Question

Use the quantum particle wavefunctions for the kinetic energy levels in a one
dimensional box to qualitatively demonstrate that the classical probability distribution
(any value of x is equally allowed) is obtained for particles at high temperatures.

Answer 1

The time-independent Schrodinger equation for a particle of mass m moving in one direction with energy E is

          ??22md2?(x)dx2+V(x)?(x)=E?(x)                              (1)

     ? is the reduced Planck Constant where ?=h2?

· m is the mass of the particle

· ?(x)

• is the stationary time-independent wavefunction
• V(x) is the potential energy as a function of position
• E is the energy, a real number

This equation can be modified for a particle of mass m free to move parallel to the x-axis with zero potential energy (V = 0 everywhere) resulting in the quantum mechanical description of free motion in one dimension:

         ??22md2?(x)dx2=E?(x)                                                  (2)

This equation has been well studied and gives a general solution of:

?(x)=Asin(kx)+Bcos(kx)                                                  (3)

where A, B, and k are constants.

wavefunction

The solution to the Schrödinger equation we found above is the general solution for a 1-dimensional system. We now need to apply our boundary conditions to find the solution to our particular system.

According to our boundary conditions, the probability of finding the particle at x=0 or x=L is zero. When x=0 sin(0)=0 and cos(0)=1; therefore, B must equal 0 to fulfill this boundary condition giving:

?(x)=Asin(kx)                                                              (4)

We can now solve for our constants (A and k) systematically to define the wavefunction.

Solving for k
Differentiate the wavefunction with respect to x:

d?dx=kAcos(kx)                                                              (5)

d2?dx2=?k2Asin(kx)                                                        (6)

Since ?(x)=Asin(kx)

, then d2?dx2=?k2?                                                                  (7)

If we then solve for k by comparing with the Schrödinger equation above, we find:

k=(8?2mEh2)1/2                                                                                                     (8)

Now we plug k into our wavefunction:

?=Asin(8?2mEh2)1/2x                                                    (9)

value of A
To determine A, we have to apply the boundary conditions again. Recall that the probability of finding a particle at x = 0 or x = L is zero.

When x = L:

0=Asin(8?2mEh2)1/2L                                                   (10)

This is only true when

(8?2mEh2)1/2L=n?                                                       (11)

where n = 1,2,3…

Plugging this back in gives us:

?=Asinn?Lx                                                             (12)

To determine A, recall that the total probability of finding the particle inside the box is 1, meaning there is no probability of it being outside the box. When we find the probability and set it equal to 1, we are normalizing the wavefunction.

?L0?2dx=1                                                                (13)

For our system, the normalization looks like:

A2?L0sin2(n?L)xdx=1                                                (14)

Using the solution for this integral from an integral table, we find our normalization constant, A:

A=2L???                                                                   (15)

Which results in the normalized wavefunction for a particle in a 1-dimensional box:

?=2L???sinn?Lx                                                   (16)

the Allowed Energies

Solving for E results in the allowed energies for a particle in a box:

En=n2h28mL2                                                                                       (17)

This is a very important result; it tells us that:

1. The energy of a particle is quantized.
2. The lowest possible energy of a particle is NOT zero. This is called the zero-point energy and means the particle can never be at rest because it always has some kinetic energy.

This is also consistent with the Heisenberg Uncertainty Principle: if the particle had zero energy, we would know where it was in both space and time.