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I am looking at the derivation presented here.

  1. The first thing I am unsure about is where the form of $\psi_0=Ae^{\frac{-m\omega x^2}{2\hbar}}$ came from.

  2. Also, is this form for all $\psi$, or just for the ground state?

  3. Also, I can't figure out where this comes from:

    To get $\langle x^2 \rangle$ and $\langle p^2 \rangle$, we use the following useful calculation. Note that $x$ acting on $\psi_0$ must give the same answer as $-\frac{p^2}{m\omega}$[...]

Any help will be much appreciated.

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3 Answers 3

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In the first part of the linked document are not giving a formal derivation of the uncertainty principle. It is giving a particular example to show the general idea. The gaussian wavefunction is chosen because it is a particularly simple and happens to exactly satisfy the lower bound $\sigma_x\sigma_p = \frac{1}{2}\hbar$. For the harmonic oscillator being considered, then only the ground state will be a gaussian. A more general derivation is given on the last page.

The relation $\frac{-i\hat{p}}{2m}\psi_0 = x \psi_0$ comes from the fact that the annihilation operator acting on the ground state gives $a_-\psi_0 = 0$ by the definition of the ground state.

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After a quick go-through over your reference, it was simply using the ground state (Yes, $\psi_o$ is just the ground state.) of the harmonic oscillator as an example to demonstrate that $\Delta x$ and $\Delta p$ really do satisfy the uncertainty principle.

For your second question, equation (8) is obtained by taking the derivative of $\psi_0$ and then rearranging. try it yourself. ^^

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The given wave function $\psi_0$ is indeed only the ground state wave function for the quantum harmonic oscillator. Wave functions of excited levels involve Hermite polynomials in general and are not that simple.

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