Let's try to clarify and prove the formula for the **cosmic variance** of the CMB power spectrum coefficients \( C_\ell \): ### The Formula You wrote: \[ \text{Cosmic variance} = \frac{2}{2\ell + 1} C_\ell^2 \] And you want to prove: \[ \langle (C_\ell - \langle C_\ell \rangle)^2 \rangle = \frac{2}{2\ell + 1} \langle C_\ell \rangle^2 \] with the hint involving 4-point correlations of the spherical harmonic coefficients \( a_{\ell m} \). --- ## Step 1: The Estimator for \( C_\ell \) The angular power spectrum estimator is: \[ \hat{C}_\ell = \frac{1}{2\ell+1} \sum_{m=-\ell}^{\ell} |a_{\ell m}|^2 \] where the \( a_{\ell m} \) are the coefficients from the spherical harmonic decomposition of the CMB sky. --- ## Step 2: Mean of \( \hat{C}_\ell \) Assume the \( a_{\ell m} \) are Gaussian random variables with zero mean and variance \( C_\ell \): \[ \langle a_{\ell m} a^*_{\ell' m'} \rangle = \delta_{\ell \ell'} \delta_{m m'} C_\ell \] So, \[ \langle \hat{C}_\ell \rangle = \frac{1}{2\ell+1} \sum_{m=-\ell}^\ell \langle |a_{\ell m}|^2 \rangle = \frac{1}{2\ell+1} \sum_{m=-\ell}^\ell C_\ell = C_\ell \] --- ## Step 3: Variance of \( \hat{C}_\ell \) We want: \[ \mathrm{Var}(\hat{C}_\ell) = \langle \hat{C}_\ell^2 \rangle - \langle \hat{C}_\ell \rangle^2 \] Compute \( \langle \hat{C}_\ell^2 \rangle \): \[ \langle \hat{C}_\ell^2 \rangle = \frac{1}{(2\ell + 1)^2} \sum_{m, m'} \langle |a_{\ell m}|^2 |a_{\ell m'}|^2 \rangle \] Expand \( |a_{\ell m}|^2 = a_{\ell m} a^*_{\ell m} \): \[ \langle |a_{\ell m}|^2 |a_{\ell m'}|^2 \rangle = \langle a_{\ell m} a^*_{\ell m} a_{\ell m'} a^*_{\ell m'} \rangle \] By Wick's theorem (for Gaussian variables): \[ \langle abcd \rangle = \langle ab \rangle \langle cd \rangle + \langle ac \rangle \langle bd \rangle + \langle ad \rangle \langle bc \rangle \] For our case: - \( a = a_{\ell m} \), \( b = a^*_{\ell m} \), \( c = a_{\ell m'} \), \( d = a^*_{\ell m'} \) So, \[ \langle a_{\ell m} a^*_{\ell m} a_{\ell m'} a^*_{\ell m'} \rangle = \langle a_{\ell m} a^*_{\ell m} \rangle \langle a_{\ell m'} a^*_{\ell m'} \rangle + \langle a_{\ell m} a_{\ell m'} \rangle \langle a^*_{\ell m} a^*_{\ell m'} \rangle + \langle a_{\ell m} a^*_{\ell m'} \rangle \langle a^*_{\ell m} a_{\ell m'} \rangle \] But - \( \langle a_{\ell m} a^*_{\ell m} \rangle = C_\ell \) - \( \langle a_{\ell m} a_{\ell m'} \rangle = 0 \) - \( \langle a_{\ell m} a^*_{\ell m'} \rangle = C_\ell \delta_{m m'} \) So \[ \langle |a_{\ell m}|^2 |a_{\ell m'}|^2 \rangle = C_\ell^2 + 0 + |C_\ell|^2 \delta_{m m'} \] --- ## Step 4: Sum Over \( m, m' \) Plug into the sum: \[ \sum_{m, m'} (C_\ell^2 + C_\ell^2 \delta_{m m'}) = (2\ell+1)^2 C_\ell^2 + (2\ell+1) C_\ell^2 \] So, \[ \langle \hat{C}_\ell^2 \rangle = \frac{1}{(2\ell+1)^2} \left[ (2\ell+1)^2 C_\ell^2 + (2\ell+1) C_\ell^2 \right] = C_\ell^2 + \frac{C_\ell^2}{2\ell+1} \] --- ## Step 5: Subtract the Mean Squared \[ \mathrm{Var}(\hat{C}_\ell) = \langle \hat{C}_\ell^2 \rangle - \langle \hat{C}_\ell \rangle^2 = \left( C_\ell^2 + \frac{C_\ell^2}{2\ell+1} \right) - C_\ell^2 = \frac{2C_\ell^2}{2\ell+1} \] --- ## Final Result \[ \boxed{ \mathrm{Var}(\hat{C}_\ell) = \frac{2}{2\ell+1} C_\ell^2 } \] This is the standard formula for **cosmic variance**. --- ## Summary You **prove** this using the properties of Gaussian random fields and Wick's theorem, as shown above. The "hint" in your question refers to how to calculate 4-point correlations (the key step above), which for Gaussian variables reduces to sums of products of 2-point functions. If you need a more detailed explanation of the hint and how it connects to Wick's theorem, let me know!