 #1
JD_PM
 1,128
 158
 Homework Statement:

Given the Lagrangian density
\begin{equation*}
\mathcal{L}=\mathcal{L}_0 + \mathcal{L}_I=\frac 1 2 \partial_{\mu} \phi \partial^{\mu} \phi  \frac{m^2}{2} \phi^2  \frac{\lambda_3}{3!} \phi^3  \frac{\lambda_4}{4!} \phi^4
\end{equation*}
Where
\begin{equation*}
\mathcal{L}_0 = \frac 1 2 \partial_{\mu} \phi \partial^{\mu} \phi  \frac{m^2}{2} \phi^2
\end{equation*}
\begin{equation*}
\mathcal{L}_I =  \frac{\lambda_3}{3!} \phi^3  \frac{\lambda_4}{4!} \phi^4
\end{equation*}
a) Derive what are the interaction vertices of this theory.
b) Determine the Feynman rules for the vertices. Hint: the combinatorics
will not be trivial.
 Relevant Equations:
 Please see below
Let us first take the Smatrix expansion (i.e. Dyson's formula)
\begin{align*}
S_{fi}&=\langle f  T \left\{ \exp\left( i\frac{\lambda_3}{3!}\int d^4 x :\phi \phi \phi (x) : + i\frac{\lambda_4}{4!}\int d^4 x :\phi \phi \phi \phi (x) : \right) \right\} i \rangle \\
&= \langle f  i \rangle + i \frac{\lambda_3}{3!} \int d^4 x \langle f :\phi \phi \phi (x): i \rangle + i\frac{\lambda_4}{4!} \int d^4 x \langle f  :\phi \phi \phi \phi (x) :  i \rangle \\
&+ \left(\frac{i\lambda_3}{3!}\right)^2\frac{1}{2!} \int d^4 x d^4 y \langle f T\left\{ : \phi \phi \phi (x) :: \phi \phi \phi (y) : \right\}i\rangle \\
&+ \left(\frac{i\lambda_4}{4!}\right)^2\frac{1}{2!} \int d^4 x d^4 y \langle f T\left\{ : \phi \phi \phi \phi (x) :: \phi \phi \phi \phi (y) :\right\} i\rangle \\
&+ \left(\frac{i\lambda_3}{3!}\right)\left(\frac{i\lambda_4}{4!}\right) \int d^4 x d^4 y \langle f T\left\{ : \phi \phi \phi \phi \phi \phi \phi (x) :\right\} i\rangle + \mathcal{O}(\lambda_3^4, \lambda_4^4)
\end{align*}
a) My guess is that there will be two interaction vertices: ##i \lambda_3## and ##i \lambda_4## corresponding to ##\phi^3## and ##\phi^4## interaction terms (respectively) in the lagrangian. They look as follows (where I used dotted lines for mesons as convention)
OK let's actually prove it. The idea I had in mind is to evaluate the first order terms in the expansion only, and check out explicitly that the factors of ##1/3!## and ##1/4!## cancel out.
For the first one
\begin{align*}
i \frac{\lambda_3}{3!} \int d^4 x \langle f :\phi \phi \phi (x): i \rangle &= \frac{\lambda_3}{3!} \int d^4 x \langle f :\left(\phi^+ + \phi^\right)\left(\phi^+ + \phi^\right)\left(\phi^+ + \phi^\right)(x): i \rangle\\
&= i \frac{\lambda_3}{3!}\int d^4 x \langle f : \phi^ \phi^ \phi^ (x): i \rangle + i \frac{\lambda_3}{3!}\int d^4 x \langle f : \phi^+ \phi^+ \phi^+ (x): i \rangle \\
&+i \frac{\lambda_3}{3!}\int d^4 x \langle f : \phi^ \phi^+ \phi^+ (x): i \rangle + i \frac{\lambda_3}{3!}\int d^4 x \langle f : \phi^+ \phi^ \phi^ (x): i \rangle \\
&+i \frac{\lambda_3}{3!}\int d^4 x \langle f : \phi^ \phi^ \phi^+ (x): i \rangle + i \frac{\lambda_3}{3!}\int d^4 x \langle f : \phi^+ \phi^+ \phi^ (x): i \rangle \\
&+ i \frac{\lambda_3}{3!}\int d^4 x \langle f : \phi^ \phi^+ \phi^ (x): i \rangle + i \frac{\lambda_3}{3!}\int d^4 x \langle f : \phi^+ \phi^ \phi^+ (x): i \rangle \\
&= i \frac{\lambda_3}{3!}\int d^4 x \langle f \phi^ \phi^ \phi^ (x) i \rangle + i \frac{\lambda_3}{3!}\int d^4 x \langle f \phi^+ \phi^+ \phi^+ (x) i \rangle \\
&+i \frac{\lambda_3}{3!} 3\int d^4 x \langle f \phi^ \phi^+ \phi^+ (x) i \rangle + i \frac{\lambda_3}{3!} 3\int d^4 x \langle f \phi^ \phi^ \phi^+ (x) i \rangle
\end{align*}
The issue I have in here is that when evaluating each term individually I do not get the factor of ##1/3!## canceled out. Let's show one computation explicitly
\begin{align*}
i \frac{\lambda_3}{3!}\int d^4 x \langle f \phi^ \phi^ \phi^ (x) i \rangle &= i \frac{\lambda_3}{3!}\left(\frac{1}{2V \omega_{\vec k}}\right)^{3/2}\langle f b^{\dagger}(\vec k)b^{\dagger}(\vec k)b^{\dagger}(\vec k) i\rangle \int d^4 x \exp\left(3k\cdot x\right) \\
&= i \frac{\lambda_3}{3!}\left(\frac{1}{2V \omega_{\vec k}}\right)^{3/2}\langle 0 b(\vec k)b(\vec k)b(\vec k) b^{\dagger}(\vec k)b^{\dagger}(\vec k)b^{\dagger}(\vec k)0\rangle \int d^4 x \exp\left(3k\cdot x\right) \\
&= i \frac{\lambda_3}{3!}\left(\frac{1}{2V \omega_{\vec k}}\right)^{3/2}\langle 00\rangle \int d^4 x \exp\left(3k\cdot x\right) \\
&= i \frac{\lambda_3}{3!}\left(\frac{1}{2V \omega_{\vec k}}\right)^{3/2} \frac{1}{3}\delta^{(4)}(k)
\end{align*}
Where I've used (I am indeed working in n.u.)
$$
\phi^(x)=\sum_{\vec k} \Big(\frac{1}{2V \omega_{\vec k}} \Big)^{1/2} b^{\dagger}(\vec k)e^{ik \cdot x}
$$
$$
\big[ b (\vec k) , b^{\dagger} (\vec k') \big]=\delta_{\vec k \vec k'}
$$
$$
\int \mathrm{d}^4 x \exp(\mathrm{i} ({k}_1{k}_2) \cdot {x})=(2 \pi)^4 \delta^{(4)}({k}_1{k}_2).
$$
$$
\langle 00\rangle=1
$$
As shown, I get no ##3!## factor canceling ##1/3!## out... What am I missing? I think once I see my mistake, b) will be easier to establish
@vanhees71 , @Gaussian97 may you have time to discuss this one?
Thank you
\begin{align*}
S_{fi}&=\langle f  T \left\{ \exp\left( i\frac{\lambda_3}{3!}\int d^4 x :\phi \phi \phi (x) : + i\frac{\lambda_4}{4!}\int d^4 x :\phi \phi \phi \phi (x) : \right) \right\} i \rangle \\
&= \langle f  i \rangle + i \frac{\lambda_3}{3!} \int d^4 x \langle f :\phi \phi \phi (x): i \rangle + i\frac{\lambda_4}{4!} \int d^4 x \langle f  :\phi \phi \phi \phi (x) :  i \rangle \\
&+ \left(\frac{i\lambda_3}{3!}\right)^2\frac{1}{2!} \int d^4 x d^4 y \langle f T\left\{ : \phi \phi \phi (x) :: \phi \phi \phi (y) : \right\}i\rangle \\
&+ \left(\frac{i\lambda_4}{4!}\right)^2\frac{1}{2!} \int d^4 x d^4 y \langle f T\left\{ : \phi \phi \phi \phi (x) :: \phi \phi \phi \phi (y) :\right\} i\rangle \\
&+ \left(\frac{i\lambda_3}{3!}\right)\left(\frac{i\lambda_4}{4!}\right) \int d^4 x d^4 y \langle f T\left\{ : \phi \phi \phi \phi \phi \phi \phi (x) :\right\} i\rangle + \mathcal{O}(\lambda_3^4, \lambda_4^4)
\end{align*}
a) My guess is that there will be two interaction vertices: ##i \lambda_3## and ##i \lambda_4## corresponding to ##\phi^3## and ##\phi^4## interaction terms (respectively) in the lagrangian. They look as follows (where I used dotted lines for mesons as convention)
OK let's actually prove it. The idea I had in mind is to evaluate the first order terms in the expansion only, and check out explicitly that the factors of ##1/3!## and ##1/4!## cancel out.
For the first one
\begin{align*}
i \frac{\lambda_3}{3!} \int d^4 x \langle f :\phi \phi \phi (x): i \rangle &= \frac{\lambda_3}{3!} \int d^4 x \langle f :\left(\phi^+ + \phi^\right)\left(\phi^+ + \phi^\right)\left(\phi^+ + \phi^\right)(x): i \rangle\\
&= i \frac{\lambda_3}{3!}\int d^4 x \langle f : \phi^ \phi^ \phi^ (x): i \rangle + i \frac{\lambda_3}{3!}\int d^4 x \langle f : \phi^+ \phi^+ \phi^+ (x): i \rangle \\
&+i \frac{\lambda_3}{3!}\int d^4 x \langle f : \phi^ \phi^+ \phi^+ (x): i \rangle + i \frac{\lambda_3}{3!}\int d^4 x \langle f : \phi^+ \phi^ \phi^ (x): i \rangle \\
&+i \frac{\lambda_3}{3!}\int d^4 x \langle f : \phi^ \phi^ \phi^+ (x): i \rangle + i \frac{\lambda_3}{3!}\int d^4 x \langle f : \phi^+ \phi^+ \phi^ (x): i \rangle \\
&+ i \frac{\lambda_3}{3!}\int d^4 x \langle f : \phi^ \phi^+ \phi^ (x): i \rangle + i \frac{\lambda_3}{3!}\int d^4 x \langle f : \phi^+ \phi^ \phi^+ (x): i \rangle \\
&= i \frac{\lambda_3}{3!}\int d^4 x \langle f \phi^ \phi^ \phi^ (x) i \rangle + i \frac{\lambda_3}{3!}\int d^4 x \langle f \phi^+ \phi^+ \phi^+ (x) i \rangle \\
&+i \frac{\lambda_3}{3!} 3\int d^4 x \langle f \phi^ \phi^+ \phi^+ (x) i \rangle + i \frac{\lambda_3}{3!} 3\int d^4 x \langle f \phi^ \phi^ \phi^+ (x) i \rangle
\end{align*}
The issue I have in here is that when evaluating each term individually I do not get the factor of ##1/3!## canceled out. Let's show one computation explicitly
\begin{align*}
i \frac{\lambda_3}{3!}\int d^4 x \langle f \phi^ \phi^ \phi^ (x) i \rangle &= i \frac{\lambda_3}{3!}\left(\frac{1}{2V \omega_{\vec k}}\right)^{3/2}\langle f b^{\dagger}(\vec k)b^{\dagger}(\vec k)b^{\dagger}(\vec k) i\rangle \int d^4 x \exp\left(3k\cdot x\right) \\
&= i \frac{\lambda_3}{3!}\left(\frac{1}{2V \omega_{\vec k}}\right)^{3/2}\langle 0 b(\vec k)b(\vec k)b(\vec k) b^{\dagger}(\vec k)b^{\dagger}(\vec k)b^{\dagger}(\vec k)0\rangle \int d^4 x \exp\left(3k\cdot x\right) \\
&= i \frac{\lambda_3}{3!}\left(\frac{1}{2V \omega_{\vec k}}\right)^{3/2}\langle 00\rangle \int d^4 x \exp\left(3k\cdot x\right) \\
&= i \frac{\lambda_3}{3!}\left(\frac{1}{2V \omega_{\vec k}}\right)^{3/2} \frac{1}{3}\delta^{(4)}(k)
\end{align*}
Where I've used (I am indeed working in n.u.)
$$
\phi^(x)=\sum_{\vec k} \Big(\frac{1}{2V \omega_{\vec k}} \Big)^{1/2} b^{\dagger}(\vec k)e^{ik \cdot x}
$$
$$
\big[ b (\vec k) , b^{\dagger} (\vec k') \big]=\delta_{\vec k \vec k'}
$$
$$
\int \mathrm{d}^4 x \exp(\mathrm{i} ({k}_1{k}_2) \cdot {x})=(2 \pi)^4 \delta^{(4)}({k}_1{k}_2).
$$
$$
\langle 00\rangle=1
$$
As shown, I get no ##3!## factor canceling ##1/3!## out... What am I missing? I think once I see my mistake, b) will be easier to establish
@vanhees71 , @Gaussian97 may you have time to discuss this one?
Thank you