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 [tex]a[/tex] and locate b/w [-a/2, a/2]. Now if we add in a delta potential [tex]-V_0\delta(x)[/tex] where [tex]V_0>0[/tex]. 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?(adsbygoogle = window.adsbygoogle || []).push({});

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 [tex]0<E<\infty[/tex]. After adding the delta potential, if we still consider the energy [tex]0<E<\infty[/tex], in the region [tex]-\epsilon < x < +\epsilon[/tex] ([tex]\epsilon[/tex] 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

[tex]\psi = \sqrt{\frac{2}{a}}\sin(\frac{2n\pi x}{a})[/tex]

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

[tex]

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

[/tex]

[tex]

\psi = E\sin(kx) + F\cos(kx), \qquad +\epsilon < x< +a/2

[/tex]

I don't know what will the wavefunction in [tex]-\epsilon < x< \epsilon[/tex] 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| ?

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# Ground state wavefunction

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