3. Schrödinger’s Wave Equation

1. 3. Schrödinger’s Wave Equation -the replacement for Newton’s Laws at the quantum scale It cannot be derived.. But it was a great guess!

2. We have seen how to make a wavepacket that represents a particle. Now we need some tools to explore the mechanics of that wavepacket.

3. Reminder about wave equations Newton’s law of motion applied to vertical motion of small section of string of length dx, between xA andxB: Vertical component of force mass x acceleration Leads to wave equation: where velocity e.g. Waves on a string Tension T, mass per unit length . T y T xA xB x

4. Assume solutions of the form: Insert into wave equation: So that: which gives: or i.e. velocity = frequency x wavelength Solutions of wave equation

5. Operators and Observables An operator, , operates on a wavefunction, , to produce an observable , as well as returning the wavefunction, unchanged. Momentum operator: Total energy operator:

6. Wave properties Momentum: p = £k Particle properties: Energy: E = £T We can “extract” the momentum out of the wavefunction by the mathematical operation: Similarly we can “extract” the energy by operating with: Schrödinger’s wave equation We have seen how to make a wave-packet out of plane waves like:

7. In “operator” form this becomes: Rearranging we getSchrödinger’s equation: (one dimensional version) Schrödinger’s equation contd….. Momentum operator: Total energy operator: Now the total energy of a particle is just the sum of the kinetic energy and the potential energy: E = p2/2m +V

8. Things to note re Schrödinger’s equation • The equation is linear in R, that is there are no terms likeR 2 or (MR/Mx)2.  • The equation is homogeneous, that is there are no terms independent of R. • Taken together these features mean that if R is a solution to the equation, then so is cR, where c is any complex number. • This implies that any linear combination of solutions is also a solution

9. Differences between Schrödinger and classical wave equations K.E. + P.E. = Total Energy Eqn. derived for total energy Eqn. derived from force First derivative wrt time Second derivative wrt time Complex (note i on RHS) Real equation: implies y must be complex y is a displacement  real

10. K.E. + P.E. = Total Energy Eqn. derived for total energy Eqn. derived from force First derivative wrt time Second derivative wrt time Complex (note i on RHS) Real equation: y is a implies y must be complex displacement  real Differences between Schrödinger and classical wave equations Reminder about complex numbers: C=A+iB with i.i = -1. Complex conjugate C*=A-iB. Square modulus |C|2= C*C |C|2 = (A-iB)(A+iB) = A2 + B2. Modulus |C| = A2 + B2

11. Interpretation of the wavefunction What does R(x, t) tell us? c.f. Young’s slit experiment: high probability of detecting particle at bright fringes, low probability at dark fringes • for light expect probability of detecting a photon to be proportional to the intensity of E.M. wave. • phase of R(x, t) cannot be important – not observable. Guess that the probability of finding a particle in the range x to x+ dx at time t is proportional to |R(x,t)|2dx: Define: P(x,t) dx.  |R(x,t)|2dx Need to normalise probabilities: (one particle!)

12. It follows that: If we normalise the wavefunction so that we can define P(x,t) = |R(x,t)|2 Normalisation of wavefunction P(x,t) dx.  |R(x,t)|2dx To get rid of proportional sign we need a constant of proportionality: P(x,t) dx. = A |R(x,t)|2dx. Eliminate A by using the normalisation:

13. Suppose time-independent potential, then we expect the total energy to be a constant. “Separation of variables”: Insert into Schrödinger equation: Divide through by : LHS RHS Only depends on x Only depends on t Time independent Schrödinger equation Time-dependent, Schrödinger equation:

14. RHS becomes: Integrating: LHS becomes: yhas a solution of the form: Probability density is independent of time: T.I.S.E. contd… The only way a function of x = a function of t, for all x,t is that each function is equal to a constant. Call the constant E (we will show later this is the total energy). This is called the time-independent Schrödinger equation

15. `Solving’ theSchrödinger equation: What do we want? Usually, the allowed energy levels. First: Assume a solution - eg u(x)=Asin(kx) (TISE) Second: Substitute solution into Schrödinger equation. That gives relationship between E (and V) and k Third:Use boundary conditions (eg u and du/dx continuous) to solve for allowed values of k (eg in terms of well size, a) Fourth:Use earlier relationship between E (and V) and k to obtain allowed values of E

16. Simple example: free particle, V(x)=0 We can make a wavepacket representing a free particle by adding together plane waves of the type In this case: if and E = £T E =£2k2 /2m This confirms that the constant Eis the total energy

17. Example of solution of T.I.S.E:Particle in a box: infinite potential well Box in one dimension with walls at –a and +a for | x | a, for | x | > a For | x | a, T.I.S.E becomes: Boundary condition: u(x) must vanish for | x | > a, since otherwise we would have an infinite potential energy!

18. Therefore it is a solution as long as: Types of solution (I) A possible solution is: u(x)=Asin(kx) Check: Insert into T.I.S.E.: Boundary condition: u(x) = 0 for | x | > a implies sin(kx) = 0 for | x | = a. True if ka=m (m integer)

19. Therefore it is a solution as long as: Types of solution (II) Another type of solution is: u(x)=Bcos(kx) Check: Insert into T.I.S.E.: Boundary condition: u(x) = 0 for | x | > a implies cos(kx) = 0 for | x | = a. True if ka=m /2 (m odd integer)

20. Summary of solutions: u(x)=Asin(kx) u(x)=Bcos(kx) ka=m (m integer) ka=n /2 (n odd integer) Equivalently: ka=n/2 (n eveninteger) n = 2, 4, 6, 8,….. n = 1, 3, 5, 7,…..

21. Energy levels for a particle in a box n = 1,2,3,4,5,6,7,8,…….  Solutions for energy are called energy eigenvalues

22. x/a x/a u6(x) u5(x) u3(x) u4(x) x/a x/a x/a x/a u1(x) u2(x) Wavefunctions for particle in a box n = 1, 3, 5, 7,….. u(x)=Bcos(n x/2a) n = 2, 4, 6, 8,….. u(x)=Asin(n x/2a ) Solutions for wavefunctions are called eigenfunctions

23. It follows that: . Similarly it can be shown that Normalised eigenfunctions: n =1, 3, 5,... n =2, 4, 6..... Values of A, B ? Normalisation Normalisation condition: In the present case we only need to integrate between –a and +a since u(x) vanishes outside this range:

24. |u4(x)|2 |u3(x)|2 x/a x/a |u2(x)|2 |u1(x)|2 x/a x/a Probability density For first four eigenfunctions for particle in a box

25. Must be an uncertainty in momentum Zero point energy The lowest energy state for a particle in a box is Why can’t the energy be zero? Remember Heisenberg uncertainty relation Particle is confined in box, so x ~ a. Since momentum cannot be zero, minimum energy must be of order

26. Consequences of Zero Point Energy - 1 Helium is the only element which remains liquid at absolute zero (unless under pressure). This is due to its large vibrational zero point energy. N.B. Zero point energy  (1/mass) of particle. Molar volume of the isotope 3He in the liquid phase is larger than that of liquid 4He since the light isotope has a larger zero point energy.

27. Consequences of Zero Point Energy - 2 Conduction electrons in a metal have very large zero point motion, with typical electron velocities of order 106 ms-1. When we pass a current through a metal we are imposing a very small drift velocity on top of this random motion. The zero point energy of a gas of “fermions” gives rise to “degeneracy pressure”. Both electrons and neutrons are fermions. Degeneracy pressure contributes to: (a) the bulk modulus of a metal - i.e. the resistance of a metal to being squashed; (b) the stability of white dwarf and neutron stars.

28. Consequences of Zero Point Energy – 3Degeneracy pressure in neutron stars and white dwarves Pressure, p = 1/3 nmc2 = 2/3n x 1/2mc2 i.e. p ~ E But E ~ a-2 Therefore p ~ a-2 i.e. ~ V-2/3

29. Spacing of energy levels • N2 molecule in a box of size 1cm cubed • Electron in a box of size 1nm cubed • c.f. Thermal energy at room temperature ~ 0.025eV

30. Quantum dots and nanocrystals Small semiconductor structures can be built to confine electrons into boxes of size 10 ~ 100nm. These behave like artificial atoms, with discrete energy levels. They have interesting potential for electro-optical devices and computer memory. Two samples of powdered CdSe, a semi-conductor, differing only in the particle size. Upper part: larger crystals, lower part smaller crystals. The colour difference is due to the difference in spacing of the energy levels of electrons trapped in the nanocrystals

31. Temporary Lecture Changes • Additional single lectures: • Tuesday 12th Feb, 13:00-14:00, Shackleton (B44) LTA (ie room 1041) • Wednesay 13th Feb, 10:00-11:00, Music (B2), LTH (ie room 2065) • No lecture on Thursday 14th Feb • or in week Monday 18th - Friday 22nd Feb. • Back to normal on Monday 25th Feb.

33. Differences between Schrödinger and classical wave equations- Complex exponential or sine/cosine form?

34. Classical wave equation Here either form works well as y is real, eg So: i.e. , thus We get exactly the same result with sines/cosines. (Generally, we can use complex form and take real part of the answer; makes treatment of phase easier.)

35. Schrödinger wave equation Complex equation Note i on RHS, so time dependence of ymust be complex, – MUST use complex notation