Suppose I have a potential of the form $$V(x) = \begin{cases} \frac{1}{2}kx^2 &\text{ if }x>0 \\ {\infty} &\text{ if }x<0 \end{cases} $$

We know for the usual HO one introduces $a$ and $a^{\dagger}$ and gets nice algebraic relations between them and eventually deduces the energy levels to be $E_n = \hbar w(n+\frac{1}{2})$ purely algebraically mainly through various commutation relations.

Now suppose I do the following: Instead of working with $V(x)$ I retain my old HO potential $\frac{1}{2}kx^2$ but restrict my Hilbert space to the set of functions vanishing for $x<0$. So denote $D$ the original Hilbert space for the regular QHO (in general it will not be all of $L^2(-\infty,\infty)$ because $p$ and $x$ and their polynomials are densely defined) and let $A$ be the set of functions in $L^2(-\infty,\infty)$ which vanish for $x<0$ and let: $$D'= D \cap A $$

My question is: Instead of working with the the Hamiltonian $H= \frac{1}{2m}P^2 + V(x)$ defined on $D$ couldn't we work with $H' = \frac{1}{2m}P^2 + \frac{1}{2}kx^2$ defined on $D'$? If indeed this is the case then to get the energy levels for $H$ (defined on $D$) we can instead use $H'$ (defined on $D'$) and use the same commutation relations and algebra as for the regular QHO (the only difference is that operators are acting on $D'$ instead of $D$ but the commutation relations are still the same). This should mean that this system would have the same energy levels as the regular QHO but it does not (although it's close). So what am I missing?

  • $\begingroup$ What are the functions in $D$ that vanish for $x < 0$? That is, what are the QHO eigenstates that don't penetrate on that side of the potential? $\endgroup$
    – Dan
    Commented Feb 20, 2022 at 18:23
  • $\begingroup$ There are non. All eigenfunctions are nonzero on both sides of the real line. $\endgroup$
    – Leonid
    Commented Feb 20, 2022 at 18:27
  • $\begingroup$ So $D^{\prime}$ contains what? $\endgroup$
    – Dan
    Commented Feb 20, 2022 at 18:43
  • $\begingroup$ Linkedf. Look at odd solutions. $\endgroup$ Commented Feb 20, 2022 at 18:48
  • 1
    $\begingroup$ There are hermiticity subtleties at the origin, so tread carefully with your expectations about creation and annihilation operators. $\endgroup$ Commented Feb 20, 2022 at 19:20


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