1. from last time: 1. show that is also a solution of the SE for the SHO, and find the energy for this state 2. Sketch the probability distribution for the SHO wavefunctions, AND the probability distribution for a classical oscillator on the same axes. 3. Ponder this: how can a particle in the n=2 state get from one side of the well to the other?

2. 2. Sketch the probability distribution for the SHO wavefunctions, AND the probability distribution for a classical oscillator on the same axes. U x 0 quantum P=  2 Classical P  v -1

3. quantum physics 5: more solutions to The Schroedinger Equation: finite potentials

4. Solving the Schroedinger equation - a recipe 1.Start by writing down the S.E. with the appropriate potential energy, e.g. for the S.H. Oscillator U = ½ kx 2. If U is not cts you may need to write down the S.E. for each distinct region where there is a different U. 2.Find a wave function which is a solution to the S.E. – this is often done by educated guessing, and there may be more than one solution. 3.Apply boundary conditions – these will often limit your values of energy. 4.Evaluate any undetermined constants (like amplitudes), e.g. by using boundary conditions, applying normalisation. 5.Check your solution, if it gives you something dodgy like P= , check for errors or start again.

5. 1D finite potential well This time consider a particle, e.g. an electron, which is confined to an finite square potential well, in 1D. This is more realistic for most situations. We can write the potential energy as: In this case our potential as finite outside the “well” so the electron has a finite probability of escaping! x 0 L U

6. We’re working with the time independent form again: and our wave function will be of the form: inside the well : but now we can’t assume B equal to zero, because the well is finite! or

7. We’re only going to consider the U > E case, because E>U is not a confined particle anyway We have two regions to worry about, to the left and the right, so lets write the two possible solutions: outside the well : or

8. Now we use the requirements that the wave function is well behaved – it must be continuous across boundaries, and so must its derivative w.r.t x… at x = 0: therefore but as x  - , the solution outside the well  unless D = 0 if the wave function is to be normalisable. so on the RHS of the well, to prevent the wave function  we require F = 0

9. then you do LOTS of algebra, using the two conditions of continuity at the boundaries, to solve for A, B, D and then to get the energy – homework problem, or next year! The main results are: -energy is again quantised inside the well -outside the well the wavefunction has an exponentially decreasing form BUT its not zero! The particle can escape even though E < U! -inside the well its standing waves again

10. Our wave functions and probability densities look like this:  P=  2 x 0 L U tunelling!

11. barriers and tunneling Imagine what would happen if we had a barrier… Classically, a particle with less energy than the barrier height could not pass the barrier. But a wave-like particle CAN, because its wavefunction extends some distance into the barrier, and beyond it. x 0 L U

12. To either side we have a “typical” wavefunction for a free particle. Note that it is complex, and could be written in terms of sin, cos instead. Inside the barrier the wavefunction is exponentially decreasing, and real. Also note that the wave vector is the same on each side. x 0 L U real!! complex!!

13. probability of tunelling Within the barrier the wavefunction is of the form: where: Recall that the probability is: So the probability of transmission: - decreases with U, m - decreases with x, i.e. barrier width - increases with C, E

14. barriers, wells, atoms and conductivity… We can consider a solid to be a collection of potential wells with barriers between them. Depending on the depth of the wells, and the wavefunctions for the electrons bound to the atoms, the material will have more or less conductivity. Quantum mechanics is the basis of solid state physics, which is the basis of our understanding of semiconductors, which is the basis of all electronics and computers! so lets look at atoms now…

15. Newton C17 Light as particle mechanics Maxwell C19 Light as wave Greek Atomic theory,C5 BC Democritus etc Rutherford then Bohr, orbitals Thompson and others, Pudding models Einstein’s solution C20 Light as particle de Broglie particles as waves Quantum physics Davisson and Germer’s confirmation Heisenberg and Pauli uncertainty, exclusion Technology Problems – spectra and Schrodinger equation Technology Problems – PE effect, Black bodies Historical development

16. So far we’ve followed the wave particle and photon side, mostly. Now lets have a look at atoms, now that we have the SE to help us understand them as systems with a potential energy well which holds electrons (Fermions) to a confined space…

17. Atoms What we knew ca. 1920: 1.Atoms are stable – they don’t (usually) fall apart. 2.Atoms are very small. 3.Atoms have electrons in them, but are electrically neutral. 4.Atoms emit and absorb radiation of discrete wavelengths. The big problem with most models was number 4 – how to account for discrete energy changes. We’re going to start with Bohr’s model because it was the first to really explain this. Sort of.

18. Bohr’s model of the atom Bohr took what was known about atoms, and made some postulates, and came up with a model. The postulates: 1.an electron in an atom moves in a circular orbit about the nucleus under the influence of the Coulomb attraction. 2.It is only possible for an electron to move in an orbit for which its orbital angular momentum is quantised as L =nħ. 3.In spite of its constant acceleration E remains constant and the electron does not radiate EM radiation and collapse. 4.EM radiation is emitted if an electron discontinuously changes from one orbital to another. The frequency of EM radiation emitted is given by  E electron =hf.

19. The model: electrons move in orbits in which the centripetal force due to the coulomb attraction keeps them in a stable circular orbital, like gravity keeping planets in orbit. We only consider a one electron atom so we don’t have to worry about interactions of electrons. So the KE is ½ the PE, and the total energy is:

20. quantising the energy by quantising L: Only certain orbits are allowed, because angular momentum is quantised: hence which means the energy is: so from our force equation we can write divide one by the other to get r without v:

21. pops out of Bohr’s model as a result of quantisation of angular momentum. This is good, because it matches the experimental observation that spectral lines from hydrogen could be fitted by the equation: and this is good stuff, and the model predicted the spectrum of hydrogen, but there wasn’t any a priori justification for the postulates…

22. for next time: Readings: T4: 37.3 – 37.4T&M5: 36.3 – 36.4 PLUS look at hyperphysics on SE in 3D and atoms Examples to do: T4: 37.1, 37.2T&M5: 36.1, 36.2 Homework problem: Calculate the first Bohr radius for: a.a hydrogen atom b.a doubly ionised Lithium atom