Quantum Mechanics Positional Operator

1. Oct 11, 2007

UbikPkd

$$\Psi_{n}(x)$$ = $$\sqrt{2/L}$$sin($$\pi$$nx/L), 0 $$\leq$$x $$\leq$$ L

$$\Psi_{n}(x)$$ = 0, x<0, x>L

$$\pi$$ is meant to be just normal, not superscript, sorry

n is an integer

show that <$$\hat{x}$$> = L/2

<$$\hat{x}$$> is the expectation value of the positional operator $$\hat{x}$$ right?

$$\hat{x}$$ $$\Psi$$= x $$\Psi$$ i think so....

$$\int$$2x/L sin $$^{2}$$($$\pi$$2nx/L) dx

which gets me L/2 - L$$^{2}$$/4n$$^{2}$$ $$\pi$$$$^{2}$$

but it is supposed to be just L/2 sigh.....

any help will be much appreciated

2. Oct 11, 2007

nrqed

Hint: Put everything inside a single tex quote! As in

$$\int_0^L ~\frac{2x}{L}~ \sin^{2}(\pi 2 n x / L) dx$$

This seems right (you used the limits to be 0 and L, right?)
Can you show your steps? You must have made a mistake in the integration.

3. Oct 11, 2007

UbikPkd

ok here goes, i was wary of doing that because, as you can see, that was my first post and it took forever!

sorry that last one should have read:

$$\int_0^L ~\frac{2x}{L}~ \sin^{2}(\pi n x / L) dx$$

$$\int_0^L ~\frac{x}{L}~- ~\frac{x}{L}~ \cos(2 \pi n x / L) dx$$

integrating by parts:

$$~\frac{x^{2}}{2L}-(~\frac{x}{L}~~\frac{L}{2n \pi }~\sin(~\frac{2 n \pi x}{L}~)- \int_0^L ~\frac{L}{2 n \pi L}~\sin~\frac{2n \pi x}{L}~dx )$$

$$\left[ ~\frac{x^{2}}{2L}-~\frac{x}{2n \pi}~\sin(~\frac{2 n \pi x}{L}~)- ~\frac{L}{4 n^{2} \pi^{2}}~\cos~\frac{2n \pi x}{L}~ \right]$$ (between L and 0)

= $$~\frac{L}{2}~ - ~\frac{L}{4 n^{2} \pi^{2}}~$$

but it should just be $$~\frac{L}{2}~$$ ?????

Last edited: Oct 11, 2007
4. Oct 11, 2007

you have basically worked out the answer, it's just that you forgot that the when zero is put in as a limit in the second part (the cosine part), it gives 1 aswell. i.e. $cos(2 \pi n) = cos(0)$, so the second part that you got disappears, leaving you with just the L/2 which is what you want.

$$<x> = ~\frac{L}{2}~ - ~\frac{L}{(2 \pi n)^2}~ + ~\frac{L}{(2 \pi n)^2}~$$

Last edited: Oct 11, 2007
5. Oct 11, 2007

UbikPkd

AHHHHHH ive done that so many times, still i haven't learnt, thank you very much, both of you!

nrqed
i read your blog, a few days ago id never done any quantum, they threw us in by asking for the wave function of the 'particle in a box' . It was a steep learning curve that lasted all night and several litres of coffee....got there in the end!

6. Oct 11, 2007