Spherical harmonics and Bra-Ket notation Let's assume i have a wave function $$\psi = N(x+y+z)e^{-r^2/a^2} $$ with $N$ and $a$ some constants. This function can be written as a sum of the spherical harmonix $Y_{1,0}$, $Y_{1,1}$ and $Y_{1,-1}$. Now lets say i want to find the probability of obtaining $L_z=\hbar$  which corresponds to the $Y_{1,1}$ of  $\psi$. According to the postulates of QM, this probability is given by:
\begin{equation}
P=|\langle 1,1|\psi\rangle|^2 .
\end{equation}
How does one go about computing this product, considering we have a state thats depends on a function of $r$ ? Should I integrate the product of $\psi$ and $Y_{1,1}$ in 3D? Or just in the angles $\theta$ and $\phi$?.
 A: Your expression
$$P=|\langle1,1|\psi\rangle|^2$$
is inconsistent, because the ket on the right side has an
$r$-dependency, and the bra on the left side has not.
So you need to be careful how to formulate the inner product.
You can factorize your wave function into a radial and an angular part,
$$\psi(r,\theta,\phi)=N(x+y+z)e^{-r^2/a^2}=\underbrace{N r e^{-r^2/a^2}}_{=R(r)}F(\theta,\phi)$$
(I leave it to you to find the angular part $F(\theta,\phi)$.)
As you noticed in your question, you can
write the angular part $F(\theta,\phi)$
as a linear combination of $Y_{1,1}$, $Y_{1,0}$ and $Y_{1,-1}$.
This will simplify the final integration.
To get the probability of the $Y_{1,1}$ part in $\psi$ you need to calculate
the inner product as an integral in 3D space between
$R(r)Y_{1,1}(\theta,\phi)$ and $\psi(r,\theta,\phi)=R(r)F(\theta,\phi)$
$$\begin{align}
P &= \left| \iiint d^3r\ (R(r)Y_{1,1}(\theta,\phi))^* R(r)F(\theta,\phi) \right|^2\\
  &= \left| \underbrace{\int_0^\infty dr\ r^2 |R(r)|^2}_{=1, \text{ because of normalization}}
     \int_0^\pi \sin\theta\ d\theta\int_0^{2\pi} d\phi\ Y_{1,1}^*(\theta,\phi)F(\theta,\phi) \right|^2 \\
  &= \left| \int_0^\pi \sin\theta\ d\theta\int_0^{2\pi} d\phi\ Y_{1,1}^*(\theta,\phi)F(\theta,\phi) \right|^2
\end{align}$$
So the 3D integral boiled down to an integral over $\theta$ and $\phi$ only.
A: The angular momentum operator is
$$L_z = I \otimes \left(-i\hbar \frac{\partial}{\partial \phi}\right)\:.$$
The tensor product refers there to the decomposition
$$L^2(\mathbb{R}^3, d^3x)= L^2([0, +\infty), r^2 dr) \otimes L^2(S^2, \sin\theta d\theta d\phi )$$
The projector onto the eigenspace with eigenvalue $\hbar$ of $L_z$ is therefore
$$P_1 = I \otimes \sum_{j\geq 1} |j,1\rangle \langle j,1|$$
(more precisely the sum is over all values $j=,0,1,\ldots$ such that $1\leq j$).
According to the general rule, if  $$\Psi(r,\theta,\phi) = N r^{-r^2/a^2}(x+y+z) = N re^{-r^2/a^2} (\cos \theta + \sin\theta \cos \phi + \sin\theta \sin \phi)$$ $$ =  N re^{-r^2/a^2} \otimes (\cos \theta + \sin\theta \cos \phi + \sin\theta \sin \phi) $$ where we used
$$x= r\sin \theta \cos \phi\:, \quad y = r \sin \theta \sin \phi\:, z = r \cos \theta \:, $$
the wanted probability is
$$||P_1 \Psi ||^2 =\langle \Psi| P_1 \Psi \rangle =\langle N r e^{-r^2/a^2}| I |N r e^{-r^2/a^2}\rangle_{L^2([0,+\infty))} $$ $$\times < (\cos \theta + \sin\theta \cos \phi + \sin\theta \sin \phi)|1,1\rangle\langle 1,1| (\cos \theta + \sin\theta \cos \phi + \sin\theta \sin \phi) \rangle_{L^2(S^2)} $$
$$= 1 \left|\int_{S^2} Y^1_1(\theta,\phi)^* (\cos \theta + \sin\theta \cos \phi + \sin\theta \sin \phi)  \right|^2 \sin\theta d\theta d\phi \tag{1}$$
As a matter of fact the sum over all possible $j>1$ includes only the value $j=1$ because the function $$\cos \theta + \sin\theta \cos \phi + \sin\theta \sin \phi $$
on $S^2$ is a linear combination of the three $Y^{j=1}_m(\theta,\phi)$ and the remaining values of $j$ give no contribution.
You see that only the angular part of the wavefunction has relevance here.
If $\Psi(r,\theta,\phi) = N r^{-r^2/a^2}(x+y+z)$, the formula
$$|\langle 1,1| \Psi \rangle|^2$$
for the wanted probability is therefore meaningless, though it appears in many texts. Actually,  a posteriori once the students got acquainted with the, let's say, jargon, it has a precise meaning, i.e.,   the last line in (1).
A: As long as $N$ is your normalisation constant so that $<\psi|\psi>=1$, and your spherical harmonics are normalised so that $<1,1|1,1>=1$ (this will usually be the case by definition), then your expression $P$ is correct.
To calculate $P$, the inner product is an integral over 3D space with the bra expression complex conjugated, usually expressed in spherical polar coordinates. So $<1,1|\psi>=\int _0^\infty dr \int_0^{2\pi} d\phi \int_0^\pi d\theta \enspace  r^2 sin(\theta) \psi(r,\theta,\phi) Y^*_{1,1}(\theta,\phi)$. Note here we have used the convention that $(r,\theta,\phi)$ are the radial distance, polar angle, and azimuthal angle respectively.
