Uncertainty product of infinite square well

[mite]

For the ground state of a particle moving freely in a one-dimensional box 0\leqx\leqL with rigid reflecting end-points, the uncertainty product (del x)(del p) is
1 h/2
2 sqrt{2}h
3 >h/2
4 h/sqrt{3}

I used (del x)^2 =<x^2>-<x>^2 and (del p)^2 =<p^2>-<p>^2
Using the wavefunction of infinite square well potential in ground state i found the the expectation value . i got
(del x)(del p)= (h/2) ( sqrt{ (1/12) -- (1/2(Pi)^2) } )
which is none of the options.
How to solve it?

 

[vela]

I got a different answer. Please show us the details of your work. It's impossible to spot where your mistake is otherwise.

 

[mite]

\Psi=\sqrt{}(2/L) Sin(\Pix/L)

<x>=\int(2/L) (Sin(\Pix/L)^2 x dx {in the limit 0 to L}
<x>=L/2

<x^2>=\int(2/L) (Sin(\Pix/L))^2 x^2 dx {in the limit 0 to L}
<x^2>=(L^2 / 3) - ( L^2 / (2 \Pi^2)

<p>=\int(2/L) Sin(\Pix/L) (-i \hbar d/dx) Sin(\Pix/L) dx {in the limit 0 to L}
<p>=0

<p^2>=\int(2/L) Sin(\Pix/L) ( - \hbar^2 d/dx (d/dx) ) Sin(\Pix/L) dx {in the limit 0 to L}
<p^2>= (\hbar^2 \Pi^2) /L^2

(del x)^2=<x^2>-<x>^2=( L^2 /12 ) - ( L^2 /( 2 \Pi^2) )
(del p)^2=<p^2>-<p>^2=(\hbar^2 \Pi^2) /L^2

(del x)(del p)= (\hbar\Pi/L) \sqrt{}(( L^2 /12 ) - ( L^2 /( 2 \Pi^2) ) )
(del x)(del p)=(h/2) \sqrt{}( (1/12) - ( 1 / ( 2 \Pi^2 ) ) )

 

[vela]

In the very last step, you made an algebra mistake when you pulled the factor of L/2π out.

<br /> \Delta x \Delta p &amp; = \frac{\hbar\pi}{L} \sqrt{\frac{L^2}{12}-\frac{L^2}{2\pi^2}} =<br /> \frac{\hbar\pi}{L} \sqrt{\frac{L^2}{(2\pi)^2}\left(\frac{1}{3}-\frac{2}{\pi^2}\right)}<br />

 

[mite]

In the question all the options are expressed in terms of " h " rather then " \hbar " so i used \hbar=h/(2\Pi) hence i got the last step

 

[vela]

Oh, I see. Your answer looks fine. It's probably just a typo in the problem where they meant \hbar instead of h.

(Not sure what I was thinking when I wrote that last radical.)

 

[mite]

Right thank you