2
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ Note that your Hamiltonian is already in position representation. Generally, it would look like $H=P^2/2m +V(X)$ and in position-representation $\langle x| H= \left(-\nabla^2/2m +V(x) \right) \langle x|$. $\endgroup$ May 26 at 14:34

1 Answer 1

3
$\begingroup$

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

$\endgroup$
1
  • 1
    $\begingroup$ Thank you very much, that really helped me a lot! $\endgroup$
    – uzizi_1
    May 27 at 8:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.