7.1 Wave functions (Page 2/22)

University physics volume 3 Page 2 / 22

Three wave functions and the square of the amplitude of the wave functions. The first wave function, psi sub zero, and its square are symmetric, positive, peaked at x = 0, and zero far from the origin. The second wave function, psi sub 1, is antisymmetric: the function is negative at negative x, positive at positive x, and zero at the origin as well as plus and minus infinity. There is a negative minimum at negative x and positive maximum at positive x. The minimum value is exactly opposite the maximum value. The square of the amplitude of the wave function is positive and symmetric about the origin, where the value is zero, with a maximum on either side of the origin. The third wave function, psi sub N, is not symmetric. It is zero at minus infinity, decreases to a negative minimum value at some x less than zero, crosses zero, still at x less than zero, and becomes positive. It reaches a positive maximum at some positive x, then decreases to zero at large x. The minimum value is smaller in magnitude than the maximum value. The square of the amplitude of the wave function is positive, with two local maxima. The first maximum is smaller and at negative x, and the second is larger and at positive x. The square of the function is zero at one point between the maxima. — Several examples of wave functions and the corresponding square of their wave functions.

If the wave function varies slowly over the interval $Δ x$ , the probability a particle is found in the interval is approximately

P (x, x + Δ x) \approx | Ψ (x, t) |^{2} Δ x .

Notice that squaring the wave function ensures that the probability is positive. (This is analogous to squaring the electric field strength—which may be positive or negative—to obtain a positive value of intensity.) However, if the wave function does not vary slowly, we must integrate:

P (x, x + Δ x) = \int_{x}^{x + Δ x} | Ψ (x, t) |^{2} d x .

This probability is just the area under the function $| Ψ (x, t) |^{2}$ between x and $x + Δ x$ . The probability of finding the particle “somewhere” (the normalization condition ) is

P (− \infty, + \infty) = \int_{− \infty}^{\infty} | Ψ (x, t) |^{2} d x = 1 .

For a particle in two dimensions, the integration is over an area and requires a double integral; for a particle in three dimensions, the integration is over a volume and requires a triple integral. For now, we stick to the simple one-dimensional case.

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.

The wave function Psi of x and t is plotted as a function of x. It is a step function, zero for x less than 0 and x greater than L, and constant for x between zero and L. — Wave function for a ball in a tube of length L .

Solution

The wave function of the ball can be written as

Ψ (x, t) = C (0 < x < L),

where C is a constant, and

Ψ (x, t) = 0

otherwise. We can determine the constant C by applying the normalization condition (we set

t = 0

to simplify the notation):

P (x = − \infty, + \infty) = \int_{− \infty}^{\infty} | C |^{2} d x = 1 .

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

P (x = 0, L) = \int_{0}^{L} | C |^{2} d x = 1 .

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

| C |^{2} \int_{0}^{L} d x = 1 .

Integration gives

C = \sqrt{\frac{1}{L}} .

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

P (x = 0, L / 2) = \int_{0}^{L / 2} {| \sqrt{\frac{1}{L}} |}^{2} d x = (\frac{1}{L}) \frac{L}{2} = 0.50 .

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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Where is the ball? (part ii)

A ball is again constrained to move along a line inside a tube of length L . This time, the ball is found preferentially in the middle of the tube. One way to represent its wave function is with a simple cosine function ( [link] ). What is the probability of finding the ball in the last one-quarter of the tube?

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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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