# What´s the importance of the normalization of the Kinetic term?

It´s usual to read in QFT books of how it is "easier" to have a canonically normalized kinetic term. So, for instance:

$${\cal L} = {1 \over 2 }\partial_{\mu} \phi \partial^{\mu}\phi - {1 \over 2 } m^2 \phi \phi - {\lambda \over 4!} \phi^4$$

is canon. And:

$${\cal L}_2 = \partial_{\mu} \phi \partial^{\mu}\phi - m^2 \phi \phi - 2 {\lambda \over 4!} \phi^4$$

is not.

Now, both of them have the same classical equations of motion, since ${\cal L}_2 = 2 {\cal L}$. Supose I just carry on quantization on the $\phi$ field as usual. The free propagator is:

$$\langle 0| T\{\phi(x_1) \phi(x_2)\} |0 \rangle = {i \over 2} \Delta_F(x_1-x_2)$$ - the 1/2 factor comes from the fact that is is now the Green function of $(\square+m^2)$ instead of ${1 \over 2 } (\square+m^2)$

If I calculate the four point function at tree level:

$$\langle 0| T\{\phi(x_1) \phi(x_2) \phi(x_3) \phi(x_4)\} |0 \rangle = \\ = -i 2 \lambda \int d^4z {\Delta_F(z-x_1) \over 2} {\Delta_F(z-x_2) \over 2} {\Delta_F(z-x_3) \over 2} {\Delta_F(z-x_4) \over 2}$$

All the "2" factors in the equation above disappear if I exchange ${\cal L}_2$ by ${\cal L}$. But, since they don't cancel (there's a $2^{-3}$ left) and this carries on to the cross section, I feel like I´m getting different results from equivalent Lagrangians.

What am I missing? Am I obliged to have the Kinetic term in a canonical normalization (so that ${\cal L}_2$ is "wrong")? If so, what conditions impose this normalization? Or, if not, and the two Lagrangians are really equivalent: how do this $2^{-3}$ disappears before becoming a catastrophic $2^6$ decrease in the cross section?

POST ANSWER EDIT

So, my take on the answer (please correct me if I got it wrong): starting from ${\cal L}$ above, if I do a field redefinition $\phi \rightarrow \sqrt{2}\phi$ I get:

$${\cal L}_Z = \partial_{\mu} \phi \partial^{\mu}\phi - m^2 \phi \phi - 4 {\lambda \over 4!} \phi^4$$

which is not ${\cal L}_2$ (in which I multiplied the whole ${\cal L}$ by 2).

In both ${\cal L}_Z$ and ${\cal L}_2$ I messed up with the normalization of the propagator, that means I will get $\langle p | \phi(0) | 0 \rangle = {1\over\sqrt{2}}$ instead of $\langle p | \phi(0) | 0 \rangle = 1$. The LSZ formula would then read:

$$\langle p_1 p_2 | S | p_3 ... p_n \rangle = ({1\over\sqrt{2}})^n (\mbox{amputaded diags.})$$

In the case of ${\cal L}_Z$, all the factors in the four point function would be:

$$\langle 0| T\{\phi(x_1) \phi(x_2) \phi(x_3) \phi(x_4)\} |0 \rangle = \\ = -i 4 \lambda \int d^4z {\Delta_F(z-x_1) \over 2} {\Delta_F(z-x_2) \over 2} {\Delta_F(z-x_3) \over 2} {\Delta_F(z-x_4) \over 2}$$

Which amputates to: $$-i 4 \lambda (2\pi)^4 \delta(\mbox{momentum})$$ and $$\langle p_1 p_2 | S | p_3 ... p_n \rangle_{{\cal L}_Z} = -i \lambda (2\pi)^4 \delta(\mbox{momentum})$$

Exactly the same as ${\cal L}$. Now, the same operation on ${\cal L}_2$ gives:

$$\langle p_1 p_2 | S | p_3 ... p_n \rangle_{{\cal L}_2} = -i {\lambda \over 2} (2\pi)^4 \delta(\mbox{momentum})$$

Showing that, field re-definitions are ok but multiplying the whole Lagrangian is not, the cross section will change by a factor 4. @user1631 said in his answer that means redefining $h$, I'll have to carry out this calculation without $\hbar =1$ to check that.

• For the equivalent title question(v1) in the context of Newtonian mechanics, see e.g. this post. – Qmechanic Aug 29 '12 at 19:19

## 2 Answers

First note that the canonical normalization actually has a factor of 1/2 in front of the kinetic and mass terms, and that the sign of the mass term should be reversed. So I'll consider the Lagrangians $\mathcal{L}=\frac{1}{2}((\partial \phi)^2-m^2\phi^2)-\frac{\lambda}{4!}\phi^4$ and $\mathcal{L}_2=(\partial \phi)^2-m^2\phi^2-\frac{2\cdot \lambda}{4!}\phi^4$. To find the relation between these two, let's make the field redefinition $\phi\mapsto \phi/\sqrt{2}$ in $\mathcal{L}_2$. This gives $\mathcal{L}_2=\frac{1}{2}((\partial \phi)^2-m^2\phi^2)-\frac{\lambda'}{ 4!}\phi^4$, where $\lambda'=\lambda/2$. So the two Lagrangians are only equivalent under a redefinition of the coupling along with a field redefinition. It's easy to check that this fixes your problem with the four-point function.

Note, however, that you can still work with $\mathcal{L}_2$ without making the field redefinition. The correlation functions won't agree in the two theories, but the S-matrices will (there's a wavefunction renormalization in the denominator of the LSZ formula.) And in the end, the S-matrix is the only thing you can measure anyway.

Edit: In response to the comment below, the missing factor is indeed related to $\langle p|\phi(x)|0\rangle$. Here's one way to compute this quantity if you only know the propagator. First note that by Lorentz invariance we have $\langle p|\phi(x)|0\rangle=\sqrt{Z}\exp(ipx)$, where $Z$ is some constant. Taking $x^0>y^0$ and inserting a complete set of momentum eigenstates, \begin{align*} \langle \phi(x)\phi(y)\rangle&=\int\frac{d^d\vec{p}}{(2\pi)^d2\omega}\langle 0|\phi(x)|p\rangle\langle p|\phi(y)|0\rangle=Z\int\frac{d^d\vec{p}}{(2\pi)^d2\omega}\exp(-ip(x-y)), \end{align*} which is just the ordinary propagator multiplied by $Z$. In your $\mathcal{L}_2$ theory the propagator is half of the original one, so $Z=1/2$ at tree level; loop corrections to the propagator correct $Z$ accordingly.

• Thanks for point out the mistakes - I corrected them in the question. As for your answer: does it have to do with the $\langle p | \phi | 0 \rangle$ that appears when you are building the LSZ formula? I think I got it but I want to be sure before including my conclusion in the question itself. Being more specific: I guess there will be a $\sqrt{2} \langle p | \phi | 0 \rangle$ in the case of ${\cal L}_2$ but I can't see how it appears without doing field redefinitions. – Forever_a_Newcomer Aug 30 '12 at 2:04
• Included my take on this on the question. Please let me know if I got it wrong. – Forever_a_Newcomer Sep 2 '12 at 14:45
• Looks good to me... and thanks for the fake internet points! – Matthew Sep 3 '12 at 2:29

Classically, multiplying the Lagrangian by a constant does nothing of course. Quantum mechanically, multiplying the Lagrangian by a constant is equivalent to changing the Planck constant. If you kept the Planck constant explicit in all calculations this would be more obvious, but people usually define it as 1 and hide it.

• Thanks, I'll have to check that keeping $\hbar$ around. I included some comments in the question. – Forever_a_Newcomer Sep 2 '12 at 14:47