# Delta potential problem - bound states problem

Homework Statement:
Find energies and states of particle that move in the potential ##V(x)=-\lambda \delta(x)##. If ##E<0##.
Relevant Equations:
Equation
##-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x)+V(x)\psi(x)=E\psi(x)##
where ##V(x)=-\lambda \delta(x)##.
I am confused here. For ##x>0## particle is free and for ##x<0## particle is free. That I am not sure how we can have bond states. If particle is in the area ##x>0## why it feel ##\delta## - potential at ##x=0##. Besides that, I know how to solve problem. But I am confused about this.
If we have the case of motion of a particle through a potential barrier, the particle is after that free. It does not feel the influence from the potential behind it. Could you please explain this?

anuttarasammyak
Gold Member
Energy eigenstate ##\Psi(x)## and the state of the particles in the area x>0 are two different states. QM does not allow such case division of ##\Psi(x)## according to position of particle.

Energy eigenstate ##\Psi(x)## and the state of the particles in the area x>0 are two different states. QM does not allow such case division of ##\Psi(x)## according to position of particle.
Of course, you need to have that
##\psi(0^+)=\psi(0^-)##
I agree. But why if the particle passed ##x=0## can not go to ##x \to\infty##. I do not understand that.

anuttarasammyak
Gold Member
Do you mean you got a damping solution ? If so what is the damping factor ?

When you solve equation for ##x>0## where ##V(x)=0## the Sch. equation will be
$$-\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2}=-|E|\psi(x)$$
so
$$\frac{d^2 \psi}{d x^2}-\frac{2m|E|}{\hbar^2}\psi(x)=0$$.
From that
##k^2=\frac{2m|E|}{\hbar^2}##
and from that for ##x>0##
$$\psi(x)=C_1e^{kx}+C_2e^{-kx}$$.
And from that for bound states we take that ##C_1=0##. Why? I understand that for bound states ##\psi(x) \to 0##, when ##x \to \infty##, but how to know that bound state even exists?

In case of quantum ##LHO## ##V(x)=\frac{1}{2}m\omega^2x^2## so ##V(x) \to \infty## when ##x \to \pm \infty## and therefore we must have bound states in the problem. How to know that in this case bound states even exist. It might be a silly question for you, but I really do not understand this.

anuttarasammyak
Gold Member
There are bound states and no bound states. Which is of your concern ?
Bound states are bounded around the well so they go down as they go far away from the well.

There are bound states and no bound states. Which is of your concern ?
I am not sure how to know that bound states exist. There are definitely some systems where bound states do not exist. And how particle know is in the bound state or not? For me is not logic that particle can be in bound state when it is the region where ##V(x)=0##.

anuttarasammyak
Gold Member
Have you solved problem of finite square well ? They have bound states and no bound states. Bound states go beyond the well wall and decay exponentially. Delta potential well is the special case of the finite well potential. If you find no problem in the finite well potential you do not have to worry about the delta potential case either.

Yes of course. For me is a bit different. Because there exists some potential outside the well. In this case, particle is quasi-free for me. We do not know where the particle is but if the particle is somewhere where ##V(x)=0## why it should not go to ##x \to \infty##.

anuttarasammyak
Gold Member
Because there exists some potential outside the well.
You mean in addition to delta potential another potential is applied in your exact problem? If only delta function is applied, it is finite well potential, i.e.
##V(x)=- \frac{1}{2a} ## for ##-a<x<a##, 0 otherwise
and after getting both bound and no bound solutions as function of a, make ##a\rightarrow +0## in the formula.

Last edited:
LagrangeEuler
Dr Transport
Gold Member
you are forgetting that the SE is second order, you need another BC.

And remember, a square well, no matter how small, always has at least one bound state. Think about it.

Sorry just what do you mean another BC? I can understand that if Sch. equation is second-order you have two particular solutions? Right?
But why at least one solution must be bounded?

hutchphd
Homework Helper
Yes of course. For me is a bit different. Because there exists some potential outside the well. In this case, particle is quasi-free for me. We do not know where the particle is but if the particle is somewhere where V(x)=0 why it should not go to x→∞.

This is incorrect. A finite negative potential which is zero outside a localized region will nearly always support bound states. These states all diminish at distance and can be normalized. You may call this quasi-free if you want, but it is not a good descriptor and you will be alone.
The bound state is localized. The probability for finding a particle far away becomes very very small.
But quantum tunneling exists as a very useful fact.

Dr Transport
Gold Member
Sorry just what do you mean another BC? I can understand that if Sch. equation is second-order you have two particular solutions? Right?
But why at least one solution must be bounded?

There is a condition for the continuity of the wave function and one for it's derivative. Those are necessary for the complete solution.

As for a bound solution, look at the solution of the square well, there is always a bound state. Again, look and think about it.

Yes in this ##V(x)=-\lambda \delta(x), \lambda>0## there is only continuation for wave function. The first derivative of the wave function has jump at ##x=0##. Ok. I understand.

But what about the case when ##\lambda <0##? In this case, it is more like a barrier? And you have not bound state in this case?

Dr Transport
Gold Member
Try integrating the SE over x.......

To my mind this is the case when ##E>0##. In that case for ##x<0## ##\psi_1(x)=C_1e^{ikx}+C_2e^{-ikx}## and for ##x>0## ##\psi_2(x)=C_3e^{ikx}+C_4e^{-ikx}##. Because of that there is no bound states.

Dr Transport
Gold Member
That is correct. and read your own original post. It clearly says that $E < 0$, so you are looking for a bound state and that requires that $\lambda > 0$ for the given form of the potential.

vela
Staff Emeritus