New answers tagged

0

I've solved it, in case anyone wants to see the solution. The problem is that the expression $\omega_n$ arises from: $$\omega_n^2 = c^2 k_n^2$$ with $$k_n = \frac{2 \pi }{a}\left(n+\frac{1}{2}\right), n= 0, \pm 1, \pm 2...$$ so the expression I wrote for $\omega_n$ has an absolute value. This means that when writing $E_0 (a)$ beginning with $0$: $$E_0(a) =\...


2

… why does one simply add the two actions and not add a coupling action? One does add coupling action. When we replace the partial derivatives that are present in the matter action in flat spacetime in Cartesian coordinates with covariant derivatives and integrate using volume form of the curved spacetime (this approach is called minimal coupling), this is ...


0

This is really just a pedantic detail, but only forces between monopoles fall off as $1/r^2$. For example the force between two electric dipoles falls as $1/r^3$, the force between two electric quadrupoles as $1/r^4$ and so on. This detail aside, the obvious example of a force that does not obey the inverse square law is the strong force. The reason this ...


1

Maybe I can say something about the relationship with Wick Rotation and temperature which might help. Ultimately, given a quantum system described by a Hamiltonian $\hat{H}$, suppose that you want to describe the system at a temperature $T$. Then, the partition function should be given by: $$Z = \text{tr} \left[ e^{-T\hat{H}} \right] = \sum_n \langle n | e^{-...


0

In units $c=\hbar=1$, it's clear the expression you found that the one presented in the book differ by an overall factor of $m$ (and a 2). In principle there's nothing wrong with this as both tensors will still be conserved and this difference can be considered a choice in normalization of the fields and/or of the stress tensor's definition. However I will ...


0

The concrete answer is actually in this post: Beta-function non-zero at classical level? In short, the $\beta$-function for the mass is just: $$\beta(m^2) = -2 m^2. \tag{1}$$ I am sure there are many ways to see that, e.g. the one described in the post. Here is another (trivial) way to check it. The renormalization group equation reads: $$\left\lbrace p \...


1

We can identify two cases here. Case 1: The gauge potential has a singularity which produces a singularity in the physical field too. Given that the physical field is invariant under gauge transformations, it follows that you won't be able to cure this by making a gauge transformation. We can make a trivial example of this by considering a $U(1)$ theory in ...


1

The question is rather old but maybe this answer is yet useful to someone. It all boils down to the rotation group. The thing is: a field is defined by how it transforms according to the universal cover of the Lorentz group ${\rm SL}(2,\mathbb{C})$. Now in the same way that the Lorentz group contains rotations its universal cover ${\rm SL}(2,\mathbb{C})$ ...


1

This is a bit late, but I hope it is helpful to people in the future: We know that the vector representation or the Lorentz group is $\left(\frac{1}{2},\frac{1}{2}\right)$, where I am labelling irreps by their spin, instead of their dimension.$^1$ We then know a general second rank Lorentz tensor must live by definition in the representation: $\left(\frac{1}{...


1

You did everything right. You're only lacking the correct interpretation. The momentum-space wave packet is (Eq (5.9) in my version of Srednicki) $$ f_1(\mathbf k) \propto \exp(-(\mathbf k- \mathbf k_1)^2/4\sigma^2) .$$ You found the position-space wave packet $$ \tilde f_1(\mathbf x) \propto \exp(-\sigma^2\mathbf x^2-i\mathbf k_1\mathbf x)$$ by Fourier ...


1

The basic idea underlying what's going on in that construction is the mathematical fact that the Fourier transform of a Gaussian centered around zero is also a Gaussian centered around zero. But what about a Gaussian centered around some other value? In the comments you wrote: Why a Gaussian at zero would have its mode expansion dominated by low-frequency (...


4

Those equations you write are not going to help you. The fields do satisfy them, however, as the comments hinted at, the commutator you are looking at comes from the canonical field quantisation of electromagnetism (QED), pedagogically derived for example here. Start with the gauge field (vector field here) $\mathbf{A}$, and find the canonical momentum from ...


0

I tend to believe that there is a mistake in the question. May you check ? Since you have a factor 1/2 in front of the mass term, you are dealing with the complex case of field. So in this case, how could you have a factor 1/2 in front of the kinematic term ? Also, your sign seems in the wrong direction. Have you really seen a book that contained your ...


1

Yes. The generalized positions $q^j$ are replaced with fields $\phi^{\alpha}({\bf x})$; Hamilton's principal function $S(q,P,t)$ is replaced with a functional $S[\phi,\Pi,t]$; while the Hamilton-Jacobi equation $$ H(q,\frac{\partial S}{\partial q},t) +\frac{\partial S}{\partial t} ~=~0$$ is replaced with $$ H(\phi,\frac{\delta S}{\delta \phi},t) +\frac{\...


3

The closest analogue of the quadratic potential density ${\cal V}(\phi)=\frac{1}{2}m^2\phi^2$ in field theory is the harmonic quadratic potential $V(q)=\frac{1}{2}kq^2$ in point mechanics. It should be stressed that the notion of mass enters very differently in field theory and in point mechanics, see e.g. this Phys.SE post.


2

The easiest way to work out your algebra, in case you haven't, and are alarmed by the missing constant term, is to rewrite your potential, eliminating the pestiferous unphysical $\mu^2$ that has caused grief to generations of students. Recalling that summations over repeated indices is implied, the potential is $$ \tfrac{\lambda}{4} (\phi_i \phi_i - v^2)^2 -\...


1

Such a theory disagrees with experiment. There would be no gravitational waves and coupling to a free electromagnetic wave would be impossible, as the trace of its EM tensor is zero.


1

Neither does it have a mass in the numerator for the $\phi^2$ term! Peskin & Schroeder just do not bother with a constant $m$ is this context. As you can see, this part introduces you to the ladder operators, in order to apply the formalism to the Klein-Gordon hamiltonian. No need to worry about $m$'s, which are irrelevant to the commutation relations ...


4

No, $F$ and the infinitesimal variation $\delta F$ have the same dimension. And $\phi$ and the infinitesimal variation $\delta \phi$ have the same dimension. But by definition of the functional/variational derivative $\frac{\delta F}{\delta\phi}$, the infinitesimal variation is $$ \delta F ~=~\color{red}{\int\!d^dx}~ \frac{\delta F}{\delta\phi}~\delta\phi. \...


0

In spite of what gauge theory may suggest, the field variable is the potential $A^\mu$, so not the force field tensor $F^{\mu\nu}$. In terms of the potential the Lagrangian is $$ \mathcal{L}_{EM}=\frac{\epsilon_0}{2} \left( \partial_\mu A_\nu \partial^\mu A^\nu - \partial_\nu A_\mu \partial^\mu A^\nu\right)~.$$ If you insert this in the EL equation you find $...


6

I think a quick brush up on the index notation and relativity will clear this up. (In what follows I'll be using units in which $c=1$.) First consider the operator $\partial_\mu$, which is the four-gradient, represented by: $$\partial_\mu \equiv \begin{pmatrix} \partial_t & \partial_x & \partial_y& \partial_z\end{pmatrix}.$$ This is an operator ...


2

Using a $(+1,-1,-1,-1)$ Minkowski metric, $$\Box\Phi=\partial_\mu\partial^\mu\Phi=\frac{1}{c^2}\frac{\partial^2\Phi}{\partial t^2}-\frac{\partial^2\Phi}{\partial x^2}- \frac{\partial^2\Phi}{\partial y^2}-\frac{\partial^2\Phi}{\partial z^2}$$ while $$\partial_\mu\Phi\partial^\mu\Phi=\frac{1}{c^2}\left(\frac{\partial\Phi}{\partial t}\right)^2-\left(\frac{\...


0

This really does need all three space dimensions. I'll set $\mu_0=\epsilon_0=1$ to tidy up; feel free to restore them. Since $\partial_\mu F^{\mu\nu}=0$, $\nabla\cdot E=0$ and $\nabla\times B=\dot{E}$. Since $\partial_\mu\tilde{F}^{\mu\nu}=0$, $\nabla\cdot B=0$ and $\nabla\times E=-\dot{B}$. So$$\ddot{E}=\nabla\times\dot{B}=-\nabla\times\nabla\times E=\Delta ...


1

The charge conjugation matrix $C$ is not an antilinear map. It is ordinary matrix with the property that $$ C\gamma^\mu C^{-1} =-(\gamma^\mu)^T. $$ (or maybe $C^{-1} \gamma^\mu C = -(\gamma^\mu)^T$. Conventions differ.).


3

Two complex scalar fields $\phi_{1}$ and $\phi_{2}$ can be rewritten as four real fields, in terms of their real and imaginary parts, $$\Phi=\sqrt{2}\left[\begin{array}{c} \Re\{\phi_{1}\} \\ \Im\{\phi_{1}\} \\ \Re\{\phi_{2}\} \\ \Im\{\phi_{2}\} \end{array}\right].$$ For the free theory, the Lagrange density is actually equal to $${\cal L}=\frac{1}{2}\partial^...


5

Here is a second way to see the correct result for taking the functional derivative of the spacetime derivative of the field, which I hope will be helpful. Recall that the definition of the functional derivative is $$ \frac{\delta\phi(y)}{\delta\phi(x)}=\delta(y-x) .$$ You know that Dirac deltas are distributions. That is, you should always think of them ...


5

Once nice way to calculate functional derivatives is to use the concept of the Gateaux derivative as follows: $$\frac{d}{d\epsilon}S[\phi+\epsilon \eta]\bigg|_{\epsilon=0} = \int d^4x\frac{\delta S}{\delta \phi} \eta$$ In your case, $$S[\phi+\epsilon \eta]= \int d^4x \ \bigg\{\frac{1}{2}\big((\partial \phi)^2 + 2\epsilon (\partial_\mu\phi)(\partial^\mu\eta) +...


3

The safest way to compute the functional derivative is to use the following prescription: \begin{equation} S[\phi + \delta \phi] = S[\phi] + \int {\rm d}^4 x \frac{\delta S}{\delta \phi}\delta \phi + O(\delta \phi^2) \end{equation} In other words, add a small perturbation to the field, and manipulate the action so it has the form of an integral times the ...


0

The standard way to fix this is to adopt the Belinfante-Rosenfeld tensor instead of the Noether tensor. This raises hairs. Why is the Noether tensor wrong? Not only the EM tensor must be symmetrized, also the angular momentum tensor had problems. Why is this Noether current also wrong? This problem must then be solved by ditching spin and writing it as an ...


3

This is the Belinfante-Rosenfeld procedure (note that it is not necessary to invoke spin currents for the free electromagnetic theory). Just to spell things out, this involves modifying the canonical energy-momentum tensor by adding a divergenceless term (an alternative method is using the Hilbert definition, see this question for the relation between the ...


0

The procedure, which makes the Noether tensor of EM symmetric, is known as Belinfante's method. You can find it in the standard textbooks, e.g. Landau's second volume, please google it, there are a lot of student notes discussing it, e,g. Blau's note.


11

Ultimately, the physical reason for doing this is that it works. There’s a fairly natural line of reasoning which leads to this procedure - that’s not a proof, because proofs don’t exist in physics, but it’s suggestive motivation. I'll run through this reasoning, breaking up the narrative with some elaboration that may be helpful. If you impose a local ...


14

Most theories do not have a conserved energy-momentum tensor, regardless of whether they are Lagrangian or not. For example you need locality and Lorentz invariance. When you have those, you can define the energy-momentum tensor via the partition function (which always exists, it basically defines the QFT): $$ \langle T_{\mu\nu}\rangle:=\frac{\delta}{\delta ...


10

Not all non-Lagrangian theories have a stress-energy tensor, an example of this is the critical point of the long-range Ising model, which can be expressed as a "defect" field theory where the action consists of two pieces integrated over spaces of different dimensionality and hence has no single Lagrangian that would describe it. See chapter 6 of &...


1

I just found an interesting analytical and highly non-trivial solution to equations (2) and (7) ! $$\tag{1} \Phi(T, X, Y, Z) = \tanh \bigl(Z + \mathcal{G}(a T - b X - c Y) \bigr), $$ where $a$, $b$ and $c$ are arbitrary constants satisfying the light like condition on the orthogonal plane of $Z$: $$\tag{2} a^2 - b^2 - c^2 = 0, $$ and $\mathcal{G}(u)$ is a ...


2

I'll focus on the scalar $O(n)$ gauge theory that was mentioned in the question. The set of observables in the gauged version of the model is not a subset of the observables in the ungauged version. Gauging the $O(n)$ symmetry does eliminate some observables, but it also introduces others. Observables in the gauged version are required to be invariant under ...


0

Well, I can't say this discussion helped me much. Do we have a mathematically sound formulation of QFT? Yes or no?


1

The original (and most fundamental) definition of a ghost field was one for which, when it is quantized, the particle-like excitations do not have positive norm. One of the first places this came up was in attempts to canonically quantize the electromagnetic field in a relativistically-invariant way. If you look at the (Feynman-gauge*) propagator for the ...


3

Just to add to the other answers, I wanted to give a perspective from ghosts found in modified gravity as opposed to quantum theory (which I think you may also be referring to as well). Here's a quote you might find useful: We should be clear about the distinction between the kind of ghost that arises in certain modified gravity models and the Faddeev-Popov ...


2

OP is asking for examples of a quasisymmetry of the action that is not a strict symmetry: Gaugesymmetries are often quasisymmetries. A global additive shift of the free Schroedinger field, cf. e.g. this Phys.SE post. A Galilean boost of the non-relativistic free particle, cf. e.g. this Phys.SE post. The quasisymmetry behind the conservation of Laplace-...


2

The general theory of the ghost structure in the Lagrangian field theory is rather a sophisticaed deal and is called Batalin–Vilkovisky (BV) formalism : https://en.wikipedia.org/wiki/Batalin%E2%80%93Vilkovisky_formalism https://ncatlab.org/nlab/show/BV-BRST+formalism I would not dive into deep details, lacking knowledge and understadning of the general ...


0

In the context of Yang-Mills, the ghosts are auxiliary fields introduced only as a gauge fixing mechanism (they have no physical motivation, as far as I am aware). They are just a tool to introduce a gauge fixing condition, since integration on the space of all distinct connections (connections not related by a gauge transformation) is very difficult. This ...


1

Let's start with what polarization means. When talking about polarization we always consider transverse waves. That is waves that oscillate perpendicular to the direction of propagation. Electro-Magnetic waves in vacuum would be an example of this*. (Of course a general vector field can have any polarization that is either time-like, longitudinal or ...


0

Are you familiar with classical electromagnetism? A field can be scalar. In which case there is a function that assigns a value to any position in space and time - this we then call field. More generally it could be a function that assigns a vector or even a tensor to any space and time position. This would be the vector(tensor) field. In some common cases ...


1

A (Lorentz-noninvariant) Lagrange density in electromagnetism proportional to $\vec{A}\cdot\vec{B}$ changes by a boundary term under a gauge transformation.* Under the change of gauge $\vec{A}\rightarrow\vec{A}+\vec{\nabla}\Lambda$, a Lagrange density $${\cal L}=-\frac{1}{4}\left(\vec{E}^{2}-\vec{B}^{2}\right)+k\vec{A}\cdot\vec{B}$$ changes by $${\cal L}\...


0

Air acts as an insulator in normal conditions because it is majorly composed of stable gases like oxygen and nitrogen. As we know both oxygen and nitrogen are highly electronegative elements and it is difficult to remove electrons from them. (This is without even considering the highly stable bonds of oxygen and nitrogen molecules.) Therefore a strong ...


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