-4
$\begingroup$

Where do these two equalities for the expectation value come from precisely? :

$$\begin{align} \langle x\rangle &= \int_{-\infty} ^\infty \Psi^* x \Psi\,\mathrm{d}x \\ \langle x^2\rangle &= \int_{-\infty} ^\infty x^2 |\Psi|^2\,\mathrm{d}x \end{align}$$

Note: $\Psi$ is the wave function, and $\Psi^*$ is the complex conjugate of $\Psi$.

And are these two equalities also true? :

$$\begin{align} \langle x\rangle &= \int_{-\infty} ^\infty x |\Psi|^2\,\mathrm{d}x \\ \langle x^2\rangle &= \int_{-\infty} ^\infty \Psi^* x^2 \Psi\,\mathrm{d}x \end{align}$$

Don't we just have the equality $\Psi^* x \Psi = x |\Psi|^2$ ?

$\endgroup$
  • 2
    $\begingroup$ They come from the definition of the inner product on the $L^2$ space in the "position representation". $\endgroup$ – Phoenix87 Jun 21 '15 at 10:12
  • 2
    $\begingroup$ $\Psi=\Psi(x) \in \mathbb{C}$ so $\Psi^*(x)x\Psi(x)=x\Psi^*(x)\Psi(x)=x|\Psi(x)|^2$ $\endgroup$ – glS Jun 21 '15 at 10:36
  • $\begingroup$ Ya I have same problem dawgs $\endgroup$ – bodacydo Aug 14 '15 at 21:14
1
$\begingroup$

Yes,both the set of equalities are true but only in the position representation where operator(x)=x.In the momentum representation,where operator(x) takes a different form they are not true

$\endgroup$
  • $\begingroup$ Could you develop the part where you say "In the momentum representation they are not true"? What are $<p>$ and $<p^2>$ equal to? $\endgroup$ – Quantum Force Jun 22 '15 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.