Tag Info

New answers tagged

1

I definitely wouldn't say this. Spontaneous symmetry breaking is when the Hamiltonian is symmetric under some transformations, but the ground state of the system is not. The ground state of the classical SHO is that the particle sits at the minimum of the potential, which is a time-independent state. In the quantum case the quantum ground state has the ...


0

The hexagonal Graphene lattice can be considered as a superposition of two identical sub-lattices set off by one one carbon-carbon bond length. As a result, it has two sets of wavevectors k,that are picked out by the lattice, inequivalent (since the two sublattices really are distinct) but otherwise identical (since it's semantics to say which sublattice is ...


0

Perhaps chiral superfluids and superconductors are also good examples. The A-phase of liquid 3-He, for instance, is known to be a TRSB superfluid with pairing $p_x + i p_y$.


1

There are various general ways to see the point. Let me give you a simple one that is still general enough (see also last comment at the end). Imagine you have a lagrangian density $L$ that is invariant under the continuous group transformation that acts linearly on the fields as $$ \phi\rightarrow \phi^\prime=U\phi=e^{i\omega^a T^a}\phi=(1+i\omega^a ...


1

Since the conceptual problem is not really coming from field theory itself, let's look at the 0D case. This case is too naive (no SSB here), but the discussion is still correct qualitatively. The partition function is $$Z(j,j^*)=\int dz dz^* e^{-V(|z|^2)+ z^*j + j^* z}, $$ where $dzdz^*$ really means $d\Re z\,d\Im z$ and $V(x)=r\, x+x^2$ where $r$ can be ...


1

Explicit symmetry breaking, i.e. adding a mass term for a particle simply implies that this particle is massive. In the case of a vector boson, there will be three degrees of freedom. The "additional" degree of freedom does not have to come from somewhere else by a certain mechanism, it is simply there from the beginning. An example would be Proca theory, ...


3

There is no contradiction since you should not be doing complex linear combination of generators that are already hermitian (indeed, you want the group transformation to be unitary). Hence, your linear combination with a complex coefficient (that brings you away from the $SU(2)$ group) doesn't imply a massless excitation.


1

It is indeed possible to break $ SU(3) $ to $ SU(2) \times U(1) $. To see that we need to check that $ SU(2) $ and $ U(1) $ are subgroups of $ SU(3) $. Its easy to see that $ SU(2) $ is a subgroup since the first three Gell-mann matrices are given by, \begin{equation} \lambda _i = \left( \begin{array}{cc} \sigma _i & 0 \\ 0 & 0 \end{array} ...


0

After reading the section put forth by @gcsantucci I think I kind of understand what was happening, but I'm eager to hear feedback on this. If you break $ SU(2) $ using a doublet you indeed break all the generators and get massive gauge bosons. What was confusing is that a linear combination of generators leaves the vacuum invariant (namely, $ \sigma _1 ...


0

I guess instead of me trying to explain, it is way better if you read section "Non Abelian Examples" of Peskin and Schroeder, in chapter 20. It explains exactly what you are asking for. It's actually the predecessor of the Standard model. Georgi and Glashow proposed this model before the SM, because they didn't know about the Z boson. So it is exactly what ...



Top 50 recent answers are included