In-class group activity 8:

Exploring solutions of the Schroedinger equation which are not stationary states

Open a Microsoft Word document to keep a log of your results and conclusions.   

A particle in an infinite square well does not have to be in an eigenstate of the energy operator.  If we measure the position of a particle in the well and find it at some position x, then right after the measurement the particle is in an eigenstate of the position operator.  Its energy is unknown, we can at most determine its average energy and the probability of measuring one of its eigenenergies in a subsequent measurement.  One of the fundamental assumptions of quantum mechanics is that the eigenstates of each operator form a complete set.  Any other acceptable wave function can be written as a linear combination of eigenstates, just as it can be expanded in a Fourier series.  Therefore right after our measurement, the particle is in a superposition of energy eigenstates.  Let us investigate one of those superpositions.

Problem:

A particle of mass m moves in one dimension in a square well with walls of infinite height a distance L apart.  The particle is known to be in a state consisting of an equal admixture of the two lowest energy eigenstates of the system.  Find the probability per unit length as a function of time that the particle will be at position x in the well.

• Solution:

  ψ(x,t) = 2^-1/2[ψ₁(x,t) + ψ₂(x,t)], where 
  ψ₁(x,t) = (2/L)^1/2sin(πx/L))exp(-(2πi/h)E₁t)  and ψ₂(x,t) = (2/L)^1/2sin(2πx/L)exp(-(2πi/h)E₂t),
  with E₁ = h²/(8mL²)  and  E₂ = h²/(2mL²),
  are solutions to the one-dimensional infinite square-well problem.
  P(x,t) = |ψ(x,t)|² is the probability per unit length of finding the particle at position x.

  |ψ(x,t)|² = (1/2)|ψ₁(x,t) + ψ₂(x,t)|²
  Plug in ψ₁(x,t) and ψ₂(x,t).
  |ψ(x,t)|² = (1/L)|sin(πx/L)exp(-(2πi/h)E₁t) + sin(2πx/L)exp(-(2πi/h)E₂t)|²
  = (1/L)(sin(πx/L)exp(-(2πi/h)E₁t) + sin(2πx/L)exp(-(2πi/h)E₂t))
  *(sin(πx/L)exp((2πi/h)E₁t) + sin(2πx/L)exp((2πi/h)E₂t))
  =( 1/L)[sin²(πx/L) + sin²(2πx/L) + sin(πx/L)sin(2πx/L)(exp(i2π(E₂-E₁)t/h) + exp(-2iπ(E₂-E₁)t/h)]
  = (1/L)[sin²(πx/L) + sin²(2πx/L) + 2sin(πx/L)sin(2πx/L)cos(2π(E₂-E₁)t/h)].

  P(x,t) = (1/L)[sin²(πx/L) + sin²(2πx/L) + 2sin(πx/L)sin(2πx/L)cos(2π(E₂-E₁)t/h)].
  P(x.t) is no longer independent of time, the probability per unit length of finding the particle at x is changing with time.

An animated plot of P(x,t)

Let us plot P(x,t) as as a function of x and t.
Let us measure distance in units of Å = 10^-10 m and let L = 1 Å.
Let the particle be an electron.
Then  E₁ = π²ħ²/(2mL²) = 6*10^-18 J = 38 eV  and E₂ = 4 E₁.
Then 2π(E₂-E₁)/h = (E₂-E₁)/ħ = 1.7*10¹⁷/s.
Let us measure time in units of 10^-18 s, then cos(2π(E₂-E₁)t/h) = cos(0.17 t).

We want to plot the function P(x,t) = sin²(πx) + sin²(2πx) + 2sin(πx)sin(2πx)cos(0.17 t).

The linked spreadsheet plots this function for from x = 0 to x = 1 at t = 0.  Open the spreadsheet.

• Into cell A2 type different values for the time.  Observe that the probability per unit length of finding the particle at position x is no longer independent of time.

• Animate the graph.

  • On Excel's menu click View, Toolbars, Forms.  (In Excel 2007, click the developer tab, insert, form controls.)  On the Forms toolbar click on the button and then on an empty cell on your spreadsheet.  A button will be inserted and an Assign Macro window will open.  Click New.  Enter the following code between the two lines that appear in the code window.

    Range("a2").Value = 0
    Do While Range("a2").Value < 100
    Range("a2").Value = Range("a2").Value + 0.1
    Worksheets(1).Calculate
    DoEvents
    Loop

    Close the code window and the Forms toolbar, and click the button to watch your animation.

Discuss your observations with your neighbors.  Do you observe oscillations?  Does this probability density better describe a particle bouncing back and forth in a well than the probability density associated with a stationary state?

Using a superposition of many stationary states we can construct wave packets and probability densities that bounce back and forth in the well in a way we expect a classical particle to bounce back and forth.  If we want these packets to be narrow, i.e. to have a small ∆x, then we need to use high energy stationary states.  The uncertainty principle requires a large ∆p and therefore a large p for a small ∆x.  The bouncing packets will not be energy eigenstates and the particle will not have a precisely defined energy.

Explore more superpositions of energy eigenstates by following the link below.

Link:  1-D Quantum Mechanics Applet