Quantum Mechanics Questions

Probability current density represent the overall drift of a particle in a particular direction. It is given by $$ \boxed{ \vec{J} = \frac{\hbar}{2im} [ \Psi^* \nabla \Psi - \Psi \nabla \Psi^* ] } $$ where, $\Psi = e^{i(kx-wt)} \\$ $ \Psi^* = e^{-i(kx-wt)} \\$ $\nabla \Psi = ik \Psi \\$ $\nabla \Psi^* = -ik \Psi^* \\$ $\implies \vec{J} = \frac{\hbar}{2im} [ \Psi^* ik \Psi - \Psi(-ik \Psi^*) ] \\$ $\implies \vec{J} = \frac{\hbar}{2im} [2ik] = \frac{\hbar k}{m}$ $\\$ The direction of $\vec{J}$ is along the direction of the momentum of the particle as $J \propto \hbar k $(=momentum). Also, $\frac{\hbar k}{m}$ has dimensions of velocity. So probability can be higher at some time and lower at another time and vice versa.

The transmission coefficient $T$, when $E<V_0$ is given by: $$ T = \bigg[ 1+ \frac{V_0^2 \sinh^2 (Ka)}{4E(V_0-E)} \bigg]^{-1} $$ $V_0$ is the potential barrier height $\\$ $a$ is the width of the barrier $\\$ $E$ is the energy of the particle $\\$ $K = \frac{\sqrt{2m(V_0-E)}}{\hbar}$ $\\$ Substituting the values on the equation, we will get T =0.373. Hence the transmission probability is around 37.3%.

(a) Threshold frequency is the minimum energy required to eject an electron from the surface of the metal. so $$ w_0 = h v_0 $$, where $w_0$ = work function, $v_0$ = threshold frequency and $h$ = planks constant. $ \\ $ $ \implies v_0 = \frac{w_0}{h} $ $ \\ $ $ \implies v_0 = \frac{2.5 \times 10 ^{-19}J}{6.6 \times 10^{-34} Js^{-1}} $ $ \\ $ $ \implies v_0 = 3.8 \times 10^{14} s^{-1}$ $ \\ $ (b) The energy of a photon is given by, $ \\ $ $ E = hv \implies E = 6.6 \times 10 ^ {-34} \times 6 \times 10^{14} = 3.9 \times 10^{-19} J \\$ this is the energy corresponding to the frequency given in the question. Now, the stopping potential is the maximum possible kinetic energy. So, $ \\ $ $ w_0 = 3.9 \times 10 ^ {-19} - 2 \times 10 ^ {-19} \\ $ $ w_0 = 1.9 \times 10 ^ {-19} J \\ $ $ w_0 = \frac{1.9 \times 10 ^ {-19}}{1.9 \times 10 ^ {-16}} eV \\ $ $ \implies w_0 = 1.2 eV $

Confined movement between two rigid walls resembles 1D potential box. Potential height of a rigid wall is infinity. Energy of particle confined in such a system is given by: $$ \boxed{E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}} $$ , where n = quantum number 1,2,3,... $\\$ m = mass of electron and L = length(width of 1D box) $\\$ Hence energy of first three allowed energy levels are $$ E_1 = \frac{ \pi^2 \hbar^2}{2mL^2}, E_2 = \frac{4 \pi^2 \hbar^2}{2mL^2}, E_3 = \frac{3 \pi^2 \hbar^2}{2mL^2} $$ And the corresponding wavelength (from De Broglie) would be: $\\$ using, $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_n}}$, so $$ \lambda_1 = \frac{h}{\sqrt{2mE_1}}, \lambda_2 = \frac{h}{\sqrt{2mE_2}}, \lambda_3 = \frac{h}{\sqrt{2mE_3}} $$ $$ m=9.1 \times 10^{-31} Kg, \hbar = 1.054 \times 10^{-34} Js, L = 10^-9 m$$

From De Broglie's wavelength we prove that $ \\ p = \frac{h}{\lambda} = \frac{hc}{\lambda c} \\$ $ \implies p = \frac{1240 eV nm}{(634nm)C} = 1.96 \frac{eV}{C} $

The angular frequency of a single harmonic oscillator is given by: $$ \tag{1} w = \sqrt{\frac{k}{m}} $$ $k$ is the force constant and $m$ is the mass of the oscillator. $\\$ Using Hooke's law, we have $$ \tag{2} F = kx $$, $k$ is the force constant and $x$ is the displacement. $\\$ For a quantum harmonic oscillator, the energy levels are quantized and given by the formula: $$ \tag{3} \boxed{E_n = (n + \frac{1}{2}) \hbar w } $$ Here, $E_n$ is the energy of the $n^{th}$ level, $\hbar$ is the reduced Planck’s constant, $w$ is the angular frequency of the oscillator, and $n$ is the quantum number which can take any non-negative integer value (0, 1, 2, …). $\\$ So, using (1) and (2), we have $$ w = \sqrt{\frac{k}{m}} = \sqrt{\frac{F}{xm}} $$ substituting the values for $F = 10^{-1}, x= 10^{-2}$ and $m=10^{-3}$, we have, $$ w = \frac{10^{-1}}{10^{-2} \times 10^{-3}} = 10^{2} \text{rad/s} $$ Substituting this value in (3), we have $$ E_0 = 5.28 \times 10^{-33} J $$, in the case of zero point energy, n=0

A wave function is a single-valued, continuous and normalized wave function. $\\$ (a) In the first case, $\tan x$ goes to $\infty$ as $x$ goes to $\infty$. So it is not an acceptable wave function $\\$ (b) sine and cosine functions are acceptable as they are single-valued, continuous and square-integrable. Similarly their linear combinations are also accepted.

For an infinite well potential the energy eigen value is given by: $$ \boxed{E_n = \frac{n^2 \hbar^2 \pi^2}{2mL^2}} $$ where n is the quantum numbers 1,2,3... $\\$ Let's take $E_1$ as energy of $\psi_1$, $E_2$ as energy of $\psi_2$ and $E_3$ as energy of $\psi_3$. So using the energy eigen value equation: $$ E_1 = \frac{ \hbar^2 \pi^2}{2mL^2} , E_2 = \frac{ 4 \hbar^2 \pi^2}{2mL^2} , E_3 = \frac{ 9 \hbar^2 \pi^2}{2mL^2} $$ Expectation value of energy is given by: $$ \langle E \rangle = \int_{-\infty}^{-\infty} \psi^* E \psi$$ $$ \langle E \rangle = \bigg(\frac{1}{\sqrt{14}} \bigg)^2 E_1 + \bigg(\frac{3}{\sqrt{14}}\bigg)^2 E_2 +\bigg(\frac{2}{\sqrt{14}}\bigg)^2 E_3 $$ $$ = \frac{73}{14} \frac{\pi^2 \hbar^2}{2mL^2}$$

From De Broglie wavelength we know $$\tag{1} \lambda = \frac{h}{mv}$$, also $$ \tag{2} L=mvr $$, using both we have $$ \tag{3} L = r \times \frac{h}{\lambda} $$ $\\$ Now applying the standard wave concept in a circular orbit, $\\$ $\implies n \lambda = 2 \pi r \\$ $$ \tag{4} \implies \lambda = \frac{2 \pi r}{n} $$, combining these we get, $\implies L = r \times \frac{n}{2 \pi r} \times n$ $\\$ $$ \boxed{ L = \frac{nh}{2 \pi} } $$

Energy of H-atom is given by, $$ \tag{1} E = \frac{-Re^2}{r} + \frac{p^2}{2m_e} $$ From HUP, $ \Delta x \Delta p \ge \hbar$, $$ \tag{2} \Delta p_x \approx \frac{\hbar}{r} $$ Put (2) in (1), $$ \tag{3} E = \frac{-ke^2}{r} + \frac{\hbar^2}{2mr^2} $$ For minimum energy, $$ \frac{dE}{dr} =0 \space \& \frac{d^2E}{dr^2} < 0$$ for $ r=r_0, \implies \frac{dE}{dr} = \frac{2ke^2}{r^2} - \frac{\hbar^2}{2mr^3} = 0 \\$ $\implies r_0 = 0.52 Å$. $\\$ Note: you can put $r_0$ in $\frac{d^2E}{dr^2}$ and find E for $r=r_0$ $\\$ $\implies E = -13.6eV$

The energy of a quantum is $E_1=hv=\frac{hc}{\lambda}.$ $$\\$$ The energy of an electron transmitted into radiation is $E_2=mc^2.$ $$\\$$ Here $E_1=E_2$ or,$\frac{hc}{\lambda}=mc^2$ or, $$ \lambda=\frac{h}{mc}=\frac{6.62\times{10}^{-34}}{9.1\times{10}^{-31}\times3\times{10}^8}=0.0244\mathrm{Å}$$

As radius of atom is about $5 \times 10^{-15}m$, so the electron to be confined inside of nucleus of uncertainty in its position should not be more than this. Hence $\Delta x = 5 \times 10^{-15}m$. Thus corresponding uncertainty in momentum is $$ \Delta p \Delta x \approx \frac{\hbar}{2} \implies \Delta p \approx \frac{\hbar}{2 \Delta x} $$ $$ \implies \frac{1.054 \times 10^{-34} Js}{10^{-14}m} = 1.054 \times 10^{-20} Kg m/s $$ Now momentum of $e^-$ inside of nucleus must be at least $p = \Delta p = 1.054 \times 10^{-20} kgm/s$. Electron with such momentum would have kinetic energy equal to $E=pc \space$(neglecting rest mass energy). Thus, $$ E =1.054 \times 10^{-20} Kgms^{-1} \times 3 \times 10^{8} ms^{-1} = 3.1514 \times 10^{-12} J = 19.6 \space MeV$$ This energy is far greater than emitted $\beta$ particle whose maxim energy was 1 MeV. Thus electron can't be part of nucleus.

The energy of n$^{th}$ state of $e^-$ in hydrogen atom is given by, $$ \boxed{ E = \frac{- 2 \pi^2 k^2 Z^2 e^4 m}{n^2 h^2} } $$ $ \implies E = \frac{-13.6eV}{n^2} \\$ $\\$ Energy of 1$^{st}$ excited state is when $n=2$: $\\$ $KE = -E \implies KE(E_k) = \frac{13.6eV}{4} \\$ $\lambda = \frac{h}{\sqrt{2mE_k}} \\$ $ \lambda = \frac{6.626 \times 10^{-34} \times 2} {\sqrt{2 \times 3.1 \times 10^{-31} \times 13.6 \times 1.6 \times 10^{-19}}}$ $\\$ $ \lambda = 6.6 Å $

We take $x$ as Hermitian operator and $\psi$ as its normalised state function. $$ (\Delta x)^2 = (x - \braket{x})^2 \\ \implies (\Delta x)^2 = x^2 + \braket{x}^2 - 2 \cdot x \cdot \braket{x} \\ \implies \braket{\psi | (\Delta x)^2 | \psi} = \braket{\psi | x^2 | \psi} + \braket{\psi | \braket{x}^2 | \psi} - 2 \braket{\psi | \cdot x \cdot \braket{x} | \psi} \\ \implies (\Delta x)^2 = \braket{x^2} + \braket{x}^2 - 2 \braket{x}^2 \\ \implies (\Delta x)^2 = \braket{x^2} - \braket{x}^2 $$

In order to explain interference effects, it is assumed that with each particle a wave function is associated. Born postulated that if a particle is described by a wave function $\Psi(x,t)$, then $|\Psi(x,t)|^2 dx$ gives the probability of finding the particle within an element $dx$ about the point $x$ at time $t$. Since the probability of finding the particle somewhere must be unity, the wave function must be normalized: $$ \int_{-\infty}^{\infty} |\Psi(x,t)|^2 dx = \int_{-\infty}^{\infty} \Psi^* \Psi dx = 1$$ Using Schrodinger equation, we have: $$ \tag{1} \frac{\partial \Psi}{\partial t} = \frac{i \hbar}{2m} \frac{\partial ^2 \Psi}{\partial x^2} - \frac{i}{\hbar} V \Psi$$ and $$ \tag{2} \frac{\partial \Psi^*}{\partial t} = -\frac{i \hbar}{2m} \frac{\partial ^2 \Psi^*}{\partial x^2} + \frac{i}{\hbar} V \Psi^*$$ where $V$ is assumed to be real function. On taking the time derivative of the normalized function, $$ \tag{3} \frac{d}{dt} \bigg[ \int_{-\infty}^{\infty} |\Psi(x,t)|^2 dx] \bigg] = \int_{-\infty}^{\infty} \frac{\partial (|\Psi(x,t)|^2)}{\partial t} dx $$ using, $|\Psi|^2 = \Psi^* \Psi$ $$ \frac{\partial |\Psi|^2}{\partial t} = \frac{\partial \Psi^* \Psi}{\partial t} = \Psi^* \frac{\partial \Psi}{\partial t} + \Psi \frac{\partial \Psi^*}{\partial t} $$, using (1) and (2), $$ \frac{\partial |\Psi|^2}{\partial t} = \frac{\partial}{\partial x} \bigg[ \frac{i \hbar}{2m} \Big( \Psi^* \frac{\partial \Psi}{\partial x} - \Psi \frac{\partial \Psi^*}{\partial x} \Big) \bigg]$$, so (3) becomes, $$ \frac{d}{dt} \int_{-\infty}^{\infty} |\Psi|^2 dx] = \frac{i \hbar}{2m} \Big( \Psi^* \frac{\partial \Psi}{\partial x} - \Psi \frac{\partial \Psi^*}{\partial x} \Big) \bigg|^{\infty}_{-\infty} $$ $$ = \frac{d}{dt} \int_{-\infty}^{\infty} |\Psi|^2 dx] = 0 $$, so, $$ \int_{-\infty}^{\infty} |\Psi|^2 dx] = \text{constant} $$ So, if $\Psi$ is normalized at $t=0$, then it is normalized at any time $t$. This property preserves the normalization of wave function. Without whihc features of Schrodinger equation would be incompatible.

Expectation value of position can be calculated using $$ \braket{x} = \braket{\Psi ^* | x|\Psi} = \int_{-\infty}^{\infty} \Psi^* x \Psi dx $$ So, using the wavefunction $\psi(x)$, $$ \braket{x} = \int_{-\infty}^{\infty} N exp (\frac{-x^2}{2a} - ikx) \space x \space N exp (\frac{-x^2}{2a} + ikx) dx $$ $$ = N^2 \int_{-\infty}^{\infty} x \space exp (\frac{-x^2}{a}) $$ As we know $x$ is an odd function [because $f(-x) = -f(x)$] and using the standard result $ \int_{-a}^{a} f(x) dx = 0 $, where $f(x)$ is an odd function. So the integrand become zero, thus, $$ \braket{x} = 0 $$ Hence the expectation value of position is zero. $\\$ Note: In Quantum Mechanics, when expectation value of an observable is zero that doesn't necessarily means that every measurement is zero, but it means that average result of the measurement over multiple measurements are zero.

From, HUP, we know that, $$ \boxed { \Delta E \Delta t \ge \frac{\hbar}{2} }$$ $\Delta E = \Delta m c^2$ $\\$ $\Delta E = (40m_e)c^2$ $\\$ (since variation in mass is $\pm20m_e$)

We need to normalize $\Psi = A e^{\gamma |x|}$. which is $$ \begin{matrix} \Psi = A e^{\gamma x} & \text{for} & x \ge 0 \\ \Psi = A e^{-\gamma x} & \text{for} & x < 0 \end{matrix}$$ Now, for $x \rightarrow \infty \implies \Psi \rightarrow \infty \\$ and for $x \rightarrow -\infty \implies \Psi \rightarrow \infty $. that is wave function is not finite in $-\infty \le x \le \infty$. Hence it is not normalizable.