# Energy-time uncertainty principle: Gaussian free particle

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 3.19.

As another example of the energy-time uncertainty relation, we can look again at the example of a travelling free particle with a Gaussian wave packet. We have already worked out most of what we need to test the uncertainty relation:

 $\displaystyle \langle x\rangle$ $\displaystyle =$ $\displaystyle \frac{l\hbar t}{m}\ \ \ \ \ (1)$ $\displaystyle \langle x^{2}\rangle$ $\displaystyle =$ $\displaystyle \frac{1+(2a\hbar t/m)^{2}+a(2\hbar lt/m)^{2}}{4a}\ \ \ \ \ (2)$ $\displaystyle \langle p^{2}\rangle$ $\displaystyle =$ $\displaystyle \hbar^{2}(a+l^{2}) \ \ \ \ \ (3)$

From this we get

 $\displaystyle \left\langle H\right\rangle$ $\displaystyle =$ $\displaystyle \frac{\left\langle p^{2}\right\rangle }{2m}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}(a+l^{2})}{2m} \ \ \ \ \ (5)$

We still need ${\left\langle H^{2}\right\rangle }$. From our previous calculations, we have the wave function:

$\displaystyle \Psi(x,t)=\left(\frac{2a}{\pi}\right)^{1/4}\frac{e^{\left(-ax^{2}+i(lx-\hbar l^{2}t/2m)\right)/(1+2i\hbar at/m)}}{\sqrt{1+2i\hbar at/m}} \ \ \ \ \ (6)$

We can calculate ${\left\langle H^{2}\right\rangle }$ by direct integration, using Maple:

 $\displaystyle \left\langle H^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{\hbar^{4}}{4m^{2}}\int_{-\infty}^{\infty}\left|\frac{d^{2}\Psi(x,t)}{dx^{2}}\right|^{2}dx\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar^{4}}{4m^{2}}\left(3a^{2}+6al^{2}+l^{4}\right) \ \ \ \ \ (8)$

As a check on this result, we can work out the units (always a good test to make sure you haven’t dropped a factor somewhere). From the original wave function, since exponents must be dimensionless, we know that ${a}$ has dimensions ${distance^{-2}}$ and ${l}$ has ${distance^{-1}}$. Planck’s constant has dimensions of ${energy\times time}$, so the expression above has overall units of ${energy^{4}\times time^{4}\times mass^{-2}\times distance^{-4}=energy^{2}}$. (Recall kinetic energy is ${mv^{2}/2}$.) It’s also worth noting that ${\left\langle H^{2}\right\rangle }$ is independent of time.

From here, we can get ${\sigma_{H}^{2}}$:

 $\displaystyle \sigma_{H}^{2}$ $\displaystyle =$ $\displaystyle \left\langle H^{2}\right\rangle -\left\langle H\right\rangle ^{2}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{a\hbar^{4}}{2m^{2}}\left(a+2l^{2}\right) \ \ \ \ \ (10)$

We also have, from above

 $\displaystyle \sigma_{x}^{2}$ $\displaystyle =$ $\displaystyle \left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m^{2}+4\left(at\hbar\right)^{2}}{4am^{2}} \ \ \ \ \ (12)$

We’re trying to show that ${\sigma_{H}\sigma_{x}\ge\frac{\hbar}{2}\frac{d\left\langle x\right\rangle }{dt}=l\hbar^{2}/2m}$ from above. So we want

 $\displaystyle \sigma_{H}^{2}\sigma_{x}^{2}$ $\displaystyle \ge$ $\displaystyle \frac{l^{2}\hbar^{4}}{4m^{2}}\ \ \ \ \ (13)$ $\displaystyle \frac{a\hbar^{4}}{2m^{2}}\left(a+2l^{2}\right)\frac{m^{2}+4\left(at\hbar\right)^{2}}{4am^{2}}$ $\displaystyle \ge$ $\displaystyle \frac{l^{2}\hbar^{4}}{4m^{2}} \ \ \ \ \ (14)$

The minimum of the LHS occurs at ${t=0}$ so if the inequality is true there, it is true always. In this case, it reduces to

 $\displaystyle a+2l^{2}$ $\displaystyle \ge$ $\displaystyle 2l^{2}\ \ \ \ \ (15)$ $\displaystyle a$ $\displaystyle \ge$ $\displaystyle 0 \ \ \ \ \ (16)$

This final condition is certainly true (it is required for the Gaussian wave form to converge at large ${x}$), so the uncertainty condition is verified.

## One thought on “Energy-time uncertainty principle: Gaussian free particle”

1. Chris Kranenberg

Pedagogical suggestion – Eq. 7 is a result of the observable being Hermitian. Showing this using the inner product notation can enlighten those not sure how the second derivative of (Psi star)(Psi) is a result of (p^2)^2.