# Free particle in momentum space

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 3.40.

Since the Hamiltonian for a free particle is ${H=p^{2}/2m}$, the Schrodinger equation in momentum space is

$\displaystyle i\hbar\frac{\partial\Phi}{\partial t}=\frac{p^{2}}{2m}\Phi \ \ \ \ \ (1)$

so the solution can be found by simply integrating with respect to ${t}$:

$\displaystyle \Phi(p,t)=e^{-ip^{2}t/2m\hbar}\Phi(p,0) \ \ \ \ \ (2)$

We looked at the travelling Gaussian wave packet in free space earlier. Its initial state in position space is

$\displaystyle \Psi(x,0)=\left(\frac{2a}{\pi}\right)^{1/4}e^{-ax^{2}}e^{ilx} \ \ \ \ \ (3)$

To find ${\Phi(p,0)}$ we use the conversion to momentum space we found earlier:

$\displaystyle \Phi(p,0)=\frac{1}{\sqrt{2\pi\hbar}}\left(\frac{2a}{\pi}\right)^{1/4}\int_{-\infty}^{\infty}e^{-ax^{2}+ilx-ipx/\hbar}dx \ \ \ \ \ (4)$

From the analysis of the travelling Gaussian packet we see that the integral is the same as that done when calculating ${\phi(k)}$ if we replace ${k}$ with ${p/\hbar}$. Therefore

$\displaystyle \Phi(p,0)=\left(\frac{2}{\pi a}\right)^{1/4}\frac{1}{\sqrt{2\hbar}}e^{-(p/\hbar-l)^{2}/4a} \ \ \ \ \ (5)$

Using 2, we have the full solution for ${\Phi(p,t)}$:

$\displaystyle \Phi(p,t)=\left(\frac{2}{\pi a}\right)^{1/4}\frac{1}{\sqrt{2\hbar}}e^{-ip^{2}t/2m\hbar}e^{-(p/\hbar-l)^{2}/4a} \ \ \ \ \ (6)$

Also

$\displaystyle |\Phi(p,t)|^{2}=\frac{1}{\hbar}\frac{1}{\sqrt{2\pi a}}e^{-(p/\hbar-l)^{2}/2a} \ \ \ \ \ (7)$

which is independent of time. (As a check, we can integrate this over all ${p}$ and verify that this integral is 1.)

We can calculate the means for momentum in the usual way:

 $\displaystyle \langle p\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\hbar}\frac{1}{\sqrt{2\pi a}}\int_{-\infty}^{\infty}pe^{-(p/\hbar-l)^{2}/2a}dp\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar l\ \ \ \ \ (9)$ $\displaystyle \langle p^{2}\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\hbar}\frac{1}{\sqrt{2\pi a}}\int_{-\infty}^{\infty}p^{2}e^{-(p/\hbar-l)^{2}/2a}dp\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar^{2}(l^{2}+a) \ \ \ \ \ (11)$

Both results agree with those in the analysis of the travelling Gaussian packet.

For the mean energy, we have

 $\displaystyle \langle H\rangle$ $\displaystyle =$ $\displaystyle \left\langle \frac{p^{2}}{2m}\right\rangle \ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}}{2m}(l^{2}+a)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\langle p\rangle^{2}}{2m}+\frac{a\hbar^{2}}{2m} \ \ \ \ \ (14)$

Referring back to the stationary Gaussian wave packet in free space, we see that ${\langle p^{2}\rangle=a\hbar^{2}}$, so the energy is the sum of that for a stationary Gaussian wave packet and the term ${\langle p\rangle^{2}/2m}$. For the travelling packet, there is a net non-zero average momentum, so ${\langle p\rangle}$ is non-zero. Thus the energy arises from the inherent energy of the wave packet, plus the kinetic energy of motion of the packet.