2
$\begingroup$

I am tasked with finding $\psi (x)$ of the following piecewise function:

enter image description here

This function obviously appears to not be differentiable, but am asked to consider it to approximate to a smooth wavefunction. I am asked to sketch this $\psi(x)$ pertaining to this function.

Firstly, I believe I (rightly) described this function with piecewise as:

$$\mid \Psi(x) \mid^2 \ = \begin{cases} 2.5*10^{19}x+\frac{1}{2*10^{-10}}, & \text{$(-2*10^{-10}) \le x \le 0$} \\ -2.5*10^{19}x+\frac{1}{2*10^{-10}}, & \text{$0 \le x \le (2*10^{-10})$} \end{cases}$$

We are then asked to assume that $\psi(x)$ is real only.

However, I feel a bit weird about my sketch for $\psi(x)$. I'll type it here. I'll also simplify the mess for my piecewise function a bit above.

$$ \psi(x) \ = \begin{cases} \sqrt{2.5*10^{19}x+\frac{1}{2*10^{-10}}}, & \text{$(-2*10^{-10}) \le x \le 0$} \\ \sqrt{-2.5*10^{19}x+\frac{1}{2*10^{-10}}}, & \text{$0 \le x \le (2*10^{-10})$} \end{cases} $$

However, due to the extremely high gradients of this function, trying to graph it ends up basically approximating $\psi(x) = \delta (x)$, Where $\delta(x)$ is the dirac delta function. However, this seems to take away the piecewise nature of this function, and, given $\psi(x)$, I can't seem to understand how normalizing it would bring me back to $\mid \Psi(x) \mid^2$ as I would need to. Perhaps the devil's in the details, and this isn't truly the dirac delta function? I just would have to state this is the function's arguments and basically draw it as the dirac delta function?

How should I approach this sketch? And how can this end up resembling $\mid \Psi(x) \mid^2$? I'm essentially not confident with my work but I'm not sure what I did wrong.

$\endgroup$
1
$\begingroup$

Your equation to your own question is correct. But take a look in a more compact form : \begin{equation} \psi(x)\boldsymbol{=} \left. \begin{cases} \sqrt{c\boldsymbol{-}c^2\boldsymbol{\vert} x\boldsymbol{\vert}\vphantom{\tfrac12}}\,, \quad x \boldsymbol{\in} \left[\boldsymbol{-}c^{\boldsymbol{-}1},\boldsymbol{+}c^{\boldsymbol{-}1}\right]\\ \hphantom{\sqrt{c\boldsymbol{-}}}0\hphantom{\sqrt{c\boldsymbol{-}}} \,, \quad x \boldsymbol{\notin} \left[\boldsymbol{-}c^{\boldsymbol{-}1},\boldsymbol{+}c^{\boldsymbol{-}1}\right] \end{cases} \right\} \,, \quad c=\boldsymbol{+}5\cdot10^9 \tag{01}\label{01} \end{equation} Note that $\:\boldsymbol{\vert}\psi(x)\boldsymbol{\vert}^2\:$ is very close to the Dirac $\:\delta-$function \begin{equation} \lim_{c\boldsymbol{\rightarrow}\boldsymbol{+}\infty}\boldsymbol{\vert}\psi(x)\boldsymbol{\vert}^2\boldsymbol{=}\delta(x) \tag{02}\label{02} \end{equation}

$================================================$

\begin{equation} \psi(x)\boldsymbol{=}\alpha^{{\boldsymbol{-}1}}\sqrt{\bigl[\theta(x\!\boldsymbol{+}\!\alpha)\!\boldsymbol{+}\!\theta(x\!\boldsymbol{-}\!\alpha)\!\boldsymbol{-}\!2\theta(x)\bigr]x\!\boldsymbol{+}\!\bigl[\theta(x\!\boldsymbol{+}\!\alpha)\!\boldsymbol{-}\!\theta(x\!\boldsymbol{-}\!\alpha)\bigr]\alpha}\,, \quad\alpha=c^{\boldsymbol{-}1} \tag{99}\label{99} \end{equation}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.