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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