Self interacting scalar field

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CAF123
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Homework Statement


A self-interacting real scalar field ##\psi(x)## is described by the Lagrangian density ##\mathcal L = \mathcal L_o + \mathcal L_I = \frac{1}{2} (\partial_{\mu}\psi)(\partial^{\mu}\psi) − \frac{1}{2}m^2\psi^2 − \frac{g}{4!}\psi^4 ## where g is a real coupling constant, and describes the interactions of spinless bosons, B. Write down the S-matrix expansion and identify the lowest-order normal-ordered term that gives rise to the two-body scattering process B(p1)+B(p2) → B(p3)+B(p4). Draw the Feynman diagram corresponding to this process and show that the Feynman amplitude is M = −ig.

Repeat the exercise but now for the Lagrangian ##\mathcal L = \mathcal L_o + \mathcal L_I = \frac{1}{2} (\partial_{\mu}\psi)(\partial^{\mu}\psi) − \frac{1}{2}m^2\psi^2− \frac{\lambda}{3!}\psi^3##.
(Answer: $$\mathcal M = −i\lambda^2\left( \frac{1}{ s − m^2} + \frac{1}{ t − m^2} + \frac{1}{u − m^2}\right))$$

Homework Equations


Generic S matrix expansion element $$S^{(n)} = \frac{(-i)^n}{n!} \int d^4 x_1 \dots \int d^4 x_n T(\mathcal H_I(x_1) \dots \mathcal H_I(x_n))$$

The Attempt at a Solution


Here, ##\mathcal L_I = -\frac{g}{4!}\psi^4 \Rightarrow \mathcal H_I = \frac{g}{4!}\psi^4## since there are no derivative terms. Then the lowest order term in the expansion would be $$S^{(1)} = -i \int d^4 x_1 T(\mathcal H(x_1)) = -i \int d^4 x_1 T(:\frac{g}{4!}\psi_1(x_1)\psi_2(x_1)\psi_3(x_1)\psi_4(x_1):)$$ I think it is correct and that the Feynman diagram is just a single vertex with four lines coming out of it (two incoming and two outgoing), so no propagators involved. This also makes the matrix element depend on one power of g so the answer makes sense but I am not sure how to get to ##\mathcal M## from the above expression for ##S^{(1)}##. It seems like I need to take to take the matrix element between initial and final states: ##|i\rangle = |B_1(p_1), B_2(p_2)\rangle## and ##|f\rangle = |B_3(p_3),B_4(p_4)\rangle## and deduce ##\langle i|S^{(1)}|f\rangle##

Thanks!
 

Answers and Replies

  • #2
CAF123
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Anyone any ideas? I think for the case where ##\mathcal L_I## is proportional to ##\psi^3##(three vertex interaction) then the lowest order diagram with two body scattering corresponds to s and t channel and hence the matrix element is proportional to the coupling squared. It's just getting to the actual matrix element from the last integral I showed above that I still need to get. Thanks!
 
  • #3
king vitamin
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What textbook are you using? You should obtain S matrix elements from time-ordered n-point functions using the LSZ formula, https://en.wikipedia.org/wiki/LSZ_reduction_formula . Have you seen this? Or does your textbook say anything about the relation between the S matrix and time-ordered correlaters?
 
  • #4
CAF123
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Hi King vitamin,
What textbook are you using? You should obtain S matrix elements from time-ordered n-point functions using the LSZ formula, https://en.wikipedia.org/wiki/LSZ_reduction_formula . Have you seen this? Or does your textbook say anything about the relation between the S matrix and time-ordered correlaters?
I have not seen the LSZ formula yet (except in books but it is not part of the course until next semester). I've seen that to calculate matrix elements from terms in the S matrix expansion that I take the term arising from the integral for a given n in the relevent equations above and compute ##\langle f|S^{(1)}|i\rangle##, where ##|i\rangle## are the two incoming bosons and ##|f\rangle## are the two outgoing bosons?

I can write mode expansions for the scalar fields ##\psi_i(x_1)## and then split these into an annihilation piece and a creation piece which act on the initial state. So for two incoming bosons to two outgoing bosons, the matrix element is something like $$\langle f| \psi_1^{-} \psi_2^{-} \psi_3^{+}\psi_4^{+}|i\rangle,$$ reading from right to left, the two incoming particles are destroyed (via ##\psi_{3,4}^{+}##) and the two outgoing particles are created (via ##\psi_{1,2}^{-}##). Is that right?

There is no set textbook but I just dip into Mandl/Shaw and Schwartz

Thanks!
 
  • #5
king vitamin
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Schwartz has an accessible derivation of the LSZ formula fairly early on. The starting point is to write the diagrams for

[tex]
\langle f| \psi_1^{-} \psi_2^{-} \psi_3^{+}\psi_4^{+}|i\rangle = \langle 0|\psi_{out1}^{+} \psi_{out2}^{+} \psi_1^{-} \psi_2^{-} \psi_3^{+}\psi_4^{+}\psi_{in1}^{-} \psi_{in2}^{-}|0\rangle
[/tex]

where I think I'm using the same notation as you (but you might want to check it). But then the LSZ formula basically tells you to chop off the incoming and outgoing propagators, so for phi^4 theory you just get the vertex. In phi^3 theory, there is no first-order diagrams, but you can expand to second order in the coupling, and after chopping off the external propagators you still have the internal propagators, which you can see in the formula you gave in your first post.
 
  • #6
CAF123
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Hi King vitamin, sorry for late reply.
Schwartz has an accessible derivation of the LSZ formula fairly early on. The starting point is to write the diagrams for

[tex]
\langle f| \psi_1^{-} \psi_2^{-} \psi_3^{+}\psi_4^{+}|i\rangle = \langle 0|\psi_{out1}^{+} \psi_{out2}^{+} \psi_1^{-} \psi_2^{-} \psi_3^{+}\psi_4^{+}\psi_{in1}^{-} \psi_{in2}^{-}|0\rangle
[/tex]

where I think I'm using the same notation as you (but you might want to check it).
I managed to get the answer but there are two things which I still have yet to understand fully:
1) What are the ##\psi_{out1}^{+} \psi_{out2}^{+}## and ##\psi_{in1}^{-} \psi_{in2}^{-}## terms you wrote above?
2) In my calculation, I had $$\langle f| S^{(1)}|i\rangle = -i\frac{g}{4!} \int d^4 x_1 \langle i| T(: \psi_1(x_1)\psi_2(x_1)\psi_3(x_1)\psi_4(x_1) :) |f \rangle$$ I think that the time ordering there somehow cancels the 4! factor at the front there, but I am not exactly sure how. Is it just a case of saying there are 4! possible time orderings giving rise to the Feynman diagram where we have one vertex and four bosonic lines attached to it?
Thanks!
 

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