User:Peter484

The main idea of series expansion method is at: http://en.wikipedia.org/wiki/User:Peter483

The QED equation is:

$$i \gamma^{\mu} \partial_{\mu} \psi -m\psi=q\gamma_\mu (A^{\mu}+B^{\mu})\psi$$

$$\partial_{\nu} \partial^{\nu} A^{\mu} -\partial^{\mu} \partial_{\nu} A^{\nu}=q\bar{\psi} \gamma^{\mu} \psi$$


 * $$ \gamma_\mu \,\!$$ are Dirac matrices;
 * $$\ \psi$$ a bispinor field of spin-1/2 particles (e.g. electron-positron field);
 * $$\bar\psi\equiv\psi^\dagger\gamma_0$$, called "psi-bar", is sometimes referred to as Dirac adjoint;
 * $$\ q $$ is the coupling constant, equal to the electric charge of the bispinor field;
 * $$\ A_\mu $$ is the covariant four-potential of the electromagnetic field generated by electron itself;
 * $$\ B_\mu $$ is the external field imposed by external source;

Using Lorentz gauge $$\partial_{\mu} A^{\mu} =0\,$$, equation for $$A^{\mu} \,$$ can be expressed as:

$$\partial_{\nu} \partial^{\nu} A^{\mu}=q\bar{\psi} \gamma^{\mu} \psi$$

Then we seperate time partial derivative from other parts, we get:

$$i\gamma^0 \partial_0 \psi=(-i\gamma^j \partial_j + m)\psi +q\gamma_\mu (A^{\mu}+B^{\mu})\psi$$

$$\partial_0 \partial^0 A^{\mu}=-\partial_j \partial^j A^{\mu}+q\bar{\psi} \gamma^{\mu} \psi$$

For simplicity, we use $$\hat{p^{\mu}} \rightarrow i\dfrac{\partial}{\partial x_{\mu}}$$ to replace operators in these two equations. And add $$\gamma^0 \,$$ to the $$\psi \,$$ equation:

$$\hat{E} \psi=(-\gamma^0 \gamma^j \hat{p_j} + \gamma^0 m)\psi +q \gamma^0 \gamma_\mu (A^{\mu}+B^{\mu})\psi$$......(1)

$${\hat{E}}^2 A^{\mu}= \hat{p}^2 A^{\mu}-q \bar{\psi} \gamma^{\mu} \psi$$......(2)

here we have used $$\hat{E}\rightarrow \hat{p_0}$$ and $$\hat{p}^2 \rightarrow \sum_j {\hat{p_j}}^2 \rightarrow -\bigtriangleup$$

Add operator $$\hat{E} \rightarrow i\dfrac{\partial}{\partial t}$$ to equation (1) and (2) and use (1)&(2) repeatedly, we will get the expression of higher order time partial derivative of $$\psi \,$$ and $$A^{\mu} \,$$

$${\hat{E}}^2\psi = (-\gamma^0 \gamma^j \hat{p_j} + \gamma^0 m) (\hat{E}\psi) +q \gamma^0 \gamma_{\mu} [\hat{E}(A^{\mu}+B^{\mu})]\psi+q\gamma^0 \gamma_{\mu} (A^{\mu}+B^{\mu}) (\hat{E}\psi)$$

$${\hat{E}}^2\psi =(-\gamma^0 \gamma^j \hat{p_j} + \gamma^0 m)[(-\gamma^0 \gamma^j \hat{p_j} + \gamma^0 m)\psi +q \gamma^0 \gamma_\mu (A^{\mu}+B^{\mu})\psi] +q \gamma^0 \gamma_{\mu} [\hat{E}(A^{\mu}+B^{\mu})]\psi+q\gamma^0 \gamma_{\mu} (A^{\mu}+B^{\mu}) [(-\gamma^0 \gamma^j \hat{p_j} + \gamma^0 m)\psi +q \gamma^0 \gamma_\mu (A^{\mu}+B^{\mu}]\psi )$$

Next we will use an abbreviation for the expression of $$\psi \,$$ and $$A^{\mu}$$ according to the order of electric charge $$q \,$$:

$${\hat{E}}^n \psi = \sum_j q^j a_{1.n.j}$$

$$ {\hat{E}}^n A_{\mu}= \sum_j q^j a_{2.n.j} $$

Then it is our job to calculate $$a_{1.n.j} \,$$ and $$a_{2.n.j} \,$$.

The expression of $$a_{1.n.0} \, $$ and $$a_{2.n.0}\,$$ is quite simple:

$$ a_{1.n.0}= (-\gamma^0 \gamma^j \hat{p_j} + \gamma^0 m)^n \psi$$

$$a_{2.n.0}\,$$ is a little complicated:

When $$n\,$$ is even:

$$a_{2.n.0} = \hat{p}^n A_{\mu} $$

When $$n\,$$ is odd:

$$a_{2.n.0} = \hat{p}^{n-1} \hat{E}A_{\mu} $$

Using $$ ( -\gamma^0 \gamma^j \hat{p_j} + \gamma^0 m )^2 = {\hat{p}}^2 +m^2$$, we can get:

When $$n\,$$ is even:

$$a_{1.n.0}= ({\hat{p}}^2 +m^2)^{\dfrac{n}{2}} \psi$$

When $$n\,$$ is odd:

$$a_{1.n.0} = ({\hat{p}}^2 +m^2)^{\dfrac{n-1}{2}} (-\gamma^0 \gamma^j \hat{p_j} + \gamma^0 m ) \psi$$

Now we can see we must divide the situation into $$n\,$$ is even and $$n\,$$ is odd. For simplicity, we want to combine the two expressions into one equation.

We notice the function $$ \dfrac{1+(-1)^n}{2} $$ and $$ \dfrac{1-(-1)^n}{2} $$.

When $$n\,$$ is even:

$$\dfrac{1+(-1)^n}{2} =1$$ and $$\dfrac{1-(-1)^n}{2} =0$$

When $$n\,$$ is odd:

$$\dfrac{1+(-1)^n}{2}=0$$ and $$\dfrac{1-(-1)^n}{2}=1$$

$$ ( -\gamma^0 \gamma^j \hat{p_j} + \gamma^0 m )^n =\dfrac{1+(-1)^n}{2} ({\hat{p}}^2 +m^2)^{\dfrac{n}{2}} +\dfrac{1-(-1)^n}{2} ({\hat{p}}^2 +m^2)^{\dfrac{n-1}{2}} (-\gamma^0 \gamma^j \hat{p_j} + \gamma^0 m ) $$

$$ a_{1.n.0} = {( a_{3.1} \sqrt{{\hat{p}}^2+m^2} )}^n ( \dfrac{1}{2} + \dfrac{-\gamma^0 \gamma^j \hat{p_j} + \gamma^0 m}{2a_{3.1} \sqrt{{\hat{p}}^2+m^2}}) \psi$$

here $$a_{3.1}=\pm 1$$.

We should keep in mind that $${\hat{p}}^2 +m^2 = -\bigtriangleup +m^2 $$. $$\sqrt{{ \hat{p} }^2+m^2} $$ has no meaning in the coordinates space. But since in the end we will transform the result into 3-dimension momentum space, we allow the use of $$\sqrt{{\hat{p}}^2+m^2}$$ temporarily. It's the same situation with operator $$\hat{p}$$ and $$\dfrac{1}{\hat{p}}$$.

Using the same method:

$$a_{2.n.0} = { (a_{3.1}\hat{p}) }^n ( \dfrac{1}{2} A_{\mu} + \dfrac{1}{2 a_{3.1} \hat{p} } \hat{E} A_{\mu}) $$

$$a_{2.n.1} = \sum_{n_1,n_2} { (a_{3.1}\hat{p}) }^{n-1-n_1-n_2} \dfrac{-1}{2 a_{3.1} \hat{p} } \{ [ \psi^{\dagger} ( \dfrac{1}{2} + \dfrac{-\gamma^0 \gamma^j \hat{p_j} - \gamma^0 m}{2a_{3.2} \sqrt{{\hat{p}}^2+m^2}}) {( a_{3.2} \sqrt{{\hat{p}}^2+m^2} )}^{n_1} ] \gamma_0 \gamma_{\mu}  {( a_{3.3} \sqrt{{\hat{p}}^2+m^2} )}^{n_2} ( \dfrac{1}{2} + \dfrac{-\gamma^0 \gamma^j \hat{p_j} + \gamma^0 m} {2a_{3.3} \sqrt{{\hat{p}}^2+m^2}}) \psi \dfrac{(n_1+n_2)!}{{n_1}!{n_2}!} \} $$

here $$a_{3.2},a_{3.3}=\pm 1$$

Then we will transform the result into 3-dimension momentum space:

$$f_{\vec{p}}(t,\vec{p})= (2\pi)^{-\dfrac{3}{2}} \int f_{\vec{x}} (t,\vec{x}) e^{-i\vec{p} \cdot \vec{x}} d\vec{x}$$

$$a_{2.n.1}=(2\pi)^{-\dfrac{3}{2}} \int \sum_{n_1,n_2} { (a_{3.1} p ) }^{n-1-n_1-n_2} \dfrac{-1}{2 a_{3.1} p }   \psi^{\dagger} \mid_{ \vec{\tau}_1 } ( \dfrac{1}{2} + \dfrac{-\gamma^0 \gamma^j {p_j} - \gamma^0 m}{2a_{3.2} \sqrt{{{p}}^2+m^2}}) \mid_{\vec{\tau}_1 } {( a_{3.2} \sqrt{{ \tau_1 }^2+m^2} )}^{n_1}  \gamma_0 \gamma_{\mu}  {( a_{3.3} \sqrt{{ (\vec{p}-\vec{\tau}_1 ) }^2+m^2} )}^{n_2} ( \dfrac{1}{2} + \dfrac{-\gamma^0 \gamma^j {p_j} + \gamma^0 m} {2a_{3.3} \sqrt{{{p}}^2+m^2}})\mid_{\vec{p}-\vec{\tau}_1} \psi \mid_{\vec{p}-\vec{\tau}_1} \dfrac{(n_1+n_2)!}{{n_1}!{n_2}!} d\vec{\tau}_1 $$

Here we have used:

$$( \dfrac{1}{2} + \dfrac{-\gamma^0 \gamma^j {p_j} - \gamma^0 m}{2a_{3.2} \sqrt{{{p}}^2+m^2}}) \mid_{\vec{\tau}_1 }=( \dfrac{1}{2} + \dfrac{-\gamma^0 \gamma^j \tau_{1.j} - \gamma^0 m}{2a_{3.2} \sqrt{{ \mid \vec{\tau}_1 \mid }^2+m^2}})$$

And we must keep in mind:

$$\psi^{\dagger} \mid_{ \vec{\tau}_1 }=(2\pi)^{-\dfrac{3}{2}} \int \psi_{x^{\mu}}^{\dagger} (t,\vec{x}) e^{-i\vec{\tau}_1 \cdot \vec{x}} d\vec{x}$$

It means $$ \psi^{\dagger} (t,\vec{p})$$ is the 3-momentum representation of $$ \psi_{x^{\mu}}^{\dagger} (t,\vec{x}) $$ instead of the conjugate of $$\psi (t,\vec{p})$$

In 3-dimension momentum space, operators in coordinate space become numbers we can easily move:

$$a_{2.n.1}=(2\pi)^{-\dfrac{3}{2}} \int \dfrac{-1}{2 a_{3.1} p }   \psi^{\dagger} \mid_{ \vec{\tau}_1 } ( \dfrac{1}{2} + \dfrac{-\gamma^0 \gamma^j {p_j} - \gamma^0 m}{2a_{3.2} \sqrt{{{p}}^2+m^2}}) \mid_{\vec{\tau}_1 }   \gamma_0 \gamma_{\mu}   ( \dfrac{1}{2} + \dfrac{-\gamma^0 \gamma^j {p_j} + \gamma^0 m} {2a_{3.3} \sqrt{{{p}}^2+m^2}})\mid_{\vec{p}-\vec{\tau}_1} \psi \mid_{\vec{p}-\vec{\tau}_1} \sum_{n_1,n_2}  { (a_{3.1} p ) }^{n-1-n_1-n_2} {( a_{3.2} \sqrt{{ \tau_1 }^2+m^2} )}^{n_1} {( a_{3.3} \sqrt{{ (\vec{p}-\vec{\tau}_1 ) }^2+m^2} )}^{n_2} \dfrac{(n_1+n_2)!}{{n_1}!{n_2}!} d\vec{\tau}_1 $$

Using summation formula of geometric progression and binomial expansion, we get the expression:

$$a_{2.n.1}= (2\pi)^{-\dfrac{3}{2}} \int \dfrac{-1}{2 a_{3.1} p }   \psi^{\dagger} \mid_{ \vec{\tau}_1 } ( \dfrac{1}{2} + \dfrac{-\gamma^0 \gamma^j {p_j} - \gamma^0 m}{2 a_{3.2} \sqrt{p^2+m^2} }) \mid_{\vec{\tau}_1 }   \gamma_0 \gamma_{\mu}   ( \dfrac{1}{2} + \dfrac{-\gamma^0 \gamma^j {p_j} + \gamma^0 m} {2a_{3.3} \sqrt{{{p}}^2+m^2}})\mid_{\vec{p}-\vec{\tau}_1} \psi \mid_{\vec{p}-\vec{\tau}_1} \dfrac{b_0^n-b_1^n}{b_0-b_1} d\vec{\tau}_1 $$

in which:

$$b_0=a_{3.1} p \,$$

$$b_1=a_{3.2} \sqrt{\tau_1^2+m^2}+a_{3.3} \sqrt{(\vec{p}-\vec{\tau}_1)^2+m^2}$$

Here we can $$b_j \,$$ "the j-order total energy".

Then we use the Tylor series expansion:

$$f(t)=\sum_{n\ge 0} (\hat{E}^n f)\mid _{t=0}\times \dfrac{(-it)^n}{n!} $$

$$A_{\mu}(t,\vec{p})=\sum_{n\ge 0} (\hat{E}^n A_{\mu})\mid _{t=0}\times \dfrac{(-it)^n}{n!} $$

Using $${\hat{E}}^n A_{\mu}= \sum_j q^j a_{2.n.j}$$, we get:

$$A_{\mu}(t,\vec{p}) =\sum_{n\ge 0} ( \sum_j q^j a_{2.n.j} )\mid _{t=0}\times \dfrac{(-it)^n}{n!} $$

$$A_{\mu}(t,\vec{p}) =\sum_j q^j [ \sum_n (a_{2.n.j}) \mid_{t=0} \times \dfrac{(-it)^n}{n!} ] $$

For simplicity, we use an abbreviation:

$$ A_{\mu.j}(t,\vec{p}) = \sum_n (a_{2.n.j}) \mid_{t=0} \times \dfrac{(-it)^n}{n!} $$

The solution of $$A_{\mu}(t,\vec{p}) $$ in 3-momentum space can be expressed as:

$$A_{\mu}(t,\vec{p}) =\sum_j q^j A_{\mu.j}(t,\vec{p}) $$

in which:

$$ A_{\mu.0}(t,\vec{p}) = \sum_n (a_{2.n.0}) \mid_{t=0} \times \dfrac{(-it)^n}{n!} $$

$$ A_{\mu.0}(t,\vec{p}) = \sum_{a_{_{3.1}}=\pm 1}\sum_n ( { (a_{3.1}{p} ) }^n ( \dfrac{1}{2} A_{\mu} + \dfrac{1}{2 a_{3.1} {p} } \hat{E} A_{\mu}) ) \mid_{t=0} \times \dfrac{(-it)^n}{n!} $$

$$ A_{\mu.0}(t,\vec{p}) = \sum_{a_{ _{3.1} }=\pm 1} ( \dfrac{1}{2} A_{\mu} + \dfrac{1}{2 a_{3.1} {p} } \hat{E} A_{\mu}) \mid_{t=0} e^{-ita_{_{3.1}}p}$$

$$A_{\mu.1}(t,\vec{p}) = \sum_{n>0} (a_{2.n.1}) \mid_{t=0} \times \dfrac{(-it)^n}{n!} $$

$$A_{\mu.1}(t,\vec{p}) = (2\pi)^{-\dfrac{_3}{^2}}  \int  \dfrac{-1}{2 a_{3.1} p }   \psi^{\dagger} \mid_{ t=0,\vec{\tau}_1 } ( \dfrac{1}{2} + \dfrac{-\gamma^0 \gamma^j {p_j} - \gamma^0 m}{2 a_{3.2} \sqrt{p^2+m^2} }) \mid_{\vec{\tau}_1 }   \gamma_0 \gamma_{\mu}   ( \dfrac{1}{2} + \dfrac{-\gamma^0 \gamma^j {p_j} + \gamma^0 m} {2a_{3.3} \sqrt{{{p}}^2+m^2}})\mid_{\vec{p}-\vec{\tau}_1} \psi \mid_{t=0,\vec{p}-\vec{\tau}_1} \dfrac{ e^{-itb_0} - e^{-itb_1} }{b_0-b_1} d\vec{\tau}_1 $$

in which:

$$b_0=a_{3.1} p \,$$

$$b_1=a_{3.2} \sqrt{\tau_1^2+m^2}+a_{3.3} \sqrt{(\vec{p}-\vec{\tau}_1)^2+m^2}$$

And for simplicity, we leave out the summation sign $$\sum_{ a_{_{3.1}}, a_{_{3.2}}, a_{_{3.3}}, =\pm 1}$$

We can get the solution of $$\psi \,$$ in the same way:

$${\hat{E}}^n \psi = \sum_j q^j a_{1.n.j} $$

$$ \psi (t,\vec{p}) = \sum_n ( \sum_j q^j a_{1.n.j} ) \mid_{t=0} \times \dfrac{(-it)^n}{n!} $$

$$ \psi (t,\vec{p}) = \sum_j q^j [ \sum_n ( a_{1.n.j} ) \mid_{t=0} \times \dfrac{(-it)^n}{n!} ]$$

Then we use an abbreviation:

$$ \psi_j (t,\vec{p}) = \sum_n ( a_{1.n.j} ) \mid_{t=0} \times \dfrac{(-it)^n}{n!} $$

$$\psi(t,\vec{p})=\sum_j q^j \psi_j (t,\vec{p}) $$

The expression of $$\psi_0 \,$$ is quite simple:

$$\psi_0(t,\vec{p})=\sum_{a_{_{3.1}}=\pm 1} e^{-it a_{_{3.1}} \sqrt{{p}^2+m^2} } ( \dfrac{1}{2} + \dfrac{-\gamma^0 \gamma^j {p_j} + \gamma^0 m}{2a_{3.1} \sqrt{ p^2+m^2}}) \psi \mid_{t=0} $$

$$\psi_1 \,$$ consists two parts:

$$\psi_1=\psi_{1.a}+\psi_{1.b} \,$$

$$ \psi_{1.a} =(2\pi)^{-\dfrac{_3}{^2}} \int ( \dfrac{1}{2} + \dfrac{-\gamma^0 \gamma^j {p_j} + \gamma^0 m}{2 a_{3.1} \sqrt{ p^2+m^2} } ) (\dfrac{1}{2} A_{\mu} +\dfrac{1}{2 a_{3.2} p } \hat{E} A_{\mu}) \mid_{t=0,\vec{\tau}_1} \gamma_0 \gamma^{\mu} ( \dfrac{1}{2} + \dfrac{-\gamma^0 \gamma^j {p_j} + \gamma^0 m}{2a_{3.3} \sqrt{ p^2+m^2}})\mid_{\vec{p}-{\vec{\tau}}_1 } \psi \mid_{t=0, \vec{p}-{\vec{\tau}}_1 } \dfrac{e^{-itb_0} - e^{-itb_1}}{ b_0-b_1 } d{\vec{\tau}}_1$$

in which:

$$b_0=a_{3.1} \sqrt{ p^2+m^2} $$

$$b_1=a_{3.2} \mid {\vec{\tau}}_1\mid + a_{3.3} \sqrt{ \mid \vec{p}-{\vec{\tau}}_1 \mid^2 +m^2}$$

$$ \psi_{1.b} =(2\pi)^{-2} \int ( \dfrac{1}{2} + \dfrac{-\gamma^0 \gamma^j {p_j} + \gamma^0 m}{2 a_{3.1} \sqrt{ p^2+m^2} } ) B^{\mu} (E,\vec{p}) \mid_{{\vec{\tau}}_1} \gamma_0 \gamma_{\mu} ( \dfrac{1}{2} + \dfrac{-\gamma^0 \gamma^j {p_j} + \gamma^0 m}{2a_{3.2} \sqrt{ p^2+m^2}})\mid_{ \vec{p}-{\vec{\tau}}_1 } \psi \mid_{t=0, \vec{p}-{\vec{\tau}}_1 } \dfrac{e^{-itb_0} - e^{-itb_1}}{ b_0-b_1 } dE d{\vec{\tau}}_1$$

in which:

$$b_0=a_{3.1} \sqrt{ p^2+m^2} $$

$$b_1=E + a_{3.2} \sqrt{ \mid \vec{p}-{\vec{\tau}}_1 \mid^2 +m^2}$$

Here we have used:

$$ B^{\mu}(E,\vec{p}) =(2\pi)^{-2} \int B_{x^{\nu}}^{\mu} (t,\vec{x}) e^{i(Et-\vec{p} \cdot \vec{x} )} dtd\vec{x} $$

$$ B^{\mu}(E,\vec{p}) $$ is the 4-momentum representation of external field $$B_{x^{\nu}}^{\mu} (t,\vec{x}) $$

And we also have used:

$${\hat{E}}^n B_{\vec{p}}^{\mu} (t,\vec{p})\mid_{t=0}=(2\pi)^{-\dfrac{_1}{^2}} \int E^n B^{\mu} (E,\vec{p}) dE $$

Higher order results can be achieved in the same way.

Now we analyse the term $$ \dfrac{e^{-itb_0} - e^{-itb_1}}{ b_0-b_1 } $$, it comes from $$\sum_{k_1,k_2} \dfrac{(-it)^{k_1+k_2+1}}{(k_1+k_2+1)!} b_{0}^{k_1} b_{1}^{k_2} $$. We can call this term "time-scalar sum up", because it contains scalars related to time.

When $$b_0=b_1 \,$$, $$ \dfrac{e^{-itb_0} - e^{-itb_1}}{ b_0-b_1 } =-it$$. Obviously it becomes infinite when $$t \rightarrow +\infty $$.

Actually:

$$ \dfrac{e^{-itb_0} - e^{-itb_1}}{ b_0-b_1 } =\dfrac{\cos (-tb_0) +i \sin (-tb_0) -\cos (-tb_1) -i \sin (-tb_1)}{b_0-b_1}$$

$$ =\dfrac{ -2 \sin \dfrac{-tb_0 -tb_1}{2} \sin \dfrac{-tb_0 +tb_1}{2} + 2i\cos\dfrac{-tb_0 -tb_1}{2} \sin \dfrac{-tb_0 +tb_1}{2} }{b_0-b_1}$$

$$=\dfrac{\sin \dfrac{-tb_0 +tb_1}{2}}{b_0-b_1} (-2 \sin \dfrac{-tb_0 -tb_1}{2}+ 2i\cos\dfrac{-tb_0 -tb_1}{2} )$$

$$= 2i \dfrac{ \sin \dfrac{-tb_0 +tb_1}{2}}{b_0-b_1} e^{-it( \dfrac{b_0 +b_1}{2} )} $$

We can see when $$b_0=b_1 \,$$, $$ \dfrac{\sin \dfrac{-tb_0 +tb_1}{2}}{b_0-b_1}=-\dfrac{t}{2}$$.

It menas the possibility in 3-momentum space will gather at the point $$b_0=b_1 \,$$.

Then we can use the relation $$\lim_{t\rightarrow +\infty} 2i \dfrac{ \sin \dfrac{-tb_0 +tb_1}{2}}{b_0-b_1} e^{-it( \dfrac{b_0 +b_1}{2} )} =-2\pi i \delta (b_0-b_1) e^{-itb_0} $$

Now we can see that when $$t\rightarrow +\infty$$, the 1st order time-scalar sum up can be expressed as:

$$ \dfrac{e^{-itb_0} - e^{-itb_1}}{ b_0-b_1 } =-2\pi i \delta (b_0-b_1) e^{-itb_0}$$

A more generalized expression is $$\sum_{n=0}^{j} \dfrac{e^{-itb_n} - \sum_{ _{k=0} }^{ ^{j-1} } \dfrac{{(-itb_n)}^k}{k!} } {\prod_{_{k\neq n}} ( b_n-b_k )}$$

The term $$ \sum_{ {k=0} }^{ {j-1} } \dfrac{{(-itb_n)}^k}{k!} $$ appears because of $$a_{1.n.j},a_{2.n.j}=0 \,$$ for $$n<j \,$$, but usually they cancel each other, so we have:

$$ \sum_{n=0}^{j} \dfrac{e^{-itb_n} - \sum_{ _{k=0} }^{ ^{j-1} } \dfrac{{(-itb_n)}^k}{k!} } {\prod_{_{k\neq n}} ( b_n-b_k )} = \sum_{n=0}^{j} \dfrac{e^{-itb_n}} {\prod_{_{k\neq n}} ( b_n-b_k )} $$.

In the same way as the $$\dfrac{e^{-itb_0} - e^{-itb_1}}{ b_0-b_1 }$$, we can derive a equation:

$$\lim_{t\rightarrow +\infty} \sum_{n=0}^{j} \dfrac{e^{-itb_n}} {\prod_{_{k\neq n}} ( b_n-b_k )} =(-2\pi i)^j e^{-itb_0}\prod_{n=1}^{j}\delta (b_{n-1}-b_n)$$

To achieve this, we make $$b_0 \sim b_j \,$$ all have an infinitesimal imaginary part and make sure $$Im(b_0)<Im(b_1) \cdots Im(b_j) $$. Then we do an integral $$\int db_1 db_2 \cdots db_j$$. In every step, we integral at a line nearby real axis and a complex plane lower half circle (which forming a closed curve). Since $$t\rightarrow +\infty$$, the integral of complex plane lower half circle becomes zero and we can use the residue theory. Because we must finish the $$j \,$$ steps of integral, the only way to get a non-zero integral result is:

Start from $$\dfrac{e^{-itb_j}} {\prod_{_{k\neq j}} ( b_j-b_k )} $$, do integral at this order:$$\int db_j db_{j-1} \cdots db_1$$. After an integral $$\int db_k$$, we can see there's $$k \,$$ residues we can use since $$Im(b_0)<Im(b_1) \cdots Im(b_k) $$. And we must choose the residue at $$b_k=b_{k-1} \,$$ and then do integral $$\int db_{k-1}$$.

Finally we can get the result:

$$\lim_{t\rightarrow +\infty} \sum_{n=0}^{j} \dfrac{e^{-itb_n}} {\prod_{_{k\neq n}} ( b_n-b_k )} =(-2\pi i)^j e^{-itb_{_0}}\prod_{n=1}^{j}\delta (b_{n-1}-b_n)$$

Sometimes we can get $$b_0=b_1=\cdots =b_j \,$$. For example:

The 2-order time-scalar sum up is $$ \dfrac{e^{-itb_0}} {(b_0-b_1)(b_0-b_2)} + \dfrac{e^{-itb_1}} {(b_1-b_0)(b_1-b_2)} +\dfrac{e^{-itb_2}} {(b_2-b_0)(b_2-b_1)} $$

If $$b_0=b_2\neq b_1$$, we simply ignore the term $$ \dfrac{e^{-itb_1}} {(b_1-b_0)(b_1-b_2)} $$, so:

$$\lim_{t\rightarrow +\infty} \dfrac{e^{-itb_0}} {(b_0-b_1)(b_0-b_2)} + \dfrac{e^{-itb_1}} {(b_1-b_0)(b_1-b_2)} +\dfrac{e^{-itb_2}} {(b_2-b_0)(b_2-b_1)} \approx -2\pi i \dfrac{e^{-itb_{_0}}}{b_0-b_1} \delta (b_0-b_2) $$

Another thing to say is that when we can't get $$b_0=b_1=\cdots =b_j \,$$, make sure $$b_0=b_{j_1}=b_{j_2} \cdots$$ with $$0<j_1<j_2 \cdots$$ becasue $$b_0 \,$$ is the only one which doesn't contain integration variables.

Calculation of QED in 3-dimension space: http://blog.sina.com.cn/u/1070440741

peter483@sina.com

peter483@sina.com.cn