**Required math: calculus **

**Required physics: **Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.44.

An interesting potential is one where we combine an infinite square well with a delta function barrier. That is,

Outside the square well, . Inside the well, we have two distinct regions, one on each side of the delta function.Inside the well, away from the delta function, the general solution is

where .

Since the potential is even, we can consider odd and even solutions separately. If we take the even solutions first, we have

Continuity of the wave function at gives the condition (the same condition at both points):

The wave function is automatically continuous at . However, we can use the same condition on the derivative of the wave function at this point that we used in analyzing the delta function well. That is

Working out the derivatives, we get

Substituting from above, we get

or, substituting

This is a transcendental equation in and can be solved graphically or numerically, once we’ve chosen a value for . If we take this to be 1, for example, a plot shows the intersections of and within a range of . Since the left side is a straight line, and the tangent is periodic, there are an infinite number of solutions.

Note that since both and are real and positive, we’re interested only in values of , so that’s what is shown in the plot.

The odd solutions are easier. In this case

Continuity at requires , so . In this case, the derivative is also automatically continuous across , so gives us no information. The only other boundary condition we can use is at , from which we get

This is the same condition as in the square well without the delta function, and we can write

We’ve written the energies in the latter form since the well’s width is , so we can see that the odd solutions give only the even-numbered energy levels. Thus it is only the odd-numbered energy levels (arising from the even wave functions) that get modified due to the delta function. The odd wave functions are not affected by the delta function since they are zero at .

It’s worth looking at the asymptotic behaviour of the even solutions as gets very small or very large.

For , from 12, we must have , which means

Thus we reclaim the odd numbered energy levels when the delta function disappears.

At the other extreme, if , we must have , which happens when

These are the same energy levels as in the square well of width , thus it seems that making the delta function infinitely strong separates the well into two individual wells of half the width.

Here are a few plots of the wave functions for various energies, courtesy of KCErb (see link in comments below).

Lowest even energy state. Note the discontinuity of the derivative at .

Lowest odd energy state. Here the delta function has no effect, since the wave function is zero at the origin.

Next even state. There’s a slight bump at .

Next odd state.

Next even state. The bump at is hardly noticeable now.

And another odd state.

Anonymousthanks for the tutorial.

can you please sketch the wave function?.

KCErbHere’s some mathematica code which will produce the even excited states. You just have to set the first variable “num” to a number near the roots. I’ve put the first few in a comment

KCErbOne last try with proper html, perhaps the moderator could delete the other two?

growescienceThanks for the contributions. You can format code in wordpress.com by using their code tag. See http://en.support.wordpress.com/code/posting-source-code/

KCErbAnd just for good measure here are the odd solutions in the same coding style as the evens if you want to put them together:

KCErbAs long as I’m using the comments box to talk to the site admin (sorry . . .) I’d be happy to get you some nice jpgs of the graphs so that you could just make them a part of the webpage. I could even put in a piecewise function that superimposes the solutions on the well.

growescienceThanks for the offer of some graphs. I’d be happy to install them on the web page.

AnonymousThere’s

an extra factor of two in the denominator of the LHS of eq. 1.

growescienceFixed now. Thanks.

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anonymiscI can’t tell you how much your material has helped me. It’s amazing that you would take the time to explain this in such thorough detail but I’m glad you did. When I read Griffiths the details just fly over my head.

gwrowePost authorGlad you find it useful.

A desperate physics studentHi there, this might seem like a stupid question but I’m going to ask it anyways. For the 1st part of the problem where we’re looking for the even solutions, why is the wave function automatically continuous at x=0? And why would B not equal to 0 for the even solution? I mean if we plug in 0=-Asin(0)+Bcos(0), wouldn’t B=0?

gwrowePost authorFrom equations 4 and 5, so the wave function is automatically continuous, no matter what is.

For an even solution, (well, it

couldbe, but it doesn’t have to be and in this case it isn’t). For an odd solution, it’s true that we need , however.A desperate physics studentTy ty, that helped so much. One last question, where did the cos (kx) go from equation 9 to equation 10

gwrowePost authorcos 0 = 1 🙂