7.1 Wave functions  (Page 2/22)

Where is the ball? (part i)

A ball is constrained to move along a line inside a tube of length L . The ball is equally likely to be found anywhere in the tube at some time t . What is the probability of finding the ball in the left half of the tube at that time? (The answer is 50%, of course, but how do we get this answer by using the probabilistic interpretation of the quantum mechanical wave function?)

Strategy

The first step is to write down the wave function. The ball is equally like to be found anywhere in the box, so one way to describe the ball with a constant wave function ( [link] ). The normalization condition can be used to find the value of the function and a simple integration over half of the box yields the final answer.

Solution

The wave function of the ball can be written as $\text{Ψ}(x,t)=C(0<x<L),$ where C is a constant, and $\text{Ψ}(x,t)=0$ otherwise. We can determine the constant C by applying the normalization condition (we set $t=0$ to simplify the notation):

This integral can be broken into three parts: (1) negative infinity to zero, (2) zero to L , and (3) L to infinity. The particle is constrained to be in the tube, so $C=0$ outside the tube and the first and last integrations are zero. The above equation can therefore be written

The value C does not depend on x and can be taken out of the integral, so we obtain

Integration gives

To determine the probability of finding the ball in the first half of the box $\left(0<x<L\right),$ we have

Significance

The probability of finding the ball in the first half of the tube is 50%, as expected. Two observations are noteworthy. First, this result corresponds to the area under the constant function from $x=0$ to L /2 (the area of a square left of L /2). Second, this calculation requires an integration of the square of the wave function. A common mistake in performing such calculations is to forget to square the wave function before integration.

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