关于希格斯机制的问答（转自繁星客栈）
小白：
请问，是什么对称性保护光子小姐不受希格斯粒子性骚扰？
sage:
You start with SU(2)x U(1), 4 generators of Lie algebra (or equivalently 4 gauge bosons). Now you have a Higgs doublet, which get's a vacuum expectation value (VEV). There are many ways to understand the symmetry breaking pattern. The easiest is to begin with 4 degrees of freedom in the complex Higgs doublet. After getting the VEV, there can only be 3 Goldstones. Gauge bosons have to eat Goldstone to become massive (for example in BCS). Therefore, only three gauge bosons can become massive after eating the Goldstones. There has to be one remains massless. We call that one photon.
小白：
谢谢Sage老师科普，我翻译一下：在弱电统一能标下，光子小姐和其它三个中间玻色子美女在su(2)×u(1)规范对称下统一成一类女人。低能下，这个对称性会自发破缼，同时产生三个哥徳斯通帅哥，四个美女三个帅哥，在一夫一妻制下必定有一个美女会独守空房，很不幸的是，这个独守空房的美女就是光子小姐。结果，吞食了三个帅哥的三个中间玻色子美女获得了质量，而光子小姐至今没有获得质量。请问Sage老师，是不是光子的u(1)对称没继续破缼，才使她没有机会吞食哥德斯通粒子，也就是说光子实际上是受u(1)对称性保护？
sage:
No or yes, depending on what you mean. Both the original SU(2) and U(1) are broken. However, there is a linear combination of the diagonal generator of SU(2) and U(1) remains unbroken. We call this one photon. The orthogonal linear combination becomes what we call Z-boson.
小白：谢谢sage老师，原来低能下su(2)和u(1)都破缼了，只是su(2)和u(1)对角项生成元的线性组合没有破缼，原来光子的纯洁性是受此线性组合对称性的保护。
但仍有两个问题不明白？
(1)既然低能下su(2)和u(1)都破缼了，为什么目前的电磁场还是u(1)对称的？
(2)由于直接吞食的是哥徳斯通粒子而不是希格斯粒子，希格斯双态粒子和哥德斯通粒子是什么关系
sage:
(1)既然低能下su(2)和u(1)都破缼了，为什么目前的电磁场还是u(1)对称的？
The broken U(1) is not E&M. I have said twice and I will say it again, we call the unbroken linear combination U(1) E&M.
(2)由于直接吞食的是哥徳斯通粒子而不是希格斯粒子，希格斯双态粒子和哥德斯通粒子是什么关系？
Again, there are 4 degrees of freedom in a complex doublet. 3 are eaten Goldstones, and the uneaten one is the Higgs boson.
This is identical to BCS (I thought you worked on condensed matter, so you should be familiar with this.). U(1) is Higgsed by the VEV of a complex scalar. There are two degrees of freedom in a complex scalar. One of them is the Goldstone (eaten), there is an uneaten one (sometimes called radial mode of the Cooper pair). That's the Higgs boson in the BCS.
小白：
谢谢sage老师，我把您的回复翻译一下，请老师看看我理解对否：(1) 实际上低能下的电磁场u(1)对称性就是一种混合对称性。(2) 希格斯复双态有4个自由度，其中被吃掉的三个就是哥德斯通粒子，剩下一个没被吃的就是希格斯玻色子，也就是说希格斯粒子与哥德斯通粒子是一母所生四兄弟。(3) BCS理论中的希格斯玻色子就是库伯对的径向模。请问以上理解对吗？
Sage老师虽然是粒子物理出身，感觉凝聚态物理比我理解得清楚，佩服啊。BCS理论中我没弄清楚的就是哥德斯通模，在BCS理论中它好像没被吃掉，请问sage老师BCS理论中的哥德斯通模最终跑到哪里去了？
sage:
BCS理论中我没弄清楚的就是哥德斯通模，在BCS理论中它好像没被吃掉，请问sage老师BCS理论中的哥德斯通模最终跑到哪里去了？
=======================================
The BCS condensate is made of Cooper pairs, which carries electric charge. In field theoretic language, a charged field gets a VEV. Therefore, E&M is Higgsed. Photon become massive, eating the Goldstone mode. Of course, all of these only happen inside the superconductor.
When photon becomes massive, it means it cannot penetrate very deep (more than the inverse of its mass) into the BCS superconductor. This is called Meissner effect.
小白：
非常非常谢谢Sage老师，您解决了我多年来的一个疑惑，感动得流泪啊，真有”朝闻道，夕死可以”的感觉，原来整个物理学是可以贯通的。
请问，是什么对称性保护光子小姐不受希格斯粒子性骚扰？
sage:
You start with SU(2)x U(1), 4 generators of Lie algebra (or equivalently 4 gauge bosons). Now you have a Higgs doublet, which get's a vacuum expectation value (VEV). There are many ways to understand the symmetry breaking pattern. The easiest is to begin with 4 degrees of freedom in the complex Higgs doublet. After getting the VEV, there can only be 3 Goldstones. Gauge bosons have to eat Goldstone to become massive (for example in BCS). Therefore, only three gauge bosons can become massive after eating the Goldstones. There has to be one remains massless. We call that one photon.
小白：
谢谢Sage老师科普，我翻译一下：在弱电统一能标下，光子小姐和其它三个中间玻色子美女在su(2)×u(1)规范对称下统一成一类女人。低能下，这个对称性会自发破缼，同时产生三个哥徳斯通帅哥，四个美女三个帅哥，在一夫一妻制下必定有一个美女会独守空房，很不幸的是，这个独守空房的美女就是光子小姐。结果，吞食了三个帅哥的三个中间玻色子美女获得了质量，而光子小姐至今没有获得质量。请问Sage老师，是不是光子的u(1)对称没继续破缼，才使她没有机会吞食哥德斯通粒子，也就是说光子实际上是受u(1)对称性保护？
sage:
No or yes, depending on what you mean. Both the original SU(2) and U(1) are broken. However, there is a linear combination of the diagonal generator of SU(2) and U(1) remains unbroken. We call this one photon. The orthogonal linear combination becomes what we call Z-boson.
小白：谢谢sage老师，原来低能下su(2)和u(1)都破缼了，只是su(2)和u(1)对角项生成元的线性组合没有破缼，原来光子的纯洁性是受此线性组合对称性的保护。
但仍有两个问题不明白？
(1)既然低能下su(2)和u(1)都破缼了，为什么目前的电磁场还是u(1)对称的？
(2)由于直接吞食的是哥徳斯通粒子而不是希格斯粒子，希格斯双态粒子和哥德斯通粒子是什么关系
sage:
(1)既然低能下su(2)和u(1)都破缼了，为什么目前的电磁场还是u(1)对称的？
The broken U(1) is not E&M. I have said twice and I will say it again, we call the unbroken linear combination U(1) E&M.
(2)由于直接吞食的是哥徳斯通粒子而不是希格斯粒子，希格斯双态粒子和哥德斯通粒子是什么关系？
Again, there are 4 degrees of freedom in a complex doublet. 3 are eaten Goldstones, and the uneaten one is the Higgs boson.
This is identical to BCS (I thought you worked on condensed matter, so you should be familiar with this.). U(1) is Higgsed by the VEV of a complex scalar. There are two degrees of freedom in a complex scalar. One of them is the Goldstone (eaten), there is an uneaten one (sometimes called radial mode of the Cooper pair). That's the Higgs boson in the BCS.
小白：
谢谢sage老师，我把您的回复翻译一下，请老师看看我理解对否：(1) 实际上低能下的电磁场u(1)对称性就是一种混合对称性。(2) 希格斯复双态有4个自由度，其中被吃掉的三个就是哥德斯通粒子，剩下一个没被吃的就是希格斯玻色子，也就是说希格斯粒子与哥德斯通粒子是一母所生四兄弟。(3) BCS理论中的希格斯玻色子就是库伯对的径向模。请问以上理解对吗？
Sage老师虽然是粒子物理出身，感觉凝聚态物理比我理解得清楚，佩服啊。BCS理论中我没弄清楚的就是哥德斯通模，在BCS理论中它好像没被吃掉，请问sage老师BCS理论中的哥德斯通模最终跑到哪里去了？
sage:
BCS理论中我没弄清楚的就是哥德斯通模，在BCS理论中它好像没被吃掉，请问sage老师BCS理论中的哥德斯通模最终跑到哪里去了？
=======================================
The BCS condensate is made of Cooper pairs, which carries electric charge. In field theoretic language, a charged field gets a VEV. Therefore, E&M is Higgsed. Photon become massive, eating the Goldstone mode. Of course, all of these only happen inside the superconductor.
When photon becomes massive, it means it cannot penetrate very deep (more than the inverse of its mass) into the BCS superconductor. This is called Meissner effect.
小白：
非常非常谢谢Sage老师，您解决了我多年来的一个疑惑，感动得流泪啊，真有”朝闻道，夕死可以”的感觉，原来整个物理学是可以贯通的。
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