Expectation Value of Momentum for a Free Particle

The expectation value of the momentum operator \hat{p} = -i\hbar \frac{\partial}{\partial x} is given by:

        
      \[ \langle p \rangle = \int_{-\infty}^{\infty} \psi^*(x,t) \left(-i\hbar \frac{\partial}{\partial x} \right) \psi(x,t) \,dx. \]

By expressing the wave function in Fourier space:

        
      \[ \psi(x,t) = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} \tilde{\psi}(p) e^{i(px - Et)/\hbar} dp, \]

The expectation value simplifies to:

        
      \[ \langle p \rangle = \int_{-\infty}^{\infty} | \tilde{\psi}(p) |^2 p \, dp. \]

Gaussian Wave Function

For a Gaussian wave packet at t = 0:

        
      \[ \psi(x,0) = \frac{1}{(2\pi \sigma_x^2)^{1/4}} e^{-\frac{x^2}{4\sigma_x^2}} e^{i k_0 x}. \]

The Fourier transform gives:

        
      \[ \tilde{\psi}(p) = \left( \frac{2\sigma_x^2}{\hbar^2 \pi} \right)^{1/4} e^{-\sigma_x^2 (p - \hbar k_0)^2 / \hbar^2}. \]

The expectation value of momentum is then:

        
      \[ \langle p \rangle = \int_{-\infty}^{\infty} p \frac{1}{\hbar \sqrt{\pi} \sigma_p} e^{-(p - \hbar k_0)^2 / \sigma_p^2} dp. \]

Since this is a Gaussian distribution centered at p = \hbar k_0, the result is:

        
      \[ \langle p \rangle = \hbar k_0. \]

Conclusion

For a Gaussian wave packet, the expectation value of momentum is simply:

        
      \[ \langle p \rangle = \hbar k_0. \]