Why can't there be an antisymmetric excited state in a delta function potential?

[dyn]

I have a potential which is zero everywhere except at -2a , -a , 0 , a , 2a on the x-axis where there is an attractive delta potential at each of the 5 points. I know there is a maximum of 5 bound states. I know there can be no nodes for |x| > 2a and a maximum of one node between each delta potential. My question is ; why can't there be one further excited state - an antisymmetric state with a node at x=0 and 4 further nodes ; one between each delta potential ?

 

[mfb]

What would happen with the first and second derivative of the antisymmetric wave function at x=0?

 

[dyn]

Antisymmetric wavefunction would have the form of sine so 1st derivative would be cos(0) =1 , 2nd derivative would be sin(0)=0. I can't see how that helps and the same would happen for the 1st and 3rd excited states as well

 

[Vanadium 50]

I suggest you sketch the wavefunctions. If you encounter difficulty, sketch the wavefunctions for one delta potential, then two, then three, etc.

 

[dyn]

I tried that. I can manage to draw an antisymmetric wavefunction that goes through x=0 and has 4 other nodes !

 

[Avodyne]

That state will have positive energy. The wave function changes "too fast" between the node at x=0 and the nodes between 0 and +/-a to have negative energy. It is also not an energy eigenstate; it will have nonzero inner product with both positive and negative energy states.

 

[The_Duck]

Antisymmetric wavefunction would have the form of sine so 1st derivative would be cos(0) =1 , 2nd derivative would be sin(0)=0. I can't see how that helps and the same would happen for the 1st and 3rd excited states as well

As Avodyne says, a wave function that looks like the sine function has positive energy so will not be bound. A bound state wave function must look like a linear combination of exponentials in regions where V=0.