Solving Schrodinger Equation: xav, (x2)av, Δx

[CornMuffin]

Homework Statement

Use the ground-state wave function of the simple harmonic oscillator to find x_av, (x²)_av, and $$\Delta x$$. Use the normalization constant $$A=(\frac{m\omega _0}{\overline{h} \pi })^{1/4}$$

Homework Equations

$$\psi (x) = Asin(kx)$$
$$(f(x))_{av} = \int ^{\infty }_{-\infty } \left| \psi (x) \right| ^2 f(x) dx$$

The Attempt at a Solution

I calculated out $$x_{av}$$ as
$$x_{av} = \int ^{\infty }_{-\infty } \left| \psi (x) \right| ^2 x dx$$
$$x_{av} = A \int ^{\infty }_{-\infty } xsin^2 (kx) dx$$
$$x_{av} = A (x^2/4 - (cos(2kx))/(8k^2) - (xsin(2kx))/(4k))\right| ^{\infty}_{-\infty}$$
$$x_{av} = 0$$

I think that is right...but I am having trouble calculating $$(x^2)_{av}$$
$$(x^2)_{av} = \int ^{\infty }_{-\infty } \left| \psi (x) \right| ^2 x^2 dx$$
$$(x^2)_{av} = A \int ^{\infty }_{-\infty } x^2 sin^2 (kx) dx$$
$$(x^2)_{av} = A (x^3/6 - (xcos(2kx))/(4k^2) - ((-1+2k^2x^2)sin(2kx))/(8k^2))\right| ^{\infty}_{-\infty}$$
But this says $$(x^2)_{av} = \infty$$

which I don't think is correct...

 

[alxm]

Wrong wave function; that's not the ground state of a harmonic oscillator.

Also, $$cos(kx)$$ as $$x\rightarrow \pm \infty$$ is undefined; so a plane wave wave function isn't normalizable over infinite space. That's an important result too, though.

 

[CornMuffin]

Wrong wave function; that's not the ground state of a harmonic oscillator.

Also, $$cos(kx)$$ as $$x\rightarrow \pm \infty$$ is undefined; so a plane wave wave function isn't normalizable over infinite space. That's an important result too, though.

Is this this correct ground state?
$$\psi (x) = (\frac{m\omega }{\pi \overline{h}})^{1/4}e^{\frac{-m\omega}{2\overline{h}}x}$$

 

[Dorilian]

That is not the wave function for the ground state, you should use the hermite polynomials.

 

[alxm]

That is not the wave function for the ground state, you should use the hermite polynomials.

The first Hermite polynomial equals 1. So yes, it's correct.

 

[grey_earl]

Is this this correct ground state?
$$\psi (x) = (\frac{m\omega }{\pi \overline{h}})^{1/4}e^{\frac{-m\omega}{2\overline{h}}x}$$

You forgot a square in the argument of the exponential, it's $$\exp(-m\omega/(2 \hbar) x^2)$$.
(and for the LaTeX: it's \hbar, not \overline h)