Quantum Mechanics: Derivation of Kinetic Energy Expectation Value without using Operators?

Question

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.

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).