Following the calculation done in section 5.3 of Schwartz's QFT and The Standard Model, the matrix element, M, should have two contributions M1 and M2.
M1=ϵ1ϵ3+ϵ1ϵ4
M2=ϵ2ϵ3+ϵ2ϵ4
However when I do these calculations I get that M1 = M2 = 1, ##|M|^{2} = |M1|^{2} + |M2|^{2} = 1+1 = 2##, which does...
So to get the partition function I do the integral ##\int \alpha E^{3} e^{(B_{0}-B)E} dE##, which substituting in ##/Delta B = B_{0} - B## is ##Z = \frac{ \alpha E^{3} e^{\Delta B E}}{\Delta B} - \frac{3 \alpha E^{2} e^{\Delta B}}{\Delta B^{2}} + \frac{6 \alpha E e^{\Delta B E}}{\Delta B ^{3}} -...
By looking adding equation 1 to equation 3 you get:
##3n_{x}-3n_{y} = 2m## which multiplying each side by -1 is ##3n_{y}-3n_{x} = -2m##. The only way for this equation to be consistent with equation 2 is for m=0.
So setting m = 0, I get that ##n_{x} = n_{y}## and ##n_{z} = 3n_{x} = 3n_{y}##...
I ran through the math and you're just going to end up with essentially the same system of equations as before, since ##\varphi## is the same thing as ##\psi## other than for some constant which doesn't affect how ##\hat{n} \cdot L## acts on it.
The magnitude of ##|\vec{n}|## doesn't equal one...
So rewriting ##\psi(\vec{r})##, ##\psi(\vec{r}) = rf(r)\sqrt{\frac{2\pi}{3}}[(i-1)Y^{1}_{1} + \frac{6}{\sqrt{2}}Y^{0}_{1} + (1+i)Y^{-1}_{1}]##
Applying ##n_{x}L_{x}## to this I get:
##n_{x}rf(r)\hbar\sqrt{\frac{2\pi}{3}}(\sqrt{2}iY^{0}_{1} + \frac{6}{\sqrt{2}}(Y^{1}_{1}+Y^{-1}_{1}))##
Applying...
First I calculated ##(\vec{n} \cdot L) \psi(r) = -i\hbar(n_{x}(3y-z)+n_{y}(z-3x)+n_{z}(x-y))f(r)## and then tried to solve for ##n_{i}## such that I get (x+y+3z)f(r), and then divide ##n_{i}## by the magnitude of ##\vec{n}## to get the unit vector and m, but when I try doing this, I get the...