Is the angular momentum operator $\hat{L_z}$ a form of projection operator

Question

$\hat{L_z}$ has the eigenvalue m$\hbar$ for a given eigenfunction with non-zero $m_l$ values. This is the component/projection of the angular momentum vector on the $z$ axis. Can i say that the $\hat{L_z}$ operator is a projection operator?

Answer 1

No, a projection operator in quantum mechanics has the specific condition that it is idempotent. That is, if $\hat P$ is a projection operator, then

$$\def\bra#1{\left\langle#1\right|}\def\ket#1{\left|#1\right\rangle} \hat P^2=\hat P $$

Intuitively, the idea is that if you project a vector along another vector, then doing the projection again shouldn't change the result. $\hat L_z$ does not have this property, except for the trivial case of system with no angular momentum, where it is the zero operator.

As an example of a projection operator associated with angular momentum, consider a spin-1/2 system, with the normalized eigenstates of $\hat S_z$ being $\ket{+}$ and $\ket{-}$ for the $+1/2$ and $-1/2$ eigenvalues, respectively. Then $\ket{+}\bra{+}$ and $\ket{-}\bra{-}$ are both projection operators, for the spin-up and spin-down states. $\hat S_z$ can be written as a sum of these projection operators like (in units where $\hbar=1$)

$$\hat S_z=\frac{1}{2}\ket{+}\bra{+}-\frac{1}{2}\ket{-}\bra{-}$$

$\hat S_z^2$ is instead given by

$$\hat S_z^2=\frac{1}{4}\ket{+}\bra{+}+\frac{1}{4}\ket{-}\bra{-}=\frac{1}{4}\hat{I}$$

and is just proportional to the identity operator.

Answer 2

For an operator to be a projection, it must be the identity operator for states in its image. That is $P$ is a projection if and only if $P(P(x)) = P(x)$ for all $x$ in the domain of $P$. Consider now the action of $L_z$ on a state $\psi$ with some angular momentum in the $z$ direction. Then $L_z \psi = \ell \psi$ where $\ell$ is the eigenvalue. Therefore $L_z(L_z(\psi)) = \ell^2\psi\neq \ell\psi = L_z(\psi)$ so $L_z$ is not a projection.