账本

feynman rules

October 14, 2010

Convention:

$f(x) = \frac{1}{2\pi} \int \mathrm{d}q f(q) e^{-iqx}$ and $f(q) = \int \mathrm{d}x f(x) e^{iqx}$ .

Consider

$\frac{i}{2}g \int_{-\infty}^{z^-} \mathrm{d}y^- A^+(y^-) = \frac{i}{2}g \int_{-\infty}^{+\infty} \mathrm{d}y^- A^+(y^-) \theta(z^--y^-)$ .

Using the identity,

$\frac{-1}{2 \pi i} \int \mathrm{d}s \frac{e^{-\frac{i}{2}(z^--y^-)s}}{s+i\epsilon} =\theta(z^--y^-)$ ,

we will have

$\frac{i}{2}g \int_{-\infty}^{z^-} \mathrm{d}y^- A^+(y^-) = -g \int \frac{\mathrm{d}^4q}{(2\pi)^4} \frac{A^+(q)}{q^+ + i\epsilon} e^{-\frac{i}{2}q^+ z^-}$ .

as well as

$\frac{i}{2}g \int^{\infty}_{z^-} \mathrm{d}y^- A^+(y^-) = -g \int \frac{\mathrm{d}^4q}{(2\pi)^4} \frac{A^+(q)}{-q^+ + i\epsilon} e^{-\frac{i}{2}q^+ z^-}$ .

Now, consider

${\bar h}(z^-) \Phi(z^-) \Phi^\dagger (0) h(0)$
$= \int \frac{\mathrm{d}q_1}{(2\pi)^4} \frac{\mathrm{d}q_2}{(2\pi)^4} e^{\frac{-i}{2}(q^+_1+q^+_2)z^-}{\bar h}(q_1) \Phi(q_2) \int \frac{\mathrm{d}q_3}{(2\pi)^4} \frac{\mathrm{d}q_4}{(2\pi)^4} \Phi^\dagger (q_3) h(q_4)$
$= \int \mathrm{d}\omega e^{\frac{-i}{2}\omega z^-} \int \prod_{i=1}^{4} \frac{\mathrm{d}^4q_i}{(2\pi)^4} \delta(\omega-q^+_1-q^+_2){\bar h}(q_1) \Phi(q_2) \Phi^\dagger (q_3) h(q_4)$

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note on renormalization

October 8, 2010

work in the $\bar{\rm{MS}}$ scheme with dimension regularization $d=4-2\epsilon$ .
First every operator with $n_i$ fields of type $i$ could be written as

${\cal O}_R = \prod_{i} Z_i^{\frac{n_i}{2}} Z^{-1}_{\cal O} \Gamma \Phi_{R,i}^{n_i}$ .

Thus, generally the correction to the operator will be of the form.

$\frac{A_1}{\bar{\epsilon}^2_{\rm UV}} + \frac{A_2}{\bar{\epsilon}_{\rm UV}} + \frac{B_1}{\bar{\epsilon}^2_{\rm IR}} + \frac{B_2}{\bar{\epsilon}_{\rm IR}} + {\rm finit \ terms} + \sum_i \frac{n_i}{2}\delta Z_i - \delta Z_{\cal O}$ .

The ultra-violet term should be canceled by the counter-term, in this way $Z_{\cal O}$ could be determined order by order as long as we know $Z_{i}$ .

Secondly, since ${\cal O}_0 = Z_{\cal O} {\cal O}_R$ and ${\cal O}_0$ is scale
independent. Thus $\mu \mathrm{d} ( Z_{\cal O} {\cal O}_R )/ \mathrm{d} \mu = 0$ .

Third, the correction to the propagator is denoted as $-i\Sigma$ , thus the full propagator is $i/({\bar p} -m -\Sigma[p]) = iR/({\bar p} - m) +{\rm analytic }$ . Once on shell, only the residual part will survive due to the LSZ reduction formula and give a contribution $\sqrt{R}$ .
The fermion self energy has the form

$\Sigma_i[p] = (A_i[p^2]+B_i[p^2]) m_i + B_i[p^2] ({\bar p} - m_i)$ .

and its counter term

$\Sigma_{ct} = -\delta Z_i({\bar p} - m_i) + \delta Z_m m_i$ .

Therefore the contribution to $R$ is given by

$\delta R_i = B_i[m_i^2] + 2m_i^2 \frac{\mathrm{d}(A_i+B_i)}{\mathrm{d}p^2}|_{p^2=m_i^2} -\delta Z_i$ .

Last, in effective theory in some cases the integrations involved will be scaleless, thus zero. Hence the operator with $n_j$ external legs of type $j$ will have corrections (all infrared divergence):

$- \sum_j \frac{n_j}{2}\delta Z_j + \sum_i \frac{n_i}{2}\delta Z_i - \delta Z_{\cal O}= - \sum_j \frac{n_j}{2}\delta Z_j - \frac{A_1}{\bar{\epsilon}^2} -\frac{A_2}{\bar{\epsilon}}$ .

This should reproduce the IR poles in the full theory.

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记录一下

July 10, 2009

有时候，因为不小心或烦，删除了有用的或还要用的文件，而且又清空了回收站（或直接删除而根本不放入回收站）。怎么办？别着急，只要你的电脑还没有运行磁盘整理，且系统完好，任何时候的文件都可以找回来。方法很简单：
1、单击“开始/运行”，输入regedit 打开注册表

2、依次展开：

HKEY_LOCAL_MACHIME/SOFTWARE/microsoft

/WINDOWS/CURRENTVERSION/EXPLORER/DESKTOP/NemeSpace

在左边空白处点击“新建”，选择“主键”，命名为“645FFO40—5081—101B—9F08—00AA002F954E”，再把右边的“默认”主键的键值设为“回收站”，退出注册表。

3、重启电脑即可见到被你删除的文件。

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notes on quantization of Gauge field II — De Witt–Faddeev–Popov Approach

January 9, 2007

There is a good reason to use De Witt–Faddeev–Popov approach in practice to quantize the gauge field. It’s easy. Compare with using Dirac brackets, this functional approach is much easier to get the Feynman rules, especially for non-Abelian case.

Consider partition function

$Z = \int D A^\alpha_\mu D \psi^{\alpha\dagger} D \psi^\alpha exp(iI[A,\psi,\psi^\dagger])$ .

Here I is the gauge invariant action: $I[A,\psi,\psi^\dagger] = I[{}^\Omega A,{}^\Omega \psi,{}^\Omega \psi^\dagger]$ . And also the measure is assumed to be gauge invariant say, $D A^\alpha_\mu D \psi^{\alpha\dagger} D \psi^\alpha = D {}^\Omega A^\alpha_\mu D {}^\Omega \psi^{\alpha\dagger} D {}^\Omega \psi^\alpha$ .However, in the gauge field case, $Z$ will not be convergent, for all those fields can be different by a gauge, and leave the physical results unchanged. In this way, the integral will count the same value over and over again hence making $Z$ diverge. So intuitively, we should expect that $Z = N \cdot z$ , where $z$ is the value of the integral for one gauge choice, and N is the number of possible gauges which is of course infinity. If we can write $Z$ as $Z = N \cdot z$ , we can just strip off the unimportant overall factor $N$ and only take care of the value of $z$ to calculate physical interesting stuffs.

The way to realize this is to insert some $\delta$ function to fix the gauge. For example if we use the Lorentz gauge, we’d better be able to insert something like $\delta(\partial^\mu A^\alpha_\mu - \lambda)$ and not change the value of $Z$ .

To proceed, we refresh our mind by reviewing a character for $\delta$ -function:

$\delta(f(x)-f(a)) = \delta(x-a)/Det|f'(a)|$ ,

which leads to $Det|f'(a)| \delta(f(x)-f(a))= \delta(x-a)$ .And the determinant $Det|f'(a)|$ is independent of $x$ .

Here we will encounter almost the same case, except for replacing $x$ by some functional. We are about to insert in $Z$ a constant $\int D \Omega Det|\Delta| \delta(f(\Omega, x)-\lambda) = 1$ , where the $f(\Omega,x)$ is some gauge choice. this obviously leaves $Z$ unchanged. The integral was over all gauges $\Omega$ . Here , $Det|\Delta|$ is called Faddeev–Popov determinant and it’s easy to show that it is independent of gauge choice $\Omega$ .
After doing so and noting those gauge independent elements, we get
$Z = \int D A^\alpha_\mu D \psi^{\alpha\dagger} D \psi^\alpha D \Omega Det|\Delta| \delta(f(\Omega, x)-\lambda) exp(iI[A,\psi,\psi^\dagger])$ .

$=\int D \Omega \int D A^\alpha_\mu D \psi^{\alpha\dagger} D \psi^\alpha Det|\Delta| \delta(f(\Omega_{fix}, x)-\lambda) exp(iI[A,\psi,\psi^\dagger])$

Now we can drop the overall factor $\int D \Omega$ and if by indroducing a fermion like field (Faddeev–Popov ghost) making

$Det|\Delta|= \int D \omega^\dagger D \omega exp(i \omega^\dagger \Delta \omega)$ ,

we get

$Z = \int D A^\alpha_\mu D \psi^{\alpha\dagger} D \psi^\alpha D \omega^\dagger D \omega \delta(f(\Omega_{fix}, x)-\lambda) exp(iI[A,\psi,\psi^\dagger]+i \omega^\dagger \Delta \omega)$ .

Finally we multiply $Z$ by some distribution function, usually guassian-type $\int D\lambda exp(- \frac{i}{2\xi} \lambda^2)$ , and use the fact that $Z$ doesn’t depend on $\lambda$ . So we can integrate out the $\delta$ -function and get
$Z = \int D A^\alpha_\mu D \psi^{\alpha\dagger} D \psi^\alpha D \omega^\dagger D \omega exp(iI[A,\psi,\psi^\dagger]-i \frac{i}{2\xi} f^2+ i \omega^\dagger \Delta \omega)$

$= \int D A^\alpha_\mu D \psi^{\alpha\dagger} D \psi^\alpha D \omega^\dagger D \omega exp(iI_{NEW} [A,\psi,\psi^\dagger, \omega^\dagger, \omega]).$
Here $I_{NEW} [A,\psi,\psi^\dagger, \omega^\dagger, \omega] = I[A,\psi,\psi^\dagger]-i \frac{i}{2\xi} f^2+ i \omega^\dagger \Delta \omega$ . The second term is the gauge fixed term, the third the ghost field.

After all, we should note here that though we choose a gauge, this $Z$ really does NOT care about what kind gauge we have chosen. That is to say after you evaluate a physical interesting thing such as cross section, the $\xi$ will be canceled out and shouldn’t show up in your final results(if it does, you make some mistakes).

Last we take the QCD as an example. We will use both Lorentz gauge( $\partial^\mu A^\alpha_\mu = 0$ ) and axial gauge( $A^\alpha_3 = 0$ ). In QCD, $A^\alpha_\mu$ transform as $A^\alpha_\mu \rightarrow A^\alpha_\mu + D_\mu \Omega^\alpha$ .

So for Lorentz gauge $f^\alpha =\partial^\mu A^\alpha_\mu$ , $\Delta = \frac{\partial f^\alpha}{\partial \Omega^\beta} = \delta_{\alpha \beta} \partial^\mu D_\mu$ .

And for axial case, $f^\alpha =A^\alpha_3$ , $\Delta = \frac{\partial f^\alpha}{\partial \Omega^\beta} = \delta_{\alpha \beta} \partial_3$ (remember that $A^\alpha_3 = 0$ ). So in the axial case, we see that the ghosts decouple and by gauge indepence, they should decouple in general.

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notes on quantization of Gauge field I — Dirac Bracket

January 4, 2007

Dirac bracket is an extension of Poisson bracket, used to deal with the Hamiltonian with constraints. Especially it is useful in quantizing gauge theories. In detail, Dirac bracket is defined as $[A,B]_D= [A,B]_P - [A, \Phi_i]_P (\Delta^{-1})^{ij} [\Phi_j,B]_P$ , where $[,]_P$ is the Poisson bracket, $\Phi$ is some constraints such as a gauge condition, and $\Delta$ is some matrix we will talk about later.

Suppose there exits a Hamiltonian $H$ with a set of constraints $\{ \Phi_i \approx 0 \}$ (Here “ $\approx$ ” means equations hold on shell, say after evaluate all the Poisson brackets then turn on the constraints conditions. We call this “weakly equal to”). For example, in E&M theory, $L = F^{\mu \nu}F_{\mu \nu}$ . So the canonical momenta are $\Pi^\mu=F^{0\mu}$ . We immediately get a set of primary constraints that $\Pi^0 \approx 0$ (we call this a set of constraints, because the equation holds on every point of space time). These constraints are needed to hold all the time, so $[\Pi^0,H] = 0$ is demanded; hence a set of secondary constraints $\partial_i \Pi^i \approx 0$ follows. There are no more constraints here, because $[\partial_i \Pi_i,H] = 0$ reduces to $0 = 0$ (trivial). If it’s a non-trivial equation, we keep on this process until we get a “ $0 = 0$ “.

According to Dirac, we decompose those constraints into two classes: the first-class constraints and the second-class constraints. The first-class constraint means its Poison bracket with all the other constraints vanishes $[\Phi_i,\Phi_j]_P \approx 0$ on the other hand, the second-class does not, which indicates that no linear combination $\sum c_i [\Phi_i,\Phi_j]_P$ vanishes. As the E&M example above, $\Pi^0$ and $\partial_i \Pi^i$ are first-class constraints.

In quantizing the gauge field, the primary first-class constraints(such as $\Pi^0 \approx 0)$ or maybe all the first-class constraints are harmless for they can be eliminated by a choice of gauge. So we took linear combination of constraints so that as many constraints can be put into first-class as possible then eliminate them by choosing gauge. After that, we left with a set of second-class constraints.

The existence of the second-class constraints indicates some degrees of freedom are not physically important. The naive idea to deal with them is just to “throw them away” and only keeping those degrees of freedom of physical importance. For example, suppose that we have constraints $p_1 \approx 0$ , $q_1 \approx 0$ , we can just take $p_1$ and $q_1$ as identically $0$ . This is a simplest example. For generalization, Dirac proposed the following scheme:

Since for second-class constraints no linear combination $\sum c_i[\Phi_i,\Phi_j]_P$ vanishes, then $Det(\Delta) \neq 0$ , where $\Delta_{ij} \equiv [\Phi_i,\Phi_j]_P$ . Hence $\Delta$ is reversible. By introducing Dirac bracket $[A,B]_D = [A,B]_P - [A,\Phi_i]_P (\Delta^{-1})^{ij}[\Phi_j,B]_P$ , we can get all commutation relations for canonical variables. And use them to quantize gauge field theory. We can check by using the $p_1 \approx 0$ , $q_1 \approx 0$ , that this scheme is the same as we throw away $p_1$ and $q_1$ .

Still use the E&M case as an example. First as we mentioned above , $\Pi^0 = 0$ and $\partial_i \Pi^i = 0$ are first class constraints. Now we propose a gauge(here we use coulomb gauge) $\partial^i A_i = 0$ . Together with $\partial_i \Pi^i=\partial_i F^{i0} = 0$ , we can eliminate $A^0$ (In this case $A^0= 0$ ). Then we left with two second-class constraints: $\partial_i \Pi^i = 0$ and the gauge condition $\partial^i A_i = 0$ , since $\Delta^{1 \bold{x},2 \bold{y}} = -\Delta^{2 \bold{y},1 \bold{x}} = [\partial^i A_i, \partial_i \Pi^i]_P = \nabla^2 \delta( \bold{x}-\bold{y})$ , Δ^1x,1y=Δ^2x,2y = 0, Det(Δ)≠0. Finally, we use the Dirac bracket we get, [A_i, Π^j]_D= iδⁱ_jδ(x-y)+i ∂²/∂_i∂_j(1/4π|x-y|). And extend this to quantum version, to quantize E&M.

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Something New

December 25, 2006

下一个project和这篇文章有关. 也是关于J/psi. 早就该看的, 不过碰上期末和假期, 一直压着. 和前一个pro不同, 这里涉及的scale仿佛多了一点. 先在这里标记一下.

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Weinberg’s QFT

December 19, 2006

很久没有看书(书终归是书, 还是和文献不同). 现在正好是学期末–无心向学但又不该休假，晃荡了一周, 从疯狂的期末恢复过来, 想选本书看看了.

其实也没有选择余地, 手头没什么书, 随手抄了Weinberg的量子场论, 卷二. 以前大略看过卷一, 断断续续地读了几次, 但终没有看完. 坚持最长的是在完成peskin后, 认真读了卷一的前几章, 写的的确精彩. 印象深的是他在用群论(group theory: little group and etc.)来处理粒子场:不同粒子是洛仑兹群不同表示; 零质量粒子必定只有两个自旋分量等等. 这些是别的许多场论书,包括peskin, 都不曾谈及的. 所以如果说别的场论是”高等数学”的话, Weinberg的大概算是”数分”了. 除了书中用了一些拓扑, 群论的知识外, Weinberg的场论中的推导不是一般的跳, 也许是他不被推荐给入门者的理由之一了.

今天随便看了卷二的第一章, 也颇有收获. 他对BRST symmetry的论述相当精彩. 不过, 个人不喜欢他前面的Witt-Faddeev-Popov (non-Abelian)的介绍, 完全留于数学的推导. 但这也许因为在卷一中已有精彩介绍(E&M Abelian case), 第九章, 只是我没有看过:D.

最后, 书是寝室长毕业前赠, 科大de盗版, 需阎xx签条子才能买, 很不错, 我喜欢.

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notes on projects

October 20, 2006

A brief procedure:

1. Determin the Kinematics. Match QCD to SCET + NRQCD, basically by writing down the gauge-invariant operators (color singlet or octect) and doing the non-relativistic expansion for quark spinors and etc. in QCD case then matching the results ,to get the hard coefficient at hard-scale. A tree level work(but NOT easy at all).

2. Calculate the scattering amplitude at that (hard) scale : factorize, optical theorem. So a loop integral may be needed for singlet case.

3. Use Renormalizatoin Group to run the coefficent to some other scale, by calculating the “vertex” corrections. Also need to take operator-mixing into consideration at this step, but it may seem small in my case. Bunch of integrals here, damned hard piece!!!

4. Interpolating: combine the end-point-result with result away from that. Maybe some structure functions are needed. No one knows whether these functions are physically correct or not. So maybe, two minus mak e a plus….

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Feynman(test embedding Youtube)

October 20, 2006

“all kinds of interesting questions which a science knowledge only adds to the excitement and mystery and the awe of a flower. It only adds. I don’t understand how it subtracts.”

我对理科的兴趣也许真的越来越少了, 不过这不能说明数学啦, 物理啦枯燥无聊; 相反地, 这表现理科是多么有趣:)—–你想, 通过12年”乏味的”初等数理教育, 另外经过4年”惨无人道”的本科高等数理教育, 接着是这些年的研究生生涯, 我依然对理科存有那么一丝丝幻想, 并没有完全泯灭. 可见,”Science”是多么有趣, 且令人激动.

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gnuplot, briefly II plot

October 14, 2006

gnuplot最重要的当然是作图拉.:) 譬如experimental data存在”calibration.dat”里:

the1st, 4th, 2nd,5th columns分别是dL , L, dL的实验误差, L的实验误差. 把L作为x, dL作为y来plot, 并且带上两者的实验误差, 用命令: plot “calibration.dat” usi 4:1:5:2 w xyerr title “L v.s dL” , 其中”usi 4:1:5:2″代表用第四,一,二,五列的数据. (默认x,y,xerr,yerr) “w xyerr”表示用xyerror bar, “title …”用来作legend.

若打算一次作多种图比如在上面的例子里再作gaussian函数图像, 就用,\来分隔 :

plot “calibration.dat” usi 4:1:5:2 w xyerr title “L vs dL” ,\

300*exp(-0.01*(x-200)**2) title “gaussian distribution”

当然, 可以把所需要作的流程, 如set term, plot “file”之类写在一个文件里面,如plot.那么直接在linux下输入gnuplot plot就行了. !!

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账本

feynman rules

note on renormalization

记录一下

notes on quantization of Gauge field II — De Witt–Faddeev–Popov Approach

notes on quantization of Gauge field I — Dirac Bracket

Something New

Weinberg’s QFT

notes on projects

Feynman(test embedding Youtube)

gnuplot, briefly II plot

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