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\textbf{\large{Indian Institute of Information Technology, Allahabad}}
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\textbf{Assignment 04 Hints}: Engineering Physics [Section: C]
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\begin{enumerate}

\item[\textbf{1}.] $ E_{n_{x},n_{y},n_{z}}=(n_{x}^{2}+n_{y}^{2}+n_{z}^{2})\frac{\pi^{2}\hbar^{2}}{2ma^{2}}=E=\frac{14\pi^{2}\hbar^{2}}{2ma^{2}} $\\
$n_{x}^{2}+n_{y}^{2}+n_{z}^{2}=14 \quad \Rightarrow \quad n_{x}=1, n_{y}=2, n_{z}=3$\\
$ \psi(x,y,z)=(\frac{2}{a})^{3/2}\sin\frac{\pi x}{a}\sin\frac{2\pi y}{a}\sin\frac{3\pi z}{a} $

\item[\textbf{2}.] $[x,p^{n}_{x}]=i\hbar np_{x}^{n-1}$\\
Proceed using induction method, considering one of the basic axioms that for $n=1$ the relation becomes $[x,p_{x}]=i\hbar$ and accept it true. Now we'll assume that given relation is true for $n$ and show that it also holds for $(n+1)$ i.e. $[x,p_{x}^{n+1}]=i\hbar (n+1)p_{x}^{n}$ 

\item[\textbf{3}.] Normalise the wave function, $ \int_{-\infty}^{\infty}|\psi(x,0)|^{2}dx=1 \Rightarrow A=\frac{1}{\sqrt{2a}} $\\
Now, \qquad $ \phi(k)=\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2a}}\int_{-a}^{a}e^{-ikx}dx \Rightarrow \frac{1}{\sqrt{\pi a}}\frac{sin(ka)}{k} $\\
Therefore, \qquad $ \psi(x,t)=\frac{1}{\sqrt{2\pi^{2}a}}\int_{-\infty}^{\infty}\frac{\sin(ka)}{k}e^{i(kx-\frac{\hbar k^{2}}{2m}t)}dk $

\item[\textbf{4}.] 
\begin{align*}
Ae^{ikx}+Be^{-ikx} =& A(\cos(kx)+i\sin(kx))+B(\cos(kx)-i\sin(kx))\\
=& (A+B)\cos(kx)+i(A-B)\sin(kx)\\
=& C\cos(kx)+D\sin(kx)
\end{align*}
with $C=A+B$ and $D=i(A-B)$\\
Now using, 
\begin{align*}
C\cos(kx)+D\sin(kx) =& \frac{1}{2}(C-iD)e^{ikx}+\frac{1}{2}(C+iD)e^{-ikx}\\
=& Ae^{ikx}+Be^{-ikx}
\end{align*}
with $A=\frac{1}{2}(C-iD)$ and $B=\frac{1}{2}(C+iD)$

\item[\textbf{5}.] $ J = \frac{i\hbar}{2m}\Big(\psi\frac{\partial\psi^{*}}{\partial x}-\psi^{*}\frac{\partial\psi}{\partial x}\Big)\Rightarrow \frac{\hbar k}{m}|A|^{2} $
 
\item[\textbf{6}.] The Schrodinger equation,\quad $ \psi^{''}(x)+\frac{2m}{\hbar^{2}}(E-V)\psi(x)=0 \quad ; \quad (E<0) $\\
Set, \quad $ U_{0}=\frac{2mV_{0}}{\hbar^{2}} \quad ; \quad k=\frac{\sqrt{2m|E|}}{\hbar} $\\
Therefore, expression becomes,\qquad $\psi^{''}(x)-k^{2}\psi(x)+U_{0}\delta(x)\psi(x)=0  $\\
At $x\neq 0$, Schrodinger has solutions, 
$$ \psi(x)\sim e^{-kx} \quad for \quad x<0 \qquad and \qquad \psi(x)\sim e^{kx} \quad for \quad x>0 $$
Using the boundary condition the energy is $-E=\frac{\hbar^{2}k^{2}}{2m}=\frac{mV_{0}^{2}}{2\hbar^{2}}$

\item[\textbf{7}.] There are both incident and reflected waves from the right and we must have $E>V_{0}$ as particles incident from right. The Schrodinger equation for $x>0$ region,
$$ \psi^{''}(x)+k_{1}^{2}\psi(x)=0 \qquad ; \quad k_{1}=\frac{\sqrt{2m(E-V_{0})}}{\hbar} $$
has sol, \qquad $ \psi(x)=e^{-ik_{1}x}+Re^{ik_{1}x} $\\
There is only transmitted wave in $x<0$ region,
$$ \psi^{''}(x)+k_{2}^{2}\psi(x)=0 \qquad ; \quad k_{2}=\frac{\sqrt{2mE}}{\hbar}$$
has sol of the form, \qquad $ \psi(x)=Se^{-ik_{2}x} $\\
Using the boundary condition for the wave function to be continuous and the continuity of the the first derivative of the wave function at $x=0$, The reflection ($R$) and transmission ($T$) coefficients are, 
$$ |R|^{2}=\Big|\frac{k_{1}-k_{2}}{k_{1}+k_{2}}\Big|^{2} \qquad |T|^{2}=1-|R|^{2} $$
Replace the expression for $k_{1}$ and $k_{2}$. 

\item[\textbf{8}.] Substituting the solution in the given wave equation, we get,\\
$$ -\frac{\omega^{2}}{c^{2}}=-k^{2}-\frac{m^{2}c^{2}}{\hbar^{2}} $$
multiplying by $c^{2}\hbar^{2}$ and writing $\hbar\omega=E$ and $k\hbar=p$,
$$ E^{2}=c^{2}p^{2}+m^{2}c^{4} $$

\item[\textbf{9}.] $$ E^{2}=c^{2}p^{2}+m^{2}c^{4}=c^{2}p^{2}\Big(1+\frac{m^{2}c^{4}}{c^{2}p^{2}}\Big) $$
Phase velocity, $ v_{p}=\frac{E}{p}=c\Big(1+\frac{m^{2}c^{4}}{c^{2}p^{2}}\Big)^{1/2} $\\
Since $\lambda=\frac{h}{p}$,
$$ v_{p}=c\Big(1+\frac{m^{2}c^{2}\lambda^{2}}{h^{2}}\Big) $$

\item[\textbf{10}.] Fermi energy, $$ E_{f}=\frac{h^{2}}{8m}\Big(\frac{3n}{\pi V}\Big)^{2/3}=6.8 \times 10^{-19}J $$
Now, $E_{total}=KE+PE=E_{f}$. Since PE is zero, $E_{f}=KE$\\
$$ v=\sqrt{\frac{2E_{f}}{m}}=1.2\times 10^{6} m/s $$

\item[\textbf{11}.]
\begin{align*}
\psi(x) =& A\sin(kx)+\frac{A\sin(ka)}{(e^{iKa}-\cos(ka))}\cos(kx)\\
=& C[\sin(kx)+e^{-iKa}\sin{k(a-x)}] \quad ; \quad C \equiv \frac{Ae^{iKa}}{(e^{iKa}-\cos(ka))}
\end{align*}

\end{enumerate}
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