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A massless real scalar admits the expansion $$ \phi(t,\mathbf{x}) = \int \frac{d^3\mathbf{p}}{(2\pi)^{3/2} \sqrt{2|\mathbf{p}|}} \bigg( e^{ - i |\mathbf{p}| t + i \mathbf{p} \cdot \mathbf{x} } a_{\mathbf{p}} + e^{+ i |\mathbf{p}| t - i \mathbf{p} \cdot \mathbf{x} } a_{\mathbf{p}}^{\ast} \bigg) $$ which when quantized has ladder operators which satisfy $[a_{\mathbf{p}},a_{\mathbf{k}}^{\ast}] = \delta^{(3)}(\mathbf{p} - \mathbf{k})$ as well as $[a_{\mathbf{p}},a_{\mathbf{k}}] =[a_{\mathbf{p}}^{\ast},a_{\mathbf{k}}^{\ast}] = 0$. These are expanded in terms of a plane-wave basis.

Is it possible to expand the field in a basis of angular momentum states? The Klein-Gordon equation in spherical coordinates is a product of spherical Bessel functions and spherical harmonics which makes me think these will be involved. Is there some set of different ladder operators $b_{|\mathbf{p}|,\ell,m}$ corresponding to various angular momentum states (i.e. probably being labelled by $|\mathbf{p}|$, $\ell$ and $m$ and so on)?

Schematically, I imagine something like $$ \phi(t,\mathbf{x}) \sim \int d|\mathbf{p}| \sum_{\ell,m} \bigg( e^{ - i |\mathbf{p}| t} j_{\ell}(|\mathbf{p}r|) Y_{\ell,m}(\theta,\phi) b_{|\mathbf{p}|,\ell,m} + \text{h.c.} \bigg) $$

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You can expand any object in any basis you want, this is precisely the point of bases. Whether a given basis is useful or not is an entirely different question. In flat space $\mathbb R^n$ the basis of plane waves is the most convenient one because it diagonalizes translations, which is a massive simplification. I cannot think of any situation where your spherical-harmonics basis is more useful than the plane-wave basis, but it is certainly a valid expansion.

Indeed, the integrand $$ f(x,\vec p):=\frac{1}{(2\pi)^{3/2}\sqrt{2|\vec p|}}e^{ipx}a_{\vec p} $$ is a certain function of $\vec p$, and you can express it in spherical coordinates and expand it in spherical harmonics: $$ f(x,r,\theta,\phi)=\sum_{\ell,m}q_{\ell,m}(x,r)Y_{\ell,m}(\theta,\phi) $$ where $$ q_{\ell,m}(x,r):=\int f(x,r,\theta,\phi)Y_{\ell,m}(\theta,\phi)^*\mathrm d\Omega $$

You can plug this into your plane-wave expansion for $\phi$, $$ \phi(x)=\int f(x,\vec p)\mathrm d\vec p+\text{c.c} $$ and simplify the $\theta,\phi$ integration somewhat but, again, this is in virtually any situation a step back, unless you have a very specific application in mind.

The plane-wave expansion is the most convenient expression for this problem. Other expansions are mathematically valid but almost invariably useless.


When spacetime is not flat, you generically do not have translations anymore, in which case there is no point in diagonalizing them -- they are not a symmetry of your problem. In general you want to expand in the simplest, largest (abelian) isometry subgroup. If you have a curved manifold with spherical symmetry but no translation symmetry, then it definitely starts making sense expanding in spherical harmonics. And this is indeed a common technique in QFT in curved spacetime. But this goes beyond the original question in the OP so I don't have much more to say here.

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    $\begingroup$ Situations with rotational symmetry but no transnational symmetry aren't that rare. I have used this type of expansion when considering impurities in a many body system $\endgroup$ – By Symmetry Jan 31 at 22:02
  • $\begingroup$ @AccidentalFourierTransform That makes sense. Just to confirm: It is wrong to assume that the ladder operators are different here, correct? Meaning $b = a$ in my question above? $\endgroup$ – QuantumEyedea Feb 1 at 2:50
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    $\begingroup$ @QuantumEyedea $a$ and $b$ are linearly related, but the precise coefficient requires evaluating the integrals carefully, which I have not done. There might be e.g. factors of $2\pi$ or so. $\endgroup$ – AccidentalFourierTransform Feb 1 at 15:14

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