I Quantum Mechanics, The Delta-Function potential

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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

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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

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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 ?
 
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Yes, could E be less than V_max ?
 

BvU

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V max is zero. E is less, yes: a bound state
 
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Ok, but E never can be less than ## V_{min}## for any ##V(x)##, right??
 

BvU

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Correct.
 

vanhees71

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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.
 

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