Physlib.QuantumMechanics.OneDimension.HarmonicOscillator.Completeness

Let $f: \mathbb{R} \to \mathbb{C}$ be a function belonging to the Hilbert space $L^2(\mathbb{R}, \mathbb{C})$ (denoted by `MemHS f`). For any $n \in \mathbb{N}$, the product of the $n$-th eigenfunction of the one-dimensional harmonic oscillator $\psi_n(x)$ and $f(x)$ is integrable with respect to the Lebesgue measure.

Let $f: \mathbb{R} \to \mathbb{C}$ be a square-integrable function belonging to the Hilbert space $L^2(\mathbb{R}, \mathbb{C})$ (denoted by `MemHS f`). For any natural number $n$, the function $$x \mapsto H_n\left(\frac{x}{\xi}\right) f(x) \exp\left(-\frac{x^2}{2\xi^2}\right)$$ is integrable with respect to the Lebesgue measure, where $H_n$ denotes the $n$-th physicist's Hermite polynomial and $\xi$ is the characteristic length scale of the one-dimensional harmonic oscillator.

Let $f: \mathbb{R} \to \mathbb{C}$ be a square-integrable function belonging to the Hilbert space $L^2(\mathbb{R}, \mathbb{C})$. For any polynomial $P(x)$ with integer coefficients, the function $$x \mapsto P\left(\frac{x}{\xi}\right) f(x) \exp\left(-\frac{x^2}{2\xi^2}\right)$$ is integrable with respect to the Lebesgue measure, where $\xi$ is the characteristic length scale of the one-dimensional harmonic oscillator.

Let $f: \mathbb{R} \to \mathbb{C}$ be a square-integrable function belonging to the Hilbert space $L^2(\mathbb{R}, \mathbb{C})$ (denoted by `MemHS f`). For any natural number $r$, the function $$x \mapsto x^r f(x) \exp\left(-\frac{x^2}{2\xi^2}\right)$$ is integrable with respect to the Lebesgue measure, where $\xi$ is the characteristic length scale of the one-dimensional harmonic oscillator.

Let $f: \mathbb{R} \to \mathbb{C}$ be a function in the Hilbert space associated with a one-dimensional harmonic oscillator. If $f$ is orthogonal to all eigenfunctions $\psi_n$ of the harmonic oscillator (i.e., the Hilbert space inner product $\langle \psi_n, f \rangle$ is zero for all $n \in \mathbb{N}$), then for any natural number $n$, the integral over the real line of the product of the $n$-th eigenfunction and $f$ is zero: $$\int_{-\infty}^{\infty} \psi_n(x) f(x) \, dx = 0$$

Let $f: \mathbb{R} \to \mathbb{C}$ be a function in the Hilbert space $L^2(\mathbb{R})$ associated with a one-dimensional harmonic oscillator. If $f$ is orthogonal to all eigenfunctions $\psi_n$ of the harmonic oscillator (i.e., $\langle \psi_n, f \rangle = 0$ for all $n \in \mathbb{N}$), then for any natural number $n$, the integral over the real line of the product of the $n$-th physicist's Hermite polynomial $H_n(x/\xi)$, the function $f(x)$, and the Gaussian factor $\exp(-x^2 / (2\xi^2))$ is zero: $$\int_{-\infty}^{\infty} H_n\left(\frac{x}{\xi}\right) f(x) \exp\left(-\frac{x^2}{2\xi^2}\right) \, dx = 0$$ where $\xi$ is the characteristic length scale of the harmonic oscillator.

Let $f: \mathbb{R} \to \mathbb{C}$ be a square-integrable function in the Hilbert space $L^2(\mathbb{R})$ (denoted by `MemHS f`). If $f$ is orthogonal to all eigenfunctions $\psi_n$ of the one-dimensional harmonic oscillator (i.e., $\langle \psi_n, f \rangle = 0$ for all $n \in \mathbb{N}$), then for any polynomial $P \in \mathbb{Z}[x]$ with integer coefficients, the following integral over the real line vanishes: $$\int_{-\infty}^{\infty} P\left(\frac{x}{\xi}\right) f(x) \exp\left(-\frac{x^2}{2\xi^2}\right) \, dx = 0$$ where $\xi$ is the characteristic length scale of the harmonic oscillator.

Let $f: \mathbb{R} \to \mathbb{C}$ be a square-integrable function in the Hilbert space $L^2(\mathbb{R})$ (denoted by `MemHS f`). If $f$ is orthogonal to all eigenfunctions $\psi_n$ of the one-dimensional harmonic oscillator (i.e., $\langle \psi_n, f \rangle = 0$ for all $n \in \mathbb{N}$), then for any natural number $r$, the following integral over the real line vanishes: $$\int_{-\infty}^{\infty} x^r f(x) \exp\left(-\frac{x^2}{2\xi^2}\right) \, dx = 0$$ where $\xi$ is the characteristic length scale of the harmonic oscillator.

Let $f: \mathbb{R} \to \mathbb{C}$ be a square-integrable function in the Hilbert space $L^2(\mathbb{R})$ (denoted by `MemHS f`). If $f$ is orthogonal to all eigenfunctions $\psi_n$ of the one-dimensional harmonic oscillator (i.e., $\langle \psi_n, f \rangle = 0$ for all $n \in \mathbb{N}$), then for any real number $c$, the following integral vanishes: $$\int_{-\infty}^{\infty} e^{icx} f(x) \exp\left(-\frac{x^2}{2\xi^2}\right) \, dx = 0$$ where $i$ is the imaginary unit and $\xi$ is the characteristic length scale of the harmonic oscillator.

Let $f: \mathbb{R} \to \mathbb{C}$ be a square-integrable function in the Hilbert space $L^2(\mathbb{R})$ (denoted by `MemHS f`). If $f$ is orthogonal to every eigenfunction $\psi_n$ of the one-dimensional harmonic oscillator (that is, $\langle \psi_n, f \rangle = 0$ for all $n \in \mathbb{N}$), then the Fourier transform $\mathcal{F}$ of the function $f(x) \exp\left(-\frac{x^2}{2\xi^2}\right)$ is identically zero: $$\mathcal{F} \left( f(x) \exp\left(-\frac{x^2}{2\xi^2}\right) \right) = 0$$ where $\xi$ is the characteristic length scale of the harmonic oscillator.

Let $f: \mathbb{R} \to \mathbb{C}$ be a square-integrable function in the Hilbert space $L^2(\mathbb{R})$ (denoted by `MemHS f`). Suppose that $f$ is orthogonal to every eigenfunction $\psi_n$ of the one-dimensional quantum harmonic oscillator, such that $\langle \psi_n, f \rangle = 0$ for all $n \in \mathbb{N}$. Assuming Plancherel's theorem holds—specifically that for any function $g \in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$, the $L^2$-norm of its Fourier transform $\|\mathcal{F} g\|_2$ is equal to its $L^2$-norm $\|g\|_2$—then $f$ is the zero element in the Hilbert space $L^2(\mathbb{R})$.

Let $f$ be a vector in the Hilbert space $\mathcal{H} = L^2(\mathbb{R})$. Suppose that for every $n \in \mathbb{N}$, $f$ is orthogonal to the $n$-th eigenfunction $\psi_n$ of the one-dimensional quantum harmonic oscillator, such that $\langle \psi_n, f \rangle = 0$. Given that Plancherel's theorem holds—specifically, that for any function $g \in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$, the $L^2$-norm of its Fourier transform $\|\mathcal{F} g\|_2$ is equal to its $L^2$-norm $\|g\|_2$—then $f$ is the zero vector in the Hilbert space.

Assume Plancherel's theorem holds, stating that for any function $g \in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$, the $L^2$-norm of its Fourier transform $\|\mathcal{F} g\|_2$ is equal to its $L^2$-norm $\|g\|_2$. Under this assumption, the topological closure of the complex linear span of the set of eigenfunctions $\{\psi_n\}_{n \in \mathbb{N}}$ of the one-dimensional quantum harmonic oscillator is equal to the entire Hilbert space $L^2(\mathbb{R})$ (denoted as $\top$).