Timeline for What is so special about spontaneous symmetry breaking? (time reversal example)
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 21, 2017 at 19:58 | comment | added | GaragePhys | The complete title is 'An introduction to quantum field theory', it's the standard reference for QFT. | |
| Apr 21, 2017 at 19:57 | comment | added | GaragePhys | I think the discussion in Peskin should be quite accesible, he proves Goldstones theorem on one page in chapter 11.1 page 351. | |
| Apr 21, 2017 at 18:06 | comment | added | user140255 | I understand what you mean! Also, do you know where I can find a comprehensive explanation of what Goldstone bosons are (considering I'm from condensed matter and not familiar with field theory formalism) | |
| Apr 21, 2017 at 5:36 | comment | added | GaragePhys | $$\left| x \right\rangle\rightarrow R \left| x \right\rangle \text{ with } R = \begin{pmatrix} cos(\alpha) & sin(\alpha) & 0 \\ -sin(\alpha) & cos(\alpha) & 0 \\ 0& 0& 1 \end{pmatrix} \approx \alpha \underbrace{\begin{pmatrix}0 & 1 & 0 \\ -1 &0 & 0 \\ 0& 0& 1 \end{pmatrix}}_{ = T} + \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0& 0& 1 \end{pmatrix}$$ Here $\left| x \right\rangle$ is clerly invariant under $R$, which is expressed by $T \left| x \right\rangle= 0$ Here you'd find more about it: \url{en.wikipedia.org/wiki/Symmetry_in_quantum_mechanics}. | |
| Apr 21, 2017 at 5:36 | comment | added | GaragePhys | No, that's not a typo, what you mean by the symbol $T$ is the generator of the symmetry, which is an infinitesimal version of it. For example if you have rotations in 3d space acting on the vector: $\left| x \right\rangle = \begin{pmatrix} 0\\ 0\\x\end{pmatrix}$ Let's consider the transformation: | |
| Apr 21, 2017 at 0:20 | comment | added | user140255 | Thanks for your answer! I am not sure if it's a typo, do you mean $ T \left g \right \rangle \neq \left g \right \rangle$ instead? Because I can't see any reason why it could give 0. Take the case of an hamiltonian symmetric under parity like an harmonic oscillator. Then applying the parity operator on the ground state gives back the ground state not 0. | |
| Apr 20, 2017 at 22:19 | history | answered | GaragePhys | CC BY-SA 3.0 |