Simplifying an expression given (on-shell & off-shell) conditions

[JD_PM]

TL;DR Summary

I want to understand how to get Eq. 8.71a in Mandl & Shaw

I was studying how to compute an unpolarized cross-section (QFT Mandl & Shaw, second edition,https://ia800108.us.archive.org/32/items/FranzMandlGrahamShawQuantumFieldTheoryWiley2010/Franz%20Mandl%2C%20Graham%20Shaw-Quantum%20Field%20Theory-Wiley%20%282010%29.pdf) and came across the following expression

$$Z:=16 \{ 2(f_1p)(f_1p') -f_1^2(pp')+m^2[-4(pf_1)+4f_1^2]+m^2 [(pp')-4(f_1 p')] +4m^4 \} \tag 1$$

Where:

$$f_1:=p+k \tag 2$$

We're also given the on-shell and off-shell conditions:

$$p^2 = p'^2=m^2, \ k^2=k'^2=0, \ pk=p'k', \ pk'=p'k \tag 3$$

Plugging $(2)$ into $(1)$ I get (I include all steps so that we can find the mistake)

$$Z=16\{2(p+k)p(p+k)p'-(p+k)^2(pp')+m^2[-4p(p+k)+4(p+k)^2]+m^2[(pp')-4(p+k)p']+4m^4\} \tag 4$$

Expanding out $(4)$ I get

$$Z=16\{ 2p^3p'+2kpkp'+2p^2kp'+2kp^2p'-p^3p'-2pkpp'-k^2pp'-4m^2p^2-4m^2pk+4m^2p^2+8m^2pk+4m^2k^2+m^2pp'-4m^2pp'-4m^2kp'+4m^4\} \tag 5$$

Now it is about simplifying $(5)$. I'll go step by step.

I first used $k^2=k'^2=0$, $p^2 = p'^2=m^2$ conditions to get

$$Z=16\{ 2p^3p'+2kpkp'+2p^2kp'+2kp^2p'-p^3p'-2pkpp'-4m^2pk+8m^2pk+m^2pp'-4m^2pp'-4m^2kp'+4m^4\} \tag 6$$

Then I simplified $Xm^2kp'$, $Ym^2pp'$ terms and got

$$Z=16\{ +4m^2(pk) -2m^2pp'+4m^4+2kpkp'-2pkpp'\} \tag 6$$

Mmm but this is not the provided solution

$$Z=32\{m^4+m^2(pk)+(pk)(pk')\}$$

At least I got $m^4$ and $m^2(pk)$ right but I got the extra term $-2m^2pp'$ and did not get $(pk)(pk')$. Actually, if the term $-p^3p'$ in $(5)$ were to be positive, $Xm^2kp'$ would cancel out but it is not the case.

Where did I get wrong then?

Any help is appreciated.

Thank you.

PS: I asked it https://math.stackexchange.com/questions/3702158/simplifying-an-expression-given-certain-on-shell-off-shell-conditions but got no answer.

 

[Dr.AbeNikIanEdL]

I don't have the time to go through all of this, but note that it is very confusing (possibly also to yourself) that you do not make clear which vectors are multiplied in a scalar product and which scalar products are then just multiplied as numbers. For example, your fourth term in (4), $kp^2p^\prime$, if I identified it correctly, should read $(k\cdot p)(p\cdot p^\prime)$, which is not equal to $k(p^2)p^\prime = m^2 (k\cdot p^\prime)$. I suggest to go through the calculation again making sure to mark scalar product between vectors and products between numbers separately.

 

[nrqed]

I don't have the time to go through all of this, but note that it is very confusing (possibly also to yourself) that you do not make clear which vectors are multiplied in a scalar product and which scalar products are then just multiplied as numbers. For example, your fourth term in (4), $kp^2p^\prime$, if I identified it correctly, should read $(k\cdot p)(p\cdot p^\prime)$, which is not equal to $k(p^2)p^\prime = m^2 (k\cdot p^\prime)$. I suggest to go through the calculation again making sure to mark scalar product between vectors and products between numbers separately.

Dr AbeNikIanEdl is absolutely right. I had the same reaction when reading your post. One must indicate which four-vectors are contracted and the most used convention is to use a dot symbol. Otherwise something like $pkpk'$ is ambiguous. But I notice that Mandl and Shaw do not use a dot, which is probably why you started writing things this way. However, note that Mandl and Shaw still indicate clearly what four vectors are contracted, although it might be easy to miss. Note that whenever two four-vectors are contracted, they are placed inside a parenthesis. For example they might write $(pk)(pk')$. So they still show clearly which ones are contracted, but the dot notation is more widely used now.

 

[JD_PM]

Dr AbeNikIanEdl is absolutely right. I had the same reaction when reading your post. One must indicate which four-vectors are contracted and the most used convention is to use a dot symbol. Otherwise something like $pkpk'$ is ambiguous. But I notice that Mandl and Shaw do not use a dot, which is probably why you started writing things this way. However, note that Mandl and Shaw still indicate clearly what four vectors are contracted, although it might be easy to miss. Note that whenever two four-vectors are contracted, they are placed inside a parenthesis. For example they might write $(pk)(pk')$. So they still show clearly which ones are contracted, but the dot notation is more widely used now.

Oh I did not know that the dot notation was more widely used, thanks for informing.

So let's go step by step

Using the dot notation on $(1)$ we get

$$Z:=16 \{ 2(f_1 \cdot p)(f_1 \cdot p') -(f_1 \cdot f_1)(p \cdot p')+m^2[-4(p \cdot f_1)+4(f_1 \cdot f_1)]+m^2 [(p \cdot p')-4(f_1 \cdot p')] +4m^4 \} \tag 1$$

Now let's work out the term $(f_1 \cdot p)(f_1 \cdot p')$ as an example; I get

$$(f_1 \cdot p)(f_1 \cdot p')=\Big((p+k)\cdot p\Big)\Big((p+k)\cdot p'\Big)=\Big(m^2+p\cdot k\Big)\Big(p\cdot p' + p'\cdot k\Big)$$ $$=m^2(p\cdot p')+(p\cdot k)(p'\cdot k)+m^2(p'\cdot k)+(p\cdot k)(p\cdot p')$$

Is this the way to go then?

 

[Dr.AbeNikIanEdL]

Looks good to me now.