2
$\begingroup$

I have set myself the task of studying the kinetic energy $T$ in a quantum mechanical system. For the latter, I use the simple case of the Hydrogen atom in the $n=1$ state. Then the wave function is (essentially) given by $\Psi = \exp(-r)$. The kinetic energy operator is a (negative) constant times the Laplace operator.

The calculation of the first moment of the kinetic energy is straightforward. Let the operator $T$ work on $\Psi$, multiply with $\Psi^{*}$ (which in the present case is equal to $\Psi$), multiply by the spherical shell $r^2dr$ and integrate from zero to infinity. The result is the well-known Rydberg energy 13.6 eV.

Problems arise when one attempts to calculate the second and higher moments of the kinetic energy. The reason is that it makes a difference on which of the two wave functions the operators work. This is somewhat surprising, since the kinetic energy is an observable and its operator is Hermitian. Nevertheless, the Hermitian rule $(\phi^{*}, T \psi) = (T \phi^*, \psi)$ is in general not obeyed in the present case. One may demonstrate this empirically (by evaluating integrals), or rigorously (by performing partial integration). The latter method shows that a boundary term appears from $r=0$.

The consequence is that $(\psi^*, TT \psi)$ is unequal to $(T \psi^*, T \psi)$. My calculations show that the first integral is equal to $-3$ and the second to $+5$ (in units of Rydberg energy). For higher moments too one observes differences that are multiples of 8 between different representations.

Of course one can use common sense and/or physical intuition. Subtracting from the second moment the square of the first moment yields the variance. This should be positive definite. Hence the second representation must be the right one, since it yields a positive variance ($+5-1 = +4$) whereas the first one is negative ($-3-1 = -4$). Unfortunately, this type of reasoning becomes more involved for the third and higher moments.

So my questions are: How should one deal with these ambiguous results, associated with Hermitian effects between different representations? Are there strict rules in QM on which representation to use in the calculation of the higher moments of the kinetic energy?

$\endgroup$
3
$\begingroup$

Sorry it is impossible that, if both $\psi, \phi$ belong to the domain of a self-adjoint operator $A$ the identity $$\langle \psi| A \phi \rangle = \langle A\psi| \phi \rangle $$ fails. The point is that your function $T\psi$, where $\psi({\bf x}):= e^{-r}$, does not belong in turn to the domain of the self-adjoint operator $T$ so that: $$\langle T\psi| T \psi \rangle = \langle T(T\psi)| \psi \rangle \quad\quad\quad \mbox{(false!)}$$ does not make sense!

In general even if $\psi$ belongs to the domain of a self-adjoint operator $A$, there is no reason why it must also belong to the domain of powers of that operators $A^n$, so the associated momenta simply do not exist (when they are defined in the most elementary way). You are probably dealing with one of these cases.

To be sure I should perform explicit computations using also the fact that $T = - \frac{\hbar^2}{2m}\Delta$, where $\Delta$ is the standard Lapalcian operator on $C^2$ functions, is not the true self-adjoint kinetic energy operator. The true operator is the unique self-adjoint exstesion of $- \frac{\hbar^2}{2m}\Delta$ when its domain is, for instance $C_0^\infty(R^3)$ or, equivalently ${\cal S}(R^3)$ or some other domains where the regularity at $r=0$ is made weaker to include functions as $e^{-r}$.

$\endgroup$
0
$\begingroup$

I have finally found a satisfactory solution to the problem I posed above !!

Careful examination of the boundary terms that arise at r=0 indicates that the problem of non-self-adjointness of T is associated with the presence of odd powers of r in the expansion of the wave function around r=0. This is an indication that the "true" wave function has only even terms in its expansion! In fact, this result makes a lot of sense, both mathematically and physically.

To get rid of the odd terms of r in Psi, I chose to adjust the wave function in a small region around the origin (0, epsilon). One can then calculate the higher moments of the kinetic energy operator. Finally taking the limit of epsilon -> 0, the results indeed satisfy the self-adjointness criterium. The computation itself is far from easy. It involves many terms with higher order derivatives of the delta function, which are generated at r=epsilon, the boundary between the thin interface and the bulk region.

In conclusion I can say that the standard solution to the SE, Psi = exp(-r), is merely an approximation of the correct wave function. Sometimes this approximation works well, and sometimes it doesn't. In the latter case, one has to use a (mathematically) more appropriate version of the wave function to get the right results.

Perhaps I shall write a short article about my little project.

$\endgroup$
  • $\begingroup$ "Careful examination of the boundary terms that arise at r=0 indicates that the problem of non-self-adjointness of T is associated with the presence of odd powers of r in the expansion of the wave function around r=0. This is an indication that the "true" wave function has only even terms in its expansion! In fact, this result makes a lot of sense, both mathematically and physically." If you took only even powers, you would lose main part of the exponential decay around $r=0$. This would most probably lead to a function that is not an eigenfunction of the Schr. equation. $\endgroup$ – Ján Lalinský Feb 8 '14 at 17:01
  • $\begingroup$ "In conclusion I can say that the standard solution to the SE, Psi = exp(-r), is merely an approximation of the correct wave function. " I do not see any valid reason for such conclusion. What do you mean by "correct wave function"? Often one is interested in the eigenfunctions of the Hamiltonian, angular momentum $\hat{L}_z$ and square of ang. momentum $\hat{L^2}$, which behave as $\exp(-r/r_0)$. Self-adjointness of $\hat{T}$ is usually not very important. $\endgroup$ – Ján Lalinský Feb 8 '14 at 17:07
  • $\begingroup$ The adjustment is only required in a very small region around the origin. At the length scale of the Bohr radius, the wave function still exhibits the usual exponential decay. The demand that the new wave function satisfies the SE, implies that the potential V is slightly different from the Coulomb point potential. $\endgroup$ – M. Wind Feb 8 '14 at 17:10
  • $\begingroup$ My approach is theoretical. It is perturbation theory. Thus the final results are independent of the value of epsilon. It would interesting to know whether there is experimental confirmation that at the proton length scale there are indeed corrections to the Coulomb point charge potential. $\endgroup$ – M. Wind Feb 8 '14 at 17:23
  • $\begingroup$ At what length scale is the modification significant? If this is greater than cca $10^{−15}~$m, it would be in contradiction with the scattering experiments and EM theory, since protons appear to have similar size. $\endgroup$ – Ján Lalinský Feb 8 '14 at 17:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.