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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.

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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).

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  • $\begingroup$ 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
  • $\begingroup$ Can anyone explain why Cory's initial definition of kinetic energy doesn't work? It seems perfectly reasonable that 1/2 m <v>^2 would be kinetic energy formulated with quantum expectation value <v>, and hypothetically it would follow the Work-Kinetic-Energy relationship where "the change in the particle's kinetic energy between two neighboring points on its path is equal to the work done by the net force as it moves between the two points" $\endgroup$
    – dturn805
    Mar 19 at 1:32
  • $\begingroup$ @dturn805 because $\langle v\rangle^2\ne \langle v^2\rangle$. The latter is the average of positive numbers so necessarily positive. The latter is the average of a bunch of vectors (or in 1d signed quantities) and proportional to $\langle p\rangle$, which is $0$ in any eigenstate of $H$ for instance whereas $\langle p^2\rangle$ is not. $\endgroup$ Mar 19 at 12:34

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