Physlib.QuantumMechanics.OneDimension.GeneralPotential.Basic

For any complex numbers $a_1, a_2 \in \mathbb{C}$ and any differentiable functions $\psi_1, \psi_2: \mathbb{R} \to \mathbb{C}$, the one-dimensional momentum operator $\hat{p}$ is linear, satisfying: \[ \hat{p}(a_1 \psi_1 + a_2 \psi_2) = a_1 \hat{p} \psi_1 + a_2 \hat{p} \psi_2 \] where the momentum operator is defined as $(\hat{p}\psi)(x) = -i \hbar \frac{d\psi}{dx}(x)$.

For any complex scalars $a_1, a_2 \in \mathbb{C}$ and functions $\psi_1, \psi_2: \mathbb{R} \to \mathbb{C}$, suppose that $\psi_1$ and $\psi_2$ are differentiable, and their images under the momentum operator, $\hat{p}\psi_1$ and $\hat{p}\psi_2$, are also differentiable. Then the square of the one-dimensional momentum operator $\hat{p}^2$ is linear, satisfying: \[ \hat{p}(\hat{p}(a_1 \psi_1 + a_2 \psi_2)) = a_1 \hat{p}(\hat{p}\psi_1) + a_2 \hat{p}(\hat{p}\psi_2) \] where the momentum operator $\hat{p}$ is defined by $(\hat{p}\psi)(x) = -i \hbar \frac{d\psi}{dx}(x)$.

Given a potential function $V: \mathbb{R} \to \mathbb{R}$ and a wave function $\psi: \mathbb{R} \to \mathbb{C}$, the potential operator is defined by the pointwise multiplication $(V\psi)(x) = V(x)\psi(x)$ for all $x \in \mathbb{R}$.

For any potential function $V: \mathbb{R} \to \mathbb{R}$, complex scalars $a_1, a_2 \in \mathbb{C}$, and wave functions $\psi_1, \psi_2: \mathbb{R} \to \mathbb{C}$, the potential operator $\hat{V}$ is linear. That is, it satisfies the identity $$\hat{V}(a_1 \psi_1 + a_2 \psi_2) = a_1 \hat{V}\psi_1 + a_2 \hat{V}\psi_2$$ where the action of the potential operator is defined by $(\hat{V}\psi)(x) = V(x)\psi(x)$.

The Schrödinger operator for a one-dimensional quantum mechanical system with mass $m$ and potential function $V: \mathbb{R} \to \mathbb{R}$ maps a wave function $\psi: \mathbb{R} \to \mathbb{C}$ to a new function $\mathbb{R} \to \mathbb{C}$ defined by the expression \[ x \mapsto -\frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2}(x) + V(x)\psi(x) \] where $\hbar$ is the reduced Planck constant and $\frac{d^2\psi}{dx^2}$ denotes the second derivative of $\psi$. Formally, the operator is defined as the sum of the kinetic energy term $\frac{1}{2m} \hat{p}^2 \psi$ (where $\hat{p}$ is the momentum operator) and the potential energy term $V \psi$.

For any complex scalars $a_1, a_2 \in \mathbb{C}$ and wave functions $\psi_1, \psi_2: \mathbb{R} \to \mathbb{C}$, suppose that $\psi_1$ and $\psi_2$ are differentiable and that their images under the momentum operator, $\hat{p}\psi_1$ and $\hat{p}\psi_2$, are also differentiable. Then the one-dimensional Schrödinger operator $\hat{H}$ is linear, satisfying: \[ \hat{H}(a_1 \psi_1 + a_2 \psi_2) = a_1 \hat{H}\psi_1 + a_2 \hat{H}\psi_2 \] where the Schrödinger operator $\hat{H}$ for a system with mass $m$ and potential $V$ is defined as $(\hat{H}\psi)(x) = -\frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2}(x) + V(x)\psi(x)$, and the momentum operator $\hat{p}$ is defined as $(\hat{p}\psi)(x) = -i \hbar \frac{d\psi}{dx}(x)$.