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:

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

$a_{2.n.1}=\sum _{n_{1},n_{2}}{(a_{3.1}{\hat {p}})}^{n-1-n_{1}-n_{2}}{\dfrac {-1}{2a_{3.1}{\hat {p}}}}\{[\psi ^{\dagger }({\dfrac {1}{2}}+{\dfrac {-\gamma ^{0}\gamma ^{j}{\hat {p_{j}}}-\gamma ^{0}m}{2a_{3.2}{\sqrt {{\hat {p}}^{2}+m^{2}}}}}){(a_{3.2}{\sqrt {{\hat {p}}^{2}+m^{2}}})}^{n_{1}}]\gamma _{0}\gamma _{\mu }{(a_{3.3}{\sqrt {{\hat {p}}^{2}+m^{2}}})}^{n_{2}}({\dfrac {1}{2}}+{\dfrac {-\gamma ^{0}\gamma ^{j}{\hat {p_{j}}}+\gamma ^{0}m}{2a_{3.3}{\sqrt {{\hat {p}}^{2}+m^{2}}}}})\psi {\dfrac {(n_{1}+n_{2})!}{{n_{1}}!{n_{2}}!}}\}$

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

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

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

$a_{2.n.1}=(2\pi )^{-{\dfrac {3}{2}}}\int \sum _{n_{1},n_{2}}{(a_{3.1}p)}^{n-1-n_{1}-n_{2}}{\dfrac {-1}{2a_{3.1}p}}\psi ^{\dagger }\mid _{{\vec {\tau }}_{1}}({\dfrac {1}{2}}+{\dfrac {-\gamma ^{0}\gamma ^{j}{p_{j}}-\gamma ^{0}m}{2a_{3.2}{\sqrt {{p}^{2}+m^{2}}}}})\mid _{{\vec {\tau }}_{1}}{(a_{3.2}{\sqrt {{\tau _{1}}^{2}+m^{2}}})}^{n_{1}}\gamma _{0}\gamma _{\mu }{(a_{3.3}{\sqrt {{({\vec {p}}-{\vec {\tau }}_{1})}^{2}+m^{2}}})}^{n_{2}}({\dfrac {1}{2}}+{\dfrac {-\gamma ^{0}\gamma ^{j}{p_{j}}+\gamma ^{0}m}{2a_{3.3}{\sqrt {{p}^{2}+m^{2}}}}})\mid _{{\vec {p}}-{\vec {\tau }}_{1}}\psi \mid _{{\vec {p}}-{\vec {\tau }}_{1}}{\dfrac {(n_{1}+n_{2})!}{{n_{1}}!{n_{2}}!}}d{\vec {\tau }}_{1}$

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:

$a_{2.n.1}=(2\pi )^{-{\dfrac {3}{2}}}\int {\dfrac {-1}{2a_{3.1}p}}\psi ^{\dagger }\mid _{{\vec {\tau }}_{1}}({\dfrac {1}{2}}+{\dfrac {-\gamma ^{0}\gamma ^{j}{p_{j}}-\gamma ^{0}m}{2a_{3.2}{\sqrt {{p}^{2}+m^{2}}}}})\mid _{{\vec {\tau }}_{1}}\gamma _{0}\gamma _{\mu }({\dfrac {1}{2}}+{\dfrac {-\gamma ^{0}\gamma ^{j}{p_{j}}+\gamma ^{0}m}{2a_{3.3}{\sqrt {{p}^{2}+m^{2}}}}})\mid _{{\vec {p}}-{\vec {\tau }}_{1}}\psi \mid _{{\vec {p}}-{\vec {\tau }}_{1}}\sum _{n_{1},n_{2}}{(a_{3.1}p)}^{n-1-n_{1}-n_{2}}{(a_{3.2}{\sqrt {{\tau _{1}}^{2}+m^{2}}})}^{n_{1}}{(a_{3.3}{\sqrt {{({\vec {p}}-{\vec {\tau }}_{1})}^{2}+m^{2}}})}^{n_{2}}{\dfrac {(n_{1}+n_{2})!}{{n_{1}}!{n_{2}}!}}d{\vec {\tau }}_{1}$

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

$a_{2.n.1}=(2\pi )^{-{\dfrac {3}{2}}}\int {\dfrac {-1}{2a_{3.1}p}}\psi ^{\dagger }\mid _{{\vec {\tau }}_{1}}({\dfrac {1}{2}}+{\dfrac {-\gamma ^{0}\gamma ^{j}{p_{j}}-\gamma ^{0}m}{2a_{3.2}{\sqrt {p^{2}+m^{2}}}}})\mid _{{\vec {\tau }}_{1}}\gamma _{0}\gamma _{\mu }({\dfrac {1}{2}}+{\dfrac {-\gamma ^{0}\gamma ^{j}{p_{j}}+\gamma ^{0}m}{2a_{3.3}{\sqrt {{p}^{2}+m^{2}}}}})\mid _{{\vec {p}}-{\vec {\tau }}_{1}}\psi \mid _{{\vec {p}}-{\vec {\tau }}_{1}}{\dfrac {b_{0}^{n}-b_{1}^{n}}{b_{0}-b_{1}}}d{\vec {\tau }}_{1}$

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:

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

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

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

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

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

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:

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

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

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

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

$A_{\mu .1}(t,{\vec {p}})=(2\pi )^{-{\dfrac {_{3}}{^{2}}}}\int {\dfrac {-1}{2a_{3.1}p}}\psi ^{\dagger }\mid _{t=0,{\vec {\tau }}_{1}}({\dfrac {1}{2}}+{\dfrac {-\gamma ^{0}\gamma ^{j}{p_{j}}-\gamma ^{0}m}{2a_{3.2}{\sqrt {p^{2}+m^{2}}}}})\mid _{{\vec {\tau }}_{1}}\gamma _{0}\gamma _{\mu }({\dfrac {1}{2}}+{\dfrac {-\gamma ^{0}\gamma ^{j}{p_{j}}+\gamma ^{0}m}{2a_{3.3}{\sqrt {{p}^{2}+m^{2}}}}})\mid _{{\vec {p}}-{\vec {\tau }}_{1}}\psi \mid _{t=0,{\vec {p}}-{\vec {\tau }}_{1}}{\dfrac {e^{-itb_{0}}-e^{-itb_{1}}}{b_{0}-b_{1}}}d{\vec {\tau }}_{1}$

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:

$\psi _{0}(t,{\vec {p}})=\sum _{a_{_{3.1}}=\pm 1}e^{-ita_{_{3.1}}{\sqrt {{p}^{2}+m^{2}}}}({\dfrac {1}{2}}+{\dfrac {-\gamma ^{0}\gamma ^{j}{p_{j}}+\gamma ^{0}m}{2a_{3.1}{\sqrt {p^{2}+m^{2}}}}})\psi \mid _{t=0}$

$\psi _{1}\,$ consists two parts:

$\psi _{1}=\psi _{1.a}+\psi _{1.b}\,$

$\psi _{1.a}=(2\pi )^{-{\dfrac {_{3}}{^{2}}}}\int ({\dfrac {1}{2}}+{\dfrac {-\gamma ^{0}\gamma ^{j}{p_{j}}+\gamma ^{0}m}{2a_{3.1}{\sqrt {p^{2}+m^{2}}}}})({\dfrac {1}{2}}A_{\mu }+{\dfrac {1}{2a_{3.2}p}}{\hat {E}}A_{\mu })\mid _{t=0,{\vec {\tau }}_{1}}\gamma _{0}\gamma ^{\mu }({\dfrac {1}{2}}+{\dfrac {-\gamma ^{0}\gamma ^{j}{p_{j}}+\gamma ^{0}m}{2a_{3.3}{\sqrt {p^{2}+m^{2}}}}})\mid _{{\vec {p}}-{\vec {\tau }}_{1}}\psi \mid _{t=0,{\vec {p}}-{\vec {\tau }}_{1}}{\dfrac {e^{-itb_{0}}-e^{-itb_{1}}}{b_{0}-b_{1}}}d{\vec {\tau }}_{1}$

in which:

$b_{0}=a_{3.1}{\sqrt {p^{2}+m^{2}}}$

$b_{1}=a_{3.2}\mid {\vec {\tau }}_{1}\mid +a_{3.3}{\sqrt {\mid {\vec {p}}-{\vec {\tau }}_{1}\mid ^{2}+m^{2}}}$

$\psi _{1.b}=(2\pi )^{-2}\int ({\dfrac {1}{2}}+{\dfrac {-\gamma ^{0}\gamma ^{j}{p_{j}}+\gamma ^{0}m}{2a_{3.1}{\sqrt {p^{2}+m^{2}}}}})B^{\mu }(E,{\vec {p}})\mid _{{\vec {\tau }}_{1}}\gamma _{0}\gamma _{\mu }({\dfrac {1}{2}}+{\dfrac {-\gamma ^{0}\gamma ^{j}{p_{j}}+\gamma ^{0}m}{2a_{3.2}{\sqrt {p^{2}+m^{2}}}}})\mid _{{\vec {p}}-{\vec {\tau }}_{1}}\psi \mid _{t=0,{\vec {p}}-{\vec {\tau }}_{1}}{\dfrac {e^{-itb_{0}}-e^{-itb_{1}}}{b_{0}-b_{1}}}dEd{\vec {\tau }}_{1}$

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