Quantum mechanics: Normalization and the width of distribution

[Kentaxel]

1. Homework Statement [/b]

Normalize the following wave function, obtain the corresponding function in position-space (fourier transform) and find the width of the distribution in the x variable.

Homework Equations

\phi(p_x) =<br /> \begin{cases}<br /> 0, &amp; \;\; |p_x-p_0| &gt; \gamma \\<br /> C, &amp; \;\; |p_x-p_0| \leq \gamma<br /> \end{cases}

3. The Attempt at a Solution

Now the normalization of the initial wave function pops out the relation*

C=\frac{1}{\sqrt{2\gamma}}

So for so good. Using this result to obtain a corresponding wave function in position space (1dim x-space) i am able to obtain the function

\Psi (x) = \sqrt{\frac{\hbar}{\pi \gamma}} e^{i x p_0 / \hbar} \frac{\sin(\gamma x / \hbar)}{x}

Which in fact normalizises to unity as expected, suggesting I'm on the right track. Now however i encounter problems. In trying to find the width of the distribution in the variable x (i.e Δx), I'm not sure if my calculatins are incorrect or if my interpretation of the problem is incorrect or if my solution is actually fine but the problem it self is incredibly complicated. My approach is as follows:

I want to know for what value of x the amplitude of the probability density i.e |ψ(x)|^2 has droped to e^(-1) times its maximum value right? Thus i simply calculate where the function obtains it's maximum value using the derivative with respect to x and setting it equal to zero, or so i thought. It turns out however that this yields a slightly uncomfortable equation To calculate (though by no means impossible) this fact however leads me to believe its not the right way to go about solving the problem. I push on, and find the maximum to be at x=0 and |ψ(0)|^2 = γ/(∏*h-bar). Thus i want to know for what x |ψ(x)|^2 = γ/(e*∏*h-bar) right? This expression however seems to yield a result only solvable with numerical methods which is not what i expect and does not give the nice representation of the width Δx that i want but rather*

\frac{\hbar}{\pi \gamma} \frac{\sin^2(\gamma x / \hbar)}{x^2} = \frac{\gamma}{e \pi \hbar}

The problem is that I am supposed to show that the result leads to the uncertanity relation between the different distrubutions, ΔxΔp ≥ h-bar, which with this result seems impossible to me.

 

[genericusrnme]

Why not use the most common form of the standard deviation?

\sigma _x = \sqrt{\langle x^2 \rangle - \langle x \rangle ^2}

(I'm assuming what you mean by width of distrobution is the same thing I mean by standard deviation)

 

[Kentaxel]

Why not use the most common form of the standard deviation?

\sigma _x = \sqrt{\langle x^2 \rangle - \langle x \rangle ^2}

(I'm assuming what you mean by width of distrobution is the same thing I mean by standard deviation)

So how would i actually apply this to my problem? The initial task was to show that the uncertainity principle comes out of the relation between the wavefunction in position-space and momentum-space. This seems to me like using another definition to obtain the desired result.

By width of distribution i mean the value of x where the probability density has dropped to e^(-1) of it's maximum amplitude as discribed in the text.