While having a deep respect and a good understanding of quantum mechanics, I have serious reservations about the specific form of the kinetic energy operator as it appears in the 3D Schroedinger equation.
Let us assume we solve the time-independent Schroedinger equation in r-representation for some system. We can now multiply the equation by the complex conjugate of the wave-function, $\Psi^*(r)$. The resulting equation expresses that at the local level (in each volume element) an energy balance holds: $V(r) + T(r) = E = constant$. However the kinetic energy term is peculiar, as it can attain negative values, which goes against physical intuition.
This raises the question whether the kinetic energy term represents the true (average) kinetic energy at position $r$, or perhaps something else.
How is the kinetic energy operator derived? Two approaches:
Kinetic energy is best understood in k-representation. That is because in QM momentum is intimately related to wave factor. So start with $\Psi(r)$. Next perform Fourier transformation to obtain $\Phi(k)$. Then $\Phi^*(k)\Phi(k)\, k^2 \,dk$ is the probability density. All properties of the kinetic energy follow from this probability density. To return to $r$-representation, one may use Plancherel's theorem. This theorem says that a product of two wave functions integrated over r-space is equal to that in $k$-space. We can apply Plancherel to the product of $\Phi(k)^*$ and $\Phi(k)\,k^2$, and this leads to the familiar expression for the kinetic energy in $r$-representation. However, there is a weakness in this argument: Plancherel is only applicable at the state-level. There is no reason at all to assume that the result is (physically) meaningful at a local level. Indeed, a Fourier transform can easily lead to a result which is negative for a range of values. Furthermore, there is no standard way of applying Plancherel's theorem. For example if one considers the kinetic energy squared, there are different ways of distributing the powers of $k$ to the two wave functions. After Fourier transformation each choice leads to a different density for the kinetic energy squared in $r$-representation.
Similar to the 1D case, we can postulate that an outgoing spherical wave $\Psi(r) = (1/r)exp(ikr)$ and an incoming spherical wave $\Psi(r) = (1/r)exp(-ikr)$ are two examples of a wave function with a well-defined kinetic energy. Which can be obtained by applying the operator $-(1/r)(d^2/dr^2)(r)$ to $\Psi(r)$. Multiplication by $r^2\,\Psi^*(r)$ indeed yields a constant kinetic energy density. This seems straightforward. However, one should note that there is an alternative method: $r^2\times (1/r)(d/dr)(r\Psi^*) \times (1/r)(d/dr)(r\Psi)$. This leads to precisely the same result. Which also means that one may consider a linear combination of the two densities. When we work with a specific wave function, it is not a priori clear to me which kinetic energy density is the right one.
I have shown that both method 1 and 2 lead to ambiguity in the definition of the kinetic energy operator and in the resulting kinetic energy density. However as we know in the Schroedinger equation a unique choice is made. I would like to understand the physical/mathematical motivation and justication of this choice.