Write a plane-wave function \(\psi(\vec{r}, t)\) for a nonrelativistic free particle of mass \(m\) moving in three dimensions with momentum p, including correct time dependence as required by the Schrödinger equation. What is the probability density associated with this wave?

The general solution for the plane-wave function satisfying the Schrödinger equation is given by: \(\psi(\vec{r},t) = A e^{\frac{i}{\hbar} (\vec{p}\cdot\vec{r} - Et)}\) where A is a normalization constant, \(\vec{r}\) is the position, \(\vec{p}\) is the momentum, E is the energy, and t is the time.

We are given the momentum p and mass m. The energy E for a nonrelativistic free particle with momentum p is given by: \(E = \frac{p^2}{2m}\) Substituting the values of E and p into the plane-wave function: \(\psi(\vec{r},t) = A e^{\frac{i}{\hbar} (\vec{p}\cdot\vec{r} - \frac{p^2}{2m}t)}\)

The probability density, \(|\psi(\vec{r},t)|^2\), is given by the square of the absolute value of the wave function: \(|\psi(\vec{r},t)|^2 = |\psi(\vec{r},t)| * |\psi^*(\vec{r},t)|\) where \(\psi^*(\vec{r},t)\) is the complex conjugate of the wave function. For the plane-wave function, the complex conjugate is given by: \(\psi^*(\vec{r},t) = A^* e^{-\frac{i}{\hbar} (\vec{p}\cdot\vec{r} - \frac{p^2}{2m}t)}\) Now, calculating the probability density: \(|\psi(\vec{r},t)|^2 = A A^* e^{\frac{i}{\hbar} (\vec{p}\cdot\vec{r} - \frac{p^2}{2m}t)} * e^{-\frac{i}{\hbar} (\vec{p}\cdot\vec{r} - \frac{p^2}{2m}t)} = |A|^2\) The probability density of a free particle with momentum p is constant and given by \(|A|^2\). To obtain the correct normalization, we need to integrate this probability density over all space and set it equal to one, which will give us the value of the normalization constant A.