When dealing with vector fields in his QFT book, Schwartz writes the classical field in terms of a basis which I don't know how he is getting.

He first introduces the Proca Lagrangian

$$\mathcal{L}=-\dfrac{1}{4}F_{\mu\nu}F^{\mu\nu}+\dfrac{1}{2}m^2A_\mu A^\mu.$$

The equations of motion are then $(\Box + m^2)A_\mu = 0$ and $\partial^\mu A_\mu = 0$.

He then says:

Let us now find explicit solutions to the equations of motion. We start by Fourier transforming our classical fields. Since $(\Box+m^2)A_\mu=0$, we can write any solution as

$$A_\mu(x)=\sum_i \int \dfrac{d^3 p}{(2\pi)^3}\tilde{a}_i(p)\epsilon^i_\mu(p) e^{ipx}, \quad p_0=\omega_p=\sqrt{|\mathbf{p}|^2+m^2}$$

for some basis vectors $\epsilon^i_\mu(p)$. For example, we could trivially take $i=1,2,3,4$ and use four vectors $\epsilon^i_\mu(p)=\delta^i_\mu$ in this decomposition. Instead, we want a basis that forces $A_\mu$ to automatically satisfy also its equation of motion $\partial^\mu A_\mu=0$. This will happen if $p^\mu \epsilon_\mu^i(p) =0$. For any fixed $4$-momentum $p^\mu$ with $p^2=m^2$, there are three independent solutions to this equation given by three $4$-vectors $\epsilon_\mu^i(p)$, necessarily $p^\mu$ dependent, which we call polarization vectors. Thus we only have to sum over $i=1,2,3$. We conventionally normalize the polarizations by $\epsilon_\mu^\ast \epsilon^\mu =-1$.

I don't understand this at all. Taking the Fourier transform of $\Box A_\mu + m^2 A_\mu = 0$ and denoting $\hat{A}_\mu$ the Fourier transform we have

$$A_\mu(x)=\int \dfrac{d^3 p}{(2\pi)^3} (a_\mu(p) e^{-ipx}+a_\mu^\ast(p)e^{ipx}).$$

This is done by taking the three-dimensional Fourier transform, getting the equation $\partial_t^2\hat{A}+\omega_p^2 \hat{A}=0$, realizing that $\hat{A}_\mu= a_\mu(p) e^{-i\omega_p t}+b_\mu(p) e^{i\omega_p t}$ and finally using the condition that $A_\mu$ is real so that $b_\mu(p)=a_\mu^\ast(-p)$ which leads directly to the formula above.

There is no $\epsilon^i_\mu(p)$ anywhere. Nor can I see why there should be. The sum has only two terms anyway.

So I'm really missing something here. How is this result properly derived?

  • $\begingroup$ Your solution is the trivial solution they speak about where $\epsilon = \delta $, you should think about $\epsilon $ as the basis of coordinate, so it tell us at which direction The field is oscillating (polarization) and because of the second equation of motion, we get that the direction of polarization must be orthogonal to that of momentum, so we can forget about one direction in our calculations (the momentum direction) , that's why it's better to write the field in terms of these basis (polarization) vectors. $\endgroup$
    – Ismasou
    May 29, 2017 at 6:30
  • $\begingroup$ If it was classical physics, you'll write the field in terms of $e_x e_y e_z$. $\endgroup$
    – Ismasou
    May 29, 2017 at 6:36

1 Answer 1


You're almost there. Given $$ A_\mu(x)=\int \dfrac{d^3 p}{(2\pi)^3} (a_\mu(p) e^{-ipx}+a_\mu^\dagger(p)e^{ipx}). $$ you may pick any set of four linearly independent vectors $\{\epsilon_\mu^1,\epsilon_\mu^2,\epsilon_\mu^3,\epsilon_\mu^4\}$ and expand the $a_\mu$ in terms of them: $$ a_\mu=\sum_{i=1}^4a_i \epsilon_\mu^i $$ for some coefficients $a_i$. The "trivial basis" mentioned by S. is $\epsilon^i_\mu\equiv\delta^i_\mu$, in which case this expression becomes $$ a_\mu=\sum_{i=1}^4a_i \delta_\mu^i=a_\mu $$ i.e. the coefficients $a_i$ are just the cartesian components of $a_\mu$. In principle, this is a valid basis, but we can do better. For one thing, in this basis the transversality condition $p\cdot a=0$ is not immediate to implement.

If we choose a basis that changes with $p$, such that $$ p\cdot\epsilon^1=p\cdot\epsilon^2=p\cdot\epsilon^3=0,\qquad \epsilon^4=p/m $$ then we may write, as before, $$ a_\mu=\sum_{i=1}^4a_i \epsilon_\mu^i $$ but now the condition $p\cdot a$ is equivalent to $a_4=0$, so that in effect $$ a_\mu=\sum_{i=1}^3a_i \epsilon_\mu^i $$

With this, $$ A_\mu(x)=\sum_{i=1}^3\int \dfrac{d^3 p}{(2\pi)^3} (a_i \epsilon_\mu^i e^{-ipx}+a^\dagger_i \epsilon_\mu^{i*} e^{ipx}). $$ which is the final expression for a free Proca field.

  • $\begingroup$ I think I understand the point. The field $A$ is a covector field, so that $A(x)= A_\mu(x) dx^\mu$. When we Fourier transform we get another covector field $a_\mu(p) dp^\mu$. Thus $a_\mu$ is expressed in the canonical basis, but we can switch to another basis $\epsilon^i(p)$, with $\epsilon^i(p) = \epsilon^i_\mu(p) dp^\mu$. We can therefore find out that $a_\mu(p) = \tilde{a}_i(p) \epsilon^i_\mu(p)$. Just why would we do this? Is it to choose a basis that already imposes the condition on the $A_\mu$ field that we want, namely $\partial^\mu A_\mu = 0$? So we end up restricting to a subspace? $\endgroup$
    – Gold
    May 29, 2017 at 23:21
  • $\begingroup$ @user1620696 yes, that is exactly what's going on. The new basis is convenient because it makes the transversality condition $\partial\cdot A=0$ trivial to implement. In the old basis you would have to carry the condition $p\cdot a=0$ along in all your calculations. $\endgroup$ May 30, 2017 at 16:47
  • $\begingroup$ Sorry to comment on an older post, but your last expression is a sum of a positive and negative mode which is what I would expect. However the solution given in Schwartz (as quoted in the original question) has only a positive mode $e^{ipx}$. What happened to the negative mode? $\endgroup$
    – CBBAM
    Apr 5 at 7:06

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