Question

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

Short answer

Answer: The probability density associated with the wave function for a nonrelativistic free particle of mass \(m\), moving in three dimensions with momentum \(\vec{p}\) is independent of time, position, and momentum, and is given by the squared magnitude of the wave function's constant, \(\rho(\vec{r},t) = |A|^2\).

Step-by-step solution

Write the wave function for a free particle in 3 dimensions

To start, we recall that a wave function for a free particle in three dimensions can be written in the following form: $$\Psi(\vec{r}, t) = A e^{i(\vec{k}\cdot\vec{r} - \omega t)}$$ where \(A\) is an arbitrary constant, \(\vec{k}\) is the wave vector, \(\vec{r}\) is the position vector, \(\omega\) is the angular frequency, and \(t\) is the time.

Relate the wave vector and angular frequency to momentum and energy

In this step, we're going to relate the wave vector \(\vec{k}\) and angular frequency \(\omega\) to the particle's momentum vector \(\vec{p}\) and its energy. For a nonrelativistic free particle, the energy can be expressed as: $$E=\frac{\vec{p}^2}{2m}$$ The momentum can be related to the wave vector as follows: $$\vec{p}=\hbar \vec{k}$$ and the energy can be related to the angular frequency: $$E=\hbar \omega$$ We can substitute the expression for momentum into the energy equation: $$\hbar \omega = \frac{\hbar^2 \vec{k}^2}{2m}$$ Solving for \(\omega\), we get: $$\omega=\frac{\hbar \vec{k}^2}{2m}$$ Now, we have the necessary relations to complete the wave function.

Write the wave function with mass and momentum terms

Now that we have the relations between wave vector, angular frequency, momentum, and energy, we can write the wave function for the nonrelativistic free particle as: $$\Psi(\vec{r}, t) = A e^{i(\frac{\vec{p}}{\hbar}\cdot\vec{r} - \frac{\vec{p}^2 t}{2m\hbar})}$$

Find the probability density

With the wave function in hand, we can determine the probability density, which is given by the square of the magnitude of the wave function: $$\rho(\vec{r},t) = |\Psi(\vec{r},t)|^2$$ Substituting the wave function expression into the probability density formula, we get: $$\rho(\vec{r},t) = |A|^2 e^{i(\frac{\vec{p}}{\hbar}\cdot\vec{r} - \frac{\vec{p}^2 t}{2m\hbar})}e^{-i(\frac{\vec{p}}{\hbar}\cdot\vec{r} - \frac{\vec{p}^2 t}{2m\hbar})}=|A|^2$$ This means that the probability density is independent of time, position, and momentum, and is simply equal to the squared magnitude of the wave function's constant \(A\).