Quantum Mechanics, The Delta-Function potential

[eliUCV]

Hi, I have a question, why when we study the Delta-Function Potencial we can treat with $E < V$, since the following relation says

$\frac{d^2 \psi}{dx^2} = \frac{2m}{\hbar^2} (V - E) \psi$

And do not allow it? or it is just $E < V_{min}$ ?

 

[BvU]

Hello UCV, !

What about the sign of V ?

 

[eliUCV]

Thanks :D
$V(x) = - \alpha \delta(x)$ where $\alpha > 0$
I think it is important to say that it is time-independent Schrodinger's Equation

 

[BvU]

Yep. Exactly matches the treatise in the link. One bound state.
For $\ V(x) = - \alpha \delta(x) \$ there is no $V_{min}\$ so no $E < V_{min}\$; did you mean $E < V_{max}$, in this case 0 ?

 

[eliUCV]

Yes, could E be less than V_max ?

 

[BvU]

V max is zero. E is less, yes: a bound state

 

[eliUCV]

Ok, but E never can be less than $V_{min}$ for any $V(x)$, right??

 

[BvU]

Correct.

 

[vanhees71]

Since in
$$\hat{H}=\frac{1}{2m} \hat{p}^2 + V(\hat{x}),$$
the first term ("kinetic energy") it's a positive definite operator. For the very artificial potential in question, somewhat sloppily speaking the potential has a minimum at $x=0$ (namely "$-\infty$") there. The energy eigenvalues thus must all be $>-\infty$, which is hardly a constraint.

To get the eigenvalues you have to solve the time-independent Schrödinger equation,
$$-\frac{\hbar^2}{2m} u_E''(x) -\alpha \delta(x) u_E(x)=E u_E(x).$$
For bound states you need $E<0$, but for $E<0$ the only solution must be of the form
$$\psi(x)=A \exp(-k|x|), \quad k>0.$$
Now
$$\psi'(x)=-A k \mathrm{sign}(x) \exp(-k |x|)$$
and
$$\psi''(x)=[A k^2 -2 A k \delta(x)] \exp(-k |x|).$$
Plugging this into the Schrödinger equation we get (cancelling the common A exp factor)
$$-\frac{\hbar^2}{2m} [k^2 - 2 k \delta(x)]-\alpha \delta(x)=E.$$
So the only bound-state solution is to make the $\delta$ distribution contribution vanish by choosing
$$\frac{\hbar^2}{m} k = \alpha \;\Rightarrow \; k=\frac{m \alpha}{\hbar^2}.$$
From this we get
$$E=-\frac{\hbar^2}{2m} k^2 = -\frac{m \alpha^2}{2 \hbar^2}.$$
Of course there are plenty of scattering solutions with $E>0$ too.