Quantum Mechanics Positional Operator

[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

 

[nrqed]

$$\Psi_{n}(x)$$ = $$\sqrt{2/L}$$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

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.

 

[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 $$~\frac{L}{2}~$$ ?

 

[ppyadof]

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.

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

 

[UbikPkd]

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.

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

AHHHHHH I've 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!

 

[ppyadof]

AHHHHHH I've 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!

That's ok. Only too glad to help. Steep learning curves are the best in the long run, take my word for it :P