Turning delay power into a bandpower without losing the signal

It builds on the previous chapter's delay transform and the conversions $X$ and $Y$ . Wavenumbers are in $h\,\mathrm{Mpc^{-1}}$ and dimensionless power in $\mathrm{mK^2}$ .

From two visibilities to a power spectrum

HERA cross-multiplies two delay-transformed visibilities, $\tilde V_1$ and $\tilde V_2$ , taken from data with independent noise, and normalizes the product:

\hat P(k_\perp,k_\parallel)=\frac{X^2Y}{\Omega_{pp}B}\,\tilde V_1\tilde V_2^{*}.

Here $\Omega_{pp}$ is the sky integral of the squared beam and $B$ is the effective bandwidth. HERA Collaboration 2022, equation 12 In this series $X$ converts angle to transverse distance and $Y$ converts frequency to line-of-sight distance, as in the previous chapter. The paper's equations 13 and 14 attach the letters the other way round; the reading guide in this series shows how to check which is which from the units.

The units confirm the form. $\tilde V$ is in $\mathrm{mK\,sr\,Hz}$ . Multiplying two of them by $X^2$ in $(h^{-1}\mathrm{Mpc})^2$ per steradian and $Y$ in $h^{-1}\mathrm{Mpc}$ per hertz, then dividing by steradians and hertz, leaves $\mathrm{mK^2}\,h^{-3}\mathrm{Mpc^3}$ .

Using two independent copies matters. The product of one visibility with itself would add the noise power to every mode. The cross product has no noise bias, so its noise averages to zero, and individual estimates can come out negative. HERA Collaboration 2022, §4.1

The dimensionless power spectrum

Cosmologists usually quote

\Delta^2(k)=\frac{k^3P(k)}{2\pi^2},

the contribution to the brightness variance per logarithmic interval of $k$ , in $\mathrm{mK^2}$ . HERA Collaboration 2022, equation 15 Its square root is a typical brightness contrast on that scale.

HERA then averages its cylindrical $(k_\perp,k_\parallel)$ estimates in shells of constant $|k|$ . It weights each cell by the inverse square of its expected noise power and gives zero weight to every cell with $k_\parallel$ below the horizon plus the buffer. HERA Collaboration 2022, §4.4

Window functions

A reported bandpower is not the power at one wavenumber. The taper and the binning mix nearby modes, so the estimates $\hat{\mathbf p}$ relate to the true bandpowers $\mathbf p$ through a window matrix:

\hat{\mathbf p}=\mathbf W\,\mathbf p .

HERA Collaboration 2022, equation 18 Each row of $\mathbf W$ sums to one, so a flat spectrum passes through unchanged. HERA chose a diagonal normalization, which keeps error bars small but lets neighbouring windows overlap. HERA Collaboration 2022, §4.1

Figure 18 of the paper shows the windows for Band 2, Field 1. Their 16th and 84th percentiles sit $0.031\,h\,\mathrm{Mpc^{-1}}$ from the median, and the lowest window has no response below $0.128\,h\,\mathrm{Mpc^{-1}}$ because of the buffer. HERA Collaboration 2022, Fig. 18 and footnote 7 To reduce overlap HERA reports every other native mode. Even so, in the public covariance the correlation between neighbouring native rows is $0.25$ to $0.33$ , as computed in the method guide of this series.

A model is compared fairly only after it passes through the same window:

\hat m_\alpha=\sum_\beta W_{\alpha\beta}\,m_\beta .

Worked example: a model through HERA's window

Take the Lab's heated history and its density-only prediction at $z=7.93$ , described below. At the first HERA bandpower, $k=0.192\,h\,\mathrm{Mpc^{-1}}$ :

The model evaluated at the bin centre gives $15.49\,\mathrm{mK^2}$ .
Weighted by the window row gives $15.87\,\mathrm{mK^2}$ , 2.5 % higher, because the window reaches up to larger $k$ where $\Delta^2$ is rising.
For the cold history the same step moves $2893$ to $2964\,\mathrm{mK^2}$ .

Here the window changes the answer by a few percent. It matters more for a spectrum with a sharp feature, and it always matters for consistency.

From a bandpower to an upper limit

HERA has not detected the signal. It turns each bandpower into a 95 % upper limit by adding twice the error bar to the estimate, after setting negative estimates to zero:

U=\max\left(\hat\Delta^2,0\right)+2\sigma .

HERA Collaboration 2022, §4.4 The reason is physical. A real sky signal can only add positive power; noise can push an estimate either way.

At $k=0.192$ in Band 2, Field 1 the public file gives $\hat\Delta^2=211.7\,\mathrm{mK^2}$ and $\sigma=367.3\,\mathrm{mK^2}$ , so $U=946.4\,\mathrm{mK^2}=(30.76\,\mathrm{mK})^2$ , the paper's headline limit at $z=7.9$ . At $k=0.320$ the estimate is $-516.0$ with $\sigma=1229.8$ ; it is set to zero and $U=2459.7\,\mathrm{mK^2}=(49.60\,\mathrm{mK})^2$ .

Signal loss

Every step that changes the data can remove part of the hydrogen. HERA lists four identified losses and corrects the power spectra for each. HERA Collaboration 2022, §3.11, Table 2

**Table 4.1** Identified signal losses, each corrected in HERA's final power spectra.
Step	Power loss, Band 1	Power loss, Band 2
Absolute calibration	9 %	8 %
Cross-coupling filtering	3 %	1 %
Time averaging	1 %	1 %
Redundant averaging	1 %	1 %

The paper states that these losses do not depend on scale or on the reionization model. A companion paper, still in preparation when HERA published, ran simulated skies through nearly the whole pipeline and recovered a realistic model for $k\ge0.192\,h\,\mathrm{Mpc^{-1}}$ . HERA Collaboration 2022, §5

The standard test is signal injection. Add a known copy of the signal, scaled by $A$ , run the identical treatment $T$ and measure what comes out:

f=\frac{P\left[T(\mathrm{fg}+A\,s)\right]-P\left[T(\mathrm{fg})\right]}{A^2\,P(s)} .

If $f=1$ nothing was lost. Making $A$ large keeps cross terms between foreground and signal from hiding or faking a loss. The Lab computes $f$ over the modes the treatment uses, in $0.05\le k\le0.6\,h\,\mathrm{Mpc^{-1}}$ .

Worked example: avoidance against a smooth fit

Wedge avoidance only discards modes; it does not change the ones it keeps. In the Lab default record the recovered fraction is $f=0.999$ .

Subtracting a third-order polynomial in frequency from each visibility is different. A least-squares fit is linear in the data, so it removes the part of the hydrogen that happens to look smooth, along with the foreground. In the Lab the recovered fraction falls to $f=0.823$ . Worse, in this schematic sky the third-order fit leaves foreground residuals $5.8\times10^{4}$ times the hydrogen power in the modes it uses. Losing 18 % of the signal would not even buy a clean measurement. Real analyses that subtract foregrounds use more careful models, but they still need an injection test.

Comparing a model with the limit

The Lab's prediction keeps only density fluctuations:

\Delta^2_{21}(k)=\bar T_{\mathrm b}^2\,(1+\mu^2)^2\,\Delta^2_m(k),

with $\Delta^2_m$ the linear matter power spectrum and $\mu$ the cosine between $\mathbf k$ and the line of sight. It is the redshift-space power of Furlanetto, Oh and Briggs with the isotropic part set equal to the density, which means no spin-temperature or ionization fluctuations. Furlanetto, Oh and Briggs 2006, §4.1, equation 107 The matter spectrum uses the Eisenstein-Hu no-wiggle transfer function. Eisenstein and Hu 1998 HERA's kept modes have $k_\perp\ll k_\parallel$ , so $\mu\approx1$ and the factor is $4$ . Fluctuations in spin temperature, coupling and ionization are left out, so this is not a full model.

Passed through the windows, the heated history ( $\bar T_{\mathrm b}=16.2\,\mathrm{mK}$ ) gives $0.0168$ of the limit at $k=0.192$ . The cold history ( $\bar T_{\mathrm b}=-221\,\mathrm{mK}$ ) gives $3.13$ times the limit there and $2.06$ times it at $0.320$ . Even this minimal model of fully coupled, unheated gas at $z=7.9$ is excluded. HERA's companion analysis reached a related conclusion: under standard galaxy astrophysics, the limits disfavour scenarios in which the gas is never heated much above the temperature it reaches by cooling alone. HERA Collaboration 2022, §1

Lab tasks

Open the 21 cm cosmic dawn Lab.

Read the injected signal recovered in the default state. Switch the foreground treatment to the smooth fit and read it again. Then raise the order of the fit under the advanced settings and say how the recovered fraction moves.
Choose the cold-hydrogen preset. In the HERA chart, find the bandpowers that the windowed model exceeds; there should be two at $z=7.9$ and none at $z=10.4$ . The ratio readout should show $3.13$ .
Stay on that preset and, under the advanced settings, raise the temperature added by heating by one step, to $10\,\mathrm K$ . The ratio falls to about $0.02$ . In this density-only model, warming the gas from $1.7$ to about $12\,\mathrm K$ is enough to pass the limit, because the brightness drops from $-221$ to about $-19\,\mathrm{mK}$ .

Exercises

Check that $X^2Y\,|\tilde V|^2/(\Omega_{pp}B)$ has units of $\mathrm{mK^2}\,h^{-3}\mathrm{Mpc^3}$ when $X$ is in $h^{-1}\mathrm{Mpc}$ per radian and $Y$ in $h^{-1}\mathrm{Mpc}$ per hertz.
At $k=0.448$ in Band 2, Field 1, $\hat\Delta^2=5974.1\,\mathrm{mK^2}$ and $\sigma=3441.1\,\mathrm{mK^2}$ . Find $U$ and its square root.
A window row has weights $0.2,0.6,0.2$ on three modes where a model is $10$ , $20$ and $40\,\mathrm{mK^2}$ . Find the windowed value and compare it with the centre value.
An injection test gives $f=0.82$ . By what factor must the measured power, and hence the limit, be corrected?
Why may a single bandpower estimate be negative, while the upper limit derived from it may not?

Explained answers

Exercise 1. $|\tilde V|^2$ is in $\mathrm{mK^2\,sr^2\,Hz^2}$ . Multiply by $X^2$ , in $h^{-2}\mathrm{Mpc^2\,sr^{-1}}$ , and $Y$ , in $h^{-1}\mathrm{Mpc\,Hz^{-1}}$ , to get $\mathrm{mK^2\,sr\,Hz}\,h^{-3}\mathrm{Mpc^3}$ . Dividing by $\Omega_{pp}$ in steradians and $B$ in hertz leaves $\mathrm{mK^2}\,h^{-3}\mathrm{Mpc^3}$ .

Exercise 2. The estimate is positive, so $U=5974.1+2\times3441.1=12856.3\,\mathrm{mK^2}$ , and $\sqrt U=113.39\,\mathrm{mK}$ . This matches the published value.

Exercise 3. $0.2\times10+0.6\times20+0.2\times40=22\,\mathrm{mK^2}$ , 10 % above the centre value of $20$ . A spectrum that rises with $k$ is pulled up by a symmetric window.

Exercise 4. Only 82 % of the true power survives, so the measured power must be divided by $0.82$ , multiplied by $1.22$ . Correcting for loss always raises the estimate and the limit; ignoring the loss would make the limit look stronger than it is.

Exercise 5. Cross-multiplying two independent noisy copies gives a product whose noise has zero mean, so the noise can push one estimate below zero. A real sky signal only contributes positive power, so the truth is at least zero. Replacing a negative estimate by zero before adding $2\sigma$ keeps the limit from being tighter than the data justify.

Try it in the Lab

21 cm cosmic dawn and radio foregrounds: Switch to the smooth fit and compare the injected signal recovered with wedge avoidance.