I have a rather silly question I am afraid… I am just getting to know the bra-ket-notation and still think I did not quite get it…
I want to compute a certain term, which contains the braket notation and I don't really know how to proceed. For $H=-\frac{\hbar^2}{2m}\nabla^2+V$ being the Hamiltonian of the Schrödinger equation for one particle and $P(q)=\textbf{1}_{[q-\varepsilon/2,\, q+\varepsilon/2]}$ the characteristic function on the interval $[q-\varepsilon/2,\, q+\varepsilon/2]$
\begin{align}
\text{Im}^+\langle \psi \mid P(q')HP(q)\mid \psi\rangle &= \text{Im}^+\langle \psi \mid \textbf{1}_{[q'-\varepsilon/2,\, q'+\varepsilon/2]}\left(\frac{-\hbar^2}{2m}\nabla^2+V\right)\textbf{1}_{[q-\varepsilon/2,\, q+\varepsilon/2]}\mid \psi\rangle\\
&= \frac{-\hbar^2}{2m} \text{Im}^+\langle \psi \mid \textbf{1}_{[q'-\varepsilon/2,\, q'+\varepsilon/2]}\left(\nabla^2+V\right)\textbf{1}_{[q-\varepsilon/2,\, q+\varepsilon/2]}\mid \psi\rangle\\
&= \, ? \\
\end{align}
I don't really know how to proceed since I don't really understand how the operators in the middle part of the bra and ket act on $\psi$?
On an easier note how would one compute
\begin{align}
\langle\psi\mid P(q)\mid \psi\rangle = \langle\psi\mid \textbf{1}_{[q-\varepsilon/2,\, q+\varepsilon/2]}\mid \psi\rangle = \, ?
\end{align}
I am sorry if this question is stupid.. I just would be very thankful for any help!
Best Answer
For an interval $\Delta$ we can define $P(\Delta)$ in the position representation (using the bra-ket notation) as follows:
$$ \langle x|P(\Delta) := \mathbf 1_{\Delta}(x) \, \langle x| \quad, \tag{1}$$
where $\mathbf1_\Delta$ the characteristic function. Note that
$$P(\Delta) = \mathbb I\, P(\Delta) = \int \mathrm dx\, |x\rangle\langle x|\, P(\Delta)\overset{(1)}{=} \int \mathrm dx\, \mathbf 1_{\Delta}(x)\, |x\rangle\langle x| =\int_\Delta \mathrm dx\, |x\rangle\langle x| \quad . \tag{2}$$
Here, $\mathbb I$ denotes the identity operator of the corresponding Hilbert space and we've used the (formal) completeness relation of the position eigenstates.
Eventually, this yields
$$P(\Delta)|\psi\rangle \overset{(2)}{=}\int_{\Delta} \mathrm dx\, |x\rangle \langle x|\psi\rangle = \int_{\Delta} \mathrm dx\, |x\rangle \,\psi(x) \quad \tag{3}$$
and hence $$\langle \psi|P(\Delta)|\psi\rangle = \int_{\Delta} \mathrm dx\, |\psi(x)|^2 \quad . \tag{4}$$