I originally asked this on the physics Stack Exchange site, but perhaps it could be more easily answered here.

Given the definition of the correlation function for CMB temperature fluctuations as

$$ C\left(\theta\right) = \left\langle \frac{\delta T}{T}\left(\hat{n}_1\right) \frac{\delta T}{T}\left(\hat{n}_2\right) \right\rangle_{\hat{n}_1\cdot \hat{n}_2 = \cos\theta} ,$$

I should be able to simplify it to

$$ C\left(\theta\right) = \frac{1}{4\pi} \sum_{l=0}^\infty (2l + 1) \, C_l \, P_l\left(\cos\theta\right) $$

(where $P_l \left(x\right)$ are the Legendre polynomials) by decomposing the temperature fluctuations into spherical harmonics like this

$$ \frac{\delta T}{T} = \sum_{l=0}^\infty \sum_{m=-l}^l a_{lm} Y_{lm}. $$

I think the first step of this procedure should look like this

$$ C\left(\theta\right) = \left\langle \sum_{l_1=0}^\infty \sum_{m_1=-l_1}^{l_1} a_{l_1 m_1} Y_{l_1 m_1}\left(\hat{n}_1\right) \sum_{l_2=0}^\infty \sum_{m_2=-l_2}^{l_2} a_{l_2 m_2} Y_{l_2 m_2}\left(\hat{n}_2\right) \right\rangle_{\hat{n}_1\cdot \hat{n}_2 = \cos\theta} .$$

I understand that the spherical harmonics can be written in the form

$$ Y_{lm}(\theta,\phi) \propto P_{lm} \left(\cos\theta\right) e^{i m \phi} $$

(where $P_{lm}(x)$ are the associated Legendre polynomials) and that $C_l$ should come out as

$$ C_l = \frac{1}{2l + 1} \sum_{m=-l}^l a_{lm} a_{l-m} $$

(though I could be off on this last piece). However, I am unsure of the mathematical steps involved in simplifying the four sums down to one. What identities, properties, or other insights will allow me to make this simplification?


  • $\begingroup$ Would Mathematics be a better home for this question? If you decide you would like it migrated, you can raise a moderator flag. $\endgroup$
    – rob
    Commented Jun 20, 2018 at 16:05
  • $\begingroup$ Wherever it is answered most easily, I suppose. It is possible I set things up naively, in which case this physics page might be more suitable, but maybe the math page could better address spherical harmonic manipulations. Should I also ask it there and then keep whichever yields better results? $\endgroup$
    – Grayscale
    Commented Jun 20, 2018 at 16:31
  • $\begingroup$ We could leave it here for a few days to see if you get any physics insight; that's probably better than cross-posting. But I'll leave the decision up to you. $\endgroup$
    – rob
    Commented Jun 20, 2018 at 20:09
  • $\begingroup$ Fixed some errors. $\endgroup$
    – Grayscale
    Commented Jun 20, 2018 at 22:01
  • $\begingroup$ After further reading, I think part of the solution may be that there are two relevant kinds of averages. One is the cosmic mean, which is the average of the results obtained by other observers in many points in space for given directions. However, since we cannot actually measure the cosmic mean, instead we do a second kind of average, summing over all directions on the sky from a single vantage point. So I think there may be more than just math involved here. @rob $\endgroup$
    – Grayscale
    Commented Jun 20, 2018 at 22:02

1 Answer 1


In order to arrive at the correlation function in terms of the Legendre polynomials you begin by noting that

\begin{align} \left\langle a_{l_1 m_1}a^*_{l_2 m_2}\right\rangle = C_{l_1}\,\delta_{l_1l_2}\,\delta_{m_1m_2} \end{align}

where $\delta_{xy}$ are Kronecker deltas. (Note that, this only holds if the random fields described by the $a_{l m}$, i.e. in this case the CMB, is statistically homogeneous and isotropic!)

After using expansion into spherical harmonics (your third equation), you'll find the correlation function to read

\begin{align} C(\theta) = \left\langle\frac{\delta T}{T}(\hat n_1)\frac{\delta T}{T}(\hat n_2)\right\rangle = \left\langle\sum_{l_1}\sum_{l_2}\sum_{m_1}\sum_{m_2}a_{l_1 m_1}Y_{l_1m_1}(\hat n_1)\,a_{l_2m_2}Y_{l_2m_2}(\hat n_2)\right\rangle \end{align}

For real fields (and the CMB temperature is a real field) one can write \begin{align} a^*_{lm} = (-1)^ma_{l-m} \quad\text{and}\quad Y^*_{lm}(\hat n) = (-1)^mY_{l-m}(\hat n). \end{align}

This allows you to use $\sum_{m_2} a_{l_2m_2}Y_{l_2m_2}(\hat n_2)=\sum_{m_2} a^*_{l_2m_2}Y^*_{l_2m_2}(\hat n_2)$ such that the correlation function reads

\begin{align} C(\theta) &= \left\langle\sum_{l_1}\sum_{l_2}\sum_{m_1}\sum_{m_2}a_{l_1 m_1}a^*_{l_2m_2}\,Y_{l_1m_1}(\hat n_1)Y^*_{l_2m_2}(\hat n_2)\right\rangle =\\ &= \sum_{l_1}\sum_{l_2}\sum_{m_1}\sum_{m_2}\left\langle a_{l_1 m_1}a^*_{l_2m_2}\right\rangle\,Y_{l_1m_1}(\hat n_1)Y^*_{l_2m_2}(\hat n_2) =\\ &= \sum_{l_1}\sum_{l_2}\sum_{m_1}\sum_{m_2} C_{l_1}\,\delta_{l_1l_2}\,\delta_{m_1m_2} \,Y_{l_1m_1}(\hat n_1)Y^*_{l_2m_2}(\hat n_2) =\\ &= \sum_{l_1}C_{l_1}\,\sum_{m_1} Y_{l_1m_1}(\hat n_1)Y^*_{l_2m_2}(\hat n_2) \end{align}

Finally, use the relation \begin{align} P_l(\cos\theta) = \frac{4\pi}{2l+1}\sum_m Y_{lm}(\hat n_1)Y_{lm}(\hat n_2) \end{align} to find your quoted result (your 2nd equation).

  • $\begingroup$ Thanks! Where exactly did the fact that we are averaging over $\hat{n}_1 \cdot \hat{n}_2 = \cos\theta$ come in though? $\endgroup$
    – Grayscale
    Commented Aug 14, 2019 at 16:09
  • $\begingroup$ (1/3) The fact that the final expression is independent of direction (either $\hat n_1$ or $\hat n_2$) is inherent in above definition of the power spectrum $C_\ell$; it simply doesn't allow for anything direction-dependent! (cf. my remark about assumed isotropy.) $\endgroup$
    – Bas
    Commented Aug 15, 2019 at 17:01
  • $\begingroup$ (2/3) As you can see, the average $\langle ... \rangle$ in the two-point correlation function $\langle \frac{\delta T}{T}(\hat n_1)\frac{\delta T}{T}(\hat n_2)\rangle$ slips into the sums such that you have to evaluate $\langle a_{\ell_1m_1}a_{\ell_2m_2}\rangle$. The remaining part of the expression (the sum over spherical harmonics) is evaluated using the orthogonality of the spherical harmonics. Here, you see explicitly that functions dependent on directions $\hat n_1$ and $\hat n_2$ are related to a function dependent only on their product $\hat n_1\cdot\hat n_2$. $\endgroup$
    – Bas
    Commented Aug 15, 2019 at 17:03
  • $\begingroup$ (3/3) The average over the $a_{\ell m}$ is the only part left which could carry information on a direction dependence. But this is what we -- by assumption -- fixed to be related to the power spectrum $C_\ell$ which does not know about direction. $\endgroup$
    – Bas
    Commented Aug 15, 2019 at 17:05
  • $\begingroup$ I interpreted your question as "where does the direction dependence vanish". To be precise, we do not average over $\cos\theta$ anywhere! If we averaged over any function of $\theta$, the $\theta$-dependence would vanish in the final result. But in fact, we see that $C(\theta)$ is explicitly dependent on $\theta$. $\endgroup$
    – Bas
    Commented Aug 15, 2019 at 17:07

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