Coefficients in expansions of a vector potential

[Haorong Wu]

TL;DR Summary

Why the coefficients in expansions of a vector potential are different in different papers?

I have seen two expansions of a vector potential,

$$\mathbf A=\sum_\sigma \int \frac{d^3k}{(16 \pi^3 |\mathbf k|)^{1/2}} [\epsilon_\sigma(\mathbf k) \alpha_\sigma (\mathbf k) e^{i \mathbf k \cdot \mathbf x}+c.c.],$$
and
$$\mathbf A=\sum_\sigma \int \frac{d^3k}{ (2 \pi)^3(2 |\mathbf k|)^{1/2}} [\epsilon_\sigma(\mathbf k) \alpha_\sigma (\mathbf k) e^{i \mathbf k \cdot \mathbf x}+c.c.],$$
where $\epsilon_\sigma$ are polarizations, $\alpha_\sigma (\mathbf k)$ are amplitudes.

My problem is the two coefficients $(2 \pi)^{3/2}$ and $(2 \pi)^3$ in these two expressions, respectively. Why they are different?

I do not remember where I have read something about it, that it is related to the definition of Fourier transform.

Thanks for any hints.

 

[Dale]

The constant coefficients don’t matter. They are only there for convenience. They are different because the two different authors disagreed on which was most convenient.

 

[vanhees71]

It's just a question of conventions. You have as many normalization conventions in the Fourier mode decomposition of (quantum) fields you have authors of textbooks and papers (or even more ;-)).

 

[Meir Achuz]

Physicists usually define the FT as $f(x)=\int...F(k)$,and mathematicians
use $f(x)=(2\pi)^{-1/2}\int$.

 

[vanhees71]

Well, there are many types of physicists and all define their FTs differently. Sometimes the convention changes even in different subfields of physics.

E.g., in non-relativistic quantum mechanics one usually defines the FT between position and momentum (or rather "wave-vector") representation in a symmetric way, i.e., as unitary transformations from $\mathrm{L}^2 \rightarrow \mathrm{L}^2$ (for 1D motion):
$$\psi(x)=\int_{-\infty}^{\infty} \mathrm{d} k \frac{1}{\sqrt{2 \pi}} \tilde{\psi}(k) \exp(\mathrm{i} k x) \; \Leftrightarrow \; \tilde{\psi}(k)=\int_{-\infty}^{\infty} \mathrm{d} x \frac{1}{\sqrt{2 \pi}} \psi(x) \exp(-\mathrm{i} k x).$$
In relativistic (Q)FT most physicists use
$$\psi(x)=\int_{-\infty}^{\infty} \mathrm{d} k \frac{1}{2 \pi} \tilde{\psi}(k) \exp(\mathrm{i} k x) \; \Leftrightarrow \; \tilde{\psi}(k)=\int_{-\infty}^{\infty} \mathrm{d} x \psi(x) \exp(-\mathrm{i} k x).$$
For the FT wrt. time vs. angular frequency you have the same conventions but with the opposite signs in the exponentials. That's because one usually has to solve wave equations and likes to have $k$ the direction of the wave's phase velocity and not $-k$.

In many engineering texts this signs in the exponential are reversed ;-)).

In short, it's a mess, and one must be careful when reading texts to make sure to figure out, which convention is used.

 

[Haorong Wu]

Thank you all! It is clear now.

 

[Delta2]

The different constants shouldn't trouble you a lot. Unless of course the constants present in formulas have dimensions. In this case they are plain numbers, no dimensions involved.