New answers tagged symmetry-breaking
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It depends.
In the two Higgs doublets models, there are two cases:
The two doublets are independent
There is a cross potential term between these doublets a la: $\phi_1^2\phi_2^2$
For case 1, the two VEVs are independent.
For case 2, the two VEVs are are coupled to each other. But it is hard to tell which is inducing which.
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In short yes.
I believe this idea arose when people were thinking of the symmetry breaking from a bigger symmetry group (for example some hypothetical GUT scale group), down to the standard model as the universe cooled down.
Morally speaking, the same story of the Higgs mechanism applies.
That is we have a theory with symmetry group $G$ in which the vacuum ...
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@mike stone 's answer is impeccable, but requires emphasizing that the answer therefore depends on the representation of the Higgs multiplet under that group.
For most of the groups used in physics, this was done in a legendary paper Group theory of the spontaneously broken gauge symmetries, Ling-Fong Li (1974), Phys Rev D9 1723, routinely used as a "...
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The strongest motivation for initially equal amounts of matter and anti-matter actually comes not from the standard model but from cosmology - there is a staggeringly large number of photons in the Universe, relative to baryons$^1$. It's possible to arrive at this conclusion a couple of different ways. One is to look at the cosmic microwave background, which ...
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The mathematical meaning of a singularity is that we cannot define mathematical quantities. Since we cannot define mathematical quantities, we have no idea of the physics in the initial singularity. Anything we assume is therefore a speculation. We cannot know whether the universe was created out of nothing, and, if so, whether it was created with a matter-...
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You're supposed to think of $\mu^2$ as a parameter, and there's no need to consider if it $>0$ or $<0$. You proceed by minimising the potential and then seeing that the nature of vacuum/vacua is different for $\mu^2>0$ and $\mu^2<0$. As @CosmasZachos mentioned in the comments, it is certainly a function of $T$. The exact function can be ...
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It is broken to whichever subgroup of the original group that leaves the VEV unchanged.
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Have a look here :
Physicists have observed quantized states of matter under the influence of gravity for the first time. Valery Nesvizhevsky of the Institute Laue-Langevin and colleagues found that cold neutrons moving in a gravitational field do not move smoothly but jump from one height to another, as predicted by quantum theory. The finding could be ...
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The magnetic field is not affected by your transformation: it is a pseudovector, and it does not change sign upon spatial inversions. Since neither the magnetic field nor the particle's velocity is changed by the reflection, the right-hand rule predicts the same direction for the force both before and after the reflection.
When in doubt, it is typically ...
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The magnetic field also changes sign because, in the mirror image of the solenoid creating the field, the windings turn in the opposite direction to those in our world.
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The Witten index in a supersymmetric theory counts the number of bosonic ground states minus the number of fermionic ground states. The reason for this is that the trace sums over all states and for each excited bosonic state, we also get an excited fermionic state of the same energy, due to supersymmetry, which results in a cancellation.
The reason for $\...
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Let's say you have a certain gauge theory, i.e., a QFT with gluons plus, perhaps, some other matter fields. Take for example one of such matter fields $\phi$. In general $\phi$ will couple to the gluons, i.e., it will interact with them. In fact, it is possible that $\phi$ only interacts with some of the gluons. Let's give these objects some names.
Take the ...
answered Jun 24 at 11:57
AccidentalFourierTransform
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Hint: Scale the quantum fluctuations $\eta$ with a factor $\sqrt{\hbar}$, i.e. put $$\phi~=~v+\sqrt{\hbar}\eta.$$ This makes it easier to see that the extra terms vanish as $\hbar\to 0$. See also e.g. this related Phys.SE post.
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Looking at Chapter 6 of Altland-Simons book, you can read about the non interacting Bose gas and the weakly interacting Bose gas, and I think that the anser might be in those pages.
In my opinion the key point is that the non interacting system is pathological in the sense that there is no solution that minimizes the action, and therefore "expanding the ...
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