Given the potential find the eigenfunction

[Ashphy]

Homework Statement

Given the potential find the eigenfunction

Relevant Equations

$$V(x)=\begin{cases}0; x>0\\ \infty;x<0 \end{cases}$$

Hi, this was one of the oral exam questions my teacher asked so i tried to solve it. Consider y>0 the energy spectrum here is continuous and non degenerate while for y<0 the spectrum is discrete and non degenerate because E<0.
for y>0 i thought of 2 cases
case 1 there is no wave function for x<0 because of infinite potential so the general solution must be $$\psi(x)=Ae^{ikx}+Be^{-ikx}$$ then i apply continuity of the function and continuity of the derevative to finde theat A=-B such $$\psi(x)=A(e^{ikx}-e^{-ikx})=-2iAsinkx$$ but this is not normalizable since the integral is divergent so i consider the case 2 such i have an oscillating wave function for x>0 and an exponentially decreasing function for x<0 and then i go ahead and find C (constant associated to the real exponential for x<0) and B in function of A but then again it is not normalizable. What am i doing wrong? if anyone could please tell me if there is a better approach to the problem it would be really helpful, thank you

 

[PeroK]

I assume you haven't seen anything like this before? Non-normalizable solutions for a free particle?

 

[Ashphy]

no i have not, i just recently started studying potentials and as of my class notes i have written that acceptable wave eigenfunctions must be normalizable, is it not so?

 

[hutchphd]

Because plane-wave states are not properly normalisable we employ the trick of normalising them in a large (relative to potential range) cubic box of side L with periodic boundary conditions. We then take the limit L→∞ at the end of the calculation. Of course we no longer have a strictly continuous spectrum of eigenstates, except in the limit.
Note that this is equivalent to solving the problem of a well and letting one wall "go out to infinity". One becomes used to such chicanery because of its utility.

 

[PeroK]

no i have not, i just recently started studying potentials and as of my class notes i have written that acceptable wave eigenfunctions must be normalizable, is it not so?

To be viable physical solutions, they must be normalizable. What you've found is only part of the story. To normalize such wavefunctions you have to bundle them up into a wave-packet.

That said, the non-normalizable eigenfunctions can still be useful for studying quantum behaviour - such as reflection and transmission coefficients.

You didn't do anything wrong, by the way. This should have been covered in your course.

 

[Ashphy]

To be viable physical solutions, they must be normalizable. What you've found is only part of the story. To normalize such wavefunctions you have to bundle them up into a wave-packet.

That said, the non-normalizable eigenfunctions can still be useful for studying quantum behaviour - such as reflection and transmission coefficients.

You didn't do anything wrong, by the way. This should have been covered in your course.

thank you for that, i may be missing notes, is there any good book where i can find more?

 

[Ashphy]

Because plane-wave states are not properly normalisable we employ the trick of normalising them in a large (relative to potential range) cubic box of side L with periodic boundary conditions. We then take the limit L→∞ at the end of the calculation. Of course we no longer have a strictly continuous spectrum of eigenstates, except in the limit.
Note that this is equivalent to solving the problem of a well and letting one wall "go out to infinity". One becomes used to such chicanery because of its utility.

Thank you! i would have never thought.
so in this case do i just take my solutions for the potential well,with $$ \psi=\sqrt{2/L}sin(\frac{n\pi x}{L}) \\L\to\infty $$ doesn't it make $\psi=0$ ?

 

[hutchphd]

Typically when you need to calculate a real world number (one that corresponds to a measurable quantity) the answer will not depend explicitly on the "size of the box" and the limit doesn't matter. We are imperfect beings.........

 

[PeroK]

thank you for that, i may be missing notes, is there any good book where i can find more?

I like Griffiths:

https://www.abebooks.co.uk/servlet/...lok_DwYlApnJtjz0YZQ2C5LFJVsM5J1MaAnfmEALw_wcB

 

[Ashphy]

I like Griffiths:

https://www.abebooks.co.uk/servlet/...lok_DwYlApnJtjz0YZQ2C5LFJVsM5J1MaAnfmEALw_wcB

thank you again!

 

[Ashphy]

Typically when you need to calculate a real world number (one that corresponds to a measurable quantity) the answer will not depend explicitly on the "size of the box" and the limit doesn't matter. We are imperfect beings.........

following your tip i tried to solve it like so: Let the general solution to a finite well from 0 to L be $\psi=Bsin(kx)+Acos(kx)$ then i apply the periodic condition such $$\begin{cases}\psi(0)=\psi(L)\\ \psi '(0)=\psi '(L)\end{cases},$$ Solving the system gives $$\begin{cases}B=A\frac{1-coskL}{sinkL}\\coskL=n\pi\end{cases}$$ so i find the condition $k=\frac{n\pi}{L};\;E=\frac{n^2\pi^2\hbar^2}{2mL^2}$ which seem reasonable but then if i substitute k in B and i find an indeterminate form of $\frac{0}{0}$ so i thought i'd put B=0 (i feel that's wrong) and find $\psi=Acos(\frac{n\pi x}{L})$ that gives $\psi=A$ for $L\to\infty$ which seems wrong. Is this reasonable? am i approaching it correctly?