Standard Deviation of Position and Momentum

Classical Mechanics Level 5

Let $\psi(x) = {\pi}^{-1/4} {e}^{-{x}^{2}/2}.$

First calculate ${\sigma}_{x} = \sqrt{\left<{x}^{2}\right> - {\left<x\right>}^{2}}$ .

Then calculate ${\sigma}_{p} = \sqrt{\left<{p}^{2}\right> - {\left<p\right>}^{2}}$ .

If ${\sigma}_{x}{\sigma}_{p} = a\hbar$ , evaluate $a$ .

Image Credit: Wikipedia Fermilab, U.S. Department of Energy

The answer is 0.5.

1 solution

Matt DeCross
May 10, 2016

The given wavefunction is a Gaussian, which is a minimum uncertainty state -- it saturates the uncertainty principle, so the constant $a$ is $\frac12 = 0.5$ . One can also explicitly compute the expectation values, but this is tedious.

Why does the Gaussian function minimize uncertainty? I tried calculating expectation values for momentum but I kept getting zero. Might have to do with how I wrote <p^2>.

M. V. G. - 1 year, 11 months ago

For sure <p^2> should not be zero, although <x> and <p> are. The Fourier transform of a Gaussian is also a Gaussian, so the variance can't be zero in either position or momentum space. I think I probably just used my knowledge that coherent states of the simple harmonic oscillator are minimum uncertainty states, and that these states are also Gaussian, above.

Matt DeCross - 1 year, 10 months ago

Standard Deviation of Position and Momentum

Image Credit: Wikipedia Fermilab, U.S. Department of Energy

The answer is 0.5.

1 solution

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