1. The One-Dimensional Finite Square W ell defined over from x = –a to x = a, so ‘width’ of well is 2a [different definition from Griffiths 1d∞SW  evenness/oddness] also put well bottom at zero energy: different from Griffiths too there will be at least one (even) ground state, but only a finite number of bound states for the moment, assume E 0 let’s go after the even solution(s): the constancy of V  cosine inside and exponential outside well the modest (FINITE) jump in V at the well edges [x = –a, x = a] implies that the wavefunction’s slope is continuous at those places the wavefunction must always be continuous

2. One-Dimensional Finite Square W ell boundary condition on  (x): finite at x = ±∞  A = 0 evenness of solution  A > = B < := A continuous at x = –a   < (–a) =  E (–a)  Ae –  a = Dcos(ka) continuous at x = a  same thing due to evenness of solution!! boundary condition on d  /dx: continuous at x = a  same thing boundary condition on d  /dx: continuous at x = –a divide these two equations and multiply by a: new notation:

3. think of the L and R sides as functions of z L side: ordinary tan or cot function times z, which looks about the same (except for first odd case) R side: a circle of radius z 0 centered at the origin where the two graphs cross are the solutions Graphical solution to One-Dimensional Finite Square W ell here is case z 0 ≈ 4.5 E E O z

4. Some illuminating limits and special cases if V 0 is huge (V 0 >>E ) the situation is essentially the ∞SW, and z 0 is very large too: the ‘circle’ is far up, and the intersections of it with the tan/cot functions occur at z ≈ n  if z 0 ≤  /2, there is only one solution (the even ground state)  if the well is very (z 0 <<  /2), shallow, then the sole bound state occurs for very small z the condition becomes, inserting the Taylor expansion for tan z (which is actually overkill but worth doing anyway…) Thus the sole ground state has floated to the very top of the well

5. it’s not surprising that the ground state is less than halfway up Estimating the energy for a special case in particular, for z 0 =  /2, we have almost the lowest odd state too obviously E 2 = V 0 but what is E 1 ??

6. new physics: a continuum of free (or scattering) states the wavenumber will be position-dependent in an obvious way a different way of thinking: allow the wavicle to enter from the left, and be both reflected and transmitted by the well therefore, to the left of the well and inside the well we have both L and R waves, but to the right of the well only an R wave Consider the case E > V 0

7. Impose boundary conditions and get A > in terms of A < 4 equations and 5 unknowns: express the other 4 in terms of the ‘outgoing’ amplitude A >  multiply (II) by sin(ℓa) and (IV) by cos(ℓa) and add  C; do visa versa and subtract  D… analogously, use (I) and (III) and double angle formulas to get nice result for outgoing amplitude in terms of incoming amplitude

8. wavenumber is the same to left and to right of the well, so no worries about differing wave speeds (affects probability flux) transmission probability is (amplitude ratio) 2 ; reciprocal is tidier From this result, get the real-valued transmission probability we get note

9. Graph of transmission probability as a function of E/V 0 graph is a little off: T := 0 right up to E/V 0 = 1 T becomes unity when there are an integral number of half- wavelengths insides the well’s width; to get T = 1 

10. certain values of E for which the transmission is perfect! Implications of this result reflection coefficient for reference only, we get (don’t copy this down!)