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

${\sigma }_x=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),{\sigma }_y=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),{\sigma }_z=\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right)$

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

$H=-\text{boldsymbol}\mu \cdot \mathbf{B}=-\frac{{\mu }_{0}B}{2}\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$

and the Schrödinger equation is

$i\hslash \frac{d}{dt}\left(\begin{array}{c}a(t)\\ b(t)\end{array}\right)=-\frac{{\mu }_{0}B}{2}\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{c}a(t)\\ b(t)\end{array}\right)$

where $\left(\begin{array}{c}a(t)\\ b(t)\end{array}\right)$ is the wave function of the particle at time t . Initially we have

$\left(\begin{array}{c}a(0)\\ b(0)\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ i\end{array}\right)$

and so the solution is

$\left(\begin{array}{c}a(t)\\ b(t)\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{c}\exp \left(i\frac{{\mu }_{0}Bt}{2\hslash }\right)\\ i\exp (-i\frac{{\mu }_{0}Bt}{2\hslash })\end{array}\right)$

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

${|\frac{1}{\sqrt{2}}(11)\left(\begin{array}{c}a(t)\\ b(t)\end{array}\right)|}^2=\frac{1}{4}{|\exp \left(i\frac{{\mu }_{0}Bt}{2\hslash }\right)+i\exp (-i\frac{{\mu }_{0}Bt}{2\hslash })|}^2=\frac{1}{2}(1+\sin \left(\frac{{\mu }_{0}Bt}{\hslash }\right))$

Similarly, the probability of the particle being in the state s_y = -1/2 at time t is $\frac{1}{2}(1-\sin \left(\frac{{\mu }_{0}Bt}{\hslash }\right))$

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:

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

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

Using the formula:

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

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

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

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

Calculation:

The wave function is:

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

Its angular part is:

$\psi (\theta ,\varphi )=K^{\prime }(\cos \varphi \sin \theta +\sin \varphi \sin \theta +2\cos \theta )$

Using identities:

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

Substituting and simplifying:

$\psi (\theta ,\varphi )=K^{\prime }[\frac{1}{2}(e^{i\varphi }+e^{-i\varphi })\sin \theta +\frac{1}{2i}(e^{i\varphi }-e^{-i\varphi })\sin \theta +2\cos \theta ]$

This becomes:

$\psi (\theta ,\varphi )=\sqrt{\frac{1}{8\pi }}[-\frac{1}{2}(1-i)\sqrt{\frac{8\pi }{3}}{Y}_1^1+\frac{1}{2}(1+i)\sqrt{\frac{8\pi }{3}}{Y}_1^{-1}+2\sqrt{\frac{4\pi }{3}}{Y}_1^0]$

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

$\sqrt{⟨{\overrightarrow{L}}^2⟩}=\sqrt{l(l+1)}\hslash =\sqrt{2}\hslash$

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,

•  $L^2|\varphi >=6{\hslash }^2|\varphi >$
• $L_z|\varphi >=2\hslash |\varphi >$

Here L is angular momentum and $\hslash$ 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

• $<\varphi |L^2|\varphi >=<\varphi |L_x^2|\varphi >+<\varphi |L_y^2|\varphi >+<\varphi |L_z^2|\varphi >$

• $<\varphi |L^2|\varphi >=<\varphi |L_x^2|\varphi >+<\varphi |L_y^2|\varphi >+<\varphi |L_z^2|\varphi >$

Using, $<\varphi |\varphi >=1$, 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.

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

Explanation:

• $H=\frac{p_x^2}{2m}+\frac{p_y^2}{2m}+\frac{1}{2}m{\omega }^2{x}^2+2m{\omega }^2{y}^2$
• $H=\frac{p_x^2}{2m}+\frac{p_y^2}{2m}+\frac{1}{2}m{\omega }^2{x}^2+\frac{1}{2}m(2\omega {)}^2{y}^2$
• ${\omega }_x=\omega$, ${\omega }_y=2\omega$
• $E=(n_x+\frac{1}{2})\hslash {\omega }_x+(n_y+\frac{1}{2})\hslash {\omega }_y$

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

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

• $\frac{27}{2}\hslash \omega$$=(n_x+\frac{1}{2})\hslash {\omega }_x+(n_y+\frac{1}{2})\hslash {\omega }_y$
• Substitute values of ${\omega }_x$ and ${\omega }_y$ in terms of $\omega$
• $\frac{27}{2}\hslash \omega$$=(n_x+\frac{1}{2})\hslash \omega +2(n_y+\frac{1}{2})\hslash \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}\hslash \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:

$⟨x_1{x}_2⟩=\int d{x}_1\int d{x}_2$ ${\psi }_0(x_1){\psi }_1(x_2)x_1{x}_2{\psi }_0(x_1){\psi }_1(x_2)$,

• 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 $⟨x_1{x}_2⟩=⟨x_1⟩⟨x_2⟩$, which simplifies the integrations to:

$⟨x_1⟩=\int d{x}_1{\psi }_0(x_1)x_1{\psi }_0(x_1)$ , and $⟨x_2⟩=\int d{x}_2{\psi }_1(x_2)x_2{\psi }_1(x_2)$.

• Calculation of these integrals gives: $⟨x_1⟩=\frac{a}{2}$, and $⟨x_2⟩=\frac{a}{2}$.
• 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 $x_2$: $⟨x_1{x}_2⟩=⟨x_1⟩⟨x_2⟩=\frac{a^2}{4}$

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 $l=0$
• $R_{20}=N(1-\frac{r}{2a})e^{-\frac{r}{2a}}$
• 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=2a$r=2a$ r=2a r=2a $r=2a$  is zero.
• So, graph 1 satisfies this condition.

So, the correct answer is Graph-1.

Explanation:

 $E_n=n{E}_0$(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 $E_i=2\times 0+2\times E_0+2\times 2E_0+2\times 3E_0+-------+2\times (\frac{N-2}{2})E_0$
• $E_i=2E_0[1+2+3+-----------+\frac{N-2}{2}]$
• 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^2{E}_0}{4}-\frac{N{E}_0}{2}$

Case-2-Final ground state energy after polarization

After polarization, only one electron filled in the state.

• $E_f=1\times 0+1\times E_0+1\times 2E_0+1\times 3E_0+...+1\times (N-1)E_0$
• $E_f=E_0[1+2+3+-----------+(N-1)]$
• $\sum [1+2+3+-------N]=\frac{N(N+1)}{2}$
• $\sum [1+2+3+-------+(N-1)=\frac{N(N-1)}{2}$
• $E_f=\frac{N^2{E}_0}{2}-\frac{N{E}_0}{2}$

The change in ground state energy is $E_f-E_i=\frac{N^2{E}_0}{2}-\frac{N{E}_0}{2}-\frac{N^2{E}_0}{4}+\frac{N{E}_0}{2}=\frac{N^2{E}_0}{4}$

So, the correct answer is $\frac{1}{4}{N}^2{E}_0$.

Concept:

The momentum operator is given by

p = - ih $\frac{\partial }{\partial x}$

where h is the Plank constant.

Calculation:

e^{$\frac{iaP}{h}$} |x>

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

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

= |x> - a∇|x> + $\frac{1}{2!}$ (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:

$\hat{H}=-\frac{{\hslash }^2}{2m}\frac{d^2}{d{x}^2}$

2. Expectation Value of Energy:

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

〈 E 〉 = ${\int }_0^1{\psi }^{\ast }(x)\hat{H}\psi (x)dx$

Calculation -

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

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

Simplifying:

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

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

4. Expectation Value Integral:
   
〈 E 〉 = ${\int }_0^1\psi (x)(-\frac{{\hslash }^2}{2m}\frac{d^2}{d{x}^2}\psi (x))dx$

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

${\psi }_n(x)=\sqrt{2}\sin (n\pi x)$ with energy eigenvalues $E_n=\frac{n^2{\pi }^2{\hslash }^2}{2m}.$

Comparing the given wavefunction:

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

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:

$E_1=\frac{{\pi }^2{\hslash }^2}{2m},E_2=\frac{4{\pi }^2{\hslash }^2}{2m}$

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 $\frac{4{\pi }^2{\hslash }^2}{5m}$

Therefore, the correct answer is (2).

Explanation:

Let us reconsider the system of equations:

$\frac{dA}{dt}=-i\omega B$ and $\frac{dB}{dt}=i\omega A.$

By differentiating the first equation again, $\frac{d²A}{dt²}=-i\omega \frac{dB}{dt}.$

• Substituting the second equation into this results in $\frac{d²A}{dt²}=-\omega ²A.$
• 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 $\langle A{⟩}_{\psi }(t)=0$ 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 $\langle B{⟩}_{\psi }(0)=i,$ we need to multiply cosh(ωt) by i.
• So the time-evolved expectation value is $\langle A{⟩}_{\psi }(t)=\langle \psi |A(t)|\psi ⟩=\langle \psi |Csinh(\omega t)|\psi ⟩=sinh(\omega t)$

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, $\psi =\sqrt{\frac{2}{L}}cos\frac{\pi x}{L}$ 

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

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,

• $E_1^{(1)}=\underset{\frac{-L}{2}}{\overset{\frac{+L}{2}}{\int }}[\sqrt{\frac{2}{L}}cos\frac{\pi x}{L}{]}^2.ϵcos\frac{\pi x}{L}dx$
• $E_1^{(1)}=\frac{2ϵ}{L}\underset{\frac{-L}{2}}{\overset{\frac{+L}{2}}{\int }}co{s}^3\frac{\pi x}{L}dx$

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

•  $E_1^{(1)}=\frac{2ϵ}{L}\times 2\underset{0}{\overset{\frac{+L}{2}}{\int }}{\cos }^3\frac{\pi x}{L}dx$

Using the trignometric formula for $co{s}^3\frac{\pi x}{L}$,

• $co{s}^3\frac{\pi x}{L}=\frac{[cos(\frac{3\pi x}{L})+3cos(\frac{\pi x}{L})]}{4}$
• Substituting this value, we get,
• $E_1^{(1)}=\frac{4ϵ}{L}\underset{0}{\overset{\frac{+L}{2}}{\int }}\frac{[cos(\frac{3\pi x}{L})+3cos(\frac{\pi x}{L})]}{4}$
• $E_1^{(1)}=\frac{ϵ}{L}[(\frac{Sin\frac{3\pi x}{L}}{3\pi /L}){|}_0^{L/2}+3(\frac{Sin\frac{\pi x}{L}}{\pi /L}){|}_0^{L/2}$

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

So, the correct answer is $\frac{8ϵ}{3\pi }$.

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 $\frac{\partial G}{\partial q_j}$

= - ϵ p

q' - q = δ q_i 

= ϵ $\frac{\partial G}{\partial p_j}$

= ϵ p

Now G = qp

- ϵ $\frac{\partial G}{\partial q}$ = - ϵ p = δ p

ϵ $\frac{\partial G}{\partial p}$ = - ϵ q = δ q

The correct answer is option (2).