- #1
KFC
- 488
- 4
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?
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| ?
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| ?