Fourier transformation of the Wavefunction in QM

[B4cklfip]

Homework Statement

Given is the function phi(x) = A_0 for -L≤x≤L and phi(x) = 0 otherwise.
The task is to first sketch phi(x) as function of x. Then to calculate the fourier-transformation and sketch phi(k) as function of kL. Also I have to compare accessible Broads.

Relevant Equations

phi(k)=1/sqrt(2*pi) integral_-inf_inf (dx exp(i*k*x)*phi(x))

Hello Physics Forum,

I am not sure what to to in this task, because the wavefunction is only given as A_0. Maybe someone can explain it to me.

Thanks in Advance,
B4ckflip

 

[pasmith]

It says in your problem statement that $$\phi(x) = \begin{cases} 0 & x < -L \\<br /> A_0 & -L \leq x \leq L \\<br /> 0 & x > L\end{cases}.$$ So what is $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{ikx} \phi(x)\,dx?$$

 

[B4cklfip]

\phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{ikx} \phi(x)\,dx is the Fouriertransformation of \phi(x). It changes the dependence of the wavefunction from position x to momentum p.

 

[pasmith]

And how would you do the integral, given that $\phi(x)$ is defined piecewise?

 

[B4cklfip]

Thanks for the hint, I now tried to solve it and got following result:

$$\tilde{\phi}(x) = \begin{cases} 0 & x < -L \\ \frac{A_0}{\sqrt{2pi}k} \cdot 2sin(kL)\ & -L \leq x \leq L \\ 0 & x > L\end{cases}$$

I have integrated from -L to L for the second interval. Is it correct ?
And how can I sketch specially phi(x) ?

 

[vela]

Thanks for the hint, I now tried to solve it and got following result:

$$\tilde{\phi}(x) = \begin{cases} 0 & x < -L \\ \frac{A_0}{\sqrt{2pi}k} \cdot 2sin(kL)\ & -L \leq x \leq L \\ 0 & x > L\end{cases}$$

I have integrated from -L to L for the second interval. Is it correct?

No, it's not correct. Note that you integrated with respect to $x$, so $x$ doesn't appear in the final result once you plug the limits in. $\tilde{\phi}$ is not a function of $x$ but of $k$.