1. 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
e.g. Waves on a string
Tension T, mass per unit length . T y T xA xB x
Newton’s law of motion applied to vertical motion of small section of string of length dx, between xA and xB :
Vertical component of force mass x acceleration
If theta is angle between horizontal and string, upperward force = Tsin(theta) = T y/x
Leads to wave equation:
where velocity 3

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

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. Schrödinger’s wave equation
We have seen how to make a wave- packet out of plane waves like:
Wave properties
Energy: E =  
Momentum: p = k
Particle properties:
We can “extract” the momentum out of the wavefunction by the mathematical operation:
Similarly we can “extract” the energy by operating with:
Discuss operators, eg for momentum or energy. The operator operates on the wavefunction to produce the observable – see end of waves course 6

7. 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
In “operator” form this becomes:
Rearranging we get Schrödinger’s equation:
(one dimensional version)

8. Things to note re Schrödinger’s equation
The equation is linear in , that is there are no terms like  2 or ( /x)2. 
The equation is homogeneous, that is there are no terms independent of .
Taken together these features mean that if  is a solution to the equation, then so is c, 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 must be complex y is a displacement  real

10. Differences between Schrödinger and classical wave equations
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 must be complex displacement  real
K.E P.E. = Total Energy
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 (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 (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 |(x,t)|2 dx:
Define: P(x,t) dx.  |(x,t)|2 dx
Need to normalise probabilities:
(one particle!)

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

13. Time independent Schrödinger equation
Time-dependent, Schrödinger equation:
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

14. T.I.S.E. contd… has a solution of the form:
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).
RHS becomes:
Integrating:
LHS becomes:
This is called the time-independent Schrödinger equation
has a solution of the form:
Probability density is independent of time:

15. `Solving’ the Schrödinger 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 =  
E =  2k2 /2m
This confirms that the constant E is the total energy

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

19. Types of solution (II) Another type of solution is: u(x)=Bcos(kx)
Check:
Insert into T.I.S.E.:
Therefore it is a solution as long as:
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 even integer)
n = 2, 4, 6, 8,…..
n = 1, 3, 5, 7,…..
So with combination of cos and sin solutions, every value of n is covered by bottom formula 20

21. Energy levels for a particle in a box
Solutions for energy are called energy eigenvalues

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

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

24. Probability density
|u1(x)|2
|u2(x)|2
|u3(x)|2
|u4(x)|2
x/a
..ie where are you most likely to find the particle
For first four eigenfunctions for particle in a box 24

25. 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.
Must be an uncertainty in momentum
Since momentum cannot be zero, minimum energy must be of order
Stop at the question. Note that Emin doesn’t have to be exactly same as E1 – it is an uncertainty relationship 25

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.
Larger by factor 4/3 corresponding to larger zero point motion (what about hydrogen…?) – The confining dimension a is also relevant 26

27. Consequences of Zero Point Energy - 2
Conduction electrons in a metal have very large zero point motion, with typical electron velocities of order 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.
Read Section 3.4.3 27

28. Consequences of Zero Point Energy – 3 Degeneracy 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
Read Section 3.4.3 28

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 must be complex,
– MUST use complex notation