Physical Chemistry
Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
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Chapter 11, Problem 11.19E
Interpretation Introduction

Interpretation:

The value of px for Ψ0 and Ψ1 of a harmonic oscillator is to be calculated. Whether the calculated values make sense or not is to be stated.

Concept introduction:

In quantum mechanics, the wavefunction is given by Ψ. The wavefunction contains all the information about the state of the system. The wavefunction is the function of the coordinates of particles and time. The square of the probability function, |Ψ|2, relates to the probability density.

Expert Solution & Answer
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Answer to Problem 11.19E

The value of px for Ψ0 and Ψ1 of a harmonic oscillator is 0. The average momentum of harmonic oscillator is zero because the movement of mass takes place back and forth in both directions and momentum is a vector quantity. Therefore, the calculated value makes sense.

Explanation of Solution

The general wavefunction of harmonic oscillator is expressed as,

Ψ=(απ)14(12nn!)12Hn(α1/2x)eαx2/2 …(1)

Substitute the value of n=0 in equation (1).

Ψ0=(απ)14(120×0!)12H0(α1/2x)eαx2/2=(απ)14H0(α1/2x)eαx2/2

Substitute the value of n=1 in equation (1).

Ψ1=(απ)14(1211!)12H1(α1/2x)eαx2/2=(α4π)14H1(α1/2x)eαx2/2

The value of px is calculated by the formula.

px=+ΨixΨdx …(2)

Substitute the value of Ψ0 in equation (2).

px=+((απ)14H0(α1/2x)eαx2/2)ix((απ)14H0(α1/2x)eαx2/2)dx

From Table 11.1, substitute the value of H0(α1/2x)=1 in the above equation.

px=(απ)12+(eαx2/2)(ix(eαx2/2))dx=i(απ)12+(eαx2/2)(2αx2eαx2/2)dx=ia32π12+xeαx2dx…(3)

The above equation show that eαx2 is an even function whereas x is an odd function. Therefore, the overall function is odd.

The integration of odd function going from to+ and centered at 0 is expressed as,

+f(x)dx=0

Therefore, equation (3) becomes,

px=0

Substitute the value of Ψ1 in equation (2).

px=+((α4π)14H1(α1/2x)eαx2/2)ix((α4π)14H1(α1/2x)eαx2/2)dx

From Table 11.1, substitute the value of H1(α1/2x)=2α1/2x in the above equation.

px=(α4π)12+(2α1/2xeαx2/2)ix(2α1/2xeαx2/2)dx=i4α3/22π1/2+(xeαx2/2)x(xeαx2/2)dx=i4α3/22π1/2+xeαx2/2[eαx2/2αx2eαx2/2]dx=i4α5/22π1/2+(xeαx2αx3eαx2)dx

The above equation shows that eαx2 is an even function whereas x3 and x are odd functions. Therefore, the overall function is odd.

The integration of odd function going from to+ and centered at 0 is expressed as,

+f(x)dx=0

Therefore, equation (3) becomes,

px=0

Hence, the value of px for Ψ0 and Ψ1 of a harmonic oscillator is 0.

The average momentum of harmonic oscillator is zero because the movement of mass takes place back and forth in both directions and momentum is a vector quantity.

Conclusion

The value of px for Ψ0 and Ψ1 of a harmonic oscillator is 0. The average momentum of harmonic oscillator is zero because the movement of mass takes place back and forth in both directions and momentum is a vector quantity. Therefore, the calculated value makes sense.

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