- #1

JD_PM

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- 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 S-matrix 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

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 0|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} \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 0|0\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 0|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} \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 0|0\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