# Interpretations

## Interpretation of the Wavefunction

Wavefunction $$\Psi(\vec{x},t)$$, w/ Probability Density $$\rho(\vec{x},t)=|\Psi(\vec{x},t)|^2$$, $$e\rho(\vec{x},t)$$ gives the charge density.

$$\int_{V_0} d\vec{x}\rho(\vec{x},t) =$$ probability to find the particle in the volume.

Probability flux: $$\vec{j}(\vec{x},t)=-\frac{i\hbar}{2m}(\Psi^*\vec{\nabla}\Psi-(\vec{\nabla}\Psi^*)\Psi) = -\frac{i\hbar}{2m}2i\Im\left[\Psi^*\vec{\nabla}\Psi\right] = \frac{\hbar}{m}\Im\left[\Psi^*\vec{\nabla}\Psi\right]$$.

$$\int_\mathbb{R}d\vec{x} j(\vec{x},t) = \frac{\hbar}{m}\Im\int_\mathbb{R}d\vec{x}\Psi(\vec{x},t)^*\vec{\nabla}\Psi(\vec{x},t) = \frac{1}{2m}\int_\mathbb{R}d\vec{x}\Psi^*(\hat{P}\Psi) + (\hat{P}^*\Psi^*)\Psi = \frac{1}{m}\Re\langle \hat{P}\rangle_t = \frac{1}{m}\langle\hat{P}\rangle_t$$.

### How is the probability density and flux related?

$$\frac{\partial\rho}{\partial t}=\frac{\partial}{\partial t}\Psi^*\Psi = \frac{i}{\hbar}\hat{H}\Psi^*\Psi - i\hbar\Psi^*\hat{H}\Psi = \frac{i}{\hbar}\left[\frac{-\hbar^2}{2m}(\vec{\nabla^2}\Psi^*)\Psi + (\hat{V}\Psi^*)\Psi - \Psi^*\frac{\hbar^2}{2m}(\vec{\nabla}^2\Psi) - \Psi^*\hat{V}\Psi\right] = \frac{i\hbar}{2m}\left(\Psi^*\vec{\nabla}^2\Psi - \Psi\vec{\nabla}^2\Psi\right) = -\frac{\hbar}{m}\Im(\Psi^*\vec{\nabla}^2\Psi)$$.

The divergence of the probability flux is, $$\vec{\nabla}\cdot\vec{j}(\vec{x},t) = -\frac{i\hbar}{2m}2i\Im(\Psi^*\vec{\nabla}^2\Psi) = -\frac{\partial}{\partial t}\rho$$.

### Continuity Equation - Probability Conservation Law

$$\frac{\partial}{\partial t}\rho + \vec{\nabla}\cdot\vec{j} = 0$$.

Created: 2023-06-25 Sun 02:35

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