# Eigenstates of half Harmonic Oscillator [closed]

This might be a stupid question so pardon me! If I am looking for energy eigenstates to the 1D quantum problem such that there is an infinite barrier at $x<0$ and for $x>0$ the potential is $kx^2$, what would it be like? Please prove that your set of eigenstates forms a complete set.

• Heres a not very subtle hint: if there is an infinite wall at x=0 what condition must you impose on the wavefunction there? Furthermore, which wavefunctions that come up in solving the normal harmonic oscillator obey this condition? Showing completeness should then be trivial. The statement of completeness comes down to saying that you have an infinite set of orthonormal functions that you can expand in. For more background and rigor I'd read up on Sturm-Liouville theory. Jun 23 '13 at 2:07
• The "half harmonic oscillator" is also discussed in this Phys.SE answer. Jun 23 '13 at 15:58

The eigenstates will be the odd eigenstates of the harmonic oscillator. (This follows from the requirement that $\psi(0)=0$.)

If we accept that the eigenstates for the harmonic oscillator form a complete set for functions on $\mathbb{R}$, it then follows that the odd eigenstates form a complete set for odd functions on $\mathbb{R}$. Since any function on the half space can be represented by restricting an odd function on $\mathbb{R}$ to the half space, it must follow that the odd eigenstates form a complete set on the half space.

• "Since any function on the half space can be represented by restricting an odd function on R to the half space" gosh! I feel stupid...thanks! Jun 23 '13 at 3:06
• @NoobRevB: Glad to help. Jun 23 '13 at 3:10
• Nice, concise answer! Jun 23 '13 at 21:16
• Though I had forgotten the argument why odd states are complete for half harmonic oscillator, I was actually after a general problem. Let me know what you guys think. So if I place my infinite wall at an arbitrary $x$(not necessarily at x=0) there is no obvious way to find the eigenstates and eigenvalues right? And even if someone were to give me an infinite set of eigenstates there is no obvious way to argue that its complete? Its the parity symmetry which saves the day for us in the original question or am I missing something? Jun 23 '13 at 21:52
• @NoobRevB: Yes, this trick does not work for your more general problem. In this case it would probably be best to follow David Meltzer's advice and turn to Sturm-Liouville theory. Jun 23 '13 at 22:38

A Schroedinger (time independent) equation with a potential of the form $$V(x)=\begin{cases} 0\quad(x<0)\\ \frac12m\omega^2 x^2\quad (x>0) \end{cases}$$ will necessarily have $$\psi(x)=0$$ as solution for $$x<0$$, because the infinite potential barrier prevents the negative x region from being accessed.

For the positive part, the Schroedinger equation is the same of the Simple Harmonic Oscillator problem, so to find the solutions we can use the same reasoning used for the original problem, having care to replace the boundary condition for $$x\rightarrow\infty$$ with the requirement that the wavefunction be continuous in $$x=0$$, i.e.: $$\lim_{x\rightarrow 0^+}\psi(x)=0$$

We recall that the solutions to the original SHO problem were $$|n\rangle\in\mathcal H_{SHO}$$ such that: $$\psi_n(x)\equiv \langle x|n\rangle=\frac1{\sqrt{x_0}\sqrt[4]\pi}\frac1{\sqrt{2^n n!}}H_n(x/x_0)e^{-x^2/2x_0^2}$$ where $$x_0=\sqrt{\frac\hbar{m\omega}}$$ is the characteristic length of the SHO, and the $$H_n(x/x_0)$$ are the Hermite polynomials, whose parity is defined by $$n$$ in the following way: $$H_n(0)=\begin{cases} 0\quad(\text{n odd})\\ (-1)^{n/2}\frac{n!}{\left(\frac n2\right)!}\quad (\text{n even}) \end{cases}$$

So, we conclude that only the odd solutions of the previous problem can satisfy the new boundary conditions, so are the best candidates to build the new ones, and we can define new eigenfunctions $$|\tilde n\rangle$$ identifying them with $$|2n+1\rangle$$ up to a constant, and in the right portion of the domain: $$\tilde\psi_n(x)\equiv \langle x|\tilde n\rangle=\begin{cases} 0\quad(x<0)\\ \alpha \psi_{2n+1}(x)\quad(x>0)\end{cases}$$ The coefficient $$\alpha$$ is determined by the condition of normalization: $$\int_{-\infty}^\infty dx |\tilde\psi_n(x)|^2=|\alpha|^2\int_{0}^\infty dx |\psi_{2n+1}(x)|^2=|\alpha|^2\frac{1}2$$ from which we need to set $$\alpha=\sqrt2$$ (choosing it real by convention), so that finally

$$\tilde\psi_n(x)=\begin{cases} 0\quad(x<0)\\ \sqrt2 \psi_{2n+1}(x)\quad(x>0)\end{cases}$$

While the Hilbert space of the SHO problem was spanned by the odd and even eigenfunctions, but in the half oscillator (HHO in the following) the Hilbert space is spanned only by the odd ones: in particular, the ladder operators $$a,a^\dagger$$ as they were defined in the SHO, cannot belong to the operator algebra of the new Hilbert space, as their action would bring a vector $$|2n+1\rangle\in\mathcal H_{HHO}$$ into a $$|2n\rangle$$ or a $$|2n+2\rangle$$ which $$\notin\mathcal H_{HHO}$$. I don't know if a suitable redefinition exists, but maybe this topic has been already thoroughly studied and the answer is well known.

To answer to the last question: the half oscillator is a regular Sturm Liouville problem, exactly as the SHO was, so its solutions (at least their restriction to the positive $$x$$ axis) are guaranteed to be orthogonal and complete (ref., for example Arfken pag. 386, $$7^{th}$$ edition). But please note that they will not be a complete set for all continuous functions, defined on $$\mathbb R$$: they are a complete set only for the subspace of Hilbert space $$L^2[0,\infty)$$ formed by the functions $$f(x)$$ such that $$f(0)=0$$.