Question

Sorry, there is quite a bit to this problem, I got #1, but I'm having trouble with #2.

#1

A) Write down the Time Dependent Schrodinger equation for an electron trapped in a one-dimensional infinite potential well of 130 pm wide.

B) What is the potential energy function in this case?

C) What are the first four stationary state energies?

D) What are the first four time dependent stationary state wave functions?

E) What are the first four stationary state probability distributions for the electron?

F) What frequency photon will excite an electron, trapped in this well in its ground state, to jump up to the fourst stationary state?

#2

Using the information from #2:

A) What is the average position of the electron in its ground state?

B) What is the uncertainty of its position in this state?

C) What is the minimum uncertainty of its momentum in this state? (Use the exact form of the Heisenberg Uncertainty Relationships)

Answer 1

Solutions to the Schroedinger equation for a free particle of mass m, (-?²/(2m))?²?(x,t)/?x² = i???(x,t)/?t, are the plane waves ?(x,t) = exp(i(kx - ?t)), and wave packets build from those plane waves. |?(x,t)|² represents the probability per unit length of finding the particle at a time t at position x.

How do we make predictions about a particle that is not free, but is confined?

Stationary states

The Schroedinger equation for a particle moving in one dimension through a region where its potential energy is a function of position has the form

(-?²/(2m))?²?(x,t)/?x² + U(x)?(x,t) = i???(x,t)/?t.

We are often interested in finding the eigenstates of the energy operator i??/?t, i.e. we are interested in finding the wave functions of particles whose energy can be predicted with certainty. For an eigenfunction of the energy operator we have

i???(x,t)/?t = E?(x,t),    or    ??(x,t)/?t = (-iE/?)?(x,t).

To within a constant the function ?(x,t) is equal to its own derivative with respect to t. The exponential function behaves this way.  
Recall that d(exp(at))/dt = a exp(at). Also recall that we are taking a partial derivative with respect to t, and, as far as this derivative is concerned, x is treated like a constant.

Therefore the solution to the above equation is

?(x,t) = ?(x) exp(-iEt/?).

For eigenfunctions of the energy operator i??/?t the Schroedinger equation becomes time independent.

(-?²/(2m))?²?(x)/?x² + U(x)?(x) = E?(x).

The operator   (-?²/(2m))?²/?x² + U(x) is called the Hamiltonian operator H, and the time-independent Schroedinger equation is often abbreviated as

H?(x) = E?(x).

If we find the possible solutions ?(x) of the time-independent Schroedinger equation for a particular potential energy function U(x), then we can obtain the corresponding wave functions ?(x,t) by just multiplying ?(x) by exp(-iEt/?), where E is the appropriate eigenvalue for each solution.

The wave functions of particles whose energy can be predicted with certainty are of the form ?(x,t) = ?(x) exp(-iEt/?).

The probability density is

|?(x,t)|² = ?(x) exp(-iEt/?) ?^*(x) exp(iEt/?) = ?(x)?^*(x) = |?(x)|² .

The probability of finding the particle at a particular position x for a particle with well defined energy is independent of time. It does not change with time. The wave functions of particles with well defined energy are therefore often called stationary states.

Energy and position are incompatible measurements. Knowing the energy of a particle prevents us from knowing its position and tracking it. A stationary-state wave function contains no information about the position of the particles as a function of time. All we get is probabilities. We can certainly make a measurement and determine the position of a particle at a particular time, but then we loose the information about its energy.

The time-independent Schroedinger equation involves only real variables and can have real solutions. The complex part of the solution ?(x,t) is then contained in the function exp(-iEt/?).

Assume we want to solve the Schroedinger equation (-?²/(2m))?²?(x)/?x² + U(x)?(x) = E?(x).
Dividing both sides of the equation by -?²/(2m) and defining k₁² = 2mE/?², k₀(x)² = 2mU(x)/?², and k(x)² = k₁² - k₀(x)² we can simplify the notation.

(-?²/(2m))?²?(x)/?x² + U(x)?(x) = E?(x).

?²?(x)/?x² - (2mU(x)/?²)?(x) + (2mE/?²)?(x) = 0.

?²?(x)/?x² - k₀(x)²?(x)) + k₁²?(x) = 0.

?²?(x)/?x² + (k₁² - k₀(x)²)?(x) = 0,
or
?²?(x)/?x² + k(x)²?(x) = 0,

Let us solve this equation for the "infinite square well". We assume U(x) = 0 for x = 0 to L, and U(x) = infinite everywhere else. A particle cannot penetrate a region with infinite potential energy, there is no chance that we can find it there, and its wave function in that region is zero. We put the particle in a one-dimensional box, out of which it has no chance of escaping.

In the region from x = 0 to x = L the potential energy U(x) = 0. The particle can freely move inside the box. Therefore k₀(x) = 0 and k(x)² = k₁². Possible wave functions for the particle must satisfy the equation

?²?(x)/?x² + k²?(x) = 0,

and they must be zero at x = 0 and x = L, because the wave function must be continuous and it is zero outside the region from x = 0 to x = L.

Real solutions of the Schroedinger equation which are zero at x = 0 are ?(x) = Asin(kx), since ?²sin(kx)/?x² = -k²sin(kx) and sin(0) = 0. For these solutions to be zero at x = L we need

sin(kL) = 0, kL = n?, with n = 1, 2, 3, ...,

since the sine function is zero if its argument is an integer multiple of ?.

The possible values of k are k_n = n?/L, the possible values of the energy E_n = h²k_n²/(2m) are E_n = n²?²?²/(2mL²). The potential and the first five possible energy levels a particle can occupy are shown in the figure below. Arbitrary units are used.

The energy of the particle in the infinite square well is quantized. It can only take on the values E_n = n²?²?²/(2mL²). If we measure the energy we can only measure one of these eigenvalues, E_n = n²?²?²/(2mL²), n = 1, 2, 3, ... .  The confinement of the particle leads to energy quantization.
If we measure E_n, then right after the measurement the wave function of the particle is ?_n(x, t) = A_nsin(n?x/L)exp(-iE_nt/?).
For example if we measure the energy to be E₃ = 9?²?²/(2mL²), then right after the measurement the wave function of the particle is
?₃(x,t) = A₃sin(3?x/L)exp(-iE₃t/?).

A measurement changes the information an observer has about the system and therefore can change the wave function of the system.

The square of the normalized wave function |?_n(x,t)|² = |?_n(x)|² = A_n²sin²(n?x/L) is equal to the probability per unit length of finding the particle with energy E_n at position x. To normalize the wave function we have to choose A_n² = 2/L. Then ?_-?^+?|?(x,t)|²dx = 1, and the total probability of finding the particle inside the well is 1.

Example:

An electron (m = 9.109*10^-31 kg) is confined in a one-dimensional infinite square well of width L = 10 nm.   The figure below shows

• (a) the lowest five energy levels E_n and the wave functions ?_n(x) = (2/L)^1/2sin(n?x/L)

• (b) the corresponding probability density functions |?_n(x)|² = (2/L)sin²(n?x/L)

The wave functions and the probability density functions have an arbitrary magnitude (i.e. they are not normalized) and are shifted by the corresponding electron energy

The lowest possible energy of the confined particle is its ground-state energy. All other possible energies are are excited-state energies. A particle can make transitions between different energy levels when it is interacting with its environment. If a particle makes a transition from a lower energy level to a higher energy level, it has to absorb an amount of energy ?E = E_high - E_low from the outside world. It can absorb a photon with energy hf = ?E, or it can receive the required energy from another particle in a collision. If the particle makes a transition from a higher energy level to a lower energy level, it has to release an amount of energy ?E = E_high - E_low to the outside world. One way to do this is to emit a photon of energy hf = ?E.

Because the energy levels are quantized, the energies of photons involved in transitions are quantized. The particle in the well can only interact with photons of certain energies, it can only emit or absorb light of certain colors.

Electrons in atoms are not confined in a square well, but they are confined.  This confinement leads to energy quantization. If we measure the energy of an electron in an atom, we can only measure one of a set of discrete values. The energies of the electrons in atoms are quantized. Atomic spectra reveal this quantization.

Light from a hot filament contains all the colors of the rainbow. But light given off or absorbed by individual atoms consists of a unique set of colors.

Another famous experiment that reveals energy quantization in atoms is the Franck