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 Dec 18 answered Wigner's friend and quantum Zeno effect Dec 18 comment Schrodinger's equation (explanation to non physicist) @Temitope.A: Entanglement isn't obvious in anything here because I've only written single-particle wavefunctions. A two-particle wavefunction $\Psi(\vec r_1, \vec r_2)$ gives a probability $\int_{V_1}\int_{V_2}|\Psi|^2 d \vec r_1 d \vec r_2$ of detecting one particle in a region $V_1$ and a second particle in a region $V_2$. A simple solution for distinguishable particles is $\Psi(\vec r_1, \vec r_2) = \psi_1(\vec r_1) \psi_2(\vec r_2)$, and it can be shown that this satisfies all our conditions. An entangled state cannot be written so simply. (Indistinguishable particles take more care.) Dec 18 revised Schrodinger's equation (explanation to non physicist) added 1 characters in body Dec 18 comment Is a hard drive heavier when it is full? Ron, I believe you're correct, and so my answer might be misleading. I do wonder whether information entropy can be thought of as a source of mass: to order bits, we must do work by removing entropy. Such an explanation would have an order of magnitude similar to my answer. Dec 18 comment Amplitude of an electromagnetic wave containing a single photon Anna V: the quantum electromagnetic field has operators analogous to the harmonic oscillator. Instead of the Hamiltonian $H = p^2 + x^2$ (I'm dropping the units here) and energy $(\frac{1}{2} + n) \hbar \omega$, we write a Hamiltonian for each frequency $\omega$: $H = \frac{\epsilon_0}{2} E^2 + \frac{1}{2\mu_0} B^2$, where $E$ and $B$ are treated as conjugate quantum operators (just like $x$ and $p$). We find creation and annihilation operators such that $E = a^\dagger + a$ and $B = a^\dagger - a$, so the energy must be $(\frac{1}{2} + n) \hbar \omega$. $\langle E \rangle$ is well-defined. Dec 18 awarded Editor Dec 18 revised Amplitude of an electromagnetic wave containing a single photon changed "factor of 2" to "factor of 1/2" Dec 18 comment Schrodinger's equation (explanation to non physicist) One type of measurement is strong measurement, where we the experimentalists, measure some differential operator $A$, and find some particular (real) number $a_i$, which is one of the eigenvalues of $A$. (Important detail: for $A$ to be measureable, it must have all real eigenvalues.) Then, we know the wavefunction "suddenly" turns into $\psi_i$, which is the eigenfunction of $A$ whose eigenvalue was that number $a_i$ we measured. The system has lost of knowledge of the original wavefunction $\psi$. The probability of measuring $a_i$ is $|<\psi_i | \psi>|^2$. Dec 18 answered Amplitude of an electromagnetic wave containing a single photon Dec 16 answered Schrodinger's equation (explanation to non physicist) Oct 7 answered Why are some materials diamagnetic, others paramagnetic, and others ferromagnetic? Jul 21 answered The most price-efficient experimental setup involving SPDC, single-photon counting etc Jul 21 awarded Supporter Jul 10 answered Is a hard drive heavier when it is full? Jul 6 awarded Teacher Jul 6 answered Misconception about the expectation of a quantum system