The problem I'm struggling with has asked me to find the $x$-representation of the half harmonic oscillator wave function with a potential of $\frac12kx^2$. Our setup started with the WKB approximation of the full harmonic oscillator, giving us this equation to start from: $$ \int_{0}^{x_0} \sqrt{2m}\sqrt{E_n-\frac12kx^2}dx=\left(n+\frac12\right)h $$ After some solving and using $\omega = \sqrt{\frac km}$, we eventually get to the normalized solution: $$\psi(x) = \langle x|n\rangle = 2^{-n/2}(n!)^{-1/2}\left(\frac{m\omega}{\hbar\pi}\right)^{1/4}H_n\left(\sqrt{\frac{m\omega}{\hbar}}\right)e^{-\frac{m\omega}{2\hbar}x^2}$$ where $H_n$ is the appropriate Hermite polynomial.

Now, I'm not expected to know what Hermite polynomials are at this level (first semester of 400-level Quantum Mechanics), but the way the professor wants me to answer this question is to see what I understand about the full harmonic oscillator and its solution above, and normalize it for the half harmonic oscillator. Essentially, we know $ \int_{-\infty}^{\infty}{\psi^*\psi} = 1$ and we want to show that $\int_{-\infty}^{\infty}\phi^*\phi$ is also equal to 1, where $\phi$ is the half harmonic oscillator wavefunction. In other words, we want to show $$\phi(x) = C^2\int_{-\infty}^{\infty}{\psi^*\psi} = 1$$ Intuitively, I know the answer is $\frac{1}{\sqrt{2}}$, and my professor confirmed this, but how am I supposed to get to this point? I only know the beginning and the end of the solution, but not the middle steps.

  • $\begingroup$ The SHO wave functions are normalized over the entire real line. Now you need to normalize them over "half" the real line. Note that the SHO wave functions are eigenstates of parity, i.e., they're either even or odd functions. That should get you the rest of the way. $\endgroup$
    – march
    Commented Oct 30, 2022 at 17:02

1 Answer 1


I assume that by the half oscillator you mean a HO with infinite potential for $x<0$ so that $\psi(0)=0$?

If so its wavefunctions are identical with the odd ($n=1,3,5,\ldots$) wavefunctions of the ordinary HO. The even ($n=0,2,4,\ldots$) wavefunctions are not allowed as they do not vanish at the origin.

  • $\begingroup$ Are the wave functions identical? There must be a normalization factor involved, since my professor has told me that the normalization constant of $\frac{1}{\sqrt{2}}$ is the correct value to obtain. $\endgroup$
    – Weasnaw
    Commented Oct 30, 2022 at 22:18
  • $\begingroup$ The wave fuctions are identical in $x>0$, but they are zero in $x<0$. $\endgroup$
    – mike stone
    Commented Oct 31, 2022 at 12:41

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