昨晚无聊 推演了一下ksi-重整化

ksi->0 对应landau gauge

ksi->无穷 对应幺正规范 unitarity gauge

加上一些规范条件 可以直接写出总的包含矢量玻色子（vector bosons）、标量玻色子（scalar bosons）、鬼场（ghost fields）的拉格朗日量 并且要满足一般的规范条件

再直接加上费米子的拉氏量 并对所有的拉氏量取到二阶微扰 那么就可以得到各玻色子以及鬼场的质量 然后就直接写出各场的传播子delta 并且要求所有的传播子在对称破缺之前的渐进行为一致（重整化的要求） 但是还有一点 我们要消除对称性破缺时的紫外发散！！！

问题在于 我给定一个拉氏量 微扰展开到高阶 那么在消除无穷大时 我们要确保高阶也是满足同等的对称性的！！比如规范条件等等 昨晚我没有想明白这个问题

转到大家比较熟悉的S矩阵理论：S=<initial state|H_I |final state>

给定哈密顿量H 进行微扰 拆分H=H_0+H_I

H_0是自由粒子的哈密顿量 H_I 是相互作用的哈密顿量 为写出S矩阵进行这样拆分的前提是 H_0与H有完全一样的质量谱 然后把H_I 作为微扰项进行展开

问题就表述为：H_I一定要与H保持同样的各种对称性？ 怎么证明？ 一般来说相互作用会破缺对称性的 大家也许解过一些schrodinger方程 发现了这种情况 那么当延伸至场论时不知道是否会存在这个问题 Γ[x,n]=Γ_R[x,n]+Γ_∞[x,n] Γ_R是重整的 我们是否一定要求Γ_∞与Γ有同样的对称性？怎么证明？ Γ表示场的总拉氏量

我的表述可能还没有问到问题的实质...........CONTINUE THINKING........

I am not sure I understand what you are asking. But, here is what a textbook should say:

The only possible confusing parts are gauge fixing and regularization. Because of these issues, people suspected for quite a while whether gauge theory makes sense. Fadeev-Popov procedure is manifestly gauge invariant. This solves the gauge fixing problem. (The result is gauge invariant in any gauge). And with dimensional regularization, it is straightforward to see that the theory could be regulated in a gauge invariant way. Thereofore, all the counter term generated this way preserves gauge invariance.

Notice that none of these arguments refers to a particular order of perturbative expansion. Therefore, it is true to all orders of perturbation theory.

I am not sure what you are getting at in the H_0 versus H_I. The eigenstates of H=H_0+H_I are different from H_0, of course. However, we are using the spirit of perturbation theory, when H_I is small. We are asking questions such as the S-matrix, where the in-coming and out-going states are eigenstates of H_0, since they are free particles. In this case, eigenstates of H_0, i.e., free particles, are natural choices for the base on which we use our perturbation theory.

Notice that none of these arguments refers to a particular order of perturbative expansion. Therefore, it is true to all orders of perturbation theory.

I nod but want an exact proof in maths. I think we may be very careful while getting rid of infinite in counter terms.Maybe I understand wrongly,but I don't know where is it now.^_^ thanks

I am not sure what you are getting at in the H_0 versus H_I. The eigenstates of H=H_0+H_I are different from H_0, of course. However, we are using the spirit of perturbation theory, when H_I is small. We are asking questions such as the S-matrix, where the in-coming and out-going states are eigenstates of H_0, since they are free particles. In this case, eigenstates of H_0, i.e., free particles, are natural choices for the base on which we use our perturbation theory.

As I know, for writting S matrix, H_0 and H_I should satisfy some conditions.H_0 has the same mass spectrum with H; H_I is so small that we can use perturbative theory ;etc . Oh ,I realised that a not suitable example I used. Let me think in night.I must work for boss now.

thank you very much,sage.
best wishes for u.＾＿＾

I nod but want an exact proof in maths.
============================================

Then you have to formulate a rigorous question. You want the mathenatical proof of

1) Renormalizability?

2) Gauge invariance?

3) Unitarity?

4) All of them?

Then you have to formulate a rigorous question. You want the mathenatical proof of

1) Renormalizability?

2) Gauge invariance?

3) Unitarity?

4) All of them?
*************************************************
I want to understand all of them deeply,even their philosophy. ^_^
I never have traditional physics lessons in classroom but work for Chinese Nuclear Data Center in China Insititue Of Atomic Energy(I am a student ).There are no teachers who give me knowledge on quantum field theory compeletly now,especial renormalizability, Gauge invariance , unitarity and other fundmental concepts in the whole theoretical frame. I even don't know how important they are.and I don't know what is the most important in the whole theory.Maybe the gauge invariance or unitarity or other things? Which concept will be keep in advanced and fundamental theory in future ? Would you talk of it and give me some advices in quantum field theory.thanks very much.（I am reading Weinberg's books on quantum field theory for almost one year after I got here)
hawk
best regards and thanks.

Actually, those concepts like unitarity, gauge invariance, and renormalizability are pretty well understood now, after 'thooft's work.

I won't be able to explain the technical details here because it will take much too long. You are in good hands with Weinberg. On the other hand, it might hard for the fisrt time reader. If so, you might try to read Peskin first. Weinberg is for deeper understanding.

Anyway, in principle:

Unitarity is a requirement of the consistency of any theory. In gauge theories, this is not entirely obvious. From Lorentz invariance, we will write a gauge field as A_mu. However, we know that there are degrees of freedoms in this gauge field that is not physical. Therefore, for unitarity, we must require those degrees of freedom does not contribute to any physical observable. This is ensured by gauge symmetry.

A confusing point is that we must choose a gauge to do our calculation. It is not trivial to verify any calculation is in the end gauge invariant. However, Fadeev-Popov procedure shows that our theory preserve gauge symmetry with any gauge fixing, since F-P is manifestly gauge invariant (a rather technical way to see this is the BRST symmetry).

Another confusing point is that, naively, the classical gauge symmetry is not necessary translating into a full symmetry of every observable, since the full theory is only defined by a regularization+renormalization procedure. For example, the naive cut-off regularization breaks the manifest gauge symmetry (it is still secret gauge invariant in the end). However, it is possible to regulate the theory in a manifestly gauge invariant way. One of the famous example is dimensional regularization.

Yet another seemingly confusing point is that whether all these nice property is still true in a theory where gauge symmetry is spontaneously broken. The answer is that the symmetry is only broken by the vacuum (Or ther physical states does manifest gauge symmetry). However, gauge symmetry is still a good symmetry for the Lagrangian and provides all the services it suppose to do. another way of saying this is that the theory still has a gauge symmetry, although non-linearly realized.

Now, there is also the issue of renormalizability. It turns out that gauge theories are renormalizable. Gauge symmetry played an important role here in that it constrains the type of divergences we can have and relate different divergences. The definite proof is provided by 'tHooft. On the other hand, in the modern point view of effective field theory, renormalizability is not a requirement for a consistent theory.

If you go to old Fan2 Xing1, in the collection of my articles, you could see some discussion about effective field theory.

我不知道是否存在这种可能：在S矩阵的无穷微扰级数展开中，这个级数的整体满足某个对称性，但并不是展开的每一项都分别满足这种对称性。如果这样，在有限阶的微扰近似下，领头的几项主要贡献之和，可能并不满足某个对称性，于是理论本身满足某个对称性，但前面有限项之和未必满足。

不好意思，我楼上白痴了一把，实际情况是显然的：
在S矩阵的级数展开式中，若把耦合常数的幂次称为展开的阶数，且S矩阵满足某种对称性，则S矩阵展开式中每一阶所对应的一个完整的项，都会单独具有这种对称性。因为S矩阵的对称性，其实就是相关的作用量所满足的对称性；而S矩阵相当于这个作用量的指数泛函的编时乘积

我不知道是否存在这种可能：在S矩阵的无穷微扰级数展开中，这个级数的整体满足某个对称性，但并不是展开的每一项都分别满足这种对称性。如果这样，在有限阶的微扰近似下，领头的几项主要贡献之和，可能并不满足某个对称性，于是理论本身满足某个对称性，但前面有限项之和未必满足。

不好意思，我楼上白痴了一把，实际情况是显然的：
在S矩阵的级数展开式中，若把耦合常数的幂次称为展开的阶数，且S矩阵满足某种对称性，则S矩阵展开式中每一阶所对应的一个完整的项，都会单独具有这种对称性。因为S矩阵的对称性，其实就是相关的作用量所满足的对称性；而S矩阵相当于这个作用量的指数泛函的编时乘积

==========================================================
我想是可以找到这种例子的。
举个简单的例子：H（X）=f(x)+g(x).为一般起见，取H为偶函数，而f(x)与g就可以不必是偶函数。这样空间反射对称性在f,g中就被破坏了。当然在此例中我们可以认为g是无穷阶部分，而f是低阶部分。
问题是，我们有什么样的守恒定律的内在要求来避免这种情况？



Fadeev-Popov procedure is manifestly gauge invariant. This solves the gauge fixing problem. (The result is gauge invariant in any gauge).
============================================================

就像上面sage兄提到一样，虽然Fadeev-Popov量子化过程与规范的选取，以及展开式的阶数无关，但是我理解为，我们毕竟还是要选取一个规范来计算，还是要选取一个B[f]函数（虽然一般选为高斯函数以方便计算）。不知道B[f]的奇点会有什么作用不.

就像上面sage兄提到一样，虽然Fadeev-Popov量子化过程与规范的选取，以及展开式的阶数无关，但是我理解为，我们毕竟还是要选取一个规范来计算，还是要选取一个B[f]函数（虽然一般选为高斯函数以方便计算）。不知道B[f]的奇点会有什么作用不.

b[f], gauge-fixing term? which singularity you are referring to?

Yes, we need to fix a gauge. But, the fact we can fix a gauge using FP procedure showes that whatever we get is going to be gauge invariant. no matter which gauge you choose to work in.

"H_0是自由粒子的哈密顿量 H_I 是相互作用的哈密顿量 为写出S矩阵进行这样拆分的前提是 H_0与H有完全一样的质量谱 然后把H_I 作为微扰项进行展开
问题就表述为：H_I一定要与H保持同样的各种对称性？ 怎么证明？ 一般来说相互作用会破缺对称性的"
-----
H_0 的划分 不一定是要求和H 有同样的质量谱吧
只要H_0的谱能构成物理的H的完备基就可以吧 藕合常数够小
另外 场论里面按照藕合常数来划分 也保留了整体对称性 而local的对称性不增加新的内禀的守恒物理量 这样的化分 和量子力学的问题有点不一样吧

"实际情况是显然的：
在S矩阵的级数展开式中，若把耦合常数的幂次称为展开的阶数，且S矩阵满足某种对称性，则S矩阵展开式中每一阶所对应的一个完整的项，都会单独具有这种对称性。因为S矩阵的对称性，其实就是相关的作用量所满足的对称性；而S矩阵相当于这个作用量的指数泛函的编时乘积"

--- 好像不一定吧
如果是有效理论 有高阶藕合的话 DDDD这样的话 S矩阵元展开的第2,3阶本身并不是规范不变吧 只是这种差异是高价的 (和重正化不变性的情况类似) - 我感觉是这样的

如果把S写作S=Texp(iA)，其中T表示编时乘积算符，A代表作用量积分。我的意思是，如果A满足对称性，在S展开的每一阶都应该具备这种对称性。