# Question on computation with the bra-ket-notation

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!

• Note that your Hamiltonian is already in position representation. Generally, it would look like $H=P^2/2m +V(X)$ and in position-representation $\langle x| H= \left(-\nabla^2/2m +V(x) \right) \langle x|$. May 26 at 14:34

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 $$\langle \psi|P(\Delta)|\psi\rangle = \int_{\Delta} \mathrm dx\, |\psi(x)|^2 \quad . \tag{4}$$

• Thank you very much, that really helped me a lot! May 27 at 8:17