Recent content by Demystifier

Here it is tacitly assumed that ##x\gg n##. Under this approximation $$x(x-1)(x-2)....(x-(n-1)) \approx x^n$$

Not at all, it works for particles which are distinguishable. For example, one particle can be a hydrogen atom made of a proton and an electron, so that the total spin is zero. The other particle can be a helium atom, also with total spin zero. The full system is then a hydrogen atom and a...

Let me give a simple explanation how is symmetrization (or anti-symmetrization) due to indistinguishability of particles different from entanglement in a narrow sense. Consider the symmetrized state $$|\psi\rangle = \frac{1}{\sqrt{2}}[|A\rangle|B\rangle+|B\rangle|A\rangle]$$ If we think of it...

Not necessarily. https://en.wikipedia.org/wiki/Stigler%27s_law_of_eponymy

Another paper appeared today arguing that it's just postselection. https://arxiv.org/abs/2509.03127

Exactly! You should learn it first.

Which part was new to you? :oldbiggrin:

I learned that $$|e^{i\pi}|=|\pi^{ie}|=|i^{e\pi}|=1$$

With standard definition any two photons are entangled (because they are bosons so the wave function is always symmetrized), but it is not the case that you can use any two photons to violate Bell inequalities. In real experiments, the photons that violate Bell inequalities are prepared in a...

Yes, but for indistinguishable particles one can define a more narrow notion of entanglement; the operator acting on the vacuum that creates the joint state can't be factorized into a product of single-particle creation operators. Let me illustrate this by an example. Let ##a^{\dagger}(k)## be...

The paper says that photon paths are indistinguishable. It's not the same as indistinguishability of photons themselves, but they define photon states with a second-quantized formalism, which implies that photons are indistinguishable too, even if they don't say it explicitly. That being said...

Also note that even the QED vacuum, i.e., a state with zero number of photons, is highly entangled. Hence it is really impossible to have a truly non-entangled state when the notion of "entanglement" is understood in a more general sense.

The effect is explained in the paper by quantum indistinguishability. But indistinguishability in QM is implemented by symmetrization of the wave function, and mathematically symmetrization looks just as a special kind of entanglement of particles that otherwise would be distinguishable. In...

Iskandarani, I would suggest you to first study something similar but much simpler and much better understood. You can contrast the "vacuum" in classical electrodynamics with vacuum in quantum electrodynamics. The former is any configuration of EM fields for which the charge density is zero...