Quantum Mechanics MCQ Quiz - Objective Question with Answer for Quantum Mechanics - Download Free PDF

Calculation:

In the representation \((\mathbf{s}^2, s_x)\) , the spin matrices are

\(\sigma_x = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \quad \sigma_y = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_z = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \)

with eigenfunctions \(\begin{pmatrix}1 \\ 0\end{pmatrix}, \begin{pmatrix}1 \\ 1\end{pmatrix}, \begin{pmatrix}1 \\ i\end{pmatrix} \) respectively. Thus the Hamiltonian of interaction between the magnetic moment of the particle and the magnetic field is

\(H = -\boldsymbol{\mu} \cdot \mathbf{B} = -\frac{\mu_0 B}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\)

and the Schrödinger equation is

\(i\hbar \frac{d}{dt} \begin{pmatrix} a(t) \\ b(t) \end{pmatrix} = -\frac{\mu_0 B}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} a(t) \\ b(t) \end{pmatrix}\)

where \(\begin{pmatrix} a(t) \\ b(t) \end{pmatrix} \) is the wave function of the particle at time t . Initially we have

\(\begin{pmatrix} a(0) \\ b(0) \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ i \end{pmatrix}\)

and so the solution is

\(\begin{pmatrix} a(t) \\ b(t) \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} \exp\left(i \frac{\mu_0 B t}{2\hbar} \right) \\ i \exp\left(-i \frac{\mu_0 B t}{2\hbar} \right) \end{pmatrix}\)

Hence the probability of the particle being in the state s_y = +1/2 at time t is

\(\left| \frac{1}{\sqrt{2}} (1 \;\; 1) \begin{pmatrix} a(t) \\ b(t) \end{pmatrix} \right|^2 = \frac{1}{4} \left| \exp\left(i \frac{\mu_0 B t}{2\hbar} \right) + i \exp\left(-i \frac{\mu_0 B t}{2\hbar} \right) \right|^2 = \frac{1}{2} \left( 1 + \sin\left( \frac{\mu_0 B t}{\hbar} \right) \right)\)

Similarly, the probability of the particle being in the state s_y = -1/2 at time t is \(\frac{1}{2} \left( 1 - \sin\left( \frac{\mu_0 B t}{\hbar} \right) \right)\)

Calculation:

In the Schrödinger equation

d2ψ/dx2 + 2m[E - V(x)]ψ / ħ2 = 0,

we set E < 0 for a bound state as well as

k2 = 2m |E| / ħ2,    U0 = 2mV0 / ħ2,

and obtain:

d2ψ/dx2 − k2ψ − U0 δ(x)ψ = 0.

Integrating both sides over x from −ε to +ε, where ε is an arbitrarily small positive number, we obtain:

ψ′(ε) − ψ′(−ε) − k2 ∫−εε ψ dx − U0ψ(0) = 0.

With ε → 0+, this becomes:

ψ′(0+) − ψ′(0−) = U0 ψ(0).

For x ≠ 0 the Schrödinger equation has the formal solution ψ(x) ∼ exp(−k|x|) with k positive, which gives:

ψ′(x) ∼ −k |x| / x e−k|x| =  −k e−kx,        x > 0,

ψ′(x) ∼  k ekx,        x < 0 

and hence:

ψ′(0+) − ψ′(0−) = −2kψ(0) = U0ψ(0).

Thus k = −U0/2, which requires V0 to be negative. The energy of the bound state is then:

E = −ħ2k2 / 2m = −mV02 / 2ħ2

and the binding energy is:

Eb = 0 − E = mV02 / 2ħ2.

The wave function of the bound state is:

ψ(x) = A exp( (mV0 / ħ2) |x| ) = √(−mV0 / ħ2) exp(mV0 |x| / ħ2),

where the arbitrary constant A has been obtained by the normalization:

∫−∞0 ψ2 dx + ∫0∞ ψ2 dx = 1.

Concept:

The Schrödinger equation in 0 ≤ x ≤ a gives sinusoidal solutions: ψ(x) = sin(√(2mEx) / ℓ).

The Schrödinger equation in x > a gives exponentially decaying solutions for bound states: ψ(x) = Ae^{-√(2m(V₀ - E))x / ℓ}.

Boundary conditions:

ψ(0) = 0 (since V = ∞ at x < 0).

ψ(x) → 0 as x → +∞ to ensure normalizability.

Matching the wavefunction and its derivative at x = a gives the transcendental equation:

tan(√(2mEa) / ℓ) = -[E / (V₀ - E)]^1/2

Final Answer:

ψ(x) = sin(√(2mEx) / ℓ), 0 ≤ x ≤ a;

ψ(x) = Ae^{-√(2m(V₀ - E))x / ℓ}, x > a.

Calculation:

We choose the set {H, J2, Jz, L2, S2} as a complete set of commuting observables.

For the spin-orbit interaction:

\( H_{so} = A \vec{L} \cdot \vec{S} = \frac{1}{2}A(J^2 - L^2 - S^2) \)

Now for l = 2 and s = 1, the possible values of j are: \( j = 3, 2, 1 \)

Using the formula:

\( E_{so} = \frac{\hbar^2}{2} A \left[j(j+1) - l(l+1) - s(s+1)\right] \)

For \( j = 3 \): \( E_{so} = \frac{\hbar^2}{2}A[12 - 6 - 2] = 2A\hbar^2 \), degeneracy \( d = 2j+1 = 7 \)

For \( j = 2 \): \( E_{so} = \frac{\hbar^2}{2}A[6 - 6 - 2] = -A\hbar^2 \), degeneracy \( d = 5 \)

For \( j = 1 \): \( E_{so} = \frac{\hbar^2}{2}A[2 - 6 - 2] = -3A\hbar^2 \), degeneracy \( d = 3 \)

Calculation:

The wave function is:

\(\psi = Kr(\cos\phi \sin\theta + \sin\phi \sin\theta + 2\cos\theta) e^{-\alpha r}\)

Its angular part is:

\(\psi(\theta, \phi) = K'(\cos\phi \sin\theta + \sin\phi \sin\theta + 2\cos\theta)\)

Using identities:

\(\cos\phi = \frac{1}{2}(e^{i\phi} + e^{-i\phi}), \quad \sin\phi = \frac{1}{2i}(e^{i\phi} - e^{-i\phi})\)

Substituting and simplifying:

\(\psi(\theta, \phi) = K'\left[ \frac{1}{2}(e^{i\phi} + e^{-i\phi})\sin\theta + \frac{1}{2i}(e^{i\phi} - e^{-i\phi})\sin\theta + 2\cos\theta \right]\)

This becomes:

\(\psi(\theta, \phi) = \sqrt{\frac{1}{8\pi}} \left[ -\frac{1}{2}(1 - i)\sqrt{\frac{8\pi}{3}}Y_1^1 + \frac{1}{2}(1 + i)\sqrt{\frac{8\pi}{3}}Y_1^{-1} + 2\sqrt{\frac{4\pi}{3}}Y_1^0 \right] \)

This is a linear combination of spherical harmonics with l = 1. So the total angular momentum is:

\(\sqrt{\langle \vec{L}^2 \rangle} = \sqrt{l(l+1)}\hbar = \sqrt{2}\hbar \)

Hence, the correct answer is (3).

Concept:

We are using properties of angular momentum viz-\(=^2+^2+^2\) and by putting given values of \(L\) and \(L_z\) we can find value of \(L_x\).

Explanation:

 Given,

•  \(L^2|\phi>=6\hbar^2|\phi>\)
• \(L_z|\phi>=2\hbar|\phi>\)

Here L is angular momentum and \(\hbar\) is Planck's constant

Using angular momentum formula, we can write expectation values of angular momentum as

• \(=^2+^2+^2\)

Applying ket, bra operator on angular momentum operator, we get

• \(<\phi|L^2|\phi>=<\phi|L_x^2|\phi>+<\phi|L_y^2|\phi>+<\phi|L_z^2|\phi>\)

• \(<\phi|L^2|\phi>=<\phi|L_x^2|\phi>+<\phi|L_y^2|\phi>+<\phi|L_z^2|\phi>\)

Using, \(<\phi|\phi>=1\), and putting given values of \(L\) and\(L_z\) we get,

• \(6\hbar^2=++4\hbar^2\)
• Now, \(\)\(=\)

we get, \(\frac {6\hbar^2-4\hbar^2} {2}=\)

• \(=\hbar^2\)

The correct answer is \(=\hbar^2\)

Concept:

The given Hamiltonian is for the Anharmonic Oscillator. We will compare the given Hamiltonian with the equation of the Hamiltonian Anharmonic oscillator.

•  \(H=\frac{p_x^2}{2 m}+\frac{p_y^2}{2 m}+\frac{1}{2} m ω^2 x^2+2 m ω^2 y^2\)
• \(H=\frac{p_x^2}{2 m}+\frac{p_y^2}{2 m}+\frac{1}{2} m ω^2 x^2+\frac {1}{2}m (2ω)^2 y^2\)

Explanation:

• \(H=\frac{p_x^2}{2 m}+\frac{p_y^2}{2 m}+\frac{1}{2} m ω^2 x^2+2 m ω^2 y^2\)
• \(H=\frac{p_x^2}{2 m}+\frac{p_y^2}{2 m}+\frac{1}{2} m ω^2 x^2+\frac {1}{2}m (2ω)^2 y^2\)
• \(\omega_x=\omega\), \(\omega_y=2\omega\)
• \(E= (n_x+\frac{1}{2})\hbar\omega_x+(n_y+\frac{1}{2})\hbar\omega_y\)

This is the formula for energy in an oscillator in two-dimension.

Now, \(E=\)\(\frac{27}{2} \hbar \omega\)(given)

• \(\frac{27}{2} \hbar \omega\)\(= (n_x+\frac{1}{2})\hbar\omega_x+(n_y+\frac{1}{2})\hbar\omega_y\)
• Substitute values of \(\omega_x\) and \(\omega_y\) in terms of \(\omega\)
• \(\frac{27}{2} \hbar \omega\)\(= (n_x+\frac{1}{2})\hbar\omega+2(n_y+\frac{1}{2})\hbar\omega\)
• \(\frac {27}{2}=n_x+\frac{1} {2}+2n_y+1\)
• \(12=n_x+2n_y\)
• \(n_x\) should be even to satisfy the equation.
• \(n_x\)\(=0,2,4,6,8,10,12\) ; \(n_y=6,5,4,3,2,1,0\)
• \((n_x,n_y)=(0,6),(2,5),(4,4), (6,3), (8,2),(10,1), (12,0)\)
• So, the degeneracy of the energy level \(E=\)\(\frac{27}{2} \hbar \omega\) is 7.

So, the correct answer is 7.

Explanation:

• The wave function for each particle can be written in terms of the energy eigenstates of the infinite square well. The ground state and the first excited state of a particle (in a one-dimensional infinite square well of width $a$) are given by:

\(\psi_0(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right)\) , for the ground state, and

\(\psi_1(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{2\pi x}{a}\right)\) , for the first excited state.

• The expectation value of the product of the position operators \(x_1\) and \(x_2\) can be calculated with the help of the integral:

\(\langle x_1 x_2 \rangle = \int dx_1 \int dx_2\) \(\psi_0(x_1)\psi_1(x_2) x_1 x_2 \psi_0(x_1) \psi_1(x_2)\),

• Where the integration is over the range [0, a] for both \(x_1\) and \(x_2\).
• However, since the particles are distinguishable and non-interacting, we can write \(\langle x_1 x_2 \rangle = \langle x_1 \rangle \langle x_2 \rangle \), which simplifies the integrations to:

\(\langle x_1 \rangle = \int dx_1 \psi_0(x_1) x_1 \psi_0(x_1)\) , and \(\langle x_2 \rangle = \int dx_2 \psi_1(x_2) x_2 \psi_1(x_2)\).

• Calculation of these integrals gives: \(\langle x_1 \rangle = \frac{a}{2}\), and \(\langle x_2 \rangle = \frac{a}{2}\).
• So, the expectation value of the product of \(x_1\) and \( x_2\) is just the product of the expectation values of \(x_1\) and \(x_2\): \(\langle x_1 x_2 \rangle = \langle x_1 \rangle \langle x_2 \rangle = \frac{a^2}{4}\)

Solution-Option-1

Concept:

First, we will check the number of radial nodes in the graph which is given by

• Number of radial nodes \(=n-l-1\)
• Then secondly we will check the value of probability of electron at \(r=2a\).

Calculation-

• Given- \(n=2\) and \(l=0\)
• \(R_{20}=N\left(1-\frac{r}{2 a}\right) e^{-\frac{r}{2 a}}\)
• Radial nodes\(=n-l-1\) \(=2-0-1=1\)
• So, there will be only one node in the graph. Either option 1 is correct or option 2 is correct.
• Now we will check the probability of finding an electron at \(r=2a\).
• At, \(R_{20}=N(1-\frac{r}{2a})\mathrm{e}^\frac{-r}{2a}\)   
• \(P_r=|R_{20}|^2=N^2(1-\frac{r}{2a})^2\mathrm{e}^{\frac{-r}{a}}\)
• At \(r=2a\), \(P_r=0\)
• The probability of finding an electron at r=2a" id="MathJax-Element-14-Frame" role="presentation" style="position: relative;" tabindex="0">r=2ar=2a" id="MathJax-Element-270-Frame" role="presentation" style="position: relative;" tabindex="0">r=2ar=2a" id="MathJax-Element-124-Frame" role="presentation" style="position: relative;" tabindex="0">r=2ar=2a r=2a r=2a $r=2a$  is zero.
• So, graph 1 satisfies this condition.

So, the correct answer is Graph-1.

Explanation:

 \(E_n=nE_0\)(given)

Case-1-Initial ground state energy without polarization

According to Pauli Exclusion principle, only two electrons filled in one state.

• Initial ground state energy \(E_i=2\times0+2\times E_0+2\times 2E_0+2\times3E_0+-------+2\times (\frac{N-2}{2})E_0\)
• \(E_i=2E_0[1+2+3+-----------+\frac {N-2}{2}]\)
• Now, \(\sum[1+2+3+-------N]=\frac{N(N+1)}{2}\)
• \(\sum[1+2+3+-------\frac{N-2} {2}]=\frac {(\frac{N-2}{2}) (\frac {N} {2})} {2}\)
• \(E_i=2E_0\times \)\(\frac {(\frac{N-2}{2}) (\frac {N} {2})} {2}\)\(=\frac{N^2E_0}{4}-\frac{NE_0} {2}\)

Case-2-Final ground state energy after polarization

After polarization, only one electron filled in the state.

• \(E_f=1\times0+1\times E_0+1\times 2E_0+1\times3E_0+...+1\times (N-1)E_0\)
• \(E_f=E_0[1+2+3+-----------+(N-1)]\)
• \(\sum[1+2+3+-------N]=\frac{N(N+1)}{2}\)
• \(\sum[1+2+3+-------+(N-1)=\frac{N(N-1)}{2}\)
• \(E_f=\frac{N^2E_0}{2} -\frac {NE_0}{2}\)

The change in ground state energy is \(E_f-E_i=\frac{N^2 E_0} {2}-\frac {NE_0}{2}-\frac {N^2E_0} {4}+\frac {NE_0}{2}=\frac {N^2 E_0}{4}\)

So, the correct answer is \(\frac{1}{4} N^2 E_0\).

Concept:

The momentum operator is given by

p = - ih \({\partial \over \partial x}\)

where h is the Plank constant.

Calculation:

e^{\({iaP\over h}\)} |x>

= [\(\sum^{\infty}_{n=0} {1 \over n!}({iaP\over h})^n\) ]|x>

= \([\sum^{\infty}_{n=0} {1 \over n!}({iaP\over h})^n]^h \)|x>

= |x> - a∇|x> + \({1 \over 2!}\) (a∇)²|x> ... = |x-a>

X|x-a> = (x-a)|x-a>

The correct answer is an option (3).

CONCEPT: 

1. Energy Operator:

The energy (or Hamiltonian) operator in one dimension for a particle in a box is given by:

\(\hat{H} = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2}\)

2. Expectation Value of Energy:

The expectation value of energy 〈 E 〉  is given by:

〈 E 〉 = \(\int_0^1 ψ^ *(x) \hat{H} ψ(x) \, dx\)

Calculation -

1. Wavefunction:
   
\(ψ(x) = \sqrt{\frac{8}{5}} \sin(\pi x)(1 + \cos(\pi x))\)

2. Second Derivative:
   
\(\frac{d}{dx} \left( \sin(\pi x)(1 + \cos(\pi x)) \right) = \pi \cos(\pi x)(1 + \cos(\pi x)) - \pi \sin^2(\pi x)\)
   
\(\frac{d^2}{dx^2} \left( \sin(\pi x)(1 + \cos(\pi x)) \right) = -\pi^2 \sin(\pi x)(1 + \cos(\pi x)) - 2\pi^2 \cos(\pi x) \sin(\pi x) \)

Simplifying:

\(\frac{d^2}{dx^2} \left( \sin(\pi x)(1 + \cos(\pi x)) \right) = -\pi^2 \sin(\pi x) (1 + 2\cos(\pi x) + \cos^2(\pi x)) \)

3. Hamiltonian Acting on ψ(x):
   
\(\hat{H} ψ(x) = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} ψ(x) \)   

4. Expectation Value Integral:
   
〈 E 〉 = \(\int_0^1 ψ(x) \left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} ψ(x)\right) dx\)

The normalized wavefunction ψ(x) given is a linear combination of eigenfunctions of the infinite potential well.

For an infinite potential well, the eigenfunctions are

\(ψ_n(x) = \sqrt{2} \sin(n \pi x)\) with energy eigenvalues \(E_n = \frac{n^2 \pi^2 \hbar^2}{2m} .\)

Comparing the given wavefunction:

\(ψ(x) = \sqrt{\frac{8}{5}} \sin(\pi x)(1 + \cos(\pi x))\)

This is equivalent to a superposition of the first and second eigenstates.

The coefficients and normalization ensure that this wavefunction is a proper eigenstate mixture.

Energy Calculation:

By symmetry and orthogonality of the eigenfunctions, the expectation value of energy 〈 E 〉 is the weighted sum of eigenvalues:

〈 E 〉 = \(a_1^2 E_1 + a_2^2 E_2\)

Given:

\(E_1 = \frac{\pi^2 \hbar^2}{2m}, \quad E_2 = \frac{4 \pi^2 \hbar^2}{2m}\)

The weights a₁ , a₂ are found from normalization. Simplifying using known integrals, we obtain the correct weighted sum.

Finally, the result for this particular problem (via solving) yields the expectation value \(\frac{4\pi^2 \hbar^2}{5m}\)

Therefore, the correct answer is (2).

Explanation:

Let us reconsider the system of equations:

\(\frac{dA}{dt} = -iωB\) and \(\frac{dB}{dt} = iωA.\)

By differentiating the first equation again, \( \frac{d²A}{dt²} = -iω \frac{dB}{dt}.\)

• Substituting the second equation into this results in \(\frac{d²A}{dt²} = -ω² A.\)
• This differential equation is a simple harmonic one, but with a key difference: there is no negative in front of ω², leading to hyperbolic solutions.
• Specifically, we find A(t) = Csinh(ωt), for some constant C.
• Given that the expectation value \(〈A⟩_ψ(t) = 0\) at t = 0, we find \(C = 0.\)
• Thus, in general, B(t) has to be in the form of cosh(ωt), to meet the commutation relations. Finally, given that \(〈B⟩_ψ (0) = i,\) we need to multiply cosh(ωt) by i.
• So the time-evolved expectation value is \( 〈A⟩_ψ(t) = 〈ψ|A(t)| ψ⟩ = 〈ψ|Csinh(ωt)| ψ⟩ = sinh(ωt)\)

Concept: 

 Energy in first perturbation is given by

• \(E_1^{(1)}=<\psi|H^\prime|\psi>\)
• We know that, for an infinite potential well, \(\psi=\sqrt{\frac {2} {L}}cos\frac {\pi x} {L}\) 

Explanation:

Given-

• \(H^{\prime}=ϵ \cos \left(\frac{\pi x}{L}\right)\) limits from \(\frac {-L} {2}\) to \(\frac {+L} {2}\)

We know that, for an infinite potential well, \(\psi=\sqrt{\frac {2} {L}}cos\frac {\pi x} {L}\) 

• \(E_1^{(1)}=<\psi|H^\prime|\psi>\)
• \(E_1^{(1)}=\int\limits_\frac{-L} {2}^\frac{+L}{2} |\psi|^2 H^{\prime}dx\)

Now, \(\psi=\sqrt{\frac {2} {L}}cos\frac {\pi x} {L}\) and \(H^{\prime}=ϵ \cos \left(\frac{\pi x}{L}\right)\), put these values in first order energy perturbation equation, we get,

• \(E_1^{(1)}=\int\limits_\frac{-L} {2}^\frac{+L}{2} [\sqrt {\frac{2} {L}}cos\frac{\pi x}{L}]^2 .\epsilon cos\frac{\pi x} {L} dx\)
• \(E_1^{(1)}=\frac {2\epsilon} {L}\int\limits_\frac{-L} {2}^\frac{+L}{2} cos^3\frac {\pi x} {L}dx\)

Now for changing the limit from(\(\frac{-L} {2}\) to \(\frac {+L} {2}\)) to (\(0\) to \(\frac {+L} {2}\))

•  \(E_1^{(1)}=\frac {2\epsilon} {L}\times 2\int\limits_0^\frac {+L}{2}\cos^3\frac {\pi x} {L}dx\)

Using the trignometric formula for \(cos^3\frac {\pi x} {L}\),

• \(cos^3\frac{\pi x }{L}=\frac{[cos(\frac{3\pi x}{L})+3cos(\frac {\pi x}{L})]} {4}\)
• Substituting this value, we get,
• \(E_1^{(1)}=\frac {4\epsilon} {L}\int\limits_0^\frac {+L}{2}\frac{[cos(\frac{3\pi x}{L})+3cos(\frac {\pi x}{L})]} {4}\)
• \(E_1^{(1)}=\frac {\epsilon} {L} [(\frac {Sin\frac {3\pi x}{L}} {3\pi/L})|_0^{L/2} + 3(\frac {Sin\frac{\pi x}{L}} {\pi/L})|_0^{L/2}\)

• \(E_1^{(1)}=\frac {\epsilon} {L} \times\frac {L} {\pi} [\frac{1}{3}(Sin\frac{3\pi} {2}-Sin0)+3(Sin\frac {\pi}{2}-Sin0)]\)
• \(E_1^{(1)}=\frac {\epsilon} {L} [\frac{-1} {3}+3]=\frac {8\epsilon} {3\pi }\)

So, the correct answer is \(\frac{8 \epsilon}{3 \pi}\).

Concept:

A generator is an operator that acts on the wave function or quantum state vector, to produce the effect of applying a small transformation to the system.

Calculation:

q → q' = (1 + ϵ)q

p → p' = (1 - ϵ)p

If G is the generator then 

p' - p = δ p_j 

= - ϵ \({\partial G \over \partial q_j}\)

= - ϵ p

q' - q = δ q_i 

= ϵ \({\partial G \over \partial p_j}\)

= ϵ p

Now G = qp

- ϵ \({\partial G\over \partial q} \) = - ϵ p = δ p

ϵ \({\partial G\over \partial p} \) = - ϵ q = δ q

The correct answer is option (2).