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.

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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