The One-Dimensional Infinite Square W ell defined over from x = 0 to x = a, so ‘width’ of well is a potential V(x) is even about center of well x = a/2  evenness/oddness of wavefunctions about x = a/2 theme: if potential is even, lowest energy wavefunction (ground state) will be even too.. then O, E… this follows because the second derivative of a function has the same evenness/oddness as the function itself inside the well V = 0: free particle wavefunctions work outside the well V = ∞: the wavefunction must be zero the monstrous (INFINITE) jump in V at the well edges [x = 0, x = a] implies that the wavefunction’s slope is not continuous at those places the wavefunction must always be continuous, no matter what V(x)!!

Wavefunction inside the 1d ∞ Square Well rightward-moving and leftward-moving free wavicles have the same energy E, so a linear combination of them also solves the TISE with that same E choose a linear combination that satisfies the boundary conditions  (0) =  (a) = 0

Even or Odd about a/2? these are even solutions about well middle (includes ground state) they are labeled, however, by odd integers n = 1,3,5… These are odd solutions about well middle (first excited state…) (labeled by even integers) lower sign (–) let A L = A/2i upper sign (+) Let A L = A/2 Both:

Probability densities for 1d∞SW E 3 = 9 E 1 E 2 = 4 E 1 the wavefunctions are standing waves, so regardable as the sum of two oppositely-traveling wavicles with equal amplitudes wavefunctions are sinusoidal, so probability densities are sine- squared functions with integral number of cycles: area underneath is half the maximum times the width. NORMALIZE: typical value for a = 1 nm: E 1 ≈ (50 x )/(10)( )( ) ≈ 5 x J ≈ 1/3 eV [since 1 eV = x J] Hydrogen atom is about 20 times smaller (.53 Å)  E = – 13.6 eV a E 1 = h 2 /8ma 2

Things to note regarding this example walls are ‘infinitely high’ and correspond to impenetrable infinite jump in V causes a ‘kink’ in  – unusual! energy spectrum climbing like n 2 is the MOST rapid climb it can possibly be for ANY well since well ‘sides’ are ∞-ly steep! crucial helpful fact: eigenfunctions are ‘orthonormal’:

Expectation values and uncertainties I Expectation value for position should be middle of the well this tells us very little about ‘where’ the particle is requires two integrations by parts; result is this is evidently a measure of the typical ‘distance’from the center, and grows for larger n; in the limit,  a 2 /3

Expectation values and uncertainties II uncertainty in position is  x = ( – 2 ) 1/2 turning to the momentum expectations and uncertainty ground state has lowest product, which still exceeds minimum uncertainty, and the product grows with n begs the question of how to achieve minimum uncertainty

Using these orthonormal wavefunctions as building blocks: Fourier Series wavefunctions {  n }, are orthonormal, as we have seen they are complete: any odd [f(–x) = – f(x); f(0) = 0] periodic [f(x + 2a) = f(x), so period is 2a] (this is not the most general situation – stay tuned) function can be expressed as a linear combination Fourier’s Trick: to get the {c m }, multiply expansion of f(x) by  m *(x) and integrate term by term from 0 to a:

More general remarks about the Fourier series for any [odd or even or neither] periodic [f(x + 2a) = f(x), so period is 2a] function can be expressed as a linear combination conventionally, the limits are now taken to be from –a to a so the new normalization is 1/√a both cosines and sines must be included Fourier’s Trick is now very similar—the cosines and the constant are both even contributions, so