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# QUESTION ATTACHED:

QUESTION ATTACHED:

According to classical mechanics, the turning point, xtp, of an oscillator
occurs when its kinetic energy is zero, which is when its potential energy kx2
is equal to its total energy E. This equality occurs when
x2
tp = or xtp = ±
1/2
with E given by eqn 8.24. The probability of finding the oscillator stretched
beyond a displacement xtp is the sum of the probabilities ψ 2dx of finding it in
any of the intervals dx lying between xtp and infinity:
P = ∞
xtp
ψ v
2 dx
The variable of integration is best expressed in terms of y = x/α with
α = (\$2/mk)1/4. (a) Show that the turning points lie at ytp = ±(2v + 1)1/2.
(b) Go on to show that for the state of lowest energy (v = 0), ytp = 1 and the
probability is P = (1 − erf1), where the error function, erf z, is defined as
erf z = 1 − z

e−y2 dy
The values of this function are tabulated and available in mathematical
software packages.

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