Determining the BOUND STATES for the Delta function potential

In this video I will determine the wave function and energy levels of the bound states of the delta function potential.
My name is Nick Heumann, I am a recently graduated physicist. I love to teach physics, so I decided to give KZbin a try. English is not my first language, but I hope that you can understand me well enough regardless.
▬ Contents of this video ▬▬▬▬▬▬▬▬▬▬
00:00 Introducing the delta potential
05:10 Solving the Schrödinger Equation
11:30 Applying Border Conditions
15:57 Integrating the Schrödinger Equation
24:00 Determining the allowed Energy levels
28:13 Normalizing the Wavefunction

[@lilynichols5208]

Thank you for explaining why we take the integral of the Schroedinger equation!! I was quite confused when reading Griffiths, but it makes so much sense now!!

[@meghanasseshasai3407]

Very well explained, thank you so much for the video!

[@aubrylines3821]

This is so helpful thank you so much.

[@abhinavsaket1194]

good video.

[@Grellan_L]

I have one question. My lecturer says that for bound states, the sign of the curvature of the wave function has to be the same as the sign of the wavefunction itself. In your graph, the wavefunction is negative, i.e. below the x axis, yet the curvature is positive. Could you clarify that point please?

[@NickHeumannUniversity]

It's been a while since I did this video, can you timestamp the moment you are asking about please?

[@Grellan_L]

@@NickHeumannUniversity sorry yes, the graph is at the very end of the video at 31:30. Edit: I suppose what I'm asking is, should the graph be above the x-axis not below it?

[@NickHeumannUniversity]

@@Grellan_L In that graph the y axis is the energy! So if we have bound states, then E

[@Grellan_L]

@@NickHeumannUniversity yes but the wavefunction should be above the x-axis should it not? With the delta potential below it? Edit: i think I get it now, I was just confused because I'm used to seeing the wavefunction and potential superimposed with the wavefunction above the x axis and the delta potential below. Thanks for your help!

[@mcalkis5771]

Can you explain why the energy has to be negative in order for the particle to be in a bound state near the origin? If the potential at that point tends to negative infinity shouldn't any energy be considered to be less than it?

[@apoorvmishra6992]

We dont consider potential at that point. It's the potential at x= infinity (here it's 0) which determines bount or scattering states.