In quantum physics, we consider an Hilbert space $H$, i.e. we have an hermitian scalar product $(.,.)$ on $H$, and we consider self adjoint operators $T$, i.e. which satisfy $T=T^*$, where $T^*$ is the adjoint of $T$, defined by the property that $(Tx,y)=(x,T^*y)$ for every $x$ and $y$ in $H$.

Let us assume $T$ self adjoint. By definition, a complex number $\lambda$ is in the residual spectrum of $T$ if and only if $Ker(T-\lambda)=0$ (i.e. $\lambda$ is not an eigenvalue, i.e. not in the point spectrum) and $Im(T-\lambda)$ is not dense in $H$. But it is an easy computation to show that the closure of $Im(T-\lambda)$ in $H$ is $Ker(T^*-\bar{\lambda})=Ker(T-\bar{\lambda})$. So if $\lambda$ is in the residual spectrum of $T$ then $\bar{\lambda}$ is an eigenvalue of $T$. But an eingenvalue of a self-adjoint operator is real (if $Tx=\mu x$ with $x$ non-zero then $(Tx,x)=(\mu x,x)=\bar{\mu}(x,x)$ but also $(Tx,x)=(x, T^*x)=(x,Tx)=(x,\mu x)= \mu (x,x)$, thus $\bar{\mu}=\mu$ i.e. $\mu$ is real) so $\lambda$ is real and so $\lambda$ is an eigenvalue of $T$: contradiction.

Conclusion: the residual spectrum of a self-adjoint operator is empty, which explains why we do not often hear about residual spectrum in quantum mechanics courses...

Remark: in general, the operators relevant in quantum mechanics are unbounded, i.e. not continous for the norm operator, and so not defined everywhere. The preceding argument applies without assuming $T$ bounded.