Introduction to Quantum Mechanics (Coursework Sample)

INTRODUCTION TO QUANTUM MECHANICS
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1 Solution
* The function f(x) = e-i(3x+2y) will be an Eigen function if we apply the operator d2/dx2 to the function and the function is regenerated.
d2/dx2 f(x) = d/dx [d/dxe-i(3x+2y)]
= d/dx-3ie-i(3x+2y)
= -9e-i(3x+2y)
= -9f(x)
Thus, acting on the function f(x) = e-i(3x+2y) with the operator d2/dx2 regenerates the function, and the Eigen value in this case is -9.
We can say that e-i(3x+2y) is an Eigen function of the operator d2/dx2 with -9 as the Eigen value.
* f(x) = x2+y2
(1/x)(x2+y2) d/dx[f(x,y)] = (1/x)(x2+y2) d/dxx2+y2
= (1/x)(x2+y2) d/dxx2+y2 = (1/x)(x2+y2)[1/2(2x)( x2+y2)-1/2]
= (1/x)(x2+y2)[x(x2+y2)-1/2] = (1/x)(x2+y2)[x/ (x2+y2)1/2]
= (x2+y2)1/2
Thus, the function f(x) = x2+y2 is the Eigen function of the operator (1/x)(x2+y2) d/dx and the Eigen value is 1.
* f(θ)=SinθCosθ
Applying the operator Sin θddθ(Sinθddθ) + 6 Sin2θ we have;
= Sin θddθSinθCosθ(SinθddθSinθCosθ) + 6 Sin2θ
= Sin θ[CosθddθSinθ+ Sinθddθ Cosθ] {Sin θ[CosθddθSinθ+ Sinθddθ Cosθ]} + 6 Sin2θ
= Sin θ[Cos2 θ - Sin2θ] {Sin θ[Cos2 θ - Sin2θ]} + 6 Sin2θ
From this equation it is clear that the function f(θ)=SinθCosθ cannot be generated. Thus, the function is not an Eigen function of the operator
Sin θddθ(Sinθddθ) + 6 Sin2θ
2 Solution
f(x) = e-12αx^2 and f(x) = (2αx2 – 1) e-12αx^2 , -„≤x<„
To find if the two wave functions are orthogonal. We find the integral of their product over the indicated range. If it is zero then they are orthogonal CITATION Eri12 \l 1033 (Smith & Fellows, 2012).
Therefore;
-„„(2αx^2 – 1) e(-12αx^2)^2dx
= -„„(2αx^2 – 1) e1/4α^2x^4dx
=-„„[2αx2 e1/4α^2x^4 - e1/4α^2x^4]dx
=-„„2αx2 e1/4α^2x^4dx - -„„e1/4α^2x^4dx
Integrating the first part of the equation by parts;
Let u = 2αx2 and dv = e1/4α^2x^4dx
du = 4 αx dx v = 15 e1/4α^2x^4
= 25αx2 e1/4α^2x^4 - -„„4/5αx e1/4α^2x^4dx
Integrating the second part of the equation by parts;
Let u = 4/5αx and dv = e1/4α^2x^4dx
du = 4/5αdx v = 1/5 e1/4α^2x^4
= 4/25αx e1/4α^2x^4 - -„„4/25α e1/4α^2x^4dx
Integrating the second part of the equation;
4/25αx e1/4α^2x^4 - 4/125α e1/4α^2x^4
= [25αx2 e1/4α^2x^4 - 4/25αx e1/4α^2x^4 + 4/125α e1/4α^2x^4 - 1/5 e1/4α^2x^4]
= „-„+„ - „ = 0
Thus, the two wave functions are orthogonal.
3 Solution
From 0a‣Ψ(x)‣2dx = 1, we will have;
=0aA2 (Xa)41-xa2dx=1
=A40a(xa)4[1-xa]4dx=1
=0a(xa-x2)4dx=(ax)4
= a59-2a39+6a79-a29+a99
=(aA)4
A4=a-5
A = 1a54
4 Solution
For the particle moving along a circle covers some radius r from center of circle, expression for angular momentum is L= r× p CITATION Eri12 \l 1033 (Smith & Fellows, 2012)
Where p is linear momentum with operator ṕ(x, y) = ħiddxdy
Hence, operator for angular momentum is Ḽ= ṙ × ṕ, = ṙ×= ħiddxdy
This is for the particle in circular motion along xy-plane.
Applying wavefunction; Ḽ= ṙ× ħiddxdy(exp-2iФ)
Hence the position Ф is in the xy-plane
That is; Ḽ= ṙ × ħiddФ(exp-2iФ) = - ṙ2iħexp⁡(-2iФ)i, = -2iṙΨ(Ф).
The angular momentum is therefore L= -2iṙΨ(Ф)
5 Solution
For normalized wavefunction; ʃ [Ψ(Ф)]2dФ=1
* ʃ [‚(12‏)expiФ]2dФ
= ʃ 12‏exp⁡(2iФ)dФ
Assuming spherical nature of the light atom, we use spherical polar coordinates;
x = rsinÓ©cosÒ¨,
y = rsinÓ©sinÒ¨
z = rcosÓ©
dФ = r2sinөdrdөdҨ
Where r= 0 to„
ө= 0 to ‏
Ҩ= 0 to ‏
Hence the probability can be obtained from 12‏[r33×(-cosө)×exp2iҨ2i=0 after applying the respective limits.
* Applying the limits we have;
‏23‏212‏exp2iФdФ
=12‏[exp2iФ2i] With limits of 3‏2 to ‏2
This comes to;
= 14‏i[e6‏i2-e4‏i2]
= e‏i4‏i
6 Solution
For harmonic oscillator;
Zero Point Energy (ZPE) = Potential Energy V(x)
But V(x) = 0.5kf x2
The Potential Energy given above is from the relation F=ke and F= ma
Since ZPE = V(x)
ZPE = 0.5kf x2
But each mass has extension of 2.5×10725= 0.03448m
Hence for combination of two masses, total extension = 0.03445/2=0.01724m
Here the gravitational intensity is assumed to be 10N/Kg
Hence;
ZPE = 0.5×725×0.0003
= 0.10875J
Comparing the ZPE with Thermal Energy kT at 298K
Where k is the Boltzmann constant given as 1.3806503x10-23J/K
Therefore; Thermal Energy = kT
=1.3806503x10-23 J/K x 298K
= 4.114x10-21 J
On comparing, we find that the ZPE is much higher than the Thermal Energy.
This shows that the population of particles in the energy level with similar amount of energy as zero point energy is very minimal.
Converting the zero point energy to translational energy, we obtain the speed of the two masses as follows;
1.18×10-40×n2max = 0.10875
Where; nmax=‚ (0.108751.18×10-40)
= 3.0358x1019
7 Solution
* Rotational Constant is given as 11.007cm-1
Mass of 19F = 18.9984032amu
Mass of 2D = 2.0141018amu
The Bond length in 2D19F is calculated as follows;
Rotational Constant is given by B = ℏ2/2µR2
Where ℏ is the Reduced Planck’s Constant given as 1.0546 x 10-27cm2g/s
And R is the Bond length
µ, which is the reduced mass, is given by;
µ = m1m2/ m1+m2where m1 = 18.9984032amu
m2 = 2.0141018amu
µ = 18.9984032 x 2.014101818.9984032+ 2.0141018
= 38.2647221.012505 = 1.821045
From the equation of Rotational Constant, making R the subject of the equation we have R = ‚(ℏ2/2µB) = ℏ‚2µB
= 1.0546 x 10^-27‚[21.82104511.007]
= 1.0546 x 10^-27‚[3.642090.09085]
= 6.0663 x 10-28cm
* At ground state n = 0
Therefore; B0 = B – (0+1/2) α
But; B = 6.551cm-1andα = 0.183cm-1
Hence; B0 = 6.551 - ½(0.183)
= 6.551 – 0.0915 = 6.4595cm-1
Bond length; R = ℏ‚2µB0
Reduced mass is first calculated from m1 = 1.00794amu and m2 = 126.90447amu
µ = 1.00794 x 126.904471.00794+ 126.90447
= 127.9120915127.91241 = 0.9999 = 1
R = 1.0546 x 10^-27‚[26.45951] = 1.8953 x 10-27cm
And for n = 3
B3 = B – (3+1/2) α
= 6.551 – 3/2(0.183)
= 6.551 – 0.2745
= 6.2765cm-1
Bond Length; R = 1.0546 x 10^-27‚[26.27651] = 1.8952 x 10-27cm
8 Solution
* Assume the potential for bond breaking is harmonic
Then V(x) = 0.5kfx2
Where; kf = force c...