##|\psi_1\psi_2>=\cos\psi_1\cos\phi_2|HH>+\cos\psi_1\sin\psi_2|HV>+\sin\psi_1\cos\psi_2|VH>+\sin\psi_1\sin\psi_2|VV>##
Alice makes the wave function interact with a...
Let be an entangled pair of photons 1 and 2, with the same polarization. The wave function is
##|12>=\cos\psi|HH>+\sin\psi|VV>## with ##\psi## the angle of polarization. The first ##H## (##V##) in ##|HH>##
(##|VV>##) is photon 1, and the second one is photon 2.
Alice observes photon 1 with a...