I have a problem understanding the occurrence of a the relative minus signs between contributions, coming from different Feynman diagrams, involving fermions. A simple example is Bhabha scattering $e^+ e^- \to e^+ e^-$. This process can happen by scattering or annihilation. I know the heuristic argument as mentioned for example here and in many books. I'm trying to understand this by computation using the S-Matrix expansion.

Disclaimer: I will use quite sloppy notation to come as quickly as possible to my question.

We have $\lvert i\rangle = \lvert e^+ e^-\rangle = c^\dagger d^\dagger \lvert 0\rangle$ and $\langle f\rvert = \langle 0\rvert dc$

The contribution part of the second order S Matrix term is for the scattering diagram (ignoring many things)

$$ S^a \propto (\bar \Psi^- \Psi^+)_{x_1}(\bar \Psi^+ \Psi^-)_{x_2}$$

and for the annihilation diagram

$$ S^a \propto (\bar \Psi^- \Psi^-)_{x_1}(\bar \Psi^+ \Psi^+)_{x_2}$$

The corresponding amplitudes are therefore (again focusing only on the sign relevant parts)

$$\langle f\rvert S^a\lvert i\rangle \propto \langle 0\rvert cd N\{ c^\dagger c dd^\dagger\} c^\dagger d^\dagger \lvert 0\rangle$$ and $$\langle f\rvert S^b\lvert i\rangle \propto \langle 0\rvert cd N\{ c^\dagger d^\dagger d c\} c^\dagger d^\dagger \lvert 0\rangle$$

I have read the corresponding pages in quite a few books, and the standard ways to explain the minus sign are:

I That we need now to bring both terms into equal normal order (Mandl-Shaw)


II that we need to make sure that a $c$ always stands next to an $c^\dagger$ and equally for $d$, i.e. make sure a particle is always annihilated after it is created before another particle is created. (See for example (Quantum Field Theory and the Standard Model - Schwartz)

Using the anti-commutation relations between the creation and annihilation operators leads for both demands to a relative minus sign between the two contributions. My problem is understanding where the need for I or II comes from? In other words: If I follow the instructions in the textbooks I get the correct result, which is the same as if I used the heuristic rule mentioned at the beginning. Anyway I do not understand where these rules comes from.

Why do we need to bring the operators in both amplitudes into equal normal order? Or

Why do we need to annihilate a particle as soon as it was created before another particle is created?

Any help or reading suggestion would be much appreciated

  • $\begingroup$ You need to bring the states to the same order because the two contributions to the amplitude must be standardized in the same way, otherwise you would be adding apples with oranges. One may calculate the amplitude for a fixed well-defined (including the sign) initial state and final state. If your other calculation calculates a different term but the final and/or initial state differ by an odd number of exchanges of fermions, then the initial and final states have to be changed to the same form as the previous term which produces a minus sign. $\endgroup$ – Luboš Motl Oct 6 '14 at 13:10
  • $\begingroup$ Do you understand why, for anticommuting $c_i$ variables, $12 c_1 c_2 + 5 c_2 c_1 = 12 c_1 c_2 - 5 c_1 c_2 = 7 c_1 c_2$? If you do, then I can't understand how you could misunderstand the thing you are asking about. $\endgroup$ – Luboš Motl Oct 6 '14 at 13:11
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    $\begingroup$ @LubošMotl that should be an answer $\endgroup$ – David Z Oct 6 '14 at 13:12
  • $\begingroup$ @LubošMotl I do understand why $12 c_1 c_2 + 5 c_2 c_1 = 12 c_1 c_2 - 5 c_1 c_2 = 7 c_1 c_2$, anyway in the case above we are adding two amplitudes, which are c-numbers: $\langle f\rvert S^a\lvert i\rangle $ and $\langle f\rvert S^b\lvert i\rangle $ and I don't understand why we can't for example add $ <0 | cd c^\dagger d^\dagger c d c^\dagger d^\dagger |0>$ to $<0|cd c^\dagger d^\dagger d c c^\dagger d^\dagger |0> $ $\endgroup$ – jak Oct 6 '14 at 13:18
  • $\begingroup$ Thanks, @DavidZ - I wasn't quite satisfied with my remark and wanted at least one more hint what is the stumbling block... Maybe I see it now. $\endgroup$ – Luboš Motl Oct 6 '14 at 17:58

First, the second equation starting with $S^a\propto\dots $ should probably say $S^b\propto\dots$.

Now, the first two equations for the operators $S^a$ and $S^b$ which are the relevant parts of $S=S^a+S^b$ have the positive plus sign – the additional factors that are omitted don't differ by any extra sign because there is a well-defined factor (and sign) in front of the $\bar\Psi\Psi$ factor of the interaction term in the Lagrangian. You omitted the coefficient $K_a$ and $K_b$ in the two equations, respectively (some photon propagators and other things).

Without the (correct) relative sign flip, the total amplitude would be proportional to $K_a+K_b$.

To proceed (and fix the sign), it is enough to notice that in the last two displayed equations, $$ \begin{eqnarray} N\{c^\dagger c d d^\dagger\} &= c^\dagger d^\dagger c d \\ N\{c^\dagger d^\dagger dc\} &= c^\dagger d^\dagger d c \end{eqnarray} $$ where I was careful to perform an even number of fermionic fields' transpositions (the permutation of $d^\dagger$ through $c$ and $d$ in the first evaluation, or nothing in the second evaluation) so that the manipulation is independent of the question whether $(-1)$ includes the sign flips for the transpositions or not. But the last expression from $S^b$ is, because $dc=-cd$, equal to minus the first one (the $cd$ and $dc$ at the end is the only difference between the two), which is why, after the conversion of everything to the multiples of $c^\dagger d^\dagger cd $ which is still sandwiched in between the (fixed) initial and final states, produces an amplitude proportional to $K_a-K_b$, with the relative minus sign.

To answer "why I or II", I would choose "why I". The reason why we need to bring both terms into the same normal-ordered form is that we want to factorize the coefficients. But for $A_b=-A_a$ with the minus sign coming from the simple counting of permutations of the annihilation operators inside $S$ (or creation operators inside $S$), the distribution law is only possible if $A_a$ may be taken out of the parenthesis i.e. factorized, i.e. if we convert $A_b$ to $-A_a$ first: $$ K_a \cdot A_a + K_b\cdot A_b = K_a \cdot A_a - K_b\cdot A_a = (K_a-K_b)\cdot A_a$$

  • $\begingroup$ Thanks for your answer! After thinking about it for a while I think I'm now able to articulate what still bothers me: I suppose your $A_a= <f| c^\dagger d^\dagger c d |i> $ and $A_b = <f|c^\dagger d^\dagger d c |i>$?! In my understanding both result in the norm of the corresponding state. I don't understand why they are different. In other words: Whats the reson that $ \langle 0\rvert cd c^\dagger d^\dagger c d c^\dagger d^\dagger \lvert 0\rangle \neq \langle 0\rvert cd c^\dagger d^\dagger d c c^\dagger d^\dagger \lvert 0\rangle $? $\endgroup$ – jak Oct 9 '14 at 9:56
  • $\begingroup$ I don't understand how you may misunderstand these things. They differ because one of them has $cd$ somewhere inside and the other has $dc$. Because $cd=-dc$, these two matrix elements are obviously equal to minus each other, aren't they? You may only fail to see why they are minus themselves if you are completely sloppy about all the signs and all the ordering - but indeed, that's a bad starting point to become sure about similar minus sign issues. $\endgroup$ – Luboš Motl Oct 9 '14 at 10:10
  • $\begingroup$ Do you understand that $(cd)^\dagger = d^\dagger c^\dagger$ but $(cd)^\dagger \neq c^\dagger d^\dagger$, for example? It's not hard to see that one of your two final matrix elements equals $+1$ and the other is $-1$, and a minute of calculation using $(AB)^\dagger = B^\dagger A^\dagger$ and the anticommutation etc. is enough to see which is which. $\endgroup$ – Luboš Motl Oct 9 '14 at 10:10
  • $\begingroup$ Oh...Sry! After taking now a second look at what I wrote its completely obvious and I can't understand how I misunderstood it either. Of course $\langle 0\rvert cd c^\dagger d^\dagger c d c^\dagger d^\dagger \lvert 0\rangle \neq \langle 0\rvert cd c^\dagger d^\dagger d c c^\dagger d^\dagger \lvert 0\rangle$ obviously! Thanks for your patience and your help! $\endgroup$ – jak Oct 9 '14 at 10:14

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