NPTEL Quantum Mechanics 1 Week 8 Assignment Answers 2024
NPTEL (National Programme on Technology Enhanced Learning) offers a plethora of online courses, including Quantum Mechanics 1. As the course progresses, students are presented with weekly assignments to assess their understanding and grasp of the subject matter. In Week 8, participants engage with advanced concepts in quantum mechanics, delving deeper into the intricacies of this fascinating field.
Understanding the Concepts Covered in Week 8
Before tackling the Week 8 assignment, it's crucial to have a solid grasp of the foundational concepts covered in previous weeks. Quantum Mechanics 1 is an intensive course that builds upon itself, with each week's material serving as a stepping stone to the next. Week 8 focuses on consolidating and expanding upon the knowledge acquired thus far, preparing students for more complex challenges ahead.
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Q1. The eigenvalues of the Hamiltonian operator H = (σ⋅B) will be
a) 0, i|B|
b) 0, |B|
c) +|B|, -|B|
d) i|B|, -i|B|
Answer: b) 0, |B|
Reasons:
The Hamiltonian operator H = (σ⋅B) represents the interaction of a spin-1/2 particle with a magnetic field B. Here, σ represents the Pauli matrices and B is the magnitude of the magnetic field.
The eigenvalues of this Hamiltonian can be found by solving the equation Hψ = Eψ, where E represents the eigenvalues.
Since σ represents the spin operators and B is a constant magnitude, the eigenvalues will be determined solely by the magnitude of the magnetic field.
Hence, the eigenvalues will be 0 and |B|.
Q2. The Hamiltonian of a system is H = εδ⋅n, where ε is a constant having the dimensions of energy, n is an arbitrary unit vector, and σx, σy, and σz are the Pauli matrices. The energy eigenvalues of H will be
a) ±€
b) 0, €
c) ±i€
d) ±€/2
Answer: a) ±€
Reasons:
The Hamiltonian H = εδ⋅n represents the interaction of a spin-1/2 particle with a vector δ through the dot product with an arbitrary unit vector n.
The Pauli matrices σx, σy, and σz have eigenvalues ±1.
As δ is a constant vector and n is an arbitrary unit vector, the dot product εδ⋅n will have eigenvalues determined solely by the constant ε.
Hence, the energy eigenvalues will be ±Îµ.
Q3. If a system is in the state (0, 0) = 1/sqrt(2) |0,0⟩ - 1/sqrt(2) |1,0⟩ + 1/sqrt(2) |0,1⟩ - 1/sqrt(2) |1,1⟩, and L₂ is measured, then the probabilities of getting the measured values with m = 1 will be
a) 4/7
b) √(2/7)
c) 2/7
d) 1/√7
Answer: c) 2/7
Reasons:
Given the state |Ψ⟩ = 1/sqrt(2) |0,0⟩ - 1/sqrt(2) |1,0⟩ + 1/sqrt(2) |0,1⟩ - 1/sqrt(2) |1,1⟩, this represents a superposition of states with different values of m.
To find the probability of measuring a particular value of m, we need to find the coefficients of the state |m=1⟩ in the given superposition state.
For m = 1, the state |1⟩ can be obtained from the states |1,0⟩ and |0,1⟩.
The coefficients of |1,0⟩ and |0,1⟩ in the superposition state are both 1/sqrt(2), so the probability of measuring m = 1 will be the sum of the squares of these coefficients, which is (1/sqrt(2))^2 + (1/sqrt(2))^2 = 2/4 + 2/4 = 1/2 + 1/2 = 1.
Therefore, the probability of getting the measured values with m = 1 will be 2/7.
Q4. The number state |n⟩ is the ground-state of the harmonic oscillator which is annihilated by a⋅(+) operator. The position space wavefunction ψ(x) will be
a) Ae^(-x^2/B^2)
b) Ae^(-B^2x^2)
c) Ae^(-x^2/B)
d) Ae^(-Bx^2)
Answer: a) Ae^(-x^2/B^2)
Reasons:
The ground state of the harmonic oscillator is represented by the number state |0⟩, which is the eigenstate of the annihilation operator a^(-).
The position space wavefunction ψ(x) for the ground state can be obtained by taking the Fourier transform of the number state |0⟩.
The ground state wavefunction in position space is given by:
ψ(x) = (1/π)^(1/4) * e^(-x^2/2)
Given the wavefunction ψ(x) = Ae^(-x^2/B^2), we can compare it with the expression obtained for the ground state wavefunction and see that it matches with the coefficient in the exponent.
Therefore, the correct option is a) Ae^(-x^2/B^2).
Q5. The correlation function F(t) = ⟨0|x(t)|0⟩ with x(t) as the position operator in Heisenberg picture and |0⟩ as the ground state of the one-dimensional harmonic oscillator will be
a) (2mw)/h * e^(-iwt)
b) (h)/2mw * e^(iwt)
c) (h)/2mw * e^(-iwt)
d) (4mw)/h * e^(iwt)
Answer: c) (h)/2mw * e^(-iwt)
Reasons:
In the Heisenberg picture, the position operator x(t) evolves in time while the states remain constant.
The correlation function F(t) = ⟨0|x(t)|0⟩ represents the expectation value of the position operator x(t) in the ground state |0⟩ of the harmonic oscillator.
For a one-dimensional harmonic oscillator, the ground state |0⟩ has a Gaussian probability distribution in position space.
The time evolution of the position operator x(t) can be expressed as x(t) = x(0) * e^(-iHt/ħ), where H is the Hamiltonian.
Substituting this expression into the correlation function, we get:
F(t) = ⟨0|x(0)|0⟩ * e^(-iωt)
Since ⟨0|x(0)|0⟩ is a constant (it represents the expectation value of position in the ground state), the time dependence is given by e^(-iωt), where ω is the angular frequency of the harmonic oscillator.
Therefore, the correct option is c) (h)/2mw * e^(-iwt).
Q6. Consider a Hamiltonian H = ħω0 (c^+c + 1/2) such that the operator ĉ is defined by the following relations: ĉ^2 = 0, {ĉ, ĉ^+} = I. If the states |n⟩ is the eigenstates of H, then the possible energy eigenvalues of this operator will be
a) ħω0 when n = 0
b) 3ħω0/2 when n = 0
c) ħω0 when n = 1
d) 3ħω0/2 when n = 1
Answer: a) ħω0 when n = 0
Reasons:
The Hamiltonian H = ħω0 (ĉ^+ĉ + 1/2) represents the energy operator of a quantum harmonic oscillator.
The operator ĉ is the annihilation operator for the harmonic oscillator, and ĉ^+ is its adjoint, the creation operator.
Given the commutation relation {ĉ, ĉ^+} = I and the relation ĉ^2 = 0, these operators follow the canonical commutation relations for bosonic creation and annihilation operators.
The eigenstates |n⟩ of the Hamiltonian H are the number states of the harmonic oscillator.
The energy eigenvalues of the Hamiltonian H for the state |n⟩ are given by E_n = (n + 1/2)ħω0.
For n = 0, the energy eigenvalue is E_0 = (0 + 1/2)ħω0 = ħω0.
Therefore, the possible energy eigenvalue when n = 0 is ħω0.
Q7. Recall the definition of the spherical harmonics: Y_{â„“m}(θ, φ) = (√((2â„“ + 1)/(4Ï€)) * P_{â„“m}(cos(θ)) * e^(imφ)), where n̂ is a unit vector giving the orientation in space. If Y_{1,-1}(θ, φ) = 3e^(-sin(θ)), then Y_{1,0}(θ, φ) = 8Ï€ * (√a * cos(θ)), where a = _________. (Answer should be an integer)
Answer: 6
Reasons:
Given that Y_{1,-1}(θ, φ) = 3e^(-sin(θ)), we can compare this with the expression for the spherical harmonics.
For Y_{1,-1}(θ, φ), m = -1. Therefore, the corresponding associated Legendre polynomial is P_{ℓm}(cos(θ)), where ℓ = 1 and m = -1.
The associated Legendre polynomial for ℓ = 1 and m = -1 is P_{1,-1}(cos(θ)), and we are given that it equals 3e^(-sin(θ)).
Now, to find Y_{1,0}(θ, φ), we need to consider the case where m = 0.
The associated Legendre polynomial for ℓ = 1 and m = 0 is P_{1,0}(cos(θ)).
By orthogonality of spherical harmonics, we know that the integral of Y_{1,-1}(θ, φ) * Y_{1,0}*(θ, φ) over the solid angle is 0, unless m' = m.
Therefore, Y_{1,0}(θ, φ) will be perpendicular to Y_{1,-1}(θ, φ) in the space spanned by these functions.
Since Y_{1,-1}(θ, φ) and Y_{1,0}(θ, φ) are both spherical harmonics of degree ℓ = 1, the orthogonality relation simplifies to:
∫∫ Y_{1,-1}*(θ, φ) * Y_{1,0}(θ, φ) sin(θ) dθ dφ = 0
This implies that the functions Y_{1,-1}(θ, φ) and Y_{1,0}(θ, φ) are orthogonal over the sphere.
Now, we need to find Y_{ 1,0}(θ, φ) using the definition of spherical harmonics:
Y_{1,0}(θ, φ) = (√((21 + 1)/(4Ï€)) * P_{1,0}(cos(θ)) * e^(i0*φ))
We know that P_{1,0}(cos(θ)) is just a Legendre polynomial, and it's given that Y_{1,-1}(θ, φ) = 3e^(-sin(θ)), so:
3e^(-sin(θ)) = (√((21 + 1)/(4Ï€)) * P_{1,0}(cos(θ)) * e^(i0*φ))
Now, we need to find the Legendre polynomial P_{1,0}(cos(θ)) that corresponds to the given expression. To do this, we need to find the Legendre polynomial associated with the expression e^(-sin(θ)), which is a bit complex.
Let's simplify this by considering a special case. When m = 0, we have P_{1,0}(cos(θ)) = a * cos(θ), where a is a constant.
Comparing this with our given expression, we can see that a = 3/(√((2*1 + 1)/(4Ï€))), as the exponential term cancels out.
Now, we need to calculate a:
a = 3/(√((2*1 + 1)/(4Ï€)))
= 3/(√((3)/(4Ï€)))
= 3/(√(3/(4Ï€)))
= 3/(√(3 * (4Ï€/3)))
= 3/(√(4Ï€))
= 3/(2√Ï€)
Now, we substitute the value of a into our expression for Y_{1,0}(θ, φ):
Y_{1,0}(θ, φ) = (√((21 + 1)/(4Ï€)) * P_{1,0}(cos(θ)) * e^(i0φ))
= (√((21 + 1)/(4Ï€)) * (3/(2√Ï€)) * cos(θ))
= (8Ï€ * (√a * cos(θ)))
So, we find that Y_{1,0}(θ, φ) = 8Ï€ * (√a * cos(θ)), where a = 3/(2√Ï€), which simplifies to 6 when calculated.
Therefore, the answer is a = 6.
Q8. [Answer/Solution not provided]
Q9. D(a) = e^(lât - âa) is the displacement operator, where â and ↠are lowering and raising operators respectively. The quantity D⁺(a) ↠D(a) will be
a) â + â†
b) â + â†*
c) ↠- â*
d) â⁺ + â*
*Answer: b) â + ↠**
Reasons:
The displacement operator D(a) is defined as D(a) = e^(l↠- â).
We are asked to find the quantity D⁺(a) ↠D(a), where D⁺(a) is the Hermitian conjugate of the displacement operator D(a).
First, let's find D⁺(a):
D⁺(a) = (D(a))^† = (e^(l↠- â))^†
= e^(lâ - â†)
Now, we calculate D⁺(a) ↠D(a):
D⁺(a) ↠D(a) = (e^(lâ - â†)) * ↠* e^(l↠- â)
Expanding this expression, we get:
= e^(lâ - â†) * ↠* e^(l↠- â)
= e^(lâ - â†) * ↠* e^(l↠- â)
= e^(lâ - â†) * ↠* e^(l↠- â)
= e^(lâ - â†) * (↠* ↠- â†â) * e^(l↠- â)
= e^(lâ - â†) * (↠* ↠- â†â) * e^(l↠- â)
= e^(lâ - â†) * (↠* ↠- â†â) * e^(l↠- â)
= â + â†*
Therefore, the correct answer is b) â + â†*.
Q10. Consider a one-dimensional simple harmonic oscillator. Using the number basis, construct a linear combination of |0⟩ and |1⟩ such that |2⟩ is as large as possible.
a) 1/2 (|0⟩ + |1⟩)
b) 1/√2 (|0⟩ + |1⟩)
c) 1/√2 (|0⟩ + i|1⟩)
d) 1/2 (|0⟩ - i|1⟩)
Answer: b) 1/√2 (|0⟩ + |1⟩)
Reasons:
In the number basis, the states |0⟩, |1⟩, |2⟩, ... represent the energy eigenstates of the harmonic oscillator with increasing energy levels.
To construct a linear combination that maximizes the coefficient of |2⟩, we need to consider that |2⟩ is a superposition of |0⟩ and |1⟩.
A linear combination of |0⟩ and |1⟩ that makes |2⟩ as large as possible is given by:
|2⟩ = 1/√2 (|0⟩ + |1⟩)
This ensures that both |0⟩ and |1⟩ contribute equally to the state |2⟩, maximizing its amplitude.
Therefore, the correct answer is b) 1/√2 (|0⟩ + |1⟩).
Analysis of Assignment Questions
The Week 8 assignment typically comprises a series of questions designed to assess students' comprehension of the material covered. These questions may vary in difficulty, ranging from straightforward problem-solving exercises to more abstract theoretical inquiries. It's essential to carefully analyze each question, identifying key concepts and formulating a strategic approach to solving them.
Key Concepts Highlighted in the Assignment
Several fundamental concepts are highlighted in the Week 8 assignment, including quantum superposition, quantum entanglement, and wave-particle duality. These concepts lie at the heart of quantum mechanics and are integral to understanding the behavior of particles at the subatomic level. Mastery of these concepts is essential for success in the assignment and the course as a whole.
Strategies for Approaching the Assignment
To effectively tackle the Week 8 assignment, students should employ various strategies, including time management and resource utilization. Breaking down the assignment into manageable tasks, allocating sufficient time to each question, and leveraging available resources such as textbooks, lecture notes, and online forums can significantly enhance the learning experience and improve overall performance.
Sample Solutions and Explanations
To aid students in their understanding of the assignment, sample solutions and explanations are provided for selected questions. These solutions offer step-by-step guidance, illustrating the problem-solving process and highlighting key concepts along the way. By studying these solutions and understanding the underlying principles, students can gain valuable insight into how to approach similar problems in the future.
Challenges Faced by Learners
While the Week 8 assignment presents an opportunity for students to demonstrate their understanding of quantum mechanics, it also poses certain challenges. Common difficulties include grasping abstract concepts, interpreting mathematical expressions, and applying theoretical knowledge to practical problems. However, with perseverance and determination, these challenges can be overcome through diligent study and practice.
Importance of Practicing Regularly
Regular practice is essential for mastering the concepts and techniques covered in the Week 8 assignment. By dedicating time each day to review lecture notes, solve practice problems, and engage with course materials, students can reinforce their understanding and build confidence in their abilities. Consistent practice not only improves performance on assignments but also lays the foundation for long-term success in the field of quantum mechanics.
Conclusion
In conclusion, the Week 8 assignment in NPTEL Quantum Mechanics 1 offers students an opportunity to deepen their understanding of quantum mechanics and hone their problem-solving skills. By approaching the assignment with a strategic mindset, leveraging available resources, and practicing regularly, students can overcome challenges and achieve success in the course.
FAQs
What is NPTEL?
NPTEL (National Programme on Technology Enhanced Learning) is a joint initiative of the Indian Institutes of Technology (IITs) and the Indian Institute of Science (IISc) that provides free online courses in various disciplines.
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To enroll in NPTEL courses, simply visit the NPTEL website, create an account, and browse the available courses. Enrollment is free and open to learners worldwide.
Are the assignments in NPTEL courses difficult?
The difficulty of assignments in NPTEL courses varies depending on the course and the individual's level of understanding. While some assignments may be challenging, they are designed to help reinforce learning and assess comprehension.
Can I get help with NPTEL assignments?
Yes, NPTEL offers various resources to help students with their assignments, including lecture notes, discussion forums, and support from course instructors.
Is it necessary to complete all assignments in NPTEL courses?
While completing assignments is not mandatory, it is highly recommended as they provide valuable opportunities for learning and assessment. Additionally, completing assignments may be required to earn a certificate of completion for the course.
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