# N+pi-&gt;N+pi exercise in QFT

1. Dec 9, 2007

### jostpuur

[SOLVED] N+pi-&gt;N+pi exercise in QFT

The due date of this exercise was several weeks ago, but I'm still struggling with this. Since some of the QFT exercises are kind of exercises, that are probably the same all over the world, I assumed there could be a non-zero probability that somebody else has been doing this same exercise earlier.

The exercise is about collision of some nucleon N and meson pi, where nucleon is described with a fermion field and the meson with a scalar field, with an interaction Hamiltonian

$$\mathcal{H}_{\textrm{int}}(x) = -g \overline{\psi}(x)\psi(x)\phi(x).$$

We were supposed to use Feynman rules, but I though I would first check with an explicit calculation that I get the same amplitudes, as we should get with the Feynman rules, just to make sure. My problem deals with this part. The intended exercise was about what happens with the amplitudes once they are written using Feynman rules, but that's another problem then.

In N+pi -> N+pi collision we need terms that have $\psi^+$ and $\phi^+$ (for destroying initial particles), and $\overline{\psi}^-$ and $\phi^-$ (for creating final particles). The first order term in the scattering matrix cannot give these operators, so we must use the second order term. The term

$$T(\overline{\psi}_1\psi_1\phi_1 \overline{\psi}_2\psi_2\phi_2)$$

(here lower indices 1 and 2 refer to position parameters $x_1$, $x_2$), contains two contractions which give the desired operators. They are

$$:\underset{\textrm{contr.}}{\overline{\psi}_{a1}} \psi_{a1}\; \phi_1\; \overline{\psi}_{b2} \underset{\textrm{contr.}}{\psi_{b2}} \phi_2 :\; = -i S^{ba}_F(x_1-x_2) \big( \underbrace{\phi_1^-\; \overline{\psi}_{b2}^- \;\phi_2^+ \;\psi_{a1}^+}_{A} + \underbrace{\phi_2^- \;\overline{\psi}_{b2}^- \;\phi_1^+ \;\psi^+_{a1}}_{B} \;+\; \textrm{others}\big)$$

and

$$:\overline{\psi}_{a1} \underset{\textrm{contr.}}{\psi_{a1}} \phi_1 \underset{\textrm{contr.}}{\overline{\psi}_{b2}} \psi_{b2}\; \phi_2:\; = i S^{ab}_F(x_1-x_2)\big(\underbrace{\phi_1^- \;\overline{\psi}_{a1}^-\; \phi_2^+\; \psi_{b2}^+}_{C} + \underbrace{\phi_2^-\;\overline{\psi}_{a1}^-\; \phi_1^+\; \psi_{b2}^+}_{D} \;+\; \textrm{others}\big)$$

By substituting

$$S^{ab}_F(x_1-x_2) = \int\frac{d^4q}{(2\pi)^4} \frac{(\displaystyle{\not}q + m_N)_{ab}}{q^2-m_N^2} e^{-iq\cdot(x_1-x_2)}$$

and the operators $\phi^-(x)$, $\phi^+(x)$, $\overline{\psi}^-_a(x)$ and $\psi^+_a(x)$ in terms of $a^{\dagger}(q)$, $a(q)$, $a^{r\dagger}(q)$ and $a^r(q)$ (where r=1,2), I calculated the amplitude from the term A.

$$\frac{ig^2}{2} \int d^4x_1\; d^4x_2\; S^{ba}_F(x_1-x_2) \langle 0| a(k') a^{s'}(p')\; \phi^-_1\; \overline{\psi}_{b2}^-\; \phi_2^+ \;\psi_{a1}^+\; a^{\dagger}(k) a^{s\dagger}(p) |0\rangle = \cdots$$
$$\cdots = \frac{i g^2}{8(2\pi)^2} \frac{\delta^4(k'+p'-k-p)}{\sqrt{E_{k',\pi} E_{p',N} E_{k,\pi} E_{p,N}}} \frac{\overline{u}^{s'}(p') (\displaystyle{\not}k' - \displaystyle{\not} p + m_N) u^s(p)}{(k'-p)^2 - m_N^2}$$

(the slash covers the prime badly, but it is k' under the slash, and not k)

This looks good. There's same kind of terms that would have come from the Feynman rules, although I'm not sure about the constants.

The amplitude from term B is

$$\frac{ig^2}{2}\int d^4x_1\; d^4x_2\; S^{ba}_F(x_1-x_2) \langle 0| a(k') a^{s'}(p')\; \phi^-_2\; \overline{\psi}^-_{b2}\; \phi^+_1\; \psi^+_{a1}\;a^{\dagger}(k) a^{s\dagger}(p) |0\rangle = \cdots$$
$$\cdots = \frac{ig^2}{8(2\pi)^2} \frac{\delta^4(k'+p'-k-p)}{\sqrt{E_{k',\pi} E_{p',N} E_{k,\pi} E_{p,N}}} \frac{\overline{u}^{s'}(p') (-\displaystyle{\not}k' - \displaystyle{\not} p' + m_N) u^s(p)}{(k'+p')^2 - m_N^2}$$

The D term has identical diagram to the A term, but there are following differences in the expression. a and b are exchanged everywhere, $x_1$ and $x_2$ are exchanged in the operator part but not in the propagator, and there's a minus sign. We can switch a and b back to the same places as they were in the A term, but when we switch $x_1$ and $x_2$, we don't get identical expression with A, because $S^{ba}_F(x_2-x_1)\neq -S^{ba}_F(x_1-x_2)$. This is because of the $m_N$-term in the propagator, which we assume to be nonzero. So in fact A and D don't give identical amplitudes, but instead when they are summed, the $m_N$-terms cancel.

The same thing happens with B and C terms. So the total scattering amplitude (in second order approximation), which should come from the expression

$$\langle 0| a(k') a^{s'}(p') \Big(\frac{(-i)^2}{2!}\int d^4x_1\; d^4x_2\; T(\mathcal{H}_{\textrm{int}}(x_1) \mathcal{H}_{\textrm{int}}(x_2))\Big) a^{\dagger}(k) a^{s\dagger}(p)|0\rangle$$

turns out to be

$$\frac{ig^2}{4(2\pi)^2} \frac{\delta^4(k'+p'-k-p)}{\sqrt{E_{k',\pi} E_{p',N} E_{k,\pi} E_{p,N}}} \Big( \frac{\overline{u}^{s'}(p') (\displaystyle{\not}k' - \displaystyle{\not} p) u^s(p)}{(k'-p)^2 - m_N^2}\quad+\quad \frac{\overline{u}^{s'}(p') (-\displaystyle{\not}k' - \displaystyle{\not} p') u^s(p)}{(k'+p')^2 - m_N^2} \Big)$$

Now... did something go wrong? Where those mass terms supposed to cancel? In the exercise session the assistant drew the two Feynman diagrams, and wrote down corresponding amplitudes using Feynman rules, and the mass terms remained there for rest of the calculation.

I had some difficulties with the fermion contractions previously. One possibility is that something goes wrong with the contractions and propagators right in the beginning.

If the mass terms instead are supposed to cancel, how could one see it simply by drawing the diagrams and using Feynman rules?

Last edited: Dec 9, 2007
2. Dec 9, 2007

### jostpuur

Another related question: Does anyone have any words of wisdom related to the states

$$a^{\dagger}(p)|0\rangle \quad\quad\quad\quad\quad (1)$$

and

$$\sqrt{2 E_p} a^{\dagger}(p)|0\rangle \quad\quad\quad\quad\quad (2)$$

being used as initial and final states in the scattering amplitude calculations?

I used the (1) in a previous exercise that dealt with a B+B->B+B scattering with $\phi^4$ theory, where there was no propagator, and got the correct result. So it seems that I should use (1) because that is what gives correct results on our course, at least... But for example P&S use (2). Now a quick thought would be, that perhaps P&S uses correspondingly different conventions with S-matrix too, but in fact it does not. It is the same

$$T\Big(\exp\Big(-i\int d^4x\; \mathcal{H}_{\textrm{int}}(x)\Big)\Big)$$

both on our course and in the P&S book. If I used (2) in this exercise I would be getting different result, so the situation with these conventions seems slightly confusing. Do the transition amplitudes themselves have different meaning with different conventions?

3. Dec 9, 2007

### Avodyne

The m terms do not cancel. I didn't follow your argument that they should.

Conventions vary with the 2E's. They can go into the amplitude, or into the relation between the amplitude and the cross section. Any one particular book presumably gets a correct final answer for the cross section, but you have to be careful comparing different books on anything but this final answer.

4. Dec 9, 2007

### jostpuur

The A term is

$$-iS^{ba}_F(x_1-x_2) \phi_1^-\;\overline{\psi}_{b2}^-\;\phi_2^+\;\psi_{a1}^+$$

and the D term is, after x1,x2 and a,b have been switched (which can be done since they are dummy variables),

$$iS^{ba}_F(x_2-x_1) \phi_1^-\;\overline{\psi}_{b2}^-\;\phi_2^+\;\psi_{a1}^+$$

Then

$$S^{ba}_F(x_2-x_1) = \int \frac{d^4q}{(2\pi)^4} \frac{(\displaystyle{\not}q + m_N)_{ba}}{q^2-m_N^2} e^{-iq\cdot(x_2-x_1)} = -\int \frac{d^4q}{(2\pi)^4} \frac{(\displaystyle{\not}q - m_N)_{ba}}{q^2-m_N^2} e^{-iq\cdot(x_1-x_2)} \neq -S^{ba}_F(x_1-x_2)$$

They still look like canceling to me.

5. Dec 9, 2007

### jostpuur

Okey, there was a mistake. Assuming that the solution here https://www.physicsforums.com/showthread.php?t=200032 was correct, the first contraction is supposed to give

$$:\underset{\textrm{contr.}}{\overline{\psi}_{a1}} \psi_{a1}\; \phi_1\; \overline{\psi}_{b2} \underset{\textrm{contr.}}{\psi_{b2}} \phi_2 :\; = -i S^{ba}_F(x_2-x_1) \big( \underbrace{\phi_1^-\; \overline{\psi}_{b2}^- \;\phi_2^+ \;\psi_{a1}^+}_{A} + \underbrace{\phi_2^- \;\overline{\psi}_{b2}^- \;\phi_1^+ \;\psi^+_{a1}}_{B} \;+\; \textrm{others}\big)$$

The x1 and x2 are in different order in the propagator. But isn't this only making the situation worse.... now the mass terms remain, and the $\displaystyle{\not}q$ terms are canceling!

hmhm....

It looks like that my previous question about what the contraction becomes when the psi-bar is on left, did not get dealt with properly yet.

Last edited: Dec 9, 2007
6. Dec 10, 2007

### jostpuur

No no no... that was some kind of mistake again. The contractions are

$$:\underset{\textrm{contr}}{\overline{\psi}_{a1}}\psi_{a1}\phi_1 \overline{\psi}_{b2} \underset{\textrm{contr}}{\psi_{b2}}\phi_2: =-iS^{ba}_F(x_2-x_1)(\phi^-_1\;\overline{\psi}^-_{b2}\;\phi^+_2\;\psi^+_{a1} \;+\;\phi^-_2\;\overline{\psi}^-_{b2}\;\phi^+_1\;\psi^+_{a1} \;+\;\textrm{others})$$

$$:\overline{\psi}_{a1}\underset{\textrm{contr}}{\psi_{a1}}\phi_1 \underset{\textrm{contr}}{\overline{\psi}_{b2}} \psi_{b2}\phi_2: = iS^{ab}_F(x_1-x_2)(\phi^-_1\;\overline{\psi}^-_{a1}\;\phi^+_2\; \psi^+_{b2}\;+\; \phi^-_2\;\overline{\psi}^-_{a1}\;\phi^+_1\;\psi^+_{b2} \;+\;\textrm{others})$$

So that

$$S^{(2)} = \frac{(-i)^2g^2}{2!}\int d^4x_1\; d^4x_2\; T\big( \overline{\psi}(x_1)\psi(x_1)\phi(x_1)\overline{\psi}(x_2)\psi(x_2)\phi(x_2)\big)$$
$$= \frac{ig^2}{2}\int d^4x_1\; d^4x_2\;\Big( S^{ba}_F(x_2-x_1)( \underset{A}{\phi^-_1\;\overline{\psi}^-_{b2}\;\phi^+_2\;\psi^+_{a1}} \;+\;\underset{B}{\phi^-_2\;\overline{\psi}^-_{b2}\;\phi^+_1\;\psi^+_{a1}}) \;-\;S^{ab}_F(x_1-x_2)( \underset{C}{\phi^-_1\;\overline{\psi}^-_{a1}\;\phi^+_2\;\psi^+_{b2}} \;+\;\underset{D}{\phi^-_2\;\overline{\psi}^-_{a1}\;\phi^+_1\;\psi^+_{b2}})\Big)$$
$$+\textrm{something non-relevant}$$

Comparison of A and D: The a <-> b and x1 <-> x2 have been switched everywhere, so that they can be switched back and the expressions become identical. But there's a minus sign, and the entire expressions are canceling.... Okey it's just a one minus sign mistake. But where?

Last edited: Dec 10, 2007
7. Dec 10, 2007

### jostpuur

No no no no... The contractions are

$$:\underset{\textrm{contr}}{\overline{\psi}_{a1}}\psi_{a1}\phi_1 \overline{\psi}_{b2} \underset{\textrm{contr}}{\psi_{b2}}\phi_2: =-iS^{ba}_F(x_2-x_1)(-\phi^-_1\;\overline{\psi}^-_{b2}\;\phi^+_2\;\psi^+_{a1} \;-\;\phi^-_2\;\overline{\psi}^-_{b2}\;\phi^+_1\;\psi^+_{a1} \;+\;\textrm{others})$$

$$:\overline{\psi}_{a1}\underset{\textrm{contr}}{\psi_{a1}}\phi_1 \underset{\textrm{contr}}{\overline{\psi}_{b2}} \psi_{b2}\phi_2: = iS^{ab}_F(x_1-x_2)(\phi^-_1\;\overline{\psi}^-_{a1}\;\phi^+_2\; \psi^+_{b2}\;+\; \phi^-_2\;\overline{\psi}^-_{a1}\;\phi^+_1\;\psi^+_{b2} \;+\;\textrm{others})$$

taking into account the sign changes due to fermion operator reordering.

Thank's to the forum for its existence, so that I could write questions and remarks that helped me think.