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I am practising basic QFT and am having some trouble with calculating the vertex factor of an interacting theory involving two real scalar fields, $\phi_{1}$ and $\phi_{2}$.

If I create a generic Lagrangian with interaction terms:

$$\mathcal{H}_{\text{int}}=g\phi_{1}^{2}\phi_{2} + q\phi_{1}^{2}\phi_{2}^{2}$$

I want to calculate the vertex factors associated with these interactions. So to calculate the $\phi_{1}^{2}\phi_{2}$ vertex, I calculate the three-point Green's function:

$$G^{(3)}(x,y,z)=\langle\Omega|T\phi_{1}(x)\phi_{1}(y)\phi_{2}(z)|\Omega\rangle = \frac{\langle0|T\phi_{1}(x)\phi_{1}(y)\phi_{2}(z)\exp\left(-i\int_{-\infty}^{\infty}\mathcal{H}_{\text{int}}\:\mathrm{d}^{4}s\right)|0\rangle}{\langle0|T\exp\left(-i\int_{-\infty}^{\infty}\mathcal{H}_{\text{int}}\:\mathrm{d}^{4}s\right)|0\rangle}$$

We can expand the exponential involving $\mathcal{H}_{\text{int}}$ to first order in $g$ and $q$:

$$\exp\left(-i\int_{-\infty}^{\infty}\mathcal{H}_{\text{int}}\:\mathrm{d}^{4}s\right)=1-ig\int_{-\infty}^{\infty}\phi_{1}^{2}(s)\phi_{2}(s)\:\mathrm{d}^{4}s-iq\int_{-\infty}^{\infty}\phi_{1}^{2}(s)\phi^{2}_{2}(s)\:\mathrm{d}^{4}s$$

Clearly by Wick's theorem, any term without an even number of $\phi_{1}$ or $\phi_{2}$ terms will disappear, so the denominator will become:

$$\langle0|T\exp\left(-i\int_{-\infty}^{\infty}\mathcal{H}_{\text{int}}\:\mathrm{d}^{4}s\right)|0\rangle = 1-iq\int_{-\infty}^{\infty}\langle0|\phi_{1}^{2}(s)\phi_{2}^{2}(s)|0\rangle\:\mathrm{d}^{4}s = 1 - iq\int_{-\infty}^{\infty}\:\mathrm{d}^{4}s\Delta_{F}^{(1)}(0)\Delta_{F}^{(2)}(0)$$

Where $\Delta_{F}^{(1)}(z)$ is the Feynman propagator for $\phi_{i}$.

My understanding is (from this other question I asked), that terms containing a tadpole diverenge ($\Delta^{(i)}_{F}(0)$) can be set to zero using the requirement that $\langle \phi_{i}\rangle = 0$, so the denominator evaluates to $1$.

If this is correct, then all that's required is to evaluate the numerator:

$$\langle0|T\phi_{1}(x)\phi_{1}(y)\phi_{2}(z)\exp\left(-i\int_{-\infty}^{\infty}\mathcal{H}_{\text{int}}\:\mathrm{d}^{4}s\right)|0\rangle = -iq\int_{-\infty}^{\infty}\langle0|T\phi_{1}(x)\phi_{1}(y)\phi_{2}(z)\phi_{1}^{2}(s)\phi_{1}(s)|0\rangle\:\mathrm{d}^{4}s$$

We can use Wick's theorem to evaluate the VEV in the integral:

$$\langle 0 |T \phi_{1}(x)\phi_{1}(y)\phi_{2}(z)\phi_{1}^{2}(s)\phi_{2}(s)|0\rangle = \Delta_{F}^{(1)}(x-y)\Delta_{F}^{(2)}(z-s)\Delta_{F}^{(1)}(0) + 2\Delta_{F}^{(1)}(x-s)\Delta_{F}^{(1)}(y-s)\Delta_{F}^{(2)}(z-s)$$

So I believe that the vertex factor for the $\phi_{1}^{2}\phi_{2}$ interaction is simply $-2ig$?

To calculate the other vertex factor, we use the four-point Green's function:

$$G^{(4)}(x,y,z,w) = \langle \Omega | T \phi_{1}(x)\phi_{1}(y)\phi_{2}(z)\phi_{2}(w)|\Omega\rangle$$

Using the same approach as before, and recognising that the numerator is the same:

$$\langle0|T\phi_{1}(x)\phi_{1}(y)\phi_{2}(z)\phi_{2}(w)\exp\left(-i\int_{-\infty}^{\infty}\mathcal{H}_{\text{int}}\:\mathrm{d}^{4}s\right)|0\rangle = \langle 0 | T \phi_{1}(x)\phi_{1}(y)\phi_{2}(z)\phi_{2}(w)|0\rangle \\- iq\int_{-\infty}^{\infty}\langle 0 | T\phi_{1}(x)\phi_{1}(y)\phi_{2}(z)\phi_{2}(w)\phi_{1}^{2}(s)\phi_{2}^{2}(s)|0\rangle$$

Using Wick's theorem, we can see that this is:

$$\Delta_{F}^{(1)}(x-y)\Delta_{F}^{(2)}(z-w) - iq\int_{-\infty}^{\infty}\mathrm{d}^{4}s\left\{4\Delta_{F}^{(1)}(x-s)\Delta_{F}^{(1)}(y-s)\Delta_{F}^{(2)}(z-s)\Delta_{F}^{(2)}(w-s) + \Delta_{F}^{(1)}(x-y)\Delta_{F}^{(2)}(z-w)\Delta_{F}^{(1)}(0)\Delta_{F}^{(2)}(0) + 2\Delta_{F}^{(1)}(x-s)\Delta_{F}^{(1)}(y-s)\Delta_{F}^{(2)}(z-w)\Delta_{F}^{(2)}(0) + 2\Delta_{F}^{(1)}(x-y)\Delta_{F}^{(2)}(z-s)\Delta_{F}^{(2)}(w-s)\Delta_{F}^{(1)}(0)\right\}$$

Which, after removing the Tadpole divergences, gives us:

$$\Delta_{F}^{(1)}(x-y)\Delta_{F}^{(2)}(z-w) - iq\int_{-\infty}^{\infty}\mathrm{d}^{4}s\left\{4\Delta_{F}^{(1)}(x-s)\Delta_{F}^{(1)}(y-s)\Delta_{F}^{(2)}(z-s)\Delta_{F}^{(2)}(w-s)\right\}$$

Now, I believe that the first term arises from two separate non-interacting particles. Therefore we ignore it and our vertex factor is just $-4iq$, but I am not 100% sure, is this right?

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    $\begingroup$ As a rule of thumb, the factor for a vertex with e.g. two $\Phi_1$ legs and one $\Phi_2$ should be $-i \left( \frac{d^2}{d\Phi_1^2} \frac{d}{d\Phi_2} \mathcal H_{int} \right)_{\Phi_1 = \Phi_2 = 0}$, so the results look correct. $\endgroup$ – Noiralef May 16 '17 at 12:36
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The term $\phi_1^2\phi_2$ has two equivalent fields $\phi_1,\phi_1$, so the canonical normalisation is $$ \frac{1}{2!}\phi_1^2\phi_2 $$

On the other hand, the term $\phi_1^2\phi_2^2$ has two equivalent fields twice, $\phi_1,\phi_1$ and $\phi_2,\phi_2$, so the canonical normalisation is $$ \frac{1}{2!}\frac{1}{2!}\phi_1^2\phi_2^2 $$

More generally, the canonical normalisation of a term $\phi_1^a\phi_2^b$ would be $$ \frac{1}{a!}\frac{1}{b!}\phi_1^a\phi_2^b $$

Therefore, the answer is yes, your Dyson factors are correct. For more details, see this PSE post.

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