Concept:-

Perturbation Theory:

• Perturbation theory is an approximation method to find out the exact solution of any system with great accuracy.
• To break the energy correction in a particular energy level, we have to disturb the operator in a specific order.
• The energy correction is given by,

$\mathrm{\Delta }E=\int {{\mathrm{\Psi }}^{\ast }}_{\circ }{H}^I{\mathrm{\Psi }}_{\circ }d\tau$

The perturbed energy levels upto first order are given by,

​$E=E^{\left(0\right)}+E^{\left(I\right)}$

Explanation:-

For a hydrogen atom that is exposed to a perturbation

V= E.z

The orbital which will contribute to the first-order correction to the wave function will be,

​$E_1^1=\int \mathrm{\Psi }\ast \hat{H}\mathrm{\Psi }d\tau \ne 0$

For 2s, 3p_y, and 3dz2 orbitals the first-order correction to the wave function will be

​${\mathrm{E}}_1^1=⟨{\mathrm{\Psi }}_{2\mathrm{s}}|\mathrm{V}|{\mathrm{\Psi }}_{2\mathrm{s}}⟩=0$,

${\mathrm{E}}_1^1=⟨{\mathrm{\Psi }}_{3{\mathrm{p}}_{\mathrm{y}}}|\mathrm{V}|{\mathrm{\Psi }}_{3{\mathrm{p}}_{\mathrm{y}}}⟩=0$, and

 ${\mathrm{E}}_1^1=⟨{\mathrm{\Psi }}_{{3\mathrm{d}}_{\mathrm{z}}^2}|\mathrm{V}|{\mathrm{\Psi }}_{{3\mathrm{d}}_{\mathrm{z}}^2}⟩=0$

While for 2p_z orbital, the first-order correction to the wave function will be

​$E_1^1=⟨{\mathrm{\Psi }}_{2p_z}|V|{\mathrm{\Psi }}_{2p_z}⟩\ne 0$​

As the value of first-order correction to the wave function for 2pz orbital is non-zero, it will contribute to the first-order correction.

Conclusion:-

Hence, the first-order correction to the wave function comes only from the orbital 2pz

CONCEPT:

Quantum Tunneling and Potential Barrier

• A quantum particle with energy E < V₀ encountering a finite potential barrier of height V₀ and width L experiences a phenomenon called quantum tunneling.
• The behavior of the wave function ψ (x) is different in each of the three regions:
  • Region I (x < 0): Free particle, oscillatory wave function (e.g., \sin or \cos), representing incident and reflected waves.
  • Region II (0 < x < L): Inside the barrier, where E < V₀, the wave function becomes non-oscillatory, showing exponential decay  $e^{-\kappa x}$and slight transmission ($e^{\kappa x}$).
  • Region III (x > L): After the barrier, the wave function is again oscillatory but with much reduced amplitude due to tunneling effect.
• The square modulus |ψ(x)|² represents the probability density, which:
  • Oscillates before the barrier.
  • Decays exponentially within the barrier.
  • Re-emerges with smaller amplitude after the barrier (i.e., low but non-zero tunneling probability).

EXPLANATION:

• Option 2 correctly shows:
  • Oscillatory |ψ(x)|² in region I (incident and reflected wave superposition).
  • Exponential decay of probability inside the barrier (Region II), consistent with E < V₀.
  • Small amplitude oscillation in region III due to transmitted wave post-barrier.
• Other options are incorrect:
  • Option 1: Shows exponential decay beyond the barrier which is unphysical as transmission must allow oscillations post-barrier.
  • Option 3: Suggests oscillations inside the barrier, valid only for E > V₀.
  • Option 4: Shows no decay inside the barrier, violating tunneling theory.

Therefore, the correct answer is Option 2.

CONCEPT:

First-Order Perturbation Theory

• The first-order perturbation correction to the energy is given by the formula:

  E₁' = <ψ₀|V|ψ₀>

• For a particle in a one-dimensional box of length L, the ground state wavefunction is:

  ψ₀(x) = √(2/L) sin(πx/L)

• For the given perturbing potential:
  • V = V for 0 ≤ x ≤ L/2
  • V = V/3 for L/2 ≤ x ≤ L
• The first-order energy correction is obtained by integrating the wavefunction over the regions where the potential is different:
  • First region (0 ≤ x ≤ L/2) with potential V
  • Second region (L/2 ≤ x ≤ L) with potential V/3

EXPLANATION:

• The first-order correction involves splitting the total integral into two parts for the two potential regions:
  • Integral from 0 to L/2 for potential V
  • Integral from L/2 to L for potential V/3
• After performing the integrals, we get the following results:
  • Contribution from the first region: ½ V
  • Contribution from the second region: ⅙ V
• Therefore, the total first-order energy correction is:

  E₁' =  (VxL/2 + V/3 xL/2) / L = V/2 + V/3 = 4V/6 = 2V/3

Therefore, the correct answer is 2V/3.

Concept:

Kronecker Delta Function

• The Kronecker delta function, denoted as δ_nk, is a discrete function used in summation notation and matrix representations.
• It is defined as:

  δ_nk =

  • 1, if n = k
  • 0, if n ≠ k
• The function acts as an identity selector, ensuring that only terms with matching indices contribute in summations.

Explanation:

• Option 1: δ_nk = 2, n = k and δ_nk = 2, n ≠ k
  • Incorrect, because the Kronecker delta is not defined as 2 for any value of n and k.
• Option 2: δ_nk = 3, n = k and δ_nk = 3, n ≠ k
  • Incorrect, because δ_nk is defined as 1 when n = k and 0 otherwise.
• Option 3: δ_nk = 1, n = k and δ_nk = 0, n ≠ k
  • Correct, as it follows the standard definition of the Kronecker delta function.
• Option 4: δ_nk = 0, n = k and δ_nk = 1, n ≠ k
  • Incorrect, as it incorrectly reverses the conditions for δ_nk.

Therefore, the correct answer is δ_nk = 1 when n = k and δ_nk = 0 when n ≠ k.

The correct answer is The probability distributions of px, py, and pz are independent of angles θ and ϕ, and their sum leads to a constant overall probability distribution.

EXPLANATION:

The angular components of px, py, and pz are:

px = (3 / 4π) sinθ cosϕ,

py = (3 / 4π) sinθ sinϕ,

pz = (3 / 4π) cosθ.

The combined distribution px2 + py2 + pz2 is:

px2 + py2 + pz2 = (3 / 4π) (sin2θ cos2ϕ + sin2θ sin2ϕ + cos2θ).

Using trigonometric identities:

This simplifies to:

px2 + py2 + pz2 = 3 / 4π.

The result px2 + py2 + pz2 = 3 / 4π shows that the total probability distribution is constant and does not depend on θ or ϕ. This independence of angles confirms spherical symmetry.

CONCEPT:

Spherical Symmetry

• Angular Components of Probability Distributions:
• Combined Distribution:
  • cos2ϕ + sin2ϕ = 1
  • sin2θ + cos2θ = 1
• Spherical Symmetry:
•  
  • The result p²_x + p²_y + p²_z = ³/_4π shows that the total probability distribution is constant and does not depend on θ or ϕ.
  • This independence of angles confirms spherical symmetry.

CONCLUSION:

The spherical symmetry arises because the sum of the angular components of px, py, and pz yields a constant value independent of direction. Thus, the electron distribution in the ground state of the neon atom is spherically symmetrical.

CONCEPT:

First-Order Energy Correction using Perturbation Theory

• Perturbation theory is a method used to approximate the energy levels of a quantum system when a small perturbing potential is added to the Hamiltonian.
• For a particle in a 1D box, the unperturbed ground state energy and wave function are well-known, and we can use the first-order correction formula to find the energy change due to a perturbation.
• The first-order correction to the ground state energy due to a perturbation ($\lambda x$ ) is given by the expectation value of the perturbation in the ground state wave function:
  • $\mathrm{\Delta }{E}_1^{(1)}=⟨{\psi }_0|\lambda x|{\psi }_0⟩$

CALCULATION:

• The perturbation term is $\mathrm{\Delta }H=\lambda x$.
• The first-order correction to the energy is given by:
  • $\mathrm{\Delta }{E}_1^{(1)}={\int }_0^L{\psi }_0(x)\lambda x{\psi }_0(x)dx$
• For a particle in a box of length L" id="MathJax-Element-97-Frame" role="presentation" style="position: relative;" tabindex="0">L$L$ $L$ , the ground state wave function is:
  • ${\psi }_0(x)=\sqrt{\frac{2}{L}}\sin \left(\frac{\pi x}{L}\right)$
• Substituting the ground state wave function, we get:
  • $\mathrm{\Delta }{E}_1^{(1)}=\lambda {\int }_0^L\frac{2}{L}x{\sin }^2\left(\frac{\pi x}{L}\right)dx$
• Evaluating the integral, we find that it simplifies to:
  • $\mathrm{\Delta }{E}_1^{(1)}=\lambda \frac{L}{2}$

CONCLUSION:

• The first-order correction to the ground state energy due to the perturbation $\lambda x$ is:
  • Option (1): $\frac{\lambda L}{2}$

Concept:

In quantum mechanics, the second-order perturbation correction to the ground state energy of a system is given by: $E_0^{(2)}=\sum _{n\ne 0}\frac{|V_{0n}{|}^2}{E_0-E_n}$

Where:

• $E_0^{(2)}$ is the second-order correction to the ground state energy.

• V0n are the matrix elements of the perturbation between the ground state and the higher states.

• E0 and En are the unperturbed energies of the ground state and the higher states, respectively.

Explanation: 

For the second-order correction to the ground state energy, we calculate the contribution from each excited state:

• $E_0^{(2)}=\frac{|V_{10}{|}^2}{E_0-E_1}+\frac{|V_{20}{|}^2}{E_0-E_2}$

• Step 1: Identify the unperturbed energies:

  • $E_0=0\text{eV},E_1=8\text{eV},E_2=12\text{eV}$

• Step 2: The matrix elements of the perturbation (V0n) are given as:

  • $V_{10}=6\text{eV},V_{20}=8\text{eV},V_{12}=14\text{eV}$

• Step 3: Apply the second-order correction formula:

• Step 4: Calculate the contributions:

  • For the first term: $\frac{|V_{10}{|}^2}{E_0-E_1}=\frac{6^2}{0-8}=\frac{36}{-8}=-4.5\text{eV}$

  • For the second term: $\frac{|V_{20}{|}^2}{E_0-E_2}=\frac{8^2}{0-12}=\frac{64}{-12}=-5.33\text{eV}$

• Step 5: Add the contributions:

  • $E_0^{(2)}=-4.5\text{eV}+(-5.33\text{eV})=-9.83\text{eV}$

Conclusion:

The second-order correction to the ground state energy is -9.83 eV.

Concept:

Hückel Molecular Orbital (HMO) Theory is a method used to determine the electron energies of π-electrons in conjugated systems. Key points about Hückel Theory:

• Only π-electrons are considered: Hückel theory focuses solely on the π-electrons in conjugated systems, ignoring the σ-electrons as they do not contribute to conjugation or delocalization.
• Assumption of planar molecules: The theory assumes that molecules are planar, allowing p orbitals to overlap effectively to form a π-bonding framework.
• Use of linear combination of atomic orbitals (LCAO): Hückel theory uses the LCAO method to construct molecular orbitals as combinations of atomic p orbitals.
• Energy levels calculated using secular determinants: Hückel theory applies a secular determinant to solve for the energies of π-electrons, which are expressed in terms of α (Coulomb integral) and β (resonance integral).

Explanation: 

For cyclopropene, which has three carbon atoms in a conjugated system, we set up the Hückel determinant to calculate the π-electron energy levels.

• The secular determinant for a three-atom system like cyclopropene is given by:

  • $\left|\begin{array}{ccc}\alpha -E& \beta & 0\\ \beta & \alpha -E& \beta \\ 0& \beta & \alpha -E\end{array}\right|=0$

• Expanding this determinant, we get the characteristic equation:

  • $(\alpha -E)[(\alpha -E{)}^2-{\beta }^2]-{\beta }^2(\alpha -E)=0$

• Simplifying further, this equation factors into:

  • $(\alpha -E)[(\alpha -E{)}^2-2\beta (\alpha -E)+{\beta }^2]=0$

• The solutions to this equation are:

  • $E_1=\alpha +2\beta$
  • $E_2=\alpha -\beta$
  • $E_3=\alpha -\beta$

Conclusion:

The correct option is: α + 2β, α - β, α - β

Concept:

The Variational Principle in quantum mechanics is a method used to estimate the ground state energy of a system. According to this principle, the expectation value of the Hamiltonian for any trial wavefunction provides an upper bound to the ground state energy. The energy obtained from this trial wavefunction will be minimized with respect to the variational parameter(s) to approximate the lowest possible energy.

• Trial Wavefunction: A trial wavefunction is chosen based on the expected behavior of the system. Here, the trial wavefunction is $e^{-C{x}^2}$, which is commonly used for the Simple Harmonic Oscillator (SHO) due to its Gaussian form.
• Energy Expression: For the given wavefunction, the energy expression obtained is $\frac{{\hslash }^{2}C}{2m}+\frac{m{\omega }^2}{8C}$.
• Minimization: To find the minimum energy, the energy expression is minimized with respect to the variational parameter ( C ).

Explanation: 

To find the minimum value of the energy expression $E=\frac{{\hslash }^{2}C}{2m}+\frac{m{\omega }^2}{8C}$, we differentiate with respect to ( C ) and set the derivative to zero.

• Differentiate ( E ) with respect to ( C ):

  • $\frac{dE}{dC}=\frac{{\hslash }^2}{2m}-\frac{m{\omega }^2}{8C^2}$

•  $\frac{dE}{dC}=0$ at the critical point:

  • $\frac{{\hslash }^2}{2m}=\frac{m{\omega }^2}{8C^2}$

• Solve for ( C ):

  • $C=\frac{m\omega }{2\hslash }$

• Substitute $C=\frac{m\omega }{2\hslash }$ back into the energy expression to find the minimum energy:

  • $E_{\text{min}}=\frac{{\hslash }^2}{2m}\cdot \frac{m\omega }{2\hslash }+\frac{m{\omega }^2}{8}\cdot \frac{2\hslash }{m\omega }$

• Simplify the expression:

  • $E_{\text{min}}=\frac{\hslash \omega }{4}+\frac{\hslash \omega }{4}=\frac{\hslash \omega }{2}$

Conclusion:

The correct option is: $\frac{\hslash \omega }{2}$

Concept:

In quantum mechanics, the first-order correction to the energy in perturbation theory is calculated using the expectation value of the perturbing Hamiltonian ( H₁ ) with respect to the unperturbed wavefunction. This correction is given by:

$\mathrm{\Delta }{E}^{(1)}=⟨{\psi }_0|H_1|{\psi }_0⟩$

• Perturbation Hamiltonian: Here, the perturbation $H_1=P\sin \varphi$ acts on a rotating particle.
• Symmetry Consideration: Since $\sin \varphi$ is an odd function with respect to rotation by $\pi$, it changes sign under such transformation. However, the wavefunctions for a rotating particle are typically eigenfunctions of angular momentum, which are either even or odd functions.
• Expectation Value of Odd Functions: For an odd function like $\sin \varphi$, its expectation value with an even function (such as the ground state wavefunction) over a symmetric interval (e.g., ( 0 ) to (2π) is zero. This is because the integral of an odd function over a symmetric range cancels out.

Explanation: 

• The first-order energy correction is calculated as:

  • $\mathrm{\Delta }{E}^{(1)}=⟨{\psi }_0|P\sin \varphi |{\psi }_0⟩$

• Since $\sin \varphi$ is an odd function, and the unperturbed wavefunction ${\psi }_0$ is typically symmetric (even), the integral of ${\psi }_0^{\ast }\sin \varphi {\psi }_0$ over a full rotation (from ( 0 ) to (2π) will be zero:

  • ${\int }_0^{2\pi }{\psi }_0^{\ast }\sin \varphi {\psi }_{0}d\varphi =0$

• Thus, the first-order correction to the energy is zero due to the symmetry properties of the wavefunction and the perturbing term.

Conclusion:

The correct option is: 0