Suppose we have a quantum mechanical system, which is well described by its wave function in r-representation $\Psi$. We are interested in the properties of an observable, say the kinetic energy $T$. So we let the kinetic energy operator act on the wave function, multiply the result by Psi and integrate over spatial coordinates. Thus we obtain the expectation value $\langle T \rangle$. We note that this result is a property of the operator and also of the quantum state.

Now in QM text books one will often find detailed discussions of the result, in order to provide physical meaning and context. For example, in discussing the kinetic energy of the ground state of a hydrogen atom, it is pointed out that the kinetic energy $T$ is intimately related --through the Schroedinger equation-- to the electrostatic potential $V$ and the total energy $E$. Both $V$ and $E$ are operators that are multiplicative in $r$-representation. So $T$ (which at its core is represented by the Laplace differential operator), when written as $E$-$V$, may be regarded as multiplicative. Thus we get (with energy in Rydberg energy units and distance in Bohr radius units):

$$T = -1 + \dfrac{2}{r}$$

This result is then discussed in terms of "the classical turning point" $r=2$. The region $r > 2$ is forbidden in classical physics, since the kinetic energy is here negative. Yet in QM there is a non-zero probability that the electron is found at $r > 2$.

I wonder how much "truth" there is the above formula for $T$. It is suggestive to regard it as a proper function of the distance to the nucleus $r$. However, as soon as one digs a little deeper, conflicts arise. For example, one may compute the Fourier transform of the wave function. Now since the initial wave function was spherically symmetric, so is the FT. Hence in $\Phi(k)$ there are only states with positive wave factor $k$. There is simply no room for negative kinetic energies! So immediately a paradox arises when we regard kinetic energy as a distribution in $r$- or $k$-representation.

Furthermore, one may look at the higher moments of the kinetic energy operator. This leads to new problems. For example, the second moment $\langle TT\rangle$ can be computed as either $(\Psi, TT \Psi)$, or as $(T \Psi, T \Psi)$ [provided $T$ is self-adjoint when acting on these wave functions]. Of course it is satisfactory that these two integral representations lead to the same net result. However their integrands are obviously different! Therefore if the kinetic energy is distributed within the state, it can not be pin-pointed without some level of ambiguity.

I would like to hear your comments on this issue !


I think the only thing based on $\psi$ that can be satisfactorily said to be distributed in space is probability of configuration. Its density in configuration space is

$$ |\psi(\mathbf r_1, \mathbf r_2, ...,t)|^2. $$

Expression $|\psi|^2 \frac{\hat{T} \psi}{\psi}$ may look like density of kinetic energy, but this is inconsistent with the basic idea that kinetic energy is always positive (since it is given by square of real momentum). This problem can be solved by postulating $\frac{|\hat{\mathbf p}\psi|^2}{2m}$ as density of kinetic energy(it gives the same average), but there is no experimental confirmation for such relation, so it is rather speculative (nevertheless one could try to use it to interpret some results of quantum theory).

  • $\begingroup$ Thank you very much. I like your postulate for the density of the kinetic energy. I have noticed that when computing the even higher moments of the kinetic energy (after regularizing the wave function), that the symmetric representation makes the most sense. Your formula allows one to express the odd moments also in a symmetric way. Nice! The resulting density is larger or equal to zero. So the classical forbidden zone disappears. I like that! I think the forbidden zone never made much sense anyway.... $\endgroup$ – M. Wind Jul 29 '15 at 16:33
  • $\begingroup$ @MWind, you're welcome. Could you post some of your thinking regarding the higher moments? It sounds interesting... The forbidden zone of negative kinetic energy indeed is not present. This leads to another interesting feature: the total energy associated with definite point of configuration space is not everywhere equal to the Hamiltonian eigenvalue. Close to the nucleus, the energy is negative, far from it, positive. When energy function is integrated over the whole conf. space with prob. density based on the eigenfuction, the eigenvalue results. $\endgroup$ – Ján Lalinský Jul 29 '15 at 23:46

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