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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:

  1. 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.

  2. 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.

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  • $\begingroup$ Not sure if this helps, but considering a particle under a Hamiltonian consisting of kinetic energy operator plus a potential bounding it. For solution, $E$ can be negative so long as $E < V$ and the particle is contained. Otherwise free particle solution exists. I would think the kinetic energy operator is just a way to formulate that kinetic component of the Hamiltonian within Schrödinger eqn. $\endgroup$ – bleuofblue Jan 3 '17 at 6:59
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    $\begingroup$ The kinetic energy is a positive self-adjojnt operator, therefore its average on any quantum state is a positive real number. $\endgroup$ – yuggib Jan 3 '17 at 7:02
  • $\begingroup$ What exactly about the usual definition (canonical quantization of the kinetic energy as a function of the phase space) you don't like? $\endgroup$ – Prof. Legolasov Jan 3 '17 at 11:37
  • $\begingroup$ I don't understand what you mean by "the kinetic energy term represents the true kinetic energy". What is the "true" energy, if not the thing given by the kinetic term in the Hamiltonian? I think your question is worthwhile, but don't forget that QM is weird and nonintuitive and that it cannot be derived from something else; at best you can motivate the definitions, but they are basic axioms of physics. $\endgroup$ – Javier Apr 20 '17 at 17:53
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What bothers you is that the kinetic energy density can locally be negative. Let's see where this happens. By your definition in the question the kinetic energy density is $$\psi^*\hat{T}\psi = (E-V)\psi^*\psi,$$ so it is only negative when $V>E$, and so we only get negative kinetic energy density in a region where tunneling is taking place. Now you no doubt believe tunneling is a real phenomenon that quantum mechanics should describe, so why would you modify the kinetic energy operator to get rid of it?

This hopefully explained why locally negative kinetic energy is actually a feature not a bug. But the other reason you shouldn't worry about it is that the local kinetic energy is not an observable. The quantities that we measure correspond to expectation values of hermitian operators, and in the wave function representation the expectation value is integrated over the whole space. This can still get at local properties, for instance the probability that the particle is in a small region, since the projection operator onto that small region is a hermitian operator (in other words the probability density is a quantity that can be measured).

But the quantity $\psi^*T\psi$ integrated over some region is not the expectation of a hermitian operator since it is the expectation of the product of $T$ which is a function of momentum, and a projection operator onto the region of integration, which is a function of position. So $\psi^*T\psi$ being negative does not seem to be something observable, and if you want to define a new local kinetic energy density you need to make sure that it is an observable which is not so easy since it must depend on both position and momentum.

Let's see what happens according to the usual interpretation. Consider a wave function proportional to $\exp(-x^2/2)$. This locally has negative kinetic energy when $|x|>1$, which makes sense from the perspective of this being the ground state of the harmonic oscillator since to be in those regions it needs to tunnel.

Now can we measure the negative kinetic energy? First we need to find out if the particle is in the region we care about, say $x<-1$. So we measure a projection operator, after which the wave function collapses to something proportional to $\exp(-x^2/2)$ for $x<-1$ and zero everywhere else. Now if we measure expectation of kinetic energy you might think its negative since it is locally negative in the left region, and zero in the right region. But there is a singularity at $x=-1$. No matter how you regularize this you will end up getting that the contribution of kinetic energy at this singularity is positive enough that the total expectation of kinetic energy is positive. The reason I know this is that $T\propto p^2$ has only positive eigenvalues, as mentioned in the comments.

Now you could think of this as your position measurement disturbed the particle altering its kinetic energy to be positive, and say that it 'really' was negative before. But this is just the usual problem with measuring non-commuting observables in QM.

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  • $\begingroup$ Indeed I believe tunneling is a real phenomenon, and I also accept that QM provides a good description of it. But consider the Hydrogen ground state, where $E=-1$, $V(r)=-2/r$. This leads to a kinetic energy term $T(r) = 2/r-1$. This is interpreted as the electron being "free" for $r<2$ and "tunnelling" for $r>2$. This is strange, because the wave function $Psi = exp(-r)$ has no boundary or transition at $r=2$. In my opinion the term $2/r-1$ represents some sort of energy, but it can not be the (average) kinetic energy of the electron. $\endgroup$ – M. Wind Jan 3 '17 at 16:07
  • $\begingroup$ If you still want the energy to be E=T+V, for whatever new definition of T you have, you will run into the problem that tunneling can't be described if $\psi^*T\psi$ is everywhere positive. Now if you want to also redefine V to be nonlocal, or take $E\neq T+V$, good luck. But the fact that you need these modification is just as much of a 'defect' as the kinetic energy being negative. $\endgroup$ – octonion Jan 4 '17 at 6:10
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    $\begingroup$ And I am skeptical that defining a local kinetic energy is something that makes complete sense. In the usual definition T is a function of momentum only. When you talk about the expectation value of momentum integrated over an arbitrarily small region you are effectively multiplying the kinetic energy by a function of position that goes to zero outside the region you are concerned about. You need to make sure this is well defined as an observable if you want to be able to measure local kinetic energy. (Unless again you want to come up with a new interpretation of QM, in which case good luck) $\endgroup$ – octonion Jan 4 '17 at 6:17
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Kinetic energy operator of a point particle is a standard concept in orthodox non-relativistic quantum theory, it is defined by

$$ \hat{T} \psi = \frac{\hat{\boldsymbol \pi}^2}{2m}\psi $$ where $$ \hat{\boldsymbol \pi} = -i\hbar\nabla - q\mathbf A $$ and $\mathbf A$ is any vector potential correctly describing external electromagnetic field acting on the particle.

Since there is infinity of such vector potentials, the kinetic energy operator is not unique. But it does have unique form given above.

In case magnetic field vanishes, we can use $A=0$ and the operator can be simplified into

$$ \hat{T} \psi(\mathbf r) = -\frac{\hbar^2}{2m} \Delta \psi \,(\mathbf r). $$

The mentioned form of the operator can be motivated based on Schroedinger's equation or the classical relation $K.E. =\frac{(\mathbf p-q\mathbf A)^2}{2m}$ and the canonical quantization rule $p\rightarrow - i\hbar \nabla$. It was not, as far as I know, motivated by any concept of "kinetic energy density" in configuration space. There is no such thing in orthodox theory.

Such a concept can be introduced as an extension of the theory, but it is not universally recognized as useful, textbooks usually do not teach it.

That being said, it may be interesting to think about how to extend the theory so kinetic energy density of a sort (let us denote it $D(\mathbf r)$), can be introduced.

The sought density of kinetic energy should obey the condition that its integral gives total value - but what is this total value? In general case, $\psi$ function is not eigenfunction of kinetic energy operator and there is no single value of kinetic energy to be ascribed to $\psi$. We can therefore take as "total value" the expected average value of kinetic energy for $\psi$, which the standard rules define as

$$ K_{av} = \int \psi^* \frac{\hat{\boldsymbol \pi}\cdot\hat{\boldsymbol \pi}}{2m} \psi\,d^3\mathbf r $$

Obviously, the expression

$$ \psi^* \frac{\hat{\boldsymbol \pi}\cdot\hat{\boldsymbol \pi}}{2m} \psi, $$

although it obeys the stated condition, is not a good candidate for a density, since as you have observed, it can have negative value. But the lowest value that density can assume is at a point where particle either has lowest possible kinetic energy (zero) or where it never goes, or goes with zero probability. All such situations imply no lower than zero density.

Another possible definition of $D$ can be obtained from the integral condition if we recall the fact that the operators are Hermitian and thus their place in the integral can be changed:

$$ K_{av} = \int \frac{1}{2m}(\hat{\boldsymbol \pi}\psi)^* \cdot \hat{\boldsymbol \pi} \psi\,d^3\mathbf r $$

It is easy to see that the expression $$ \frac{1}{2m}(\hat{\boldsymbol \pi}\psi)^* \cdot \hat{\boldsymbol \pi} \psi $$ is positive or zero and integrates to $K_{av}$. It can therefore serve as definition of $D$.

Let us assume that this density $D$ at point $\mathbf r$ is a product of some value $K(\mathbf r)$ and probability density that particle is at this point $\psi^*(\mathbf r)\psi(\mathbf r)$. We can then express $K$:

$$ K = \frac{1}{2m}\frac{(\hat{\boldsymbol \pi}\psi)^* \cdot \hat{\boldsymbol \pi}\psi}{ \psi^*\psi} $$

This quantity is positive or zero at any point of configuration space, just as kinetic energy of particle is. For some regions, it is higher than $K_{av}$. This is as it should be, since result of averaging is always lower than some members in the sum. For hydrogen ground-state function, the highest value is attained when the electron is in the nucleus and decays to zero with increasing radial distance, which is reasonable.

Of course, all this is just mathematics and it would be hasty to think that electron in hydrogen atom has such variable kinetic energy depending on its position. But sometimes it may be useful to think this way, to get at least easily comprehensible, if not entirely accurate picture of what is happening in the atoms and molecules, or to develop another extension of the theory.

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The fact that the "kinetic energy" above attains negative values suggests that it is being misnamed.

Recall that for normalized waves, $\|\Phi\|^2=1$, Schrödinger originally interpreted the amplitude $\rho = \Phi\Phi^*$ as a continuous charge density spread over three dimensional space. Thus the term $-(\nabla^2 \Phi)\Phi^*$ should be interpreted as the "internal" or "self push-pull" energy density of the charge distribution $\rho$. The name "kinetic energy" should be reserved for a term contributed by the photon. Note that the photon is completely absent from Schrödinger Hamiltonian operator $H$.

The probabilistic interpretation of wave amplitudes, mean values of operators, random quantum jumps, as well as the principles of correspondence, uncertainty, indeterminacy and similar quantum concepts developed by Bohr, Born, Von Neumann and others, were never accepted by Schrödinger, nor by Planck, Einstein, De Broglie... To understand the origin of $-(\nabla^2 \Phi)\Phi^*$ it may therefore be useful to adopt the classical viewpoint prevalent at the time.

With best regards

Daniel Crespin

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  • $\begingroup$ Thank you very much for your contribution, in which you comment on the subject from a historic perspective. I would like to read more about the alternative view that the kinetic energy term is being misnamed, and that it is better to interpret it as an internal or self push-pull energy density. Could you please provide some references to the literature? $\endgroup$ – M. Wind Jul 28 '17 at 18:21

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