Derivation Problem: Particle in a Box

[nadeemo]

Homework Statement

The analytical expression of $$\Delta$$x$$\Delta$$p for a particle in a box is:

$$\Delta$$x$$\Delta$$p = h/2pi$$\sqrt{(n\pi)^{2} - 6}$$ / $$\sqrt{12}$$
for any quantum number, n

Homework Equations

($$\Delta$$x)$$^{2}$$ = <x$$^{2}$$> - <x>$$^{2}$$

and ($$\Delta$$p)$$^{2}$$ = <p$$^{2}$$> - <p>$$^{2}$$

$$\Psi$$ = $$\sqrt{2/L}$$ sin(nxpi/L)

The Attempt at a Solution

so i tried to find (delta x)^2 and multiplied it with (delta p)^2 and rooted it, but i still have "L" left over in my derivation which doesn't work out...

 

[genneth]

Can you show the calculations for the uncertainties in x and p? What you're doing is correct from what you've described.

 

[nadeemo]

<x^2> = integral of (x^2)(psi^2) = 2/ L integral (x^2)sin^2(nxpi/L) dx

integrate by parts from zero to L andd
<x^2> = L^2 - L + 1

<p^2> = integral (conjugate psi)(momentum operator)^2(psi)dx
<p^2> = -h/Lpi integral (conjugate psi)(d^2 psi/ dx^2)dx
after taking derivative and the integral
<p^2> = [ (h bar)(n)(pi)/(L) ] ^2

<p>^2 i found to be zero
and <x>^2 i found to be L^2 / 4

so ( delta x)^2 = L^2 - L + 1 - L^2 / 4 = 3L^2 /4 - L + 1
and (delta p)^2 = [(h bar)(n)(pi)/(L)] ^2

(delta x)(delta p) = sqrt [(3L^2/4 - L + 1)[(hbar)(n)(pi)/(L)]^2]

you can see that L is un removable...it doesn't really work out for me..

 

[pam]

psi(x) must be zero at the ends, so your -L+1 is wrong.
Check the dimensions. <x*2> has to ~L^2.

 

[nadeemo]

<x^2> = 2/L integral (x^2)sin^2 (nxpi/L) dx
= 2/L[x^2(x/2 - (L / 4npi)sin (2nxpi/L)) - integral x sin^2 (nxpi/L)] from 0 to L
= 2/L[x^2(x/2 - (L / 4npi)sin (2nxpi/L)) - x(x/2 - (L/4npi)sin (2nxpi/L)) - integral of sin^2 (nxpi/L)] from 0 to L
=2/L[x^2(x/2 - (L / 4npi)sin (2nxpi/L)) - x(x/2 - (L/4npi)sin (2nxpi/L)) - (x/2 - (L/4npi)sin (2nxpi/L] from 0 to L
= 2/L [ L^3 /2 - L^2/2 + L /2]
= L^2 - L + 1

i don't think i did anythign wrong in my integration...well clearly something is wrong

but i found this

its from -a/2 to a/2..so how do i make that form 0 to L?
its the same right?

 

[nadeemo]

so this pic basically solves my problem
its just that...
i wouldn't know how to integrate to get that in the first place
and
<x>^2 = L^2 / 4...and i can't incorporate that..
because well sqrt( <x^2> * (delta p)^2) = the answer that they want
but i need (delta x)^2(delta p)^2

(delta x)^2 = <x^2> - <x> ^2...and if i do that its no longer correct...

 

[genneth]

Things to note: $$\langle x \rangle = \langle p \rangle = 0$$ -- can you show this?

Further, you should be able to integrate $$\int x^n sin(x)\,dx$$ as an indefinite integral. Can you do so for n=1 or n=2?

 

[nadeemo]

<x> doesn't equal <p>
in my textbook they show that,
<x> = L/2 and <p> = 0..
and that integral doesn't really relate because x^n sin x...when i have (x^n)(sin^2 x)

 

[genneth]

True -- my bad when posting too quickly. But you should still be able to do those integrals by parts...

 

[nadeemo]

from my integration by parts for <x^2> i get (L^2)/3 when accoring to that equation
i should be gettin L^2[(n^2)(pi^2) - 6] / [12(n^2)(pi^2)]

a = L, but you hafta multiply it by 2/L, since the wavefunction is sqrt(2/L) sin(nxpi/L)