Healpy map2alm函数不返回预期的alm值数

Question

healpy.map2alm计算输入Healpix映射的alm值数组。

功能是

healpy.sphtfunc.map2alm(maps, lmax=None, mmax=None, iter=3, pol=True, use_weights=False, regression=True, datapath=None)`

参数lmax确定alm系数的数目。一个设置输入lmax，代码返回‘(lmax+1)**2’数的alm系数。

l=0返回a00。

l=1返回a11, a10, a1-1 + a00，即4个系数。

l=2返回a22, a21, a20, a2-1, a2-2，即5+3+1 =9系数。这是(lmax+1)**2 = 3**2 = 9系数。

但是，使用healpy，这不是我得到的输出，不管我选择的是哪个映射。

import numpy as np
import healpy as hp
nside = 16                         # healpix nside parameter

filename = "cmb_fullsky_map.fits"  # generic name for sky map

readmap = hp.read_map(filename)    # read in the map

# We then use healpy.map2alm(readmap) to generate an array of alm values

ell0 = hp.map2alm(readmap, lmax = 0)    # lmax = 0 should output a00
ell0.shape                              # 1 value
print "l=0 has " + str(len(ell0))
print ell0

ell1 = hp.map2alm(readmap, lmax = 1)    # lmax = 1 should outputs
ell1.shape                              # a11, a10, a1-1 
print "l=1 has " + str(len(ell1))       # 3 values
print ell1

ell2 = hp.map2alm(readmap, lmax = 2)    # lmax = 2 should output
ell2.shape                              # a22, a21, a20, a2-1, a2-2
print "l=2 has " + str(len(ell2))       # 5 values
print ell2

ell3 = hp.map2alm(readmap, lmax = 3)    # lmax = 3 should output 
ell3.shape                              # a33, a32, a31, a30, a3-1, a3-2, a3-3
print "l=3 has " + str(len(ell3))       # 7 values
print ell3

ell4 = hp.map2alm(readmap, lmax = 4)    # lmax = 4 should output
ell4.shape                              # a44, a43, a42, a41, a40, a4-1, a4-2,
print "l=4 has " + str(len(ell4))       # a4-3 a4-4
print ell4                              # 9 values

ell32 = hp.map2alm(readmap, lmax = 32)  # lmax = 32 should output
ell32.shape                             # (lmax+1)**2 = 33**2 = 1089 coefficients
print "l=32 has " + str(len(ell32))     # 65 values

我的输出不是我所期望的(lmax+1)**2。

l=0 has 1
[  2.39883065e-06+0.j]
l=1 has 3
[  2.39883065e-06 +0.00000000e+00j   4.49747594e-06 +0.00000000e+00j
  -5.39401197e-07 +1.07974023e-06j]
l=2 has 6
[  2.38695037e-06 +0.00000000e+00j   4.49747594e-06 +0.00000000e+00j
  -2.93559509e-05 +0.00000000e+00j  -5.39401197e-07 +1.07974023e-06j
  -1.75256654e-05 -1.57729954e-05j   1.65489261e-05 -1.41571515e-05j]
l=3 has 10
[  2.38695037e-06 +0.00000000e+00j   4.51148696e-06 +0.00000000e+00j
  -2.93559509e-05 +0.00000000e+00j   1.19817810e-05 +0.00000000e+00j
  -5.37909950e-07 +1.07783971e-06j  -1.75256654e-05 -1.57729954e-05j
  -7.17416081e-06 +9.14176685e-06j   1.65489261e-05 -1.41571515e-05j
  -2.64725172e-06 -3.09988314e-05j  -7.25462692e-06 -6.33251313e-07j]
l=4 has 15
[  2.38534350e-06 +0.00000000e+00j   4.51148696e-06 +0.00000000e+00j
  -2.93586471e-05 +0.00000000e+00j   1.19817810e-05 +0.00000000e+00j
  -1.72252485e-06 +0.00000000e+00j  -5.37909950e-07 +1.07783971e-06j
  -1.75253044e-05 -1.57728790e-05j  -7.17416081e-06 +9.14176685e-06j
  -1.30606717e-05 -4.21675061e-06j   1.65464103e-05 -1.41578955e-05j
  -2.64725172e-06 -3.09988314e-05j   1.09072312e-05 +3.22054569e-06j
  -7.25462692e-06 -6.33251313e-07j  -1.10079209e-06 +9.39495982e-07j
  -8.01588627e-06 -8.03762608e-06j]
l=32 has 561

你看到不一致了吗？

对于lmax=0，我预计总共有1. healpy输出1

对于lmax=1，我预计总共有4. healpy输出3

对于lmax=2，我预计总共有9. healpy输出6

对于lmax=3，我期望总共有16. healpy输出10

对于lmax=4，我预计总共会有25个。healpy输出15

对于lmax=32，我预计总数为1089。healpy输出561

问:为了得到正确的数组总数，我应该一起操作输出吗？

请说明map2alm如何执行此操作。

Answer 1

这是由于对称性。m=-1和m=1具有相同的转换，因此HEALPix只考虑m>=0。

例如，我们有lmax=2：

• a00
• a10
• a11
• a20
• a21
• a22

6个系数总计。

alm数组的长度预计为：mmax * (2 * lmax + 1 - mmax) / 2 + lmax + 1)。

Answer 2

我最近也遇到了这个问题。Healpix给出所有正的m值。负数可以用a(l,m1) = -a(l,-m)*找到，其中*是复共轭物。