Hot answers tagged

4

In the quantum mechanical description of any physical system, including a quantum field or a collection of interacting quantum fields, there is always one state vector – one collection of numbers (probability amplitudes) that generalizes what is referred to as the "wave function" in quantum mechanics of particles. In quantum field theory, a better name is a ...


4

@GerryHarp's answer contains the gist of the main idea, but there is a point that doesn't quite make sense: There is no such thing as bare ions crystal in solid state physics. To answer your last question: The "bare phonon frequencies" definitely do not refer to phonon frequencies in the absence of electrons. In fact, an ion crystal without electrons is ...


4

Lets start with the bare ions. You set up a linear lattice of protons where the boundaries are held in place by some means. The protons want to get as far apart from each other as possible. So the minimum energy state has the protons spaced on the line with equal distances between them. So yes, the ions form a regular lattice without the electrons. Now ...


3

Because we have observed processes where the photon number is not conserved. For example positronium can decay into 2 or 3 (or more) photons. This means that it is not possible to assign a global conserved charge to photons. For neutrinos we can assign lepton number and so far we have not observed a process that would violate total lepton number (lepton ...


2

Quick answer My question is how does the presence of nonzero $J(x)$ results in a non-trivial spacetime dependent value of $\langle 0|\phi(x)|0\rangle$? The equation $\phi(x)=\mathrm e^{-iPx}\phi(0)\mathrm e^{iPx}$ works both for $J=0$ and $J\neq 0$. Therefore, $$ \langle \phi(x)\rangle_J= {}_J\langle 0|\mathrm e^{-iPx}\phi(0)\mathrm e^{iPx}|0\rangle_J ...


2

You're right that the Feynman propagator for a spinor field is indeed $(i \gamma^\mu D_\mu - m)^{-1}$. The tricky part is interpreting exactly what the "inverse" means. It doesn't just mean that you invert the gamma matrices (although you do do that). The derivative operator is also being inverted. That is, if we define a function $G(y, x) := (i ...


2

Charged particles can't have Majorana masses of any type because they would violate the charge conservation law. The Majorana mass is really a term that is converting a particle into its antiparticle. It implies that the particle must be considered "physically indistinguishable" from its antiparticle. The Majorana mass term violates the lepton number or its ...


1

I am answering a question after some clarifications. If the gauge field maps the electron to a positron, it really connects the left-handed 2-spinor and right-handed 2-spinor in the Dirac's electron field into one multiplet. But that field is also a part of the $SU(2)_W$ doublet with the neutrinos. So the theory you are proposing wants to extend the ...


1

The eigenstates of a time-independent Hamiltonian $$ H |\psi_j \rangle = E_j|\psi_j\rangle$$ have the usual rotating-phase time dependence in the Schrödinger picture: $$ |\psi_j(t)\rangle = |\psi_j(t_0)\rangle \cdot \exp(E_j(t-t_0)/i\hbar) $$ However, your formula indicates that $H_0$ and $V$ are time-dependent. So if the state vector is an eigenvector of ...


1

Comments to the question (v3): The main fields assumption that goes into the proof of the Wick's theorem for fields $\hat{\phi}^i\in{\cal A}$ is that their (super)commutators $$[\hat{\phi}^i,\hat{\phi}^j]~\in~ Z({\cal A}) \tag{1}$$ are central elements of the operator algebra ${\cal A}$, cf. e.g. this Phys.SE post. For free fields $\hat{\phi}^i\in{\cal ...


1

The small number of "conceptually independent types of processes and calculations" is exactly a symptom of the theory's being fundamental! Even in classical physics, all calculations could have been mathematically reduced to the calculation of the final state that evolves from an initial state (or a state that is stationary etc.). In quantum mechanics, this ...


1

The unitary operator $U$ is not a real number, so $|U|=1$, whatever $|U|$ exactly means, does not imply $U\in \{+1,-1\}$ in any way. The unitary operator isn't even a complex number. It's more complicated than that. Moreover, time ordering isn't a "unitary operator". It isn't even an operator in the same sense as unitary operators in quantum mechanics. It ...


1

Almost everything from the wikipedia page you link is just false, or at best very misleading. IMHO, that page was written by someone that doesn't know anything about quantum mechanics beyond what one could find in TV documentaries. "Not even wrong" came into my mind many times as I was reading the article. In quantum physics, a quantum fluctuation (or ...



Only top voted, non community-wiki answers of a minimum length are eligible