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\textbf{\large{Indian Institute of Information Technology, Allahabad}}
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\textbf{Assignment 03 Hints}: Engineering Physics [Section: C]
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\flushleft \textbf{[1]} $ v=90 mph =40.2 ms^{-1} \qquad p_{x}=mv=4.02kgms^{-1} $\\
Uncertainty in velocity, $v=0.1mph=0.0447ms^{-1} \qquad \bigtriangleup p_{x}=m\bigtriangleup v_{x}=0.0045kgms^{-1} $\\
Now from Heisenberg uncertainty principle, $\bigtriangleup x\geq \frac{\hbar}{2\bigtriangleup p_{x}}=1.2\times 10^{-32}m$
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\flushleft \textbf{[2]} $\bigtriangleup E=10^{-33}J$
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\flushleft \textbf{[3]} $ \frac{d^{2}}{dx^{2}}\psi(x)=\Big[\frac{1}{x_{0}^{2}}-\frac{2n}{x_{0}x}+\frac{n(n-1)}{x^{2}}\Big] $\\
Substituting this in the time independent Schrodinger equation,\\

$ \frac{-\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}\psi(x)+V(x)\psi(x)=E\psi(x)\Rightarrow E-V=-\frac{\hbar^{2}}{2m}\Big[\frac{1}{x_{0}^{2}}-\frac{2n}{x_{0}x}+\frac{n(n-1)}{x^{2}}\Big]$
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Now from the given condition, $x\rightarrow \infty \Rightarrow V(x)\rightarrow 0$
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$ E=-\frac{\hbar^{2}}{2mx_{0}^{2}} \qquad \& \qquad V(x)=\frac{\hbar^{2}}{2m}\Big[\frac{n(n-1)}{x^{2}}-\frac{2n}{x_{0}x}\Big] $
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\flushleft\textbf{[4]} $ (\psi(x,0),\psi(x,0))=|A|^{2}\Sigma_{n,n'}(\frac{1}{2})^{(n+n')/2}(\psi_{n},\psi_{n'}) \Rightarrow |A|=\frac{1}{\sqrt{2}} $\\
Now for expression of $\psi(x,t)=e^{-i\hat{H}t/\hbar}\psi(x,0)$. Replace the $\hat{H}$ i.e. $E$ and $\psi(x,0)$.
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\flushleft\textbf{[5]} $ <p>=\int_{-\infty}^{\infty}\psi^{*}(x)(-i\hbar \frac{d}{dx})\psi(x)dx=0 $\\
$ <p^{2}>=\int_{-\infty}^{\infty}\psi^{*}(x)(-\hbar^{2} \frac{d^{2}}{dx^{2}})\psi(x)dx=\frac{\hbar^{2}}{4\sigma_{x}^{2}} $ \qquad Therefore, \qquad $\sigma_{p}=\frac{\hbar}{2\sigma_{x}}$
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\flushleft \textbf{[6]} Mean position of the particle, $ <x>=\int_{-\infty}^{\infty}\psi^{*}(x)x\psi(x)dx=0 $\\
Similarly Mean momentum of the particle, \qquad $ <p>=0 $
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\flushleft \textbf{[7]} For normalisation constant, $ \int_{-\infty}^{\infty}\psi^{*}(x)\psi(x)dx=1 \Rightarrow A=\sqrt{\frac{4}{L}} $\\
For probability, $ p(0,\frac{L}{8})=\int_{0}^{(L/8)}\psi^{*}(x)\psi(x)dx \approx 0.41$
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\flushleft \textbf{[8]} Normalisation constant is, $ C=\frac{2}{\sqrt{\pi}}\sqrt{(\frac{m\omega}{\hbar})^{3}} $.\\ Apply the Schrodinger equation , $ -\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}\psi+\frac{1}{2}kx^{2}\psi(x)=E\psi(x) $\\
Equate the $x^{2}$ terms and interpret the constant as energy $E$. Solving for $\alpha$,
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$ \alpha=\sqrt{\frac{km}{4\hbar^{2}}} $ \qquad $\&$ \qquad $E=\frac{3\hbar^{2}\alpha}{m}$
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\flushleft \textbf{[9]} $A\psi$ is also a solution of Schrodinger equation. Where $A$ is a normalisation constant. $\int A\psi^{*}A\psi dx=1$.\\
$ A^{2}\Big(\int_{0}^{\infty}e^{-x}e^{-x}dx+\int_{0}^{-\infty}e^{x}e^{x}dx\Big)=1\Rightarrow N=1 $ 
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Probability of finding the particle between $x=-1$ to $x=1$. i.e. $\int_{-1}^{1}\psi^{*}(x)\psi(x)dx$\\
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$ \int_{0}^{1}e^{-x}e^{-x}dx+\int_{0}^{-1}e^{x}e^{x}dx=\frac{1}{2}(e^{2}-e^{-2}) $
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\flushleft\textbf{[10]} $ <x>= \int_{0}^{1/\alpha}4\alpha^{3}x^{3}e^{-2\alpha x}dx =\frac{3}{2\alpha} \qquad \& \qquad <x^{2}>=\frac{3}{\alpha^{2}}$
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Probability,\qquad $ P=\int_{0}^{1/\alpha}4\alpha^{3}x^{2}e^{-2\alpha x}dx=0.32 $\\
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\flushleft \textbf{[11]} Probability of energy $(E_{1})$ state $1$,\qquad $ P_{1}=|c_{1}|^{2}=|\frac{-iE_{1}t/\hbar}{\sqrt{2}}|^{2}=\frac{1}{2} $\\
Probability of energy $(E_{2})$ state $2$,\qquad $ P_{2}=|c_{2}|^{2}=|\frac{-iE_{2}t/\hbar}{\sqrt{2}}|^{2}=\frac{1}{2} $
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Because $\Sigma_{n=1}^{\infty}|c_{n}|^{2}=1$, $\psi(x,t)$ is normalised. Now for finding the expectation values of the state $\psi(x,t)$,\\
$ <E>=\Sigma_{n=1}^{\infty}P_{n}E_{n}=\frac{1}{2}(E_{1}+E_{2}) $ ; \quad $ <E^{2}>=\Sigma_{n=0}^{n=\infty}P_{n}E_{n}^{2}=\frac{1}{2}(E_{1}^{2}+E_{2}^{2}) $\\
Therefore, \qquad $\bigtriangleup E=\sqrt{<E^{2}>-<E>^{2}}=\frac{1}{2}(E_{1}-E_{2})$\\

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\flushleft \textbf{[12]}
Consider the mass of baseball $1 Kg$ and apply de Broglie hypothesis. 
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