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Show that the probability of obtaining En, for a particle in a box with wave function Y(x)= -{√B. is given by |c₁|² 30x (L-x) 0

Show that the probability of obtaining En, for a particle in a box with wave function Y(x)= -{√B. is given by |c₁|² 30x (L-x) 0

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Show that the probability of obtaining En for a particle in a box with the provided wave function (see first image) is given by the equation in the first image. 

Hint: The probability of obtaining the ground-state energy E1 is determined in Example 3.4, as shown in the second provided image.

Show that the probability of obtaining E,, for a particle in a box with wave function (x) = is given by 3x(L − x) 0 < x < L elsewhere {√ EXCL_ |c₁|²= 240 поль [1-(-1)"]²
Transcribed Image Text:Show that the probability of obtaining E,, for a particle in a box with wave function (x) = is given by 3x(L − x) 0 < x < L elsewhere {√ EXCL_ |c₁|²= 240 поль [1-(-1)"]²
EXAMPLE 3.4 The wave function for a particle in the box is given by 30 {√BX(L_ 4(x) = Example 2.5 shows that is normalized. A measurement of the energy of the particle is carried out. What is the probability of obtaining E₁, the ground-state energy for the particle in a box? C1 SOLUTION To determine the probability, evaluate c₁: = [₁ v₁(x)(x) dx = [₁ √ ² sin L x(L-x) 0<x< L elsewhere Making the change of variables x/L = y, C1 √60 I² √60 2 π.Χ. L 30 15x (L - x) dx ["siny (y-2) dy 32² - 2 cos y ) -y cos y + T 4√60 I Thus the probability of obtaining E₁ is c₁² = 960/6 = 0.9986. The similarity between the ground-state wave function for the particle in the box and the parabolic wave function given in this example (see Fig. 2.11) is reflected in how close this probability is to one. Put another way, the probability of obtaining an energy other than the ground-state energy is 1 - 0.9986 = 0.0014. Caution: When c₁ is real and positive, it is easy to forget that the probability is given by cic, and not by c₁ itself. It may help to remember that c₁ (t) = c₁(0)e-it/h, so that even if c₁ (0) is real and positive, c₁ (t) is not.
Transcribed Image Text:EXAMPLE 3.4 The wave function for a particle in the box is given by 30 {√BX(L_ 4(x) = Example 2.5 shows that is normalized. A measurement of the energy of the particle is carried out. What is the probability of obtaining E₁, the ground-state energy for the particle in a box? C1 SOLUTION To determine the probability, evaluate c₁: = [₁ v₁(x)(x) dx = [₁ √ ² sin L x(L-x) 0<x< L elsewhere Making the change of variables x/L = y, C1 √60 I² √60 2 π.Χ. L 30 15x (L - x) dx ["siny (y-2) dy 32² - 2 cos y ) -y cos y + T 4√60 I Thus the probability of obtaining E₁ is c₁² = 960/6 = 0.9986. The similarity between the ground-state wave function for the particle in the box and the parabolic wave function given in this example (see Fig. 2.11) is reflected in how close this probability is to one. Put another way, the probability of obtaining an energy other than the ground-state energy is 1 - 0.9986 = 0.0014. Caution: When c₁ is real and positive, it is easy to forget that the probability is given by cic, and not by c₁ itself. It may help to remember that c₁ (t) = c₁(0)e-it/h, so that even if c₁ (0) is real and positive, c₁ (t) is not.
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