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A plot of Psi of x at t equal to zero as a function of x. The function is zero for x less than minus L over 2 and x greater than L over 2. For x between minus and plus L over 2, the functions is a cosine curve, concave down with a positive maximum at x equal to zero, and going to zero at minus and plus L over 2.
Wave function for a ball in a tube of length L , where the ball is preferentially in the middle of the tube.

Strategy

We use the same strategy as before. In this case, the wave function has two unknown constants: One is associated with the wavelength of the wave and the other is the amplitude of the wave. We determine the amplitude by using the boundary conditions of the problem, and we evaluate the wavelength by using the normalization condition. Integration of the square of the wave function over the last quarter of the tube yields the final answer. The calculation is simplified by centering our coordinate system on the peak of the wave function.

Solution

The wave function of the ball can be written

Ψ ( x , 0 ) = A cos ( k x ) ( L / 2 < x < L / 2 ) ,

where A is the amplitude of the wave function and k = 2 π / λ is its wave number. Beyond this interval, the amplitude of the wave function is zero because the ball is confined to the tube. Requiring the wave function to terminate at the right end of the tube gives

Ψ ( x = L 2 , 0 ) = 0 .

Evaluating the wave function at x = L / 2 gives

A cos ( k L / 2 ) = 0 .

This equation is satisfied if the argument of the cosine is an integral multiple of π / 2 , 3 π / 2 , 5 π / 2 , and so on. In this case, we have

k L 2 = π 2 ,

or

k = π L .

Applying the normalization condition gives A = 2 / L , so the wave function of the ball is

Ψ ( x , 0 ) = 2 L cos ( π x / L ) , L / 2 < x < L / 2 .

To determine the probability of finding the ball in the last quarter of the tube, we square the function and integrate:

P ( x = L / 4 , L / 2 ) = L / 4 L / 2 | 2 L cos ( π x L ) | 2 d x = 0.091 .

Significance

The probability of finding the ball in the last quarter of the tube is 9.1%. The ball has a definite wavelength ( λ = 2 L ) . If the tube is of macroscopic length ( L = 1 m ) , the momentum of the ball is

p = h λ = h 2 L ~ 10 −36 m / s .

This momentum is much too small to be measured by any human instrument.

An interpretation of the wave function

We are now in position to begin to answer the questions posed at the beginning of this section. First, for a traveling particle described by Ψ ( x , t ) = A sin ( k x ω t ) , what is “waving?” Based on the above discussion, the answer is a mathematical function that can, among other things, be used to determine where the particle is likely to be when a position measurement is performed. Second, how is the wave function used to make predictions? If it is necessary to find the probability that a particle will be found in a certain interval, square the wave function and integrate over the interval of interest. Soon, you will learn soon that the wave function can be used to make many other kinds of predictions, as well.

Third, if a matter wave is given by the wave function Ψ ( x , t ) , where exactly is the particle? Two answers exist: (1) when the observer is not looking (or the particle is not being otherwise detected), the particle is everywhere ( x = , + ) ; and (2) when the observer is looking (the particle is being detected), the particle “jumps into” a particular position state ( x , x + d x ) with a probability given by P ( x , x + d x ) = | Ψ ( x , t ) | 2 d x —a process called state reduction    or wave function collapse    . This answer is called the Copenhagen interpretation    of the wave function, or of quantum mechanics.

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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