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jostpuur
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[SOLVED] N+pi->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
<br /> \mathcal{H}_{\textrm{int}}(x) = -g \overline{\psi}(x)\psi(x)\phi(x).<br />
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
<br /> T(\overline{\psi}_1\psi_1\phi_1 \overline{\psi}_2\psi_2\phi_2)<br />
(here lower indices 1 and 2 refer to position parameters x_1, x_2), contains two contractions which give the desired operators. They are
<br /> :\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)<br />
and
<br /> :\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)<br />
By substituting
<br /> 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)}<br />
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.
<br /> \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<br />
<br /> \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}<br />
(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
<br /> \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')\;<br /> \phi^-_2\; \overline{\psi}^-_{b2}\; \phi^+_1\; \psi^+_{a1}\;a^{\dagger}(k) a^{s\dagger}(p) |0\rangle = \cdots<br />
<br /> \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}<br />
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
<br /> \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<br />
turns out to be
<br /> \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)<br />
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?
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
<br /> \mathcal{H}_{\textrm{int}}(x) = -g \overline{\psi}(x)\psi(x)\phi(x).<br />
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
<br /> T(\overline{\psi}_1\psi_1\phi_1 \overline{\psi}_2\psi_2\phi_2)<br />
(here lower indices 1 and 2 refer to position parameters x_1, x_2), contains two contractions which give the desired operators. They are
<br /> :\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)<br />
and
<br /> :\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)<br />
By substituting
<br /> 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)}<br />
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.
<br /> \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<br />
<br /> \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}<br />
(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
<br /> \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')\;<br /> \phi^-_2\; \overline{\psi}^-_{b2}\; \phi^+_1\; \psi^+_{a1}\;a^{\dagger}(k) a^{s\dagger}(p) |0\rangle = \cdots<br />
<br /> \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}<br />
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
<br /> \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<br />
turns out to be
<br /> \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)<br />
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:
hmhm...