To understand the nature of solutions, compare energy to potential at  Classically, there are two types of solutions to these equations Bound States are when E < V(  ). Unbound states are when E > V(  ) Quantum mechanically, there are bound and unbound states as well, with the same criteria Bound states are a little easier to understand, so we’ll do these first E > V(  ) E < V(  ) V(x)V(x) Bound and Unbound States

Wave Function Constraints 1.The wave function  (x) must be continuous 2.Its derivative must exist everywhere 3.Its derivative must be continuous* 4.Its second derivative must exist everywhere* 5.The wave function must be properly normalized † Must not blow up at  *Except where V(x) is infinite †There are exceptions and ways around this problem Normalization What if the wave function is not properly normalized? If the integral is finite, multiply by a constant to fix it Modified wave satisfies Schrödinger If the integral is infinite, it gets complicated For bound states, this is still trouble For unbound states, it is okay

The 1D infinite square well (1) Outside of the well, the wave function must vanish In remaining region, we need to solve differential equation What functions are minus their second derivative? We prefer real Boundary Conditions Must vanish at x = 0 And at x = L Where does sin vanish? Don’t worry about derivative because V(x) blows up there V(x)V(x)

The 1D infinite square well (2) mL 2 E/  2 Energy Diagram n = 1 n = 2 n = 3 n = 4

1D Infinite Square Well (3) BUT WAIT: What about normalization? To fix it: multiply by  (2/L). The most general solution is superposition of this solution

The Finite Square Well (1) We need to solve equation in all three regions; this takes work In region II, we get solutions like before No longer necessary that it vanish at the boundaries In regions I and III, we solve a different equation: If we are looking at bound state (E < V 0 ), in these regions, we get exponentials I IIIII

The Finite Square Well (2) Wave function must not blow up at  Wave function must be continuous at x =  ½L I IIIII Derivative of wave function must be continuous at x =  ½L

The Finite Square Well (2) Wave function must not blow up at  Wave function must be continuous at x =  ½L I IIIII Derivative of wave function must be continuous at x =  ½L

The Finite Square Well (3) Wave function penetrates into “forbidden” region Oscillates when E > V(x) Damps when E < V(x) Energies decreased slightly compared to infinite square well Finite number of bound states Due to finite extension and depth of potential well Energy Diagram Infinite Well n = 4 n = 3 n = 2 n = 1 Solve all these equations simultaneously Normalize the final wave function

The Harmonic Oscillator (1) At large x, the behavior is governed (mostly) by the x 2 term Now we guess Don’t want it blowing up at infinity! Our strategy: Check that this works Find more solutions and check them

The Harmonic Oscillator (2)

n = 3 n = 2 n = 1 n = 0 The Harmonic Oscillator (3) We still need to normalize it Expect other solutions to have similar behavior, at least at large x We will guess the nature of these solutions Multiply the wave function above by an arbitrary polynomial P(x) Substitute in and see if it works This will give you E n