# Justification of not quantizing small extra dimensions

When dealing with extra dimensions ($x ^\mu$ represents $4D$ spacetime and $y$ the extra dimension) we use what's known as Kaluza-Klein decomposition (basically a Fourier transform), $$\Phi ( x , y ) = \sum _n f _n (y) \phi _n (x)$$ Later when considering the action, we make the assumption that, \begin{align} & \int \! dy \; f _n ( y ) f _m ( y ) = \delta _{ m n } \\ & \int \! dy \; f _n ' ( y ) f _m ' ( y ) = M _{ n} ^2 \delta _{ m n } \end{align} where the primes indicate derivatives with respect to $y$.

What precisely do these assumptions mean? My best guess is that since we are apparently not quantizing the fields in the fifth dimension it means that we assume no particles can get excited in that dimension. Is this correct and if so why is that justified?

UPDATE: I realized that I was unclear about what confused me. I have no problems with assuming an orthonormal basis (or explicitly using a Fourier transform). What I want to understand are what does taking the field to be of the form, $\Phi ( x , y ) = \sum _n f _n (y) \phi _n (x)$ and just carrying out the integral over $y$ mean? This is not something we can do in normal QFT with infinite dimensions so it must be some assumption about what happens in the extra dimension. As I mentioned above my feeling is that we are not allowing field excitations in that direction but I'm not sure why that's obviously true.

The most general case would of course be to take the fields to be of the form $\Phi ( x , y ) = \sum _n \phi _n (x,y)$. By taking the field to be of the form, $\Phi ( x , y ) = \sum _n f _n (y) \phi _n (x)$ and just carrying out the integral over $y$, we are considering an effective field theory of the full theory. As you can clearly see the effect of the extra dimensions has been "integrated out". This physically means that we are essentially ignoring fluctuations/dynamics in the extra dimensions and replace them with their average values.

This approximation is commonly used in Kaluza-Klein reduction when you study the physics in 4d and is valid at scales much bigger than the radius of the compactified space. However, you must keep in mind that is an effective theory and you will have to consider the general ansatz I mentioned at the beginning if you want to study effects at smaller scales.

The integrals essentially follow from the fact that the Kaluza-Klein scalar is expanded in a Fourier series in terms of an orthonormal basis. This can be understood if we write down $f(y)$ explicitly:

$$f_n(y)=\exp(iny/R),$$

where n can take on values between $-\infty$ and $\infty$, and $R$ is the radius of the compactified dimension. The orthonormality equations then read

$$\int\overline{f_n(y)}f_m(y)\,dy=\delta_{mn},$$ $$\int\partial_y\overline{f_n(y)}\partial_yf_m(y)\,dy=mn/R^2\delta_{mn}.$$

For $m=n$, the latter expression reduces to $n^2/R^2$, where $p_n=n/R$ is the quantized momentum in the periodic dimension. Since the five-dimensional scalar is massless, i.e. $p^{\mu}p_{\mu}+p^2=0$, it follows that from the four-dimensional point of view there exists a particle with mass squared $M_n^2=-p^{\mu}p_{\mu}$, with $M_n^2=p_n^2=n^2/R^2$.

• Thanks for your answer. Sorry I was unclear about what actually bothered me about this. I updated my question. – JeffDror Nov 13 '14 at 1:09

Isn't the first integral equation just the orthonormality assumption that the basis functions $f_n$ must possess when trying to expand an arbitrary function using them, just like the Fourier basis $f_n(y)=e^{iny}$, etc? I am not sure where the second equation comes from (maybe you could give a reference), but assuming the same exponential basis one finds: $$f'_n(y)=inf_n(y)$$ therefore $$\int dy f'_n(y)f'_m(y) = -nm\int dy f_n(y)f_m(y) \\ =-nm \delta_{nm} = -n^2 \equiv M_n^2$$ Does this make sense or I have completely missed your point?

• Thanks for your answer. Sorry I was unclear about what actually bothered me about this. I updated my question. – JeffDror Nov 13 '14 at 1:09