I think I understand that the solution to the Schrodinger equation for the SHO is based on the Hermite polynomials (and the Guassian function). The solution set of all even Hermite polynomials are a solution to the Schrodinger equation, as well as the set of all odd Hermite polynomials.


  1. Is the set of all odd Hermite polynomials a solution because the potential $V(x)=1/2kx^{2}$ has even parity?
  2. Does $\psi^{*}\psi$ have even or odd parity, or neither?
  3. Why does $<x>=0$?

My main confusion is why the expectation value of the position results in an odd integrand, and why an odd integrand is equal to zero.


1) because the potential energy function has even symmetry, the solutions to Schrodinger's Equation results in purely even or odd solutions. Consider the Time Independent Schrodinger Equation (TISE) below

$$-\frac{\hbar^2}{2m}\nabla^2 \psi(x) + U(x)\psi(x) = E\psi(x)$$

But the potential contains the symmetry that $U(-x) = U(x)$. What this implies is that under the parity transform $$(x \implies -x)$$

$$-\frac{\hbar^2}{2m}\nabla^2 \psi(-x) + U(-x)\psi(-x) = E\psi(-x)$$ Or because of the potential's even symmetry

$$-\frac{\hbar^2}{2m}\nabla^2 \psi(-x) + U(x)\psi(-x) = E\psi(-x)$$

What this means is that both wavefunctions $\Psi(x)$ AND $\Psi(-x)$ solve this particular wave equation. Further, because the Schrodinger Equation is a linear Differential Equation, any linear combination of these two solutions must also be a solution to the equation. therefore, solutions exist such that:

$$\psi_{even}(x) = \psi(x) + \psi(-x)$$ $$\psi_{odd}(x) = \psi(x) - \psi(-x)$$

These follow from the definitions of even and odd functions which is easily verified.

2) Concerning the modulus squared, $\psi^* \psi$, if $\psi$ is always either even or odd, then when multiplying an odd function with itself you get an even function. likewise, an even function multiplied with itself will be even. Consider trivial polynomials for example,

$x^3$ is odd, but $(x^3)(x^3) = (x^3)^2 = x^6$ which is obviously even.

3) Lastly the fact that $\left<x \right> = 0$ follows from the previous statement, that $|\psi|^2$ is always even for potentials with even symmetry, like the SHO. This is because

$$\left<x \right> = \int_{- \infty} ^{\infty} x|\psi|^2dx$$

In the integrand, there is an odd function ($x$) times the definitely even modulus squared. This creates an odd integrand. For any odd integrand with symmetric limits, this integral must be zero.

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  • $\begingroup$ Thank you very much, user28823... fantastic, clear explanation of something that has been very hard for me to get a firm grasp on. I appreciate you taking the time to answer these questions completely! $\endgroup$ – curiousGeorge119 Nov 3 '13 at 21:47
  1. Because $V(x)$ is even, one can show that the eigenfunctions of $H$ are either even or odd. Let's call these eigenfunctions $\psi_n(x)$, which are odd if $n$ is odd, even if $n$ is even. One also shows that $\psi_n$ is real.

  2. Now, $|\psi_n(x)|^2=\psi^*_n (x)\psi_n (x) $ is obviously even. It's trivial if $n$ is even, and is easily shown if $n$ is odd.

  3. One then shows that $x |\psi_n(x)|^2$ is odd. Then, the integral of an odd function is zero, since for an odd function $f$, we have $\int dx f(x)=-\int dx f(-x)=-\int dy f(y)=- \int dx f(x)=0$, where I have used the fact that $f$ is odd, then changed variables from $x$ to $y=-x$, then changed the dummy variable $y$ back to $x$, which implies that the integral is equal to minus itself, and is thus zero. With that, you can show that $\langle \hat x \rangle= \langle \psi_n| \hat x |\psi_n\rangle=\int dx\, x |\psi_n(x)|^2=0$.

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