4

Your mistake is in saying that $\overline{\psi_{L,R}} = \overline{\psi}P_{L,R}$. In fact, $$ \overline{\psi_{L,R}} = (P_{L,R}\psi)^† \gamma^0 = \overline{\psi}\gamma^0 P_{L,R}^† \gamma^0$$ $$ = \overline{\psi} P_{R,L}$$ To see why that last line is true we have to expand $P_{L,R} = (I\mp \gamma^5)/2$ and note that $(\gamma^5)^† = \gamma^5$, $(\gamma^0)^2 = ...


3

Note that this is rather an opinion than a fully rigorous statement: $\qquad$ For vanishing gradients, i.e. for uniform systems (the case originally considered by Landau), the expansion of the free energy is indeed a Taylor expansion (in even powers of $m$) near the transition. However the addition of gradient terms in the case of non-uniform systems ...


3

The standard convention is to divide each term in the Lagrangian with its symmetry factor. Therefore the kinetic term for a real (complex) scalar field is with (without) a symmetry factor $\frac{1}{2}$, respectively. A complex scalar field $\phi= \frac{\phi^1+i\phi^2}{\sqrt{2}}$ can be viewed as 2 real scalar fields.


3

Nothing in your derivation is wrong. This is what people mean when they say that the Dirac operator is the "square root" of the Klein-Gordon operator. But it's not quite the Klein-Gordon equation because $\psi$ is still a Dirac spinor, i.e has four components, not one like the scalar field in the Klein-Gordon equation.


3

Yes and yes to your first two questions. Fermions are massless at sufficiently high temperature when the VEV is zero and the symmetry unbroken. They acquire mass when the universe expands and cools, the VEV becomes nonzero, and the symmetry breaks.


2

You need check invariance only on linear level, because you consider linear action. Third term is second order. Integration by parts is incorrect, because ζ is quadratic differential operator. I recommend you to start with most general quadratic action and find coefficients from diffeomorphism invariance, like in Zee book on gravity: After that, you need ...


2

OP is right: Wald's book has a sign mistake in eq. (E.1.36).


2

Periodic functions $f(t)$ (with a time period $T$) can be approximated by a Fourier series, i.e. by summing harmonic oscillations with the discrete frequencies $0, \frac{2\pi}{T}, 2\frac{2\pi}{T}, 3\frac{2\pi}{T}, \ldots$ . $$f(t)=\sum_{n=-\infty}^{+\infty} F_n e^{in\frac{2\pi}{T}t}$$ But, as you already noticed, with a Fourier series you cannot build ...


2

It looks to me like you are Taylor expanding $\mathcal{F}$ as a function of $\vec{m}$ without considering that $\vec{m}$ is a function of space. In particular, your $D$ and $D^2$ involve things like $\frac{\partial}{\partial m_x}$ but not $\frac{\partial}{\partial x}$. You are implicitly assuming that $\mathcal{F}(\vec{x})$ only depends on $\vec{m}(\vec{x})$ ...


2

You don't normalize your two-photon states properly. They are $|2\rangle = \tfrac{1}{\sqrt{2}}(a^\dagger)^2|0\rangle$ (and the same for the $c$ and $d$ modes). You are missing the $\tfrac{1}{\sqrt{2}}$. Then, your output state is $\frac{1}{2}\tfrac{1}{\sqrt{2}}(\hat{c}^{\dagger}\hat{c}^{\dagger}+\hat{d}^{\dagger}\hat{c}^{\dagger}+\hat{c}^{\dagger}\hat{d}^{\...


1

You shouldn't check the Cauchy-Riemann equations as they will not hold - that functional is not holomorphic as it contains explicit $\phi^\dagger$s. This is why you need to impose both $\frac{\delta S}{\delta \phi(x)} = 0$ and $\frac{\delta S}{\delta \phi^\dagger(x)} = 0$. In the holomorphic case the second equation would come for free. The way I make sense ...


1

the moon orbits both Earth and the Sun. Since the Earth orbits the Sun, and the moon orbits the Earth, it necessarily also orbits the Sun. These two motions of the moon almost do not interfere with one another, because its orbit about the Sun is almost identical to the Earth's orbit around the Sun (because they are approximately the same distance from the ...


1

Edit: The convention I used is apparently different than yours, although the method is still the same. What notes are you using? We use the commutation relations $$ [a_p,a^\dagger_q] = (2\pi)^3\delta^{(3)}(\vec{p}-\vec{q}), [a_p,a_q]=[a^\dagger_p,a^\dagger_q ]=0 $$ Now we start with $$ \langle k|k_1k_2\rangle=\langle 0|a_k a^\dagger_{k_1}a^\dagger_{k_2}|0\...


1

If the structure group of the theory is $G$, with Lie algebra $\mathfrak g$, then the (local) Yang-Mills fields, are locally defined differential $1$-forms that take value in the Lie algebra $\mathfrak g$. If the local gauge neighborhood is $U$, and we suppose that it is also a coordinate neighborhood, then we may write $$ A(x)=A_\mu^a(x)\mathrm dx^\mu\...


1

You can generally couple a continuum action $S_{\rm field} = \int_V \mathcal{L}(\psi(x^\mu)) d^4 x $ to a worldline action $S_{\rm particle} = \int_{\Delta \lambda} L(x^\mu(\lambda),\psi(x^\mu(\lambda))...)d\lambda$ through a delta function on the world-line $$S_{\rm tot} = \int_V \left(\mathcal{L}(\psi(x^\mu)) + \int_{\Delta \lambda} L(x^\mu(\lambda),\psi,.....


1

Note that the notion of gauge theories is much more general than, say, Yang-Mills theory. There is of course a long list of gauge theories that have gauge potentials -- e.g. in the case of Yang-Mills theory, the $A^a_{\mu}(x)$ field is the gauge potential -- but it is not a general requirement for a gauge theory.


1

a) In my notes on P&S I have a minus sign as well. Maybe this is just a typo? b) and c) P&S just take these values of the indices as an example. They assume a point charge at the origin of the physical space in the group direction $a=1$. They then consider the $A^{2i}$ gauge field in the "bubbling'' vacuum. That gauge field then interacts with $E^{...


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