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.


1 Answer 1


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).

  • $\begingroup$ 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, 2018 at 20:57
  • $\begingroup$ @cory21391 Getting $\psi$ as a function of $v$ requires a Fourier transform. $\endgroup$ Aug 25, 2018 at 21:05
  • $\begingroup$ @cory21391 I added to my original answer to give you a hint. $\endgroup$ Aug 25, 2018 at 21:24

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