Given the Hamiltonian for the the harmonic oscillator (HO) as $$ \hat H=\frac{\hat P^2}{2m}+\frac{m}{2}\omega^2\hat x^2\,, $$ the Schroedinger equation can be reduced to: $$ \left[ \frac{d^2}{dz^2}-\left(\frac{z^2}{4}+a\right)\right]\Psi=0~, $$ where $a=-\frac{E}{\hbar\omega}$, $z=\sqrt{\frac{2m\omega}{\hbar}}$. Now, the two independent solutions to this equation are the Wittaker's functions (Abramowitz section 19.3., or Gradshteyn at the beginning, where he defines the Wittaker's functions) $D_{-a-1/2}(z)$ and $D_{-a-1/2}(-z)$. Apparently, there is no constraint on the values for $a$. In Abramowitz, especially, there is written "both variable $z$ and $a$ can take on general complex values".

Therefore my first question is: Let us fix $a=i$ and let us therefore take the Wittaker's function $D_{-i-1/2}(z)$. This functions is solution of the time independent Schroedinger equation, and, therefore, is an eigenfunction of the ho hamiltonian. Since its value for the parameter $a$ is $i$, it follows that its eigenvalue $E$ must be $E=-i\hbar\omega$. However, this result is contradictory, since the hamiltonian must have only real eigenvalues, since it is hermitian. What do I do wrong?

My second question is: Since the functions $D_n(z)$ form a complete set for $n$ positive integer with zero, I can expand my function $D_{-i-1/2}(z)$ onto the basis set $D_n(z)$. $$ D_{-i-1/2}(z)=\sum_n C_n D_n(z)~. $$ But, evidently, if $D_{-i-1/2}(z)$ is itself an eigenfunction with a different eigenvalue with respect to any of the $D_n(z)$, the expansion above does not make sense. This question is somewhat correlated to the previous one. So, I believe I do something wrong which is in common to both of them.


"the hamiltonian must have only real eigenvalues, since it is hermitian." - only for physically meaningful systems. You have put $a=i$, meaning $m$ or $\omega$ is negative - or maybe in some bizarre alternate universe, $\hbar$. Or $m$ and $\omega$ could each be imaginary. Boundary conditions must be met. You are, in effect, exploring solutions that don't persist in time with a $exp(i\omega t)$ dependence, but decaying (or growing) exponentially. A quantum system's wavefunction shouldn't be allowed to do that!

  • $\begingroup$ Thanks, I agree with you. Though, I am not yet fully convinced. I mean, if it is only this, I can build an eigenfunction of the ho which has some imaginary eigenvalues, which has no physical meaning. In principle then, I could write the infinite dimensional matrix for the hamiltonian with its eigenvalues as diagonal elements. Physical or not physical eigenvalues, I must insert them, since we are speaking about pure mathematics. If it is hermitian, that must mean that the conjugate transpose is equal to itself, viz. that any eigenvalue must be real, physical or not. I am still confused... $\endgroup$ – Wizzerad May 11 '13 at 12:35
  • $\begingroup$ Moreover, if it is only this, I could not expand $D_{-i-1/2}(z)$ into the basis of the ho, which would mean that, mathematically speaking, the basis is not complete. Finally, $D_{-i-1/2}$ is an eigenfunction with positive real energy of the hamiltonian with repulsive harmonic potential, which does not seem bizarre at all to me. $\endgroup$ – Wizzerad May 11 '13 at 12:54
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    $\begingroup$ The problem is that you cannot normalize those eigenfunctions. I am sure if you were to try to calculate the proper normalization constant your result would either be zero or you would run into a divergent integrand. $\endgroup$ – Neuneck May 14 '13 at 7:08

Filippo Fratini's wavefunction $D_{-i-1/2}(z)$ is an eigenfunction of the harmonic oscillator with energy eigenvalue $E=-i$ (in units $\omega, \hbar, m$ are all unity). However, the asymptotic expansion of this wavefunction goes like, $$ D_{-i-1/2}(z) \rightarrow \exp(-z^{2}/4) \ as \ z \rightarrow \infty $$ and, $$ D_{-i-1/2}(z) \rightarrow \exp(z^{2}/4) \ as \ z \rightarrow -\infty \ . $$ The actual asymptotic expansions are in chapter XVI of Whittaker and Watson's book "A Course of Modern Analysis". So, $D_{-i-1/2}(z)$ is not physically sensible because the particle has disappeared to $z=-\infty$. This is also considered in the section 12.3 on solutions of Schroedinger's equation in part II of Morse and Feshbach's "Methods of Theoretical Physics".

An eigenvector $|E\rangle$ which solves $\hat{H}|E\rangle=E|E\rangle$ need not have a real eigenvalue $E$ if $\langle E|E\rangle=\infty$ because, $$ \langle E|H|E\rangle=E\langle E|E \rangle $$ says nothing about E since both sides are infinite.

  • $\begingroup$ Ok, thanks. I think now I understood. I would give +1 if I could. $\endgroup$ – Wizzerad May 11 '13 at 14:05
  • $\begingroup$ In Morse&Feshbach it is quite thoroughly explained. Good suggestion. He says "If we wish to have a wave function which is quadratically integrable, we must have n = 0, 1, 2, ... " which are the Hermite polynomials. The completeness of the hermite polynomial basis must be understood as "completeness for quadratically integrable wave functions". A function which is not quadratically integrable, like $D_{-i-1/2}(z)$, cannot be expanded there. Indeed, the coefficients would rise to infinity, if it could, as the wave function is not quadratically integrable. $\endgroup$ – Wizzerad May 11 '13 at 14:11
  • $\begingroup$ This "non integrability" of $D_{-i-1/2}(z)$ answers my questions but rises another question: Only bound states are quadratically integrable. Take a plane wave, that is not quadratically integrable. Indeed, following your example above, we have $\langle p|p\rangle = +\infty$. But still the states are physical and the eigenvalues real. So does that mean that "non quadratically integrable wave functions with real (complex) eigenvalues are (are not) physical?" At the end of the day I can still expand a plane wave into the ho basis set, though it is not quadratically integrable, right? $\endgroup$ – Wizzerad May 11 '13 at 14:32
  • $\begingroup$ I think your observations on $\langle p|p\rangle=\infty$ are completely correct. I suppose that a complex eigenvalue always means the wavefunction blows up at infinity so the solution is not physically interesting because the particle is at infinity. $\endgroup$ – Stephen Blake May 11 '13 at 18:43

So does that mean that "non quadratically integrable wave functions with real (complex) eigenvalues are (are not) physical?" At the end of the day I can still expand a plane wave into the ho basis set, though it is not quadratically integrable, right?

It is difficult to agree on what "physical" means, but if we ask if such functions have the same meaning as the quadratically integrable functions, the answer is No.

Quadratically integrable functions can be normalized to 1 and interpreted according to Born as giving probability of certain configuration.

Functions such as $e^{ipx/\hbar}$ do not belong into this set, so we either reject them as we rejected the functions above; or we include them, as Dirac did, but then we have to ascribe them different meaning. Many physicists take the second option, but I have always felt that then the meaning of the function $\psi$ becomes unclear and its application ambiguous. Dirac explains his point of view on this problem in his book "Principles of Quantum Mechanics" when dealing with scattering.

Beware, one cannot expand $e^{ipx}$ as a sum of standard eigenfunctions of the HO Hamiltonian! The latter functions fall of exponentially with $x$, while the function $e^{ipx}$ oscillates on the whole real line.


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