theorems:goldstones_theorem
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theorems:goldstones_theorem [2018/03/28 17:15] jakobadmin [Student] |
theorems:goldstones_theorem [2018/05/15 06:59] jakobadmin |
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| ====== Goldstone's theorem ====== | ====== Goldstone's theorem ====== | ||
| - | <tabbox Why is it interesting?> | ||
| - | <blockquote> | ||
| - | Goldstone's theorem states that whenever a continuous global symmetry is spontaneously broken, there exists a massless excitation about the spontaneously broken vacuum. Decomposing $\Phi(x)=|\Phi(x) |e^{i\rho(x)}$, $\rho$ transforms as $\rho(x) \to \rho(x) + \theta$. Hence the Lagrangian can depend on $\rho $ only via the derivative = $\partial_\mu \rho$; there cannot be any mass term for $\rho$, and it is a massless field. $\rho$ --- identified as the field which transforms inhomogeneously under the broken symmetry --- is referred to as the Goldstone boson. | ||
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| - | <cite>https://arxiv.org/pdf/1703.05448.pdf</cite> | ||
| - | </blockquote> | ||
| - | <tabbox Layman> | + | <tabbox Intuitive> |
| * For an intuitive explanation of Goldstone's theorem, see [[http://jakobschwichtenberg.com/understanding-goldstones-theorem-intuitively/|Understanding Goldstone’s theorem intuitively]] by J. Schwichtenberg | * For an intuitive explanation of Goldstone's theorem, see [[http://jakobschwichtenberg.com/understanding-goldstones-theorem-intuitively/|Understanding Goldstone’s theorem intuitively]] by J. Schwichtenberg | ||
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| - | <tabbox Student> | + | <tabbox Concrete> |
| <blockquote> | <blockquote> | ||
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| - | <tabbox Researcher> | + | <tabbox Abstract> |
| <blockquote> | <blockquote> | ||
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| <cite>https://arxiv.org/pdf/1612.00003.pdf</cite></blockquote> | <cite>https://arxiv.org/pdf/1612.00003.pdf</cite></blockquote> | ||
| - | --> Common Question 1# | + | <tabbox Why is it interesting?> |
| - | + | <blockquote> | |
| - | <-- | + | Goldstone's theorem states that whenever a continuous global symmetry is spontaneously broken, there exists a massless excitation about the spontaneously broken vacuum. Decomposing $\Phi(x)=|\Phi(x) |e^{i\rho(x)}$, $\rho$ transforms as $\rho(x) \to \rho(x) + \theta$. Hence the Lagrangian can depend on $\rho $ only via the derivative = $\partial_\mu \rho$; there cannot be any mass term for $\rho$, and it is a massless field. $\rho$ --- identified as the field which transforms inhomogeneously under the broken symmetry --- is referred to as the Goldstone boson. |
| - | --> Common Question 2# | + | <cite>https://arxiv.org/pdf/1703.05448.pdf</cite> |
| - | + | </blockquote> | |
| - | + | ||
| - | <-- | + | |
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| <tabbox Examples> | <tabbox Examples> | ||
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| - | <tabbox History> | ||
| </tabbox> | </tabbox> | ||
theorems/goldstones_theorem.txt · Last modified: 2020/04/12 15:05 by jakobadmin
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