How Does a Delta Potential Affect Bound States in a Quantum Infinite Well?

[KFC]

In QM, sometimes we will combine the delta potential and other familiar potential (like infinite potential well). And I am quite confuse with the bound state. For example, consider a 1D infinite potential well with width a and locate b/w [-a/2, a/2]. Now if we add in a delta potential -V_0\delta(x) where V_0>0. If we still want to find the bound state energy and wavefunctions, what kind of conditions of energy need to be satisfied for bound state?

That is, if there is no delta potential, the infinite potential well will confine all kinds of particle so we can find bound state of particles of energy 0<E<\infty. After adding the delta potential, if we still consider the energy 0<E<\infty, in the region -\epsilon < x < +\epsilon (\epsilon is small quantity), will I still have bound state?

In addition, I know how to write the bound state wavefunctions for the infinite potential well (without the delta potential), that is

\psi = \sqrt{\frac{2}{a}}\sin(\frac{2n\pi x}{a})

in presence of delta potential, do I have to break the wavefunction into three regions? Like

<br /> \psi = A\sin(kx) + B\cos(kx), \qquad -a/2 &lt; x&lt; -\epsilon<br />

<br /> \psi = E\sin(kx) + F\cos(kx), \qquad +\epsilon &lt; x&lt; +a/2<br />

I don't know what will the wavefunction in -\epsilon &lt; x&lt; \epsilon look like because I am quite confusing how to definite the energy of particle there, should I take energy be positive? or |E|<|V_0| ?

 

[intervoxel]

When you introduce the delta potential, we have two scenarios:

1 - V0 is very small, in which case it may be treated as a perturbation and the old result can be used in the calculations.

2 - V0 has a big value and you must start from scratch with a completely new Hamiltonian.

 

[JohnSimpson]

Typically what is done is to treat the delta function as a discontinuity at which boundary conditions much be met. You break the well into TWO regions, both of which have sinusoidal solutions for the TISE, and meet the boundary conditions that the wave function must go to zero at the ends of each side of the well, that it must be continuous at the delta function discontinuity, and that the derivatives follow the rule

psi_x(0+) - psi_x(0-) = 2*m*g/hbar^2

Where g is the "strength" of the delta potential (the coefficient). Meeting these conditions will result in a transcendental equation for the allowed energies.

As for what range of energies a bound state will exist I can't really remember, sorry

 

[olgranpappy]

Typically what is done is to treat the delta function as a discontinuity at which boundary conditions much be met. You break the well into TWO regions, both of which have sinusoidal solutions for the TISE, and meet the boundary conditions that the wave function must go to zero at the ends of each side of the well, that it must be continuous at the delta function discontinuity, and that the derivatives follow the rule

psi_x(0+) - psi_x(0-) = 2*m*g/hbar^2

Where g is the "strength" of the delta potential (the coefficient). Meeting these conditions will result in a transcendental equation for the allowed energies.

As for what range of energies a bound state will exist I can't really remember, sorry

Since what the OP is calling "bound states" are all the states, there will be an infinite number of "bound states" (i.e., states). The very lowest energy state may have an energy less than the lowest energy state of the "particle in a box" w/out the delta function, depending on the strength of the potential. I think it depends on whether or not
<br /> 1/V_0 &gt; mL/2\hbar^2\;.<br />
If so, then the transcendental equation you refer to will give energies that are shifted down a bit from the usual PIAB solutions. If not, the lowest energy solution will be down shifted from the first excited PIAB state and there will be one "missing". Anyways, what I'm saying just follows from solving (what I think is the correct equation):
<br /> (k/V_0)(\hbar^2/m)=\tan(kL/2)\;,<br />
for allowed 'k' values and then
<br /> E=\hbar^2k^2/2m<br />