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In Dirac's formulation of quantum mechanics, Suppose that $q$ represents position observable. About $|q\rangle \langle q|$: what does this operator mean? I do get that it results in an operator, but unsure of what physical meaning it has. The same with $|\psi \rangle \langle \psi |$. ($\Psi$ representing wavefunction, and the corresponding eigenstate $| \psi \rangle$.) |
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$|q\rangle \langle q|$ is an operator that maps the state $\alpha$ to $$|\alpha\rangle \mapsto |q\rangle \langle q|\alpha\rangle $$ You may easily calculate what the new state on the right hand side is. Because $\langle q|\alpha\rangle$ is a simple inner product – it is equal to $\alpha(q)$ if you express the state $\alpha$ as a wave function in the position representation – you see that the right hand side is simply, in the position representation, $\delta(x-q)\alpha(q)$. So it's an operator that maps any vector to a multiple of the delta-function wave function located at $x=q$, i.e. $\delta(x-q)$, multiplied by a constant given by $\alpha(q)$. Because I told you how the operator acts on any state, I have fully defined the operator. Similarly, $|\psi\rangle \langle \psi|$ is an operator that maps the state $\alpha$ to $$|\alpha\rangle \mapsto |\psi\rangle \langle \psi|\alpha\rangle $$ which may also be calculated for every pair of states. Every state $|\alpha\rangle$ is mapped to a multiple of the state $|\psi\rangle$ where the coefficient is given by the inner product. These operators are multiples of projection operators but they're only projection operators if the state $|s\rangle$ in $|s\rangle \langle s|$ is normalized to one. The state $|q\rangle$ is surely not normalized to one (its norm is infinite) so one shouldn't consider $|q\rangle \langle q|$ to be a projection operator. |
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It's a projection operator. It measures how much $|\psi \rangle$ a system is. |
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