1. Physics 3 for Electrical Engineering Ben Gurion University of the Negev www.bgu.ac.il/atomchipwww.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenterwww.bgu.ac.il/nanocenter Lecturers: Daniel Rohrlich, Ron Folman Teaching Assistants: Daniel Ariad, Barukh Dolgin Week 5. Quantum mechanics – Schrödinger’s equation superposition principle time-independent Schrödinger equation continuity of Ψ and its derivative probability interpretation of the wave function Ψ normalization of Ψ Sources: Feynman Lectures I, Chap. 37; Merzbacher (2 nd edition) Chap. 4; Merzbacher (3 rd edition) Chap. 3 Sects. 1-3, 5; פרקים בפיסיקה מודרנית, יחידות 6,7

2. Schrödinger’s equation In 1926, Heisenberg was 25. Schrödinger was 14 years older, and more reasonable. In that year, each of them invented a new mechanics. Schrödinger’s new mechanics was an equation for a continuous wave. Heisenberg’s new mechanics was a matrix algebra – full of “quantum jumps”. Yet they made exactly the same predictions! We begin with Schrödinger’s equation. In a few weeks, we will see how Heisenberg’s matrix algebra jumps out of it.

3. Schrödinger made a very reasonable analogy: Fundamentally, there are no “light rays”. The rule that “the angle of incidence equals the angle of reflection” is only an approximation. So is Snell’s law. Rules about focal length are fictions used by lens-makers. All these “light rays” are actually electromagnetic waves.

4. Schrödinger made a very reasonable analogy. Fundamentally, there are no “light rays”. The rule that “the angle of incidence equals the angle of reflection” is only an approximation. So is Snell’s law. Rules about focal length are fictions used by lens-makers. All these “light rays” are actually electromagnetic waves. So Schrödinger assumed that paths of material objects – trajectories and orbits – are approximations to “matter waves”. He asked, “What equation do these matter waves obey?”

5. Schrödinger first tried to find a relativistic equation for matter waves, but he failed. (Paul Dirac succeeded two years later.) Then he tried to find a non-relativistic equation.

6. Each physical system has its own version of Schrödinger’s equation!

7. Schrödinger first tried to find a relativistic equation for matter waves, but he failed. (Paul Dirac succeeded two years later.) Then he tried to find a non-relativistic equation. Each physical system has its own version of Schrödinger’s equation! Here is Schrödinger’s equation for a one-dimensional particle of mass m in a potential V(x): where we define

8. We don’t know what this equation means…but, as we shall soon see, neither did Schrödinger! So let’s explore: Assume V(x) = 0 (i.e. the particle is free) and solve for Ψ : If Ψ is of the form Ψ(x,t) = X(x) T(t), then substitution yields

9. We don’t know what this equation means…but, as we shall soon see, neither did Schrödinger! So let’s explore: Assume V(x) = 0 (i.e. the particle is free) and solve for Ψ : If Ψ is of the form Ψ(x,t) = X(x) T(t), then substitution yields

10. Try and which together imply. We recognize the Planck-Einstein law,, and de Broglie’s law,. So the constant is.

11. We found solutions to Schrödinger’s equation of the form, where. These are not all the solutions, but we obtain all the solutions by “superposing” these solutions:, where A(k) is any complex function of k and.

12. Superposition principle If Ψ 1 (x,t) and Ψ 2 (x,t) are solutions of Schrödinger’s equation, then αΨ 1 (x,t) + βΨ 2 (x,t) is another solution of Schrödinger’s equation, for any complex numbers α and β. The superposition principle follows directly from the fact that Schrödinger’s equation is always linear in Ψ:

13. Superposition principle Namely, if and then we multiply the first equation by α and the second equation by β and sum them to get the equation that satisfies.

14. Now let’s try to solve Schrödinger’s equation for V(x) ≠ 0 using the same method of separating out the dependence on t. We already know that we can write and if we substitute this Ψ(x,t) into Schrödinger’s equation, we obtain an equation for ψ(x):

15. Since we write which is called the time-independent Schrödinger equation.

16. Since we write which is called the time-independent Schrödinger equation. Question: Do solutions of the time-independent Schrödinger equation obey the superposition principle? Suppose ψ 1 (x) is a solution for E 1 and ψ 2 (x) is a solution for E 2. Does it matter if E 1 = E 2 ? What about the solutions ?

17. Time-independent Schrödinger equation Let’s try to solve the time-independent Schrödinger equation for a simple finite “square” potential well V(x): x V(x)V(x) V0V0 0 L/2 −L/2

18. Time-independent Schrödinger equation Inside: the solution is a linear combination of e ikx and e −ikx or, equivalently, of cos kx and sin kx, where k = x V(x)V(x) V0V0 0 L/2 −L/2 outside inside

19. Time-independent Schrödinger equation Outside: the solution is a linear combination of e ik′x and e −ik′x if E ≥ V 0, or a linear combination of e k′x and e −k′x if E ≤ V 0, where k′ = x V(x)V(x) V0V0 0 L/2 −L/2 outside inside

20. Continuity of ψ(x) and its derivative What about at x = ± L/2? The left side of the equation, Eψ(x) or (E − V 0 ) ψ(x), is finite; but the right side is not finite unless both ψ(x) and its derivative dψ(x)/dx are both continuous. Therefore, the solutions of the time-independent Schrödinger equation must be continuous with continuous first derivative. x V(x)V(x) V0V0 0 L/2 −L/2 outside inside

21. Continuity of ψ(x) and its derivative Let’s guess that because of the symmetry of V(x), the solutions ψ(x) are either symmetric or anti-symmetric. (Afterwards, we will check that this guess is correct.) We’ll save time because we’ll have to match the inside and outside solutions only at one point (let’s say, at x = L/2). x V(x)V(x) V0V0 0 L/2 −L/2 outside inside

22. Probability interpretation of ψ(x) But before we continue, we note that Schrödinger’s idea of a “matter wave” for electrons turned out to be totally unworkable. Just think about the electric charge of an electron: if an electron is really a matter wave, then we should be able to find bits of its charge in different places. No such bits of charge have ever been seen! On the contrary, the charge of the electron is, as far as we know, indivisible. Consider a two-slit interference experiment with electrons. If we look at the charges that hit the final screen, we always find whole electrons with charge −e, never parts of an electron.

23. Probability interpretation of ψ(x) But before we continue, we note that Schrödinger’s idea of a “matter wave” for electrons turned out to be totally unworkable. Just think about the electric charge of an electron: if an electron is really a matter wave, then we should be able to find bits of its charge in different places. No such bits of charge have ever been seen! On the contrary, the charge of the electron is, as far as we know, indivisible. Indeed, a few months after Schrödinger published his equation, Max Born gave ψ(x) and Ψ(x,t) a new interpretation: they are probability waves, and |ψ(x)| 2 = [ψ(x)]*ψ(x) is the probability density to find the electron at the point x. Likewise, |Ψ(x,t)| 2 = [Ψ(x,t)]* Ψ(x,t) is the probability density to find the electron at the point x at time t. This interpretation is still accepted today.

24. Normalization As long as electrons don’t simply disappear (conservation of charge) the probability to find the electron somewhere must be 1. The sum of all probabilities must be 1. In the case of the continuous variable x, the sum is an integral, so we have the mathematical requirement This mathematical requirement is called normalization.

25. Let’s return now to the time-independent Schrödinger equation for the finite well and focus on the case E < V 0. For this case we concluded that the outside solutions must be a linear combination of e k′x and e −k′x, where k′ = But normalization requires that the outside solution can only be e −k′|x|. x V0V0 0 L/2 −L/2 outside inside

26. We considered two types of solutions to the time-independent Schrödinger equation: symmetric and anti-symmetric. For the symmetric solution, the continuity conditions at x = L/2 are x V0V0 0 L/2 −L/2 outside inside

27. Therefore, the continuity conditions for the symmetric solution imply k tan kL/2 = k ′, which in turn implies discrete energy levels! From the definitions of k and k′ we have x V0V0 0 L/2 −L/2 outside inside and we can solve this equation graphically by setting each side to a variable y 2 ; the solutions are the intersections of the two curves y = k tan kL/2 and y 2 + k 2 =.

28. y = k tan kL/2 y 2 + k 2 = 2 x outside 0 L/2 −L/2 1 kL/2 2 outside V0V0 inside 1

29. For the anti-symmetric solution, the continuity conditions at x = L/2 are x V0V0 0 L/2 −L/2 outside inside

30. Therefore, the continuity conditions for the symmetric solution imply k cot kL/2 = –k′. From the definitions of k and k′ we have x V0V0 0 L/2 −L/2 outside inside and we can solve this equation graphically by setting each side to a variable y 2 ; the solutions are the intersections of the two curves y = k cot kL/2 and y 2 + k 2 =.

31. x 0 L/2 outside y = k cot kL/2 y 2 + k 2 = 2m −L/2 outside inside V0V0 kL/2 2 1 2 1

32. The first three energy levels: the ground state is symmetric, the next state is anti-symmetric, the next state is symmetric, etc. First excited state Ground state 0 L/2 −L/2

33. The first three energy levels: the ground state is symmetric, the next state is anti-symmetric, the next state is symmetric, etc. First excited state Ground state 0 L/2 −L/2 Classically forbidden region

34. The first three energy levels: the ground state is symmetric, the next state is anti-symmetric, the next state is symmetric, etc. As long as V 0 is finite, there are finitely many energy levels. For V 0 infinite (infinite “square” well), there are infinitely many levels, and ψ(x) vanishes outside the well (|x| > L/2, classically forbidden region). The derivative of ψ(x) is discontinuous at x = L/2 but this exception to the continuity rule is due to infinite V 0. The first three normalized wave functions and energies are

35. Conclusions: 1. There is no physical system exactly like an electron in a one-dimensional square well potential. But it is a model we can solve, and we learn from it some features of quantum mechanics and quantum behavior. We can also use it as an approximate model, e.g. of an electron in a wire segment. 2. We learn that energy levels can be quantized, i.e. energies can take discrete values. 3. We saw that symmetries of the model (here, parity) show up in the symmetries of the solutions. 4. We saw a unique quantum effect: quantum electrons can penetrate regions that are forbidden to classical electrons!