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.

    Reply

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