Physlib.QuantumMechanics.OneDimension.Operators.Commutation

For any Schwartz function $\psi \in \mathcal{S}(\mathbb{R}, \mathbb{C})$, the position operator $\hat{x}$ and the momentum operator $\hat{p}$ satisfy the following commutation relation: $$\hat{x}(\hat{p}\psi) - \hat{p}(\hat{x}\psi) = i\hbar\psi$$ where $i$ is the imaginary unit and $\hbar$ is the reduced Planck constant. Here, the position operator $\hat{x}$ acts as $(\hat{x}\psi)(x) = x\psi(x)$ and the momentum operator $\hat{p}$ acts as $(\hat{p}\psi)(x) = -i\hbar \frac{d\psi}{dx}(x)$.

For any Schwartz function $\psi \in \mathcal{S}(\mathbb{R}, \mathbb{C})$, the position operator $\hat{x}$ and the momentum operator $\hat{p}$ satisfy the relation: $$\hat{x}(\hat{p}\psi) = \hat{p}(\hat{x}\psi) + i\hbar\psi$$ where $i$ is the imaginary unit and $\hbar$ is the reduced Planck constant.

For any Schwartz function $\psi \in \mathcal{S}(\mathbb{R}, \mathbb{C})$, the composition of the momentum operator $\hat{p}$ and the position operator $\hat{x}$ satisfies the relation: $$\hat{p}(\hat{x}\psi) = \hat{x}(\hat{p}\psi) - i\hbar\psi$$ where $i$ is the imaginary unit, $\hbar$ is the reduced Planck constant, the position operator is defined by $(\hat{x}\psi)(x) = x\psi(x)$, and the momentum operator is defined by $(\hat{p}\psi)(x) = -i\hbar \frac{d\psi}{dx}$.