New answers tagged

3

If the nucleus is at position $A$ then classically it will have electric field configuration $E_A$. If the nucleus is at position $B$ then classically it will have electric field configuration $E_B$. Quantum mechanically, if the nucleus is in a superposition of positions $A$ and $B$ then the electric field will be in a superposition of field configurations $...


2

From the point of view of Wick's theorem, this should be obvious: If you can express any $N$-point correlator of operators acting on part A of the system through the corresponding two-point correlators, then this does not depend on the fact whether you trace the other part B of the system or not - that's precisely the point of the partial trace, it describes ...


1

If not, can one deduce that DR is "physically" (and not mathematically!) unbeatable regularisation method? Why do you say that dimensional regularisation is 'not mathematically [an] unbeatable regularisation method? The Connes-Kreimer renormalisation theory makes mathematical sense out of dimensional regularisation. The theory is geometric and is ...


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The most unphysical feature of DR is its infrared behavior for the running of the mass parameter, while the electron mass is infrared fixed in reality!


1

You only get everything form the two point correlator when the wavefunction can be written as a single Slater determinant, or equivalently as a single tensor product of one-particle states. The general state is a sum of many tensor product states. The conditions that an $n$-fermion state can be written a single tensor product are the Plucker relations.


2

No - coherent states form an (overcomplete) basis for the (infinite-dimensional) Hilbert space, and otherwise, any state could be characterized by its second moments. More generally, there is a 1-to-1 correspondence between physically admissible 2-point correlation matrices and Gaussian states. Thus, for any non-Gaussian state which can be characterized from ...


0

In the spirit of Fermi golden rule, transition rate (probability per time unit) is proportional to the matrix element of the transition and density of states in the final state. What you wrote is indeed true for the matrix element, but spontaneous decay rate is accompanied by the emission of a photon. In other words, it is not just you have to be able to ...


3

Phrases like "electrons jump between orbitals" or "electron doing a transition" are misleading in my opinion. The physical objects that matter are electronic states. An electronic state of a system with multiple interacting electrons, always describes all electrons at once. The next thing that is often misused are orbitals. Orbitals are ...


1

No. Just take two states with opposite 2-point correlators (like the all-zero and the all-one state). ("Opposite" in a suitable normalization.) Then, their sum has correlators zero (again, zero in a suitable normalization - but in any normalization, the correlation matrix of the maximally mixed state). Yet, it is not the maximally mixed state (...


0

You just blew the sign! Isolate the non-vanishing terms of your $$ p_{1\lambda}p_{3\sigma}\text{Tr}[\gamma^{\mu}(1-\gamma^5)(\gamma^{\lambda}+m_e)\gamma^{\nu}(1-\gamma^5)\gamma^{\sigma}] $$ linear in $\gamma^5$, $$ 2p_{1\lambda}p_{3\sigma}\text{Tr}[\gamma^5\gamma^{\mu} \gamma^{\lambda} \gamma^{\nu} \gamma^{\sigma}]\\ = 8i p_{1\lambda}p_{3\sigma} \epsilon^{\...


4

This is a general property of transitions $-$ in the Hartree-Fock formalism. If the eigenstates of both the initial and the final states of the transitions are (i) given by single Slater determinants which are (ii) made out of single-particle orbitals taken from the same basis set, then the selection rule you have derived is exact. As it happens, this ...


0

I actually believe that you are right. If the universe, described by a density matrix $\rho$ is stationary (time-translation is a symmetry of the system), then we know that $[H,\rho]=0$ or $H$ and $\rho$ are diagonal in the same basis. We know have that $\rho$ is a classical mixture of eigenstates. This can easily considered a Bolzmann-Gibbs distribution or ...


3

Dipole approximation Interaction between electrons and the electromagnetic field is a one-particle operator. The dipole approximation has to do with approximating the exact form of the electron-field coupling: first replacing the potential term by a dipole-like interaction (which is actually exact when the field is a plane wave) $$ V(\mathbf{r},t)\...


0

For a large interacting system, the energy eigenstates will be extremely bizarre states with lots of non local features. In real life, all the complexity of time evolution comes from the fact that the state itself will be a superposition of energy eigenstates.


0

The kets in QFT belong to the Fock space over the Hilbert space of single particle states in QM. In the bosonic case, this is the symmetric algebra over the Hilbert space and in the fermionic case, this is the anti-symmetric algebra.


3

First, let us consider the situation described by the OP: an atom and a EM mode that are coupled and exchanging energy (as described, for example, by the Jaynes-Cummings model). As the OP says, once you take into account both the energy in the atom and in the field the total energy is conserved. However, this does not make it any closer to an eigenstate ...


4

I start directly from the problematic commutator. You should be note that I replace $\vec{x}'\rightarrow \vec{y}$ for better readability. $$ I = \int \,d^3y\: \phi(\vec{y},t) \left[\pi(\vec{x},t) , \nabla^2_{\vec{y}} \phi(\vec{y}) \right] $$ I use the fact that the laplacian is acting on $\vec{y}$ to commute it freely with $\pi(\vec{x},t)$. $$ I = \int \,d^...


1

As was suggested by user296595, Breuer & Petruccione book chapter 3.3 Microscopic derivation is what you need. I suppose, you are interested whether a harmonic oscillator, if interacting with a bath, will remain harmonic. Here is the general summary of relevant formulas and definitions with application to a harmonic oscillator at the end. Looking at the ...


1

You need to think carefully about what space your quantities act on. $\phi$ is a quantum field (operator) acting on Hilbert space. However, by itself, $T^a$ is just an element of a Lie algebra in some representation. It acts on the appropriate representation space, not on Hilbert space. Therefore, the notation $T^a \lvert 0 \rangle$ is sloppy. If you go back ...


2

You may check "the theory of open quantum systems" by Prof. Petruccione, especially section 3.3 microscopic derivations. When a system is coupled to an environment, the dynamics of the total system+environment state will be governed by Schrödinger equation with the total Hamiltonian including system Hamiltonian $H_s$, environment Hamiltonian $H_e$, ...


2

It seems like you're confused about the structure of quantum mechanics. When we say that X is conserved in quantum mechanics, we just mean that the operator X commutes with the hamiltonian. It doesn't mean that we can't have superpositions of different eigenstates of X. It is also in general not meaningful operationally to ask whether a system is in a ...


1

I have a distinct feeling you chose a huge number of interacting particles so you won't have to model them in your mind. Choose just two states, and consider a quantum flipflop, describable by a two-vector, $|\psi(t)\rangle$. In the most general unitary evolution, your energy is time-independent, $$ \langle \psi(t)|H|\psi(t)\rangle = \langle \psi(0)| e^{...


0

For clarity one might avoid using the words "right" and "left" altogether, and talk about positive and negative chirality; positive and negative helicity. Helicity is +ve if $\sigma \cdot p > 0$ and negative if $\sigma \cdot p < 0$. For a Weyl spinor the chirality has one sign if the spinor matrix elements $\langle s| \sigma^a | s \...


4

Background Here we work in the Euclidean theory throughout. I also preface this with a disclaimer that I have been a bit lax with indices, but hopefully the message remains clear. The Ward identities in their broadest form possible are actually a statement about any infinitesimal shift in the fields, not necessarily symmetries of the action or otherwise: $$ \...


0

The integral $ \int dx$ is the volume $VT$ of space-time, $V$ being to the volume of the system and $T$ the time that one considers it for. The bubble diagram contributes to the vacuum-vacuum amplitude $$ \langle 0, {\rm out}|0 {\rm, in}\rangle =e^{-iW}= e^{-iTV{\mathcal E}} $$ where $\mathcal E$ is the ground state (vacuum) energy density. Obviously $V$ and ...


0

Scattering amplitudes, which are ultimately used to compute cross sections, are written in terms of asymptotic in and out states. The overlap of these asymptotic in and out states can be written in terms of the VEV of n point functions using LSZ, where here the n point function is in terms of the fully interacting fields in the Heisenberg Picture and the ...


0

Regarding the lack of an overall energy-momentum conserving delta function and the integral over all space (which contributes an infinite factor), you can see where this comes from when first considering the position-space diagram (I won't do this in full detail with all the factors, but you can go through it yourself): we have the usual integral over $\int ...


0

For a particle, the helicity and chirality are identical while for an antiparticle, the helicity and chirality are opposite. Why is it so?


1

QFT is difficult, unfortunately. Usually in a 1st pass in QFT, one is interested in computing different cross sections $d\sigma$. These differential cross sections can be related to the square of the scattering amplitude $\mathcal M$. For a scattering event of $A+B\rightarrow 1\ldots m$ distinguishable particles in $n$ spatial dimensions one has $$d\sigma=...


0

Lorentz indices are indeed raised and lowered with the metric tensor. But note that in QFT there are also other indices than Lorentz indices and these are raised an lowered with the appropriate tensor.


2

It's easiest to answer this in terms of conserved densities and currents. A general local conservation law has the form $$\frac{\partial \rho}{\partial t}=-\nabla\cdot\mathbf{J}$$ For the Schrodinger equation, multiplying through by the complex conjugate, we have $$\psi^\star i\hbar\partial_t\psi=-\frac{\hbar^2}{2m}\psi^\star\nabla^2\psi+V(x)\psi^\star\psi$$ ...


4

The GK "norm" $$ \langle \phi|\phi\rangle = \int d^3 x (\phi^*\partial_t \phi-(\partial_t \phi^*)\phi) $$ is time independent for solutions of KG equation, but may be negative. So whatever it is, it is not a probability. It is not difficult to show that the Schroedinger norm $$ \langle \psi|\psi\rangle = \int d^3x |\psi|^2 $$ is time independent ...


3

This answer will confirm that QFT does propose that quantum fields can be in configurations that break the energy bounds believed to hold for ordinary matter, such as the weak energy condition and dominant energy condition. I think it is misleading, however, to use the terminology "negative mass" for this situation, at least in many examples, as I ...


2

There are a number of reasons why the single particle theory of QM can't be extended into a quantum theory of fields and this is another one. Generally speaking we have to move into a picture where particles can be created and annihilated. This is the multi-particle picture. Freeman Dyson discusses some of the reasons in his Advanced QM. The reason why there ...


0

May be you should start with the paper arXiv:hep-th/0607177v2 : "Renormalizable Expansion for Nonrenormalizable Theories: I. Scalar Higher Dimensional Theories", by D.I. Kazakov, G.S. Vartanov, and look the references therein.


3

OP, you are exactly right. This statement in Schwartz's book is completely wrong, and for exactly the reason you mention. I am now going to do something naughty and paste my answer from a previous question because I think it applies here. Here I go. ... It seems to me that you got derailed by a rather unhelpful discussion in Schwartz. There's a lot going on ...


5

Consider any Lorentz-symmetric quantum field theory that has a single-particle sector. As a specific example, consider the Proca model. This is the QFT for a non-interacting quantum relativistic massive vector field. It is arguably the most straightforward example of a QFT after the free scalar model. It has a single-particle sector which is self-contained ...


3

The terms $K$ and $W$ are not kinetic and potential terms, rather $K$ is a K"ahler potential and $W$ is a superpotential. Neither term enters the Lagrangian directly, but they are used to construct it. For details see, for example, Cyril Closset's lecture notes on supersymmetry.


0

Hamiltonians are popular in non-relativistic formulations, where the fact the they are not relativistic invariants objects is not important. Lagrangians are popular in particle physics because they are Relativistic Invariants. This only facilitates everything because you do not have to be concerned about what is the new form of the Lagrangian when you are ...


1

The following may not necessarily be true $$ \langle c^{\dagger}_ic_j\rangle=\langle c^{\dagger}_jc_i\rangle $$ Actually, there is a whole branch of condensed matter models that depend on $$ \langle c^{\dagger}_ic_j\rangle \neq \langle c^{\dagger}_jc_i\rangle $$ meaning $$ \langle c^{\dagger}_ic_j\rangle \sim e^{\theta_{ij}i} $$ and $$ \langle c^{\dagger}...


0

To avoid the problem of questions answered only in the comments, I will post the following answer as Community wiki. Richard Myers helpfully points out that my question would be answered in most introductory QED texts including Peskin & Shroeder ch 4/5, Zee II.5 and II.6, Weinberg ch 8, Nair ch6/7, David Tong's notes ch 6.


3

OP seems to basically have this but is missing a complex conjugation somewhere. We want $(C^\dagger)_{ij}=C^*_{ji}$ that is: $$(C^{\dagger})_{ij} = \langle c_j^\dagger c_i \rangle^*$$ the inner product simply returns a number and so complex conjugating is equivalent to taking the adjoint of the operator in the middle: $$(C^\dagger)_{ij} = \langle (c_j^\...


2

The answer by @Accidental FourierTransform has already mentioned the instantons, but let me give a more pedestrian example and a more pedestrian view on the question. When we use a path integral to describe the motion of a single quantum particle, the saddle point approximation corresponds to what is otherwise known as quasiclassical approximation, and the ...


2

As a corollary, two Lagrangians that look vastly different from each other, producing vastly different classical equations of motion, can actually describe the same physics, if the fields in the two Lagrangians are related to each other by a field redefinition. Field redefinitions, even though they change the action (unobservable), leave the S-matrix (...


10

Yes, there are Lagrangians which have the same equations of motion which do not define the same quantum field theory. A standard example is the Yang-Mills action with a $\theta$-term: $S = \int\frac{1}{4} tr[F_A\wedge*F_A] + \theta \int tr[F_A \wedge F_A]$ The second term is a topological invariant, so contributes nothing to the equations of motion. But the ...


0

The probabilities have to do with the OUTCOME of the interaction. In general, an interaction can lead to plenty of results-each of which has a certain probability associated with it. That is, if the same interaction process was carried out, with similar external conditions, a large number of times, there'd be a different result each time, and the set of ...


2

Note that the LHS of eq. (5.56) is an operator. This implies that there is an implicitly written identity operator ${\bf 1}$ on the RHS of eq. (5.56). By sandwiching eq. (5.56) with the vacuum state, we only get $$\langle 0 | :\psi(x)\bar\psi(y):|0\rangle~=0;$$ not OP's last equation. There is an analogous situation for bosons. For more details, see e.g. ...


1

It's not true $:\psi(x)\bar\psi(y):=0$, but it is true that $\langle0|:\psi(x)\bar\psi(y):|0\rangle=0$. I just looked up Tong's notes and it looks like he works out a specific example in detail for the bosonic case at eq 3.33 that may be helpful to you.


5

I don't think this is true, $\pi_5(G)$ has little to do with anomalies, at least not in any direct way. The general statement is: triangle anomalies for a given symmetry $G$, in $d$ dimensions, are classified by the free part of $\Omega_{d+2}(G)$. Here $\Omega_n(G)$ denotes the cobordism group of $G$, namely the collection of $n$ dimensional manifolds $M_n$ ...


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