**Required math: calculus **

**Required physics: **Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.11.

The lowest two states for the harmonic oscillator are the ground state and first excited state . We can work out some mean values of various quantities using explicit integration.

Consider first the ground state . Using the substitutions and we have . We can simplify the operations considerably if we note the even and odd natures of some of the functions to be integrated. Since is even, is odd, so . To calculate , since the operator , we need the derivative . This is again an odd function, so as well.

For the mean square values, we do need to do some integrals (I’ve used software for this).

Here we used to convert the integration variable from to in the second line.

For we have which is an odd function. The square of an odd function is an even function, so is even, which means that the function to be integrated to find is again the product of an even function and an odd function, so here as well.

Considering , we calculate the derivative of which is of form for a constant , which is even. Thus to obtain we must integrate this even function multiplied by the odd function so the result is .

To get the mean square values, we do the integrals:

And for the momentum:

For using the results above, the uncertainty principle here comes out to

For , we have

The mean kinetic and potential energies can be worked out from the above results without doing any more integration. We get for and for .

for and for Adding these together to get the total energy gives for and for as it should.

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EvanJust want to check equation (2). Why isn’t it ξ/2 in the exponent?

gwrowePost authorBecause we have to square inside the integral:

LoganI believe you have a misplaced parenthesis in the integrand of eq. (4). The integrand should be (1-xi^2)*exp(xi^2/2)*exp(xi^2/2).

gwrowePost authorFixed now. Thanks.