# Search results

1. ### Nother current of given symmetry

Where did you get this from? With respect to translation, \bar{x} = x + b, all fields are invariant, i.e., \bar{\delta} \varphi_{a}(x) = 0, or \varphi_{a}(x) \to \bar{\varphi}_{a}(\bar{x}) = \varphi_{a} (x). \ \ \ \ \ (1) So \bar{\varphi}_{a}(x + b) = \varphi_{a}(x) . Expanding the LHS to...
2. ### A Reading canonical commutation relations from the action (QHE)

From the action, you read off the Lagrangian \mathcal{L} = \frac{me^{2}}{4\pi \hbar} \ \epsilon^{\mu\nu\rho} \ a_{\mu} \partial_{\nu} a_{\rho} . Now, calculate \frac{\partial \mathcal{L}}{\partial (\partial_{\sigma}a_{\tau})} = \frac{me^{2}}{2\pi \hbar} \ \epsilon^{\mu \sigma \tau} \ a_{\mu} ...
3. ### Nother current of given symmetry

I thought you will be asking me about something very important which I have hidden in the calculations (on purpose). But, unfortunately, you seem to be confused about the simple stuff. The Lagrangian (any Lagrangian) is a space-time density and, therefore, it cannot be invariant. What can be...
4. ### Nother current of given symmetry

Scale invariance requires the absence of dimension-full parameters. So, in 4 dimensions, your Lagrangian is not scale invariant because m and \mu are not dimensionless. Consider scaling down the coordinates infinitesimally as \bar{x}^{\mu} = e^{- \epsilon} x^{\mu} \approx x^{\mu} - \epsilon...
5. ### A Why does this term transform in this way?

I have already said that you can learn about this stuff from all textbooks on supersymmetry. I just know this stuff because it is my business to know it.
6. ### Vector calculus identity and electric/magnetic polarization

It is better for you to get used to tensor calculus: ( \vec{r} \times \nabla \times \vec{M})_{i} = \epsilon_{ijk}\ x_{j} \ ( \nabla \times \vec{M})_{k} = \epsilon_{ijk} \ x_{j} \ \epsilon_{kln} \ \partial_{l}M_{n} . Now if you use x_{j}\partial_{l}M_{n} = \partial_{l}(x_{j}M_{n}) -...
7. ### A Why does this term transform in this way?

(A , B) \otimes (A^{\prime} , B^{\prime}) = (A + A^{\prime} , B + B^{\prime}) \oplus (A + A^{\prime} - 1 , B + B^{\prime}) \oplus \cdots \oplus (|A - A^{\prime}| , B + B^{\prime}) \oplus (A + A^{\prime} , B + B^{\prime} - 1) \oplus \cdots \oplus (|A - A^{\prime}| , |B - B^{\prime}|) . Almost...
8. ### A Why does this term transform in this way?

Usually, I don’t talk about “active” vs “passive” things, because there is no meaningful distinction between the two and they often cause unnecessary confusions. But, yes I believe I used the “passive” description: That is what I meant by “two coordinate systems O \to O^{\prime} = \Lambda O and...
9. ### A Why does this term transform in this way?

Doing physics would be impossible in a world where dimensionless numbers (such as 0, 1 and i, i.e., the elements of the \gamma’s) depend on the observer's state of motion. This the intuitive reason for not transforming the \gamma^{\mu}. Group theoretically, the gamma matrices play vital role in...
10. ### A Why does this term transform in this way?

No, \gamma^{\mu}\partial_{\mu}\psi is not invariant. If it was you would not be able to add the non-invariant (spinor) term -m \psi to it and the Dirac equation would make no sense. Indeed, it transforms exactly like \psi: \gamma^{\mu}\partial_{\mu} \psi \to D(\Lambda ) \...
11. ### A Why does this term transform in this way?

I considered two coordinate systems (two observers) looking at the same point P which has two different coordinate values x and \bar{x} = \Lambda x as seen by the two observers (O , \psi ) and (O^{\prime} , \psi^{\prime}) respectively. So, \psi (x) \to \psi^{\prime} (\bar{x}) simply means \psi...
12. ### A Why does this term transform in this way?

No, \Lambda_{\frac{1}{2}} does not commute with \gamma^{\nu}.
13. ### A Why does this term transform in this way?

Throw away this textbook or notes that you are using. To prove that the Dirac equation is Lorentz covariant, you only need the following transformation laws: x^{\mu} \to \bar{x}^{\mu} = \Lambda^{\mu}{}_{\nu} \ x^{\nu} , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\psi (x) \to...
14. ### I Why do we need two representations of SU(3)

q^{a} \left( = u , d , s \right) \in \{ 3 \}, a 3-vector in the fundamental (or defining) representation space \{ 3 \} of \mbox{SU}(3). The 9-component tensor q^{a}q^{b} \in \{ 3 \} \otimes \{ 3 \} can be decomposed as follows q^{a}q^{b} = S^{ab} + A^{ab} , where S^{ab} = S^{ba} = \frac{1}{2}...
15. ### I Derivation of an identity for ##\partial^2_t \int T^{00}(x^i x_i)^2d^3

\left( ( x_{i}x^{i} )^{2} \right)_{,k} = 4 x_{k} \left( x_{j}x^{j} \right) .
16. ### A Gauge breaking and Faddeev-Popov ghost particles

I didn’t have to, because the whole of #37 was about the covariant quantization of the massless vector field A_{\mu}(x). I wouldn’t use the term “covariant formalism” if my QFT was that of massless scalars or spinors which I did not even mention in #37.
17. ### A Gauge breaking and Faddeev-Popov ghost particles

1) If you are in the business of nitpicking, then you should have said “... of QFT with massless vector fields ”, because the QFT of massless scalar does not require indefinite-metric. 2) By QFT, I meant gauge-invariant quantum field theories such as QED and QCD (which come naturally with...
18. ### A Gauge breaking and Faddeev-Popov ghost particles

You are absolutely right. If the vector field is massive, it is possible to separate the negative-norm covariantly because of the compact nature of the corresponding little group, SO(3). But, for a massless vector field, it is impossible to avoid negative norms without violating manifest Lorentz...
19. ### Metric tensor problem

U = y - t, \ \ \ \ V = y + t
20. ### I Is Gravity Part Of The Standard Model?

\mbox{Gravitational Potential} = \mbox{Newton} \left( 1 – \mbox{GR} ( \mbox{correction} ) – \mbox{SM} ( \mbox{correction} ) \right) , V(r) = - \frac{G_{N}mM}{r} \left( 1 - \frac{G_{N}(M + m)}{c^{2}r} - \frac{127}{30 \pi^{2}} \frac{G_{N} \hbar}{c^{3}r^{2}}\right) . That is a particle physics...
21. ### A Proof - gauge transformation of yang mills field strength

This is wrong. Do not use the same repeated indices in both factors. The correct expression is this L \equiv \mbox{Tr} \left( F^{a}_{\mu\nu}t^{a} F^{b \mu\nu} t^{b} \right). Now, F^{a}_{\mu\nu}F^{b \mu\nu} is a number, so you can pull it out of the trace: L = F^{a}_{\mu\nu}F^{b \mu\nu} \...
22. ### A Gamma - traceless

Non of your links explain the matter better and/or easier than I did in the other thread. And your second link contains wrong as well as confused statements.
23. ### A Gamma - traceless

Is it? Then why are you asking this? First: \psi^{\mu}(x) is a spinor-vector field. It describes spin-3/2 particle and its anti-particle. It is NOT a Dirac spinor (Dirac spinor field describes spin-1/2 particle and its anti-particle). Second: The divergence of \psi^{\mu}(x) IS a Dirac spinor...
24. ### A Gamma - traceless

It is. If he could not understand the reasoning in the SE website, he may not be able to understand my answer in the following https://www.physicsforums.com/threads/how-to-construct-a-spin-3-2-theory-from-the-ground-up.941219/post-5954706
25. ### A Is quantum theory a microscopic theory?

I am afraid there is no case to answer. Micor/macro worlds make sense provided we agree to use standard scales for length, time and mass. And we use well-defined experimental procedures to define those scales.
26. ### I Boundary terms for field operators

They always mean operator-valued distribution. Operator valued functions satisfying the QFT axioms do not exist. So, we really need distributions. The physical reason is the uncertainty principle: Measuring the field \varphi at a point x causes very large fluctuations of energy and momentum...
27. ### I Boundary terms for field operators

No, the field is not an operator (valued function on spacetime) because the norm \lVert \varphi (t,x) |0 \rangle \rVert^{2} is divergent. A densely defined operator on Hilbert space \mathcal{H} can be constructed by smearing the field \varphi (t,x) (which is an operator-valued distribution) with...
28. ### I Derivation of the Lorentz transformations

Indeed, given Minkowski spacetime M^{(1,3)}, one can show that the Poincare group \Pi (1,3) is its maximal symmetry group. Conversely, given \Pi (1,3), one can show (using the theory of induced representations) that M^{(1,3)} \cong \frac{\Pi (1,3)}{SO^{\uparrow} (1,3)} . That is Minkowski...
29. ### Derivatives on tensor components

Apply the Leibniz rule to \left( \frac{\partial}{\partial p_{\mu}} \frac{1}{\not\! p} \right) \not \! p \ .
30. ### A QFT Srednicki Chap 6: Weyl Ordering

That would be an insult to Weyl. There are far more to Weyl’s quantization than just symmetrizing product of operators. It led to: the theory of Pseudo differential operators also known as Weyl calculus; functional analysis on locally compact groups; deformation quantization with the star...