1
$\begingroup$

So I'm already aware of the quantum mechanical operator for momentum and how to derive the kinetic energy operator from this: $$\hat T=\frac{\hat p^2}{2m}=\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}$$ But I'm wondering how to derive the kinetic energy operator solely from the statistical definition of an expectation value.

I've successfully derived the momentum expectation value this way to find: $$\lt p\gt =-i\hbar \int_{-\infty}^{\infty} \psi ^\star \frac{\partial \psi}{\partial x} dx = \int_{-\infty}^{\infty} \biggl(\psi ^\star \biggl(\frac{\hbar}{i}\frac{\partial}{\partial x}\biggr) \psi \biggr)dx = \int_{-\infty}^{\infty} \biggl(\psi ^\star\hat p \psi\biggr) dx $$

It seems to follow the same derivation as before, namely: $$\lt T \gt = \frac{\lt p \gt^2}{2m} =\frac{-\hbar^2}{2m} \biggl( \int_{-\infty}^{\infty} \psi ^\star \frac{\partial \psi}{\partial x} dx \biggr)^2 $$ But I dont see how to manipulate this such that $\biggl( \int_{-\infty}^{\infty} \psi ^\star \frac{\partial \psi}{\partial x} dx \biggr)^2 = \frac{\partial^2}{\partial x^2}$

Any help clarifying this issue would be greatly appreciated.

$\endgroup$
1
$\begingroup$

What you are looking for is $\langle p^2\rangle$, not $\langle p\rangle^2$. This would entail an integral of the type $$ \int dx\, \psi(x) \left(\psi^{\prime\prime}(x)\right) \tag{1} $$ not $\left(\int dx \psi(x) \psi^\prime(x)\right)^2$. It is easy enough to check that both expressions are not the same as $\langle p^2\rangle$ is the average value of an everywhere non-negative quantity. On the other hand $\langle p\rangle=0$ for stationary states and so must have regions where it is negative over the integration region.

If $\psi(x)$ is a solution to the Schrodinger equation and the eigenvalue $E$ is known, one can also proceed starting from the Schrodinger equation $$ \psi^{\prime\prime}(x) = -\frac{2m}{\hbar^2} (E-V(x))\psi(x) $$ and sub for $\psi^{\prime\prime}(x)$ in (1).

$\endgroup$
  • $\begingroup$ Hmmm, that makes sense that I squared the wrong variable, but how would I go about getting it into a form where I'm integrating with respect to x? I suppose I could go with the integral of v^2 with respect to v but then I'd have to transform $\psi$ into a function of v instead of x, which I'm not sure exactly how to do $\endgroup$ – cory21391 Aug 25 '18 at 20:57
  • $\begingroup$ @cory21391 Getting $\psi$ as a function of $v$ requires a Fourier transform. $\endgroup$ – probably_someone Aug 25 '18 at 21:05
  • $\begingroup$ @cory21391 I added to my original answer to give you a hint. $\endgroup$ – ZeroTheHero Aug 25 '18 at 21:24

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.