#PDF
#Part I --- Short Notes (Q.M) (balki sir), Lec-6 to Lec-13
#(Short Notes) (Q.M) (balki sir) Lec-6
#The Schrödinger equation :--
#(Short Notes) (Q.M) (balki sir) Lec-7,8
#(Short Notes) (Q.M) (balki sir) Lec-8, 9
#(Short Notes) (Q.M) (balki sir) Lec-9,10
#Short notes (Balki sir) Lec-11
#Short notes (Balki sir) Lec-11 (2)
#Short notes (Balki sir) Lec-11,12
#Short notes (Balki sir) Lec-12,13
#Short notes (Balki sir) Lec-12,13 (2)
#Short notes (Balki sir) Lec-12,13 (3)
#Short notes (Balki sir) Lec-12,13 (4)
#Short notes (Balki sir) Lec-12,13 (5)
#Short notes (Balki sir) Lec-12,13 (contd.)
#Part II --- Short Notes (Q.M), Pages 1 to 10
PDF
To find \(\left\langle A\right\rangle\) where \(A\) can be any arbitrary operator.
\[
\left\langle A\right\rangle=\left\langle A\right\rangle(t)=\frac{\left\langle \Psi(t)\middle|A\middle|\Psi(t)\right\rangle}{\left\langle \Psi(t)\middle|\Psi(t)\right\rangle}
\]
First find \(\hat{A}\left|\Psi(t)\right\rangle\) then take inner product with unchanged original state \(\left\langle \Psi(t)\right|\).
Poisson Brackets \(\{A,B\}\) are replaced by commutator brackets
\[
\{A,B\}\longrightarrow \frac{[A,B]}{i(P.q)}
\]
Since \(\hat{A},\hat{B}\) are hermitian, \([A,B]\) should also be hermitian
\[
(AB)=(AB-BA)^{\dagger}=BA-AB
\]
To make it hermitian we divide by `\(i\)' so that \(\dfrac{[A,B]}{i}\) is hermitian.
where \(P.q\) abbreviation for physical quantity is choosen so that \(\dfrac{[A,B]}{i(P.q)}\) is dimensionless.
The equation which sits at heart of Quantum mechanical formalism of microscopic world / our world is given by
\[
i\hbar\frac{d}{dt}\left|\Psi(t)\right\rangle=\hat{H}\left|\Psi(t)\right\rangle
\]
(This is one of postulate of QM)
\[
\left|\Psi(t)\right\rangle=e^{\frac{-i\hat{H}(t-t_0)}{\hbar}}\left|\Psi(t_0)\right\rangle
\]
Say, \(_t_0=0_\). \(\left|\Psi(t)\right\rangle=e^{\frac{-i\hat{H}t}{\hbar}}\left|\Psi(0)\right\rangle\)
where we can say that the time evolution operator \(e^{\frac{-i\hat{H}t}{\hbar}}\) takes state \(\left|\Psi(0)\right\rangle\) to state \(\left|\Psi(t)\right\rangle\).
\[
\left\langle \Psi(t)\middle|\Psi(t)\right\rangle=\left\langle \Psi(0)\middle|e^{\frac{i\hat{H}t}{\hbar}}e^{\frac{-i\hat{H}t}{\hbar}}\middle|\Psi(0)\right\rangle
\]
\[
\left\langle \Psi(t)\middle|\Psi(t)\right\rangle=\left\langle \Psi(0)\middle|\Psi(0)\right\rangle
\]
\(\leftarrow\) length of vector is preserved
Operator \(e^{-i\hat{H}t/\hbar}\) has many names.
1) Time development operator
It is unitary operator.
2) Evolution operator.
\(UU^{\dagger}=U^{\dagger}U=I\)
3) Propagator
i.e. \(U^{\dagger}=U^{-1}\)
\[
U(t_2,t_0)=U(t_2,t_1)\cdot U(t_1,t_0)
\]
It follows semi-group property.
Operator `\(A\)' can be diagonalized with similarity transformation if \([A,A^{\dagger}]=0\)
If \(\hat{A}\) is Q.M. operator \(A^{\dagger}=A\) (hermitian) \((A,A^{\dagger})=0\) (always)
Operators \(\hat{A}\) & \(\hat{B}\) share common set of eigen values and eigen vectors if \([A,B]=0\)
\(\rightarrow\) If \([\hat{A},\hat{B}]\neq 0\Rightarrow\) A common set of eigen states does not exist. But they may share 1 or 2 common eigen states/values.
\(\rightarrow\) If \([A,B]=0\Rightarrow \exists\) a common set of eigen states. ex \([P^2,I]=0\)
\[
\hat{P}=\text{projection operator}=\left|\phi\right\rangle\left\langle \phi\right|
\]
\[
P^2=\left|\phi\right\rangle\left\langle \phi\middle|\phi\right\rangle\left\langle \phi\right|=P
\]
\[
[P^2,I]=[P,I]=0
\]
Every function is eigen state of `\(I\)', but not of \(\hat{P}\). Eigen set of \(I=\{\)odd \(f^n\), even \(f^n\), others\(\}\)
Eigen set of \(P=\{\)odd fun, even fun\(\}\)
Continuous basis :
We can also have continuous basis set to express state vector of hilbert space.
\(\hat{x}\) & \(\hat{p}\) can have infinite eigen values and thus infinite continuous eigen states for \(\hat{x}\) and \(\hat{p}\).
\[
\hat{x}\left|x'\right\rangle=x'\left|x'\right\rangle\qquad \hat{p}\left|p'\right\rangle=p'\left|p'\right\rangle
\]
Completeness
\[
\left(\int_{-\infty}^{\infty}\left|x\right\rangle\left\langle x\right|dx\right)\left|\Psi\right\rangle=\left|\Psi\right\rangle
\]
\[
\left(\int_{-\infty}^{\infty}\left|p\right\rangle\left\langle p\right|dp\right)\left|\Psi\right\rangle=\left|\Psi\right\rangle
\]
Eigen states of \(\hat{x}\) can be orthonormalized.
\[
\left.\begin{array}{l}\left\langle x\middle|x'\right\rangle=\delta(x-x')\\[2pt] \left\langle p\middle|p'\right\rangle=\delta(p-p')\end{array}\right\}\text{orthonormality}
\]
We can expand \(\left|\Psi\right\rangle\) in \(\{\left|x\right\rangle\}\) basis as well as in \(\{\left|p\right\rangle\}\) basis set.
\[
\left|\Psi\right\rangle=\int dx\,\left|x\right\rangle\underbrace{\left\langle x\middle|\Psi(t)\right\rangle}_{\psi(x,t)}
\]
\[
\left|\Psi\right\rangle=\int_{-\infty}^{\infty} dp\,\left|p\right\rangle\underbrace{\left\langle p\middle|\Psi(t)\right\rangle}_{\tilde{\psi}(p,t)}
\]
State of Quantum mechanical system is some Abstract vector which lives in seperable Hilbert space.
\(\left|\Psi\right\rangle\) can take any look depending upon the chosen basis set. We can also express \(\left|\Psi(t)\right\rangle\) in basis set formed by eigen states of Total Energy operator \((\hat{H})\)
\[
\hat{H}\left|\phi_n\right\rangle=E_n\left|\phi_n\right\rangle
\]
\[
\left|\Psi(t)\right\rangle=\sum_{n=0}^{\infty}c_n\left|\phi_n(t)\right\rangle;\ c_n=\left\langle \phi_n(t)\middle|\Psi\right\rangle
\]
where, \(\left|\phi_n(t)\right\rangle=e^{\frac{-i\hat{H}t}{\hbar}}\left|\phi_n(0)\right\rangle=e^{\frac{-iE_nt}{\hbar}}\left|\phi_n(0)\right\rangle\)
let us try to express \(\psi(x,t)\) in form of \(\tilde{\psi}(p)\).
\[
\psi(x,t)=\left\langle x\middle|\Psi(t)\right\rangle=\left\langle x\right|\int dp\left|p\right\rangle\left\langle p\middle|\Psi\right\rangle
\]
\[
\psi(x,t)=\int_{-\infty}^{\infty}dp\,\left\langle x\middle|p\right\rangle\tilde{\psi}(p)
\]
Similarly
\[
\tilde{\psi}(p,t)=\left\langle p\middle|\Psi(t)\right\rangle=\left\langle p\right|\int_{-\infty}^{\infty}dx\left|x\right\rangle\left\langle x\middle|\Psi\right\rangle
\]
\[
\tilde{\psi}(p,t)=\int_{-\infty}^{\infty}\left\langle p\middle|x\right\rangle\overline{\psi}(x,t)\,dx
\]
\[
\left\langle x\middle|\hat{p}\middle|\Psi\right\rangle=-i\hbar\frac{\partial}{\partial x}\left\langle x\middle|\Psi\right\rangle
\]
\[
\left\langle x\middle|\hat{p}\middle|p\right\rangle=p\left\langle x\middle|p\right\rangle=-i\hbar\frac{\partial}{\partial x}\left\langle x\middle|p\right\rangle
\]
\[
\left\langle x\middle|p\right\rangle\propto e^{+ipx/\hbar}\qquad \left\{\left\langle x\middle|[x,p]\middle|x'\right\rangle=i\hbar\left\langle x\middle|x'\right\rangle\right\}
\]
\[
\left\langle p\middle|\hat{x}\middle|\Psi\right\rangle=+i\hbar\frac{\partial}{\partial p}\left\langle p\middle|\Psi\right\rangle
\]
for (3D)
\[
\left\langle r\middle|\hat{p}\middle|\Psi\right\rangle=-i\hbar\nabla\psi(r,t)
\]
\[
\left\langle p\middle|\hat{r}\middle|\Psi\right\rangle=+i\hbar\nabla\tilde{\psi}(p,t)
\]
\[
\left\langle r\middle|p\right\rangle\propto e^{i\vec{p}\cdot\vec{r}/\hbar}\qquad \left\langle p\middle|r\right\rangle\propto e^{-i\vec{p}\cdot\vec{r}/\hbar}
\]
Schrödinger eq\(^n\) in \(\{\left|r\right\rangle\}\) basis.
\[
\left\langle r\right|i\hbar\frac{d}{dt}\left|\Psi(t)\right\rangle=\left\langle r\middle|\hat{H}\middle|\Psi(t)\right\rangle
\]
\[
i\hbar\frac{\partial}{\partial t}\psi(r,t)=\left\langle r\middle|\frac{p^2}{2m}+\hat{V}(r)\middle|\Psi(t)\right\rangle
\]
\[
i\hbar\frac{\partial}{\partial t}\psi(r,t)=\frac{-\hbar^2}{2m}\nabla^2\psi(r,t)+V(r)\psi(r,t)
\]
\(\searrow\) Not true if \(V(r)\) is Complex.
\[
\left\{\begin{array}{l}A\left|\lambda\right\rangle=\lambda\left|\lambda\right\rangle\\[2pt] \left\langle \lambda\right|A^{\dagger}=\left\langle \lambda\right|\lambda^{*}\\[2pt] \text{If }(A^{\dagger}=A)\ \left\langle \lambda\right|\hat{A}=\lambda^{*}\left\langle \lambda\right|\end{array}\right.
\]
If \(V(\hat{r}^{\dagger})=V(r)\) i.e. e.v. of \(V\) are real
Continued :
If eigen values of \(V(r^{\dagger})\) is imaginary then \(\hat{H}\) won't be hermitian and probabilities won't be conserved with time.
\[
\left\langle \Psi(t)\middle|\Psi(t)\right\rangle\neq\left\langle \Psi(0)\middle|\Psi(0)\right\rangle
\]
Such kind of complex potentials are used where if some particles are targeted to some target and target absorbs some of them.
If \(V(r^{\dagger})=V(r)\) (real potential function)
\[
i\hbar\frac{\partial}{\partial t}\psi(r,t)=\frac{-\hbar^2}{2m}\nabla^2\psi(r,t)+V(r)\psi(r,t)
\]
*Stationary states / Eigen states/vector of \(\hat{H}\) :--*
Then, \(H\left|\phi_n\right\rangle=E_n\left|\phi_n\right\rangle\) where \(\left|\phi_n\right\rangle\) is eigen vector of \(\hat{H}\) with eigen value \(E_n\).
\[
\Rightarrow i\hbar\frac{d}{dt}\left|\phi_n\right\rangle=\hat{H}\left|\phi_n\right\rangle=E_n\left|\phi_n\right\rangle
\]
\[
\underline{\left|\phi_n(t)\right\rangle=e^{-iE_nt/\hbar}\left|\phi_n(0)\right\rangle}
\]
we can see that these states do not change with time, so called stationary states.
so a general state of system \(\left|\Psi(t)\right\rangle\) can be expanded a linear combination of stationary state.
The sol\(^n\) of Schrödinger eq\(^n\)
\[
i\hbar\frac{d}{dt}\left|\Psi(t)\right\rangle=H\left|\Psi(t)\right\rangle
\]
is
\[
\left|\Psi(t)\right\rangle=\sum_{n=0}^{\infty}c_n\left|\phi_n(t)\right\rangle=\sum_{n=0}^{\infty}\underbrace{c_ne^{\frac{-iE_nt}{\hbar}}}
_{c_n(t)}\left|\phi_n(0)\right\rangle
\]
\[
\left|\Psi(t)\right\rangle=\sum_{n=0}^{\infty}c_n(t)\left|\phi_n(0)\right\rangle\qquad -\text{(1)}
\]
So our basis set \(\{\left|\phi_n(0)\right\rangle\}\) is time independent where as \(c_n(t)=\left\langle \phi_n(0)\middle|\Psi(t)\right\rangle\) is time dependent.
All Quantum Interference phenomena occur due to superposition of different energy states.
We can write stationary states in \(\{\left|r\right\rangle\}\) basis
\[
\left\langle r\middle|\phi_n(t)\right\rangle=\left\langle r\right|e^{\frac{-iE_nt}{\hbar}}\phi_n(t)\rangle
\]
\[
\phi_n(r,t)=e^{\frac{-iE_nt}{\hbar}}\phi_n(r,0)
\]
Then eq (1) becomes in \(\{\left|r\right\rangle\}\) basis.
\[
\left\langle r\middle|\Psi(t)\right\rangle=\sum_{n=0}^{\infty}c_n(t)\left\langle r\middle|\phi_n(0)\right\rangle
\]
\[
\Psi(r,t)=\sum_{n=0}^{\infty}c_n(t)\phi_n(r,0)
\]
our Schrödinger eq\(^n\) becomes,
\[
i\hbar\frac{\partial}{\partial t}\Psi(r,t)=\frac{-\hbar^2}{2m}\nabla^2\Psi(r,t)+V(r)\Psi(r,t)
\]
for stationary states,
\[
i\hbar\frac{\partial}{\partial t}\phi_n(r,t)=\frac{-\hbar^2}{2m}\nabla^2\phi_n(r,t)+V(r)\phi_n(r,t)
\]
using, \(\left\{\phi_n(r,t)=e^{-iE_nt/\hbar}\phi_n(r,0)\right\}\)
we have
\[
i\hbar\,\phi_n(r,0)e^{\frac{-iE_nt}{\hbar}}\left(\frac{-iE_n}{\hbar}\right)=He^{\frac{-iE_nt}{\hbar}}\phi_n(r,0)
\]
\[
E_n\phi_n(r,0)=H\phi_n(r,0)
\]
or
\[
\frac{-\hbar^2}{2m}\nabla^2\phi_n(r,0)+V(r)\phi_n(r,0)=E_n\phi_n(r,0)
\]
lets just write \(\phi_n(r,0)=\phi_n(r)\)
\[
\boxed{\nabla^2\phi_n(r)+\frac{2m\left(E_n-V(r)\right)}{\hbar^2}\phi_n(r)=0}
\]
This equation is used to find time inde- -pendent basis set \(\{\left\langle r\middle|\phi_n(0)\right\rangle\}\) or \(\{\phi_n(r,0)\}\) and often called as Time Independent Schrödinger eq\(^n\). (as it does not have any time dependent term).
Bohr quantization for periodic orbits :--
we know that
\[
\Delta x\,\Delta p\geq \hbar/2
\]
\[
dA=dx\,dp\sim h
\]
So \(\oint dA=nh\Rightarrow \left\{\oint P\,dx=nh\right\}\)
\begin{tikzpicture}[scale=0.8]
\draw[->] (-1.6,0)--(1.9,0) node[right]{$x$};
\draw[->] (0,-1.1)--(0,1.3) node[above]{$P$};
\draw[rotate=20] (0,0) ellipse (1.4 and 0.6);
\draw[rotate=20] (0,0) ellipse (1.0 and 0.35);
\end{tikzpicture}
\(\nearrow\) Total area is quantized of uncertainity principle.
Finding \(\left\langle A\right\rangle\) in given stationary state \(\left|\phi\right\rangle\)
\[
\left\langle A\right\rangle_n=\frac{\left\langle \phi_n\middle|\hat{A}\middle|\phi_n\right\rangle}{\left\langle \phi_n\middle|\phi_n\right\rangle}\quad \text{If }\left\langle \phi_n\middle|\phi_n\right\rangle=1
\]
then, \(\left\langle A\right\rangle=\left\langle \phi_n\middle|A\middle|\phi_n\right\rangle\)
we know,
\[
\left|\phi\right\rangle=I\left|\phi_n\right\rangle=\int\left|x'\right\rangle\left\langle x'\middle|\phi_n\right\rangledx'
\]
\[
\left\langle \phi_n\right|I=\int dx\left\langle \phi_n\middle|x\right\rangle\left\langle x\right|
\]
So \(\left\langle A\right\rangle_{n}=\left\langle \phi_n\middle|\hat{A}\middle|\phi_n\right\rangle=\int dx\,\phi_n^{*}(x)\left\langle x\right|\hat{A}\int\left|x'\right\rangle\phi_n(x')dx'\)
\[
\left\{\left\langle \hat{A}\right\rangle_n=\int dx\,\phi_n^{*}(x)\int \left\langle x\middle|\hat{A}\middle|x'\right\rangle\phi_n(x')dx'\right\}
\]
example : If \(\hat{A}=\hat{x}\)
\[
\begin{align*}
\left\langle \hat{x}\right\rangle_n&=\int dx\,\phi_n^{*}(x)\int\left\langle x\middle|\hat{x}\middle|x'\right\rangle\phi_n(x')dx'\\
&=\int dx\,\phi_n^{*}(x)\int x'\left\langle x\middle|x'\right\rangle\phi_n(x')dx'\\
&=\int dx\,\phi_n^{*}(x)\underbrace{\left(\int x'\delta(x'-x)\phi_n(x')dx'\right)}_{\text{---}\,I}
\end{align*}
\]
\[
\left\{\begin{array}{l}\left\langle x\middle|x'\right\rangle=\left\langle x'\middle|x\right\rangle^{*}=\left\langle x'\middle|x\right\rangle\ \ (\text{real})\\[2pt] \delta(x-x')=\delta(x'-x)\end{array}\right.
\]
\[
\left\langle x\right\rangle_n=\int_{-\infty}^{\infty}dx\,\phi_n^{*}(x)\,x\,\phi_n(x)
\]
as \(I=\left\{\begin{array}{ll}0 & x\neq x'\\ x\,\phi_n(x) & x=x'\end{array}\right.\)
Similarly
\[
\left\langle \hat{p}\right\rangle_n=\int dx\,\phi_n^{*}(x)\left(-i\hbar\frac{\partial}{\partial x}\left\langle x\middle|x'\right\rangle\right)\phi_n(x')dx'
\]
\[
\left\{\left\langle p\right\rangle_n=\int \phi_n^{*}(x)\left(-i\hbar\frac{\partial}{\partial x}\right)\phi_n(x)dx\right\}
\]
\[
\Delta x=\sqrt{\left\langle x^2\right\rangle-\left\langle x\right\rangle^2}
\]
\(\downarrow\) std. deviation / uncertainty in \(\hat{x}\).
\[
\overset{\text{or}}{=}\ \Delta x=\sqrt{\left\langle (x-\left\langle x\right\rangle)^2\right\rangle}
\]
Similarly
\[
\delta P=\sqrt{\left\langle p^2\right\rangle-\left\langle p\right\rangle^2}
\]
``Symmetry implies degeneracy''
More curvature \(\Rightarrow\) more energy
\begin{tikzpicture}[scale=0.72]
\draw[->] (-0.2,0)--(0,0);
\draw (0,2.4)--(4.6,2.4) node[right]{$E_3$};
\draw (0,1.2)--(4.6,1.2) node[right]{$E_2$};
\draw (0,0)--(4.6,0) node[right]{$E_1$};
\draw[thick] plot[domain=0:4.4,samples=80] (\x,{2.4+0.42*sin(180*\x/1.1)});
\draw[thick] plot[domain=0:4.4,samples=80] (\x,{1.2+0.42*sin(180*\x/2.2)});
\draw[thick] plot[domain=0:4.4,samples=100] (\x,{0+0.42*sin(180*\x/0.73)});
\node at (-0.45,1.2) {ex$=$};
\end{tikzpicture}
\((E_1>E_3>E_2)\)
This can be seen from \(\dfrac{-\hbar^2}{2m}\underset{\underset{(\text{curvature})}{\downarrow}}{\dfrac{\partial^2\phi}{\partial x^2}}=\underset{\underset{(\text{energy})}{\downarrow}}{E\phi}\)
We know that for a particle of mass \(m\) in potential \(V(r)\) [ONLY for \(V(r)\neq 0\)]
\[
\hat{H}=\frac{p^2}{2m}+V(r)
\]
\[
[H,r]\neq 0\ \text{ and }\ \left[H,\frac{p^2}{2m}\right]\neq 0 \qquad -\text{(1)}
\]
\[
[H,V(r)]\neq 0\ -\text{(2)}\ \Big|\ [V(r),r]=0\ -\text{(3)}
\]
Eq\(^n\) (1) implies that particle does not have simultaneous Total energy & K.E at a given distance `\(r\).'
Eq\(^n\) (2) \(\Rightarrow\) Notion of potential energy for given total energy is no more.
Eq\(^n\) (3) \(\Rightarrow\) we can talk for potential energy at given position \(r\).
Nodes vs Energy :-- (1-D Infinite well)
we know,
\[
\phi_n(x)=\sqrt{\tfrac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)\ \text{ for } 0\leq x\leq L
\]
we can find \(E_n\)' from eigen value equation for \(H\)'.
\[
H\left|\phi_n\right\rangle=E_n\left|\phi_n\right\rangle
\]
\[
\left\langle x\middle|H\middle|\phi_n\right\rangle=E_n\left\langle x\middle|\phi_n\right\rangle
\]
\[
\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\phi_n(x)=E_n\phi_n(x)\qquad -\text{(1)}
\]
\[
\left\{\begin{array}{l}\phi'(x)=\sqrt{\tfrac{2}{L}}\,\frac{n\pi}{L}\cos\left(\frac{n\pi x}{L}\right)\\[4pt] \phi''(x)=-\sqrt{\tfrac{2}{L}}\left(\frac{n\pi}{L}\right)^2\sin\left(\frac{n\pi x}{L}\right)\end{array}\right\}
\]
Using in eq\(^n\) (1) we have
\[
\frac{\hbar^2}{2m}\left(\frac{n^2\pi^2}{L^2}\right)\left(\phi_n(x)\right)=E_n\left(\phi_n(x)\right)
\]
\[
\boxed{E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}}\qquad (E_n\propto n^2)
\]
\begin{tikzpicture}[scale=0.62]
\node at (-0.7,2.6) {$n=1$}; \node at (2.6,2.6) {$\phi_1(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{\pi x}{L}\right)$};
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\draw[->](6.1,2.2)--(8.6,2.2);
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\node at (-0.7,0.9) {$n=2$}; \node at (2.6,0.9) {$\phi_2(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{2\pi x}{L}\right)$};
\draw[->](6.1,0.6)--(8.6,0.6); \node[right] at (7.2,1.4){\scriptsize 1-node};
\draw[thick] plot[domain=0:2,samples=60] ({6.1+\x},{0.6+0.6*sin(360*\x/2)});
\node at (-0.7,-0.9) {$n=3$}; \node at (2.6,-0.9) {$\phi_3(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{3\pi x}{L}\right)$};
\draw[->](6.1,-1.1)--(8.6,-1.1); \node[right] at (7.2,-0.3){\scriptsize 2-Nodes};
\draw[thick] plot[domain=0:2,samples=80] ({6.1+\x},{-1.1+0.6*sin(540*\x/2)});
\node at (8.2,-1.45) {\scriptsize $L$};
\end{tikzpicture}
no. of nodes \((m)=n-1\)
or \((n=m+1)\)
\[
\Rightarrow E_n=E_{m+1}=\frac{(m+1)^2\pi^2\hbar^2}{2mL^2}
\]
as numer of nodes increases, \(\phi''(x)\) contributes to \(E_n\) \(\left\{\frac{-\hbar^2}{2m}\phi''(x)=E_n\phi(x)\right\}\) and that is why energy increases.
Study of general potential :
\begin{tikzpicture}[scale=0.75]
\draw[->] (-0.3,0)--(6.3,0) node[right]{$x$};
\draw[->] (0,-1.3)--(0,3.1) node[above]{$V(x)$};
\draw[thick] plot[smooth,domain=0.2:6.0,samples=120]
(\x,{2.6*exp(-(\x-1.5)^2/0.28)-1.05*exp(-(\x-2.5)^2/0.30)+2.9*exp(-(\x-3.6)^2/0.22)+0.35*exp(-(\x-4.8)^2/1.2)});
\draw[dashed] (0,2.55)--(6.1,2.55) node[right]{$E_4$};
\draw[dashed] (0,1.85)--(6.1,1.85) node[right]{$E_3$};
\draw[dashed] (0,-0.55)--(4.6,-0.55) node[right]{$E_2$};
\draw[dashed] (0,-1.2)--(4.6,-1.2) node[right]{$E_1$ (Impossible)};
\node[below] at (2.5,-0.75) {\tiny $V_{min}$};
\draw[<->] (0.6,0.35)--(2.1,0.35); \node[below] at (1.3,0.4){\tiny width of Barrier};
\draw[<->] (3.0,0.55)--(4.2,0.55); \node[above] at (3.6,0.6){\tiny width of Barrier};
\end{tikzpicture}
\(E_1,E_2,E_3,E_4\) are arbitrary choosen possibilities of total energy of system.
(1) \(E_1<V_{min}\Rightarrow \left\langle H\right\rangle<\left\langle V\right\rangle\Rightarrow \left\langle H\right\rangle-\left\langle V\right\rangle<0\)
But \(\left\langle H\right\rangle=\left\langle \frac{p^2}{2m}\right\rangle+\left\langle V\right\rangle\)
\[
\Rightarrow \left\langle \frac{p^2}{2m}\right\rangle=\left\langle H\right\rangle-\left\langle V\right\rangle=-ve
\]
But
\[
\left\langle \frac{p^2}{2m}\right\rangle=\frac{\left\langle \psi\middle|PP\middle|\psi\right\rangle}{2m}=\frac{\left\langle \psi\middle|P^{\dagger}P\middle|\psi\right\rangle}{2m}=\underset{\underset{\text{Norm}>0}{\uparrow}}{\frac{\left\langle P^{\dagger}\psi\middle|P\psi\right\rangle}{2m}}
\]
So \((E_1<V_{min})\) is not possible.
(2) \(_V_{min}<E<0_\)\ \((E_2)\) :--
Most of the time particle will be found inside well, but \(\exists\) finite non-zero probability that it will tunnel and escape.
\begin{tikzpicture}[scale=0.7]
\draw[->] (-0.2,0)--(6.2,0);
\draw[thick] plot[smooth,domain=0.1:6.0,samples=120]
(\x,{0.75*exp(-(\x-1.3)^2/0.25)+1.25*exp(-(\x-2.6)^2/0.35)+0.9*exp(-(\x-4.0)^2/0.25)+0.35*exp(-(\x-5.2)^2/0.3)});
\node[right] at (4.6,1.0) {\tiny $\leftarrow$ wave function.};
\end{tikzpicture}
But \(\exists\) bound states as well depending upon \(E_2\).
(3) \((E>0)\)\ \((E_3\) or \(E_4)\)
since, \(E_3(\infty)\) or \(E_4(\infty)>V(\infty)\), so particle will always escape to infinity and bound state does not exist for such energy values.
Bound states in Finite square well
\begin{tikzpicture}[scale=0.72]
\draw[->] (-2.2,0)--(2.6,0);
\draw[thick] (-2.0,1.5)--(-1.0,1.5)--(-1.0,0)--(1.0,0)--(1.0,1.9)--(2.2,1.9);
\node[above] at (-1.6,1.5) {\tiny $V_1$}; \node[above] at (1.5,1.9) {\tiny $V_2$};
\node[below] at (-1.0,0) {\tiny $-L/2$}; \node[below] at (1.0,0) {\tiny $+L/2$};
\node at (-1.6,0.6) {\tiny $\varphi_1$}; \node at (-0.4,0.6) {\tiny $\varphi_2$ (II)};
\node at (1.6,0.9) {\tiny $\varphi_3$ (III)};
\end{tikzpicture}
For \(E>V_1\) & \(E>V_2\) we only have scattering states. To have Bound states we need \(E<V_1\) & \(E<V_2\)
(1) Region (I) :--
\[
\frac{d^2\varphi_1}{dx^2}-k^2\varphi_1=0\quad k^2=\frac{2m(V_1-E)}{\hbar^2}
\]
\[
\varphi_1=Ae^{-kx}+Be^{kx}
\]
as \((x\to-\infty)\) \(\varphi_1\to 0\Rightarrow (A=0)\)
\[
\underline{\varphi_1=Be^{kx}}\qquad (x<-L/2)
\]
(2) Region II :--
\begin{tikzpicture}[scale=0.6]
\draw[->] (-2.2,0)--(2.4,0);
\draw[thick] (-2.0,1.4)--(-1.0,1.4)--(-1.0,0)--(1.0,0)--(1.0,1.8)--(2.1,1.8);
\node[above] at (-1.7,1.4) {\tiny $V_1$}; \node[above] at (1.5,1.8) {\tiny $V_2$};
\node[below] at (-1.0,0) {\tiny $-L/2$}; \node[below] at (1.0,0) {\tiny $+L/2$};
\node at (0,0.6) {\tiny II};
\end{tikzpicture}
\[
\frac{d^2\varphi_2}{dx^2}+\frac{2m(E-0)}{\hbar^2}\varphi_2=0
\]
\[
(\partial^2+l^2)\varphi_2=0\qquad (l^2=2mE/\hbar^2)
\]
\[
\left(\varphi_2=ae^{ilx}+be^{-ilx}\right)
\]
or \(\varphi_2=a\sin(lx)+b\cos lx\)
or \(\left\{\varphi_2=a\sin(lx+\theta)\right\}\)
Region III :
\begin{tikzpicture}[scale=0.6]
\draw[->] (-2.2,0)--(2.4,0);
\draw[thick] (-2.0,1.4)--(-1.0,1.4)--(-1.0,0)--(1.0,0)--(1.0,1.8)--(2.1,1.8);
\node[above] at (-1.7,1.4) {\tiny $V_1$}; \node[above] at (1.4,1.8) {\tiny $V_2$};
\node at (1.6,0.9) {\tiny III};
\draw[blue] (-2.2,0.35)--(2.4,0.35) node[right]{\tiny $E$};
\end{tikzpicture}
\[
\frac{d^2\varphi_3}{dx^2}+\frac{2m(E-V_2)}{\hbar^2}\varphi_3=0
\]
Since, \(E<V_2\Rightarrow (E-V_2<0)\)
\[
(\partial^2-m^2)\varphi_3=0\ \left\{m^2=\frac{2m(V_2-E)}{\hbar^2}\right\}
\]
\[
\varphi_3(x)=ce^{mx}+de^{-mx}
\]
need \(\varphi_3\to 0\) as \(x\to+\infty\Rightarrow c=0\)
\[
\boxed{\varphi_3(x)=d\,e^{-mx}}
\]
Since \(V(x)\) is finite, we can't have \(\varphi''(x)\) as infinite, so we need continuity in \(\varphi'(x)\) as well as in \(\varphi(x)\)
Ground state : (0-nodes)
\begin{tikzpicture}[scale=0.62]
\draw[->] (-2.4,0)--(2.6,0);
\draw[thick] (-2.2,1.5)--(-1.2,1.5)--(-1.2,0)--(1.2,0)--(1.2,1.9)--(2.3,1.9);
\node[above] at (-1.9,1.5) {\tiny $V_1$}; \node[above] at (1.7,1.9) {\tiny $V_2$};
\draw[thick] plot[smooth,domain=-2.2:2.3,samples=90] (\x,{1.35*exp(-(\x)^2/1.2)});
\end{tikzpicture}
\(1^{st}\) excited state : (1-node)
\begin{tikzpicture}[scale=0.62]
\draw[->] (-2.4,0)--(2.6,0);
\draw[->] (0,-1.2)--(0,2.2) node[above]{\tiny $V(x)$};
\draw[thick] (-2.2,1.5)--(-1.2,1.5)--(-1.2,0)--(1.2,0)--(1.2,1.9)--(2.3,1.9);
\node[above] at (-1.9,1.5) {\tiny $V_1$}; \node[above] at (1.7,1.9) {\tiny $V_2$};
\draw[thick] plot[smooth,domain=-2.2:2.3,samples=110] (\x,{1.3*sin(140*\x)*exp(-(\x)^2/1.6)});
\end{tikzpicture}
\(2^{nd}\) excited state : (2-nodes)
\begin{tikzpicture}[scale=0.62]
\draw[->] (-2.6,0)--(2.8,0) node[right]{\tiny $x$};
\draw[->] (0,-1.2)--(0,2.2) node[above]{\tiny $V(x)$};
\draw[thick] (-2.4,1.5)--(-1.2,1.5)--(-1.2,0)--(1.2,0)--(1.2,1.9)--(2.5,1.9);
\node[left] at (-2.5,0.9) {\tiny $\infty\leftarrow$}; \node[right] at (2.6,0.9) {\tiny $\to\infty$};
\draw[thick] plot[smooth,domain=-2.4:2.5,samples=140] (\x,{1.2*cos(190*\x)*exp(-(\x)^2/1.8)});
\end{tikzpicture}
No. of Bound states :
(1) For \(E>V(\infty)\) : No Bound states exist
(2) _For \(E<V(\infty)\)_
\begin{tikzpicture}[scale=0.55]
\draw[->] (-2.2,0)--(2.4,0);
\draw[thick] (-2.0,1.2)--(-1.0,1.2)--(-1.0,0)--(1.0,0)--(1.0,1.6)--(2.2,1.6);
\node[above] at (-1.7,1.2) {\tiny $V_1$}; \node[above] at (1.6,1.6) {\tiny $V_2$};
\draw[<->] (-1.0,-0.5)--(1.0,-0.5) node[midway,below]{\tiny $L$};
\end{tikzpicture}
we know that,
\[
\frac{2mE_n}{\hbar^2}=k^2=\frac{n^2\pi^2}{L^2}\ \text{ for (1-D)}\infty\text{-well}
\]
\[
\left(E_n=\frac{n^2\hbar^2\pi^2}{2mL^2}\right)=n^2\alpha\quad \left(\alpha=\frac{\pi^2\hbar^2}{2mL^2}\right)
\]
\begin{tikzpicture}[scale=0.62]
\draw[->] (-0.2,0)--(7.2,0);
\foreach \x/\l in {1.2/\alpha,3.0/4\alpha,4.8/9\alpha,6.6/16\alpha}{\draw (\x,-0.12)--(\x,0.12); \node[below] at (\x,-0.12){\tiny $\l$};}
\node[above] at (0.6,0.15) {\tiny (1)}; \node[above] at (2.1,0.15) {\tiny (2)};
\node[above] at (3.9,0.15) {\tiny (3)}; \node[above] at (5.7,0.15) {\tiny (4)};
\node[below] at (3.6,-0.8) {\tiny $\longrightarrow$ Energy ($E$ of finit well)};
\end{tikzpicture}
No of Bound states
(1) \(E<\alpha\)
1
(2) \(\alpha<E<4\alpha\)
2
(3) \(4\alpha<E<9\alpha\)
3
(4) \(9\alpha<E<16\alpha\)
4
\(\vdots\)
Eigen functions of symmetric potential will be either even or odd functions
Linear Harmonic Oscillator :--
\[
H=\frac{P^2}{2m}+\tfrac{1}{2}m\omega^2x^2
\]
\[
\left(V(x)=\tfrac{1}{2}m\omega^2x^2\right)
\]
\begin{tikzpicture}[scale=0.6]
\draw[->] (-1.8,0)--(2.0,0) node[right]{\tiny $x$};
\draw[->] (0,-0.3)--(0,2.2) node[above]{\tiny $V(x)$};
\draw[thick] plot[domain=-1.4:1.4,samples=60] (\x,{\x*\x});
\end{tikzpicture}
\[
H=\hbar\omega\left(a^{\dagger}a+\frac{I}{2}\right)
\]
\[
a=\frac{x}{\sqrt{2\hbar/m\omega}}+\frac{ip}{\sqrt{2\hbar m\omega}}=\frac{x}{x_0}+\frac{ip}{p_0}
\]
\[
a^{\dagger}=\frac{x}{\sqrt{2\hbar/m\omega}}-\frac{ip}{\sqrt{2\hbar m\omega}}=\frac{x}{x_0}-\frac{ip}{p_0}
\]
where \(\left[a\,a^{\dagger}\right]=1\)
then, \(E_n=\hbar\omega\left(n+\tfrac{1}{2}\right)\) \((n=0,1,2,3,\ldots)\)
and,
\[
\phi_n(x)=A_ne^{-\left(\frac{x}{x_0}\right)^2}H_n(x/x_0)
\]
\[
\tilde{\phi}_n(p)=B_ne^{-\left(\frac{p}{p_0}\right)^2}H_n(p/p_0)
\]
we saw that
\[
\phi_n(x)=A_ne^{-(x/x_0)^2}H_n(x/x_0)
\]
are stationary states for S.H.O.
where \(H_n(x)\) are Hermite polynomials.
\[
H_n(x)=e^{x^2}\left(\frac{-d}{dx}\right)^ne^{-x^2}
\]
\[
H_0(x)=e^{x^2}e^{-x^2}=\underline{1}
\]
\[
H_1(x)=e^{x^2}(+2x)e^{-x^2}=2x
\]
\[
H_2(x)=e^{x^2}\left(-2e^{-x^2}+4x^2e^{-x^2}\right)
\]
\[
H_2(x)=-2+4x^2
\]
\(a\) --- lowering / annihilation operator
\(a^{\dagger}\) --- Raising / creation operator
$a\left
n\right\rangle=c_n\left
n-1\right\rangle$
$a^{\dagger}\left
n\right\rangle=c_n\left
n+1\right\rangle$
$\left\langle n\right
a^{\dagger}=\left\langle n-1\right
c_n^{*}$
$\left\langle n\right
a=\left\langle n+1\right
c_n^{*}$
$\left\langle n\middle
a^{\dagger}a\middle
n\right\rangle=
c_n
^2$
$\left\langle n\middle
aa^{\dagger}\middle
n\right\rangle=
c_n
^2$
$\left\langle n\middle
\hat{N}\middle
n\right\rangle=
c_n
^2$
$\left\langle n\middle
N+1\middle
n\right\rangle=
c_n
^2$
\(c_n=\sqrt{n}\)
$
c
=\sqrt{n+1}$
So,
\[
a\left|n\right\rangle=\sqrt{n}\left|n-1\right\rangle
\]
\[
a^{\dagger}\left|n\right\rangle=\sqrt{n+1}\left|n+1\right\rangle
\]
Some useful results
\[
1)\ \sum_{n=1}^{n}n=1+2+3+\cdots+n=\frac{n(n+1)}{2}
\]
\[
2)\ \sum_{n=1}^{n}n^2=1^2+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}
\]
\[
3)\ \sum_{n=1}^{n}n^3=1^3+2^3+3^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2
\]
\[
=\left(\textstyle\sum n\right)^2
\]
\[
4)\ \sum_{n=1}^{n}(2n)^3=2^3\sum_{n=1}^{n}n^3=8\left(\frac{n(n+1)}{2}\right)^2=8\left(\textstyle\sum n\right)^2
\]
\[
5)\ (x+y)^4=\left(\tfrac{4!}{4!}\right)x^4+\left(\tfrac{4!}{4!}\right)y^4+\left(\tfrac{4!}{2!2!}\right)x^2y^2
\]
\[
+\left(\tfrac{4!}{1!3!}\right)xy^3+\left(\tfrac{4!}{3!1!}\right)x^3y
\]
Ground State of SHO :
\[
a\left|0\right\rangle=0
\]
\[
\Rightarrow\ \left\langle x\right|a\left|0\right\rangle=0
\]
\[
\left\langle x\middle|\frac{\hat{x}}{x_0}+\frac{i\hat{p}}{\hbar}\middle|0\right\rangle=0
\]
\[
\frac{1}{x_0}\left\langle x\middle|\hat{x}\middle|0\right\rangle+\frac{i}{p_0}\left\langle x\middle|\hat{p}\middle|0\right\rangle=0
\]
\[
\frac{x}{x_0}\phi_0(x)+\frac{i}{p_0}\left(-i\hbar\frac{\partial}{\partial x}\phi_0(x)\right)=0
\]
Solve to get \(\phi(x)\) :
\[
\left\{\frac{x}{x_0}\phi_0(x)+\frac{\hbar}{p_0}\frac{\partial\phi_0(x)}{\partial x}=0\right\}
\]
\[
\left\{\begin{array}{l}\phi_0(x)=\left\langle x\middle|0\right\rangle\\ \phi_1(x)=\left\langle x\middle|1\right\rangle\\ \phi_2(x)=\left\langle x\middle|2\right\rangle\\ \phi_n(x)=\left\langle x\middle|n\right\rangle\end{array}\right.
\]
To find \(\phi_1(x)=\left\langle x\middle|1\right\rangle\)
\[
a\left|n\right\rangle=\sqrt{n}\left|n-1\right\rangle
\]
\[
a^{\dagger}\left|n\right\rangle=\sqrt{n+1}\left|n+1\right\rangle
\]
So, \(a^{\dagger}\left|0\right\rangle=1\left|1\right\rangle\)
\[
\left\langle x\right|a^{\dagger}\left|0\right\rangle=\left\langle x\middle|1\right\rangle=\phi_1(x)
\]
\[
\left\langle x\middle|\frac{x}{x_0}+\frac{ip}{p_0}\middle|0\right\rangle=\phi_1(x)
\]
\[
\left\{\frac{x}{x_0}\phi_0(x)+\frac{i}{p_0}\left(-i\hbar\frac{\partial}{\partial p}\phi_0(x)\right)=\phi_1(x)\right\}
\]
\(\phi_1(x)\) will turn out to be normalized.
Some results :
\[
1)\ \left\langle x\right\rangle_n=\left\langle n\middle|\hat{x}\middle|n\right\rangle=\left\langle n\middle|\frac{x_0(a+a^{\dagger})}{2}\middle|n\right\rangle=0
\]
\[
2)\ \left\langle p\right\rangle_n=\left\langle n\middle|\hat{p}\middle|n\right\rangle=\left\langle n\middle|\frac{p_0}{2i}(a-a^{\dagger})\middle|n\right\rangle=0
\]
\[
\begin{align*}
3)\ \left\langle x^2\right\rangle_n&=\left\langle n\middle|\hat{x}^2\middle|n\right\rangle=\left\langle n\middle|\frac{x_0^2}{4}(a+a^{\dagger})^2\middle|n\right\rangle\\
&=\left\langle n\middle|\frac{x_0^2}{4}(aa^{\dagger}+a^{\dagger}a)\middle|n\right\rangle\\
&=\frac{x_0^2}{4}\left\langle n\middle|N+1+N\middle|n\right\rangle
\end{align*}
\]
\[
\left\langle x^2\right\rangle_n=\left(n+\tfrac{1}{2}\right)\frac{x_0^2}{2}=\frac{1}{2}\left(\frac{2\hbar}{2m\omega}\right)\left(n+\tfrac{1}{2}\right)
\]
\[
\begin{align*}
4)\ \left\langle p^2\right\rangle_n&=\left\langle n\middle|p^2\middle|n\right\rangle=\left\langle n\middle|\frac{p_0^2}{4i^2}(a-a^{\dagger})^2\middle|n\right\rangle\\
&=\frac{-p_0^2}{4}\left\langle n\middle|-aa^{\dagger}-a^{\dagger}a\middle|n\right\rangle\\
&=\frac{p_0^2}{4}\left\langle n\middle|2N+1\middle|n\right\rangle=\hbar m\omega\left(n+\tfrac{1}{2}\right)
\end{align*}
\]
*Eigen-states of \((a^{\dagger})\) :*
\[
a^{\dagger}\left|\lambda\right\rangle=\lambda\left|\lambda\right\rangle
\]
where \(\left|\lambda\right\rangle\) can be expanded in \(\{\left|n\right\rangle\}\) basis :-- \(\left|\lambda\right\rangle=\sum\limits_{n=0}^{\infty}c_n\left|n\right\rangle\)
\[
a^{\dagger}\left|\lambda\right\rangle=\sum_{n=0}^{\infty}c_na^{\dagger}\left|n\right\rangle=\sum_{n=0}^{\infty}c_n\sqrt{n+1}\left|n+1\right\rangle\ -\text{(1)}
\]
also, \(a^{\dagger}\left|\lambda\right\rangle=\lambda\left|\lambda\right\rangle=\sum\limits_{n=0}^{\infty}\lambda c_n\left|n\right\rangle\quad -\text{(2)}\)
\[
\Rightarrow \sum_{n=0}^{\infty}c_n\lambda\left|n\right\rangle=\sum_{n=0}^{\infty}c_n\sqrt{n+1}\left|n+1\right\rangle
\]
which gives,
\[
\lambda\left(c_0\left|0\right\rangle+c_1\left|1\right\rangle+c_2\left|2\right\rangle+\cdots\right)
\]
\[
=c_0\left|1\right\rangle+c_1\sqrt{2}\left|2\right\rangle+c_2\sqrt{3}\left|3\right\rangle+\cdots
\]
\[
\Rightarrow \lambda c_0=0,\ \lambda c_1=c_0,\ \lambda c_2=c_1\sqrt{2},
\]
\[
\cdots\Rightarrow c_0=0=c_1=c_2=c_3\cdots
\]
\[
\Rightarrow \left|\lambda\right\rangle\ \text{d.n.e.}
\]
*Eigen states of \(a\) :*
\[
\hat{a}\left|\alpha\right\rangle=\alpha\left|\alpha\right\rangle
\]
where, \(\left|\alpha\right\rangle=\sum\limits_{n=0}^{\infty}c_n\left|n\right\rangle\)
\[
\Rightarrow a\left|\alpha\right\rangle=\sum_{n=0}^{\infty}c_na\left|n\right\rangle=\sum_{n=1}^{\infty}c_n\sqrt{n}\left|n-1\right\rangle
\]
also, \(a\left|\alpha\right\rangle=\alpha\left|\alpha\right\rangle=\sum\limits_{n=0}^{\infty}\alpha c_n\left|n\right\rangle\)
\[
\Rightarrow \sum_{n=1}^{\infty}c_n\sqrt{n}\left|n-1\right\rangle=\sum_{n=0}^{\infty}\alpha c_n\left|n\right\rangle
\]
\[
\Rightarrow \left(c_1\left|0\right\rangle+c_2\sqrt{2}\left|1\right\rangle+c_3\sqrt{3}\left|2\right\rangle+\cdots\right)
\]
\[
=\alpha\left(c_0\left|0\right\rangle+c_1\left|1\right\rangle+c_2\left|2\right\rangle+\cdots\right)
\]
\[
\Rightarrow \begin{array}{ll}c_1=\alpha c_0 & \\ c_2\sqrt{2}=\alpha c_1 & c_2=\frac{1}{\sqrt{2}}\alpha^2c_0\\ c_3\sqrt{3}=\alpha c_2 & c_3=\frac{1}{\sqrt{3\cdot 2}}\alpha^3c_0\\ \vdots & \\ c_n\sqrt{n}=\alpha c_{n-1} & c_n=\frac{1}{\sqrt{n!}}\alpha^nc_0\end{array}
\]
\[
\Rightarrow \left|\alpha\right\rangle=\sum_{n=0}^{\infty}c_n\left|n\right\rangle=\sum_{n=0}^{\infty}\frac{1}{\sqrt{n!}}\alpha^nc_0\left|n\right\rangle
\]
\[
\left\{\left|\alpha\right\rangle=c_0\sum_{n=0}^{\infty}\frac{\alpha^n\left|n\right\rangle}{\sqrt{n!}}\right\}
\]
\[
\left\langle \alpha\middle|\alpha\right\rangle=1\Rightarrow |c_0|^2\sum_{n=0}^{\infty}\frac{\left(|\alpha|^2\right)^n}{n!}=1
\]
\[
|c_0|^2e^{|\alpha|^2}=1\Rightarrow \left(|c_0|=e^{\frac{-|\alpha|^2}{2}}\right)
\]
So
\[
\left|\alpha\right\rangle=e^{\frac{-|\alpha|^2}{2}}\sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}}\left|n\right\rangle
\]
\[
a^{\dagger}\left|n\right\rangle=\sqrt{n+1}\left|n+1\right\rangle
\]
\[
\left|1\right\rangle=a^{\dagger}\left|0\right\rangle\qquad \left|2\right\rangle=\frac{1}{\sqrt{2}}a^{\dagger}\left|1\right\rangle=\frac{(a^{\dagger})^2}{\sqrt{2}}\left|0\right\rangle
\]
\[
\left|3\right\rangle=\frac{1}{\sqrt{3}}a^{\dagger}\left|2\right\rangle=\frac{(a^{\dagger})^3}{\sqrt{3!}}\left|0\right\rangle\quad \vdots
\]
\[
\left(\left|n\right\rangle=\frac{1}{\sqrt{n!}}\left(a^{\dagger}\right)^n\left|0\right\rangle\right)
\]
using, we get,
\[
\left|\alpha\right\rangle=e^{\frac{-|\alpha|^2}{2}}\sum_{n=0}^{\infty}\frac{\alpha^n\left(a^{\dagger}\right)^n}{n!}\left|0\right\rangle
\]
\[
\left|\alpha\right\rangle=e^{\frac{-|\alpha|^2}{2}}\left(\sum_{n=0}^{\infty}\frac{\left(\alpha a^{\dagger}\right)^n}{n!}\right)\left|0\right\rangle
\]
So
\[
\left\{\left|\alpha\right\rangle=e^{\frac{-|\alpha|^2}{2}}\,e^{\alpha a^{\dagger}}\left|0\right\rangle\right\}
\]
\(\underset{\text{operator}}{\nwarrow}\)
Normalized eigen states of lowering operator.
Note : All numbers of complex plane \((\alpha)\) are eigen values of anhilation operator.
States \(\left|\alpha\right\rangle\) are not orthonormal and forms over-complete space of states.
\[
\left(\int d\alpha\ \left|\alpha\right\rangle\left\langle \alpha\right|=\pi I\right)
\]
\[
\begin{align*}
\left\langle x\middle|\alpha\right\rangle&=\left\langle x\right|\sum c_0\frac{\alpha^n}{\sqrt{n!}}\left|n\right\rangle\\
&=\left\langle x\right|e^{\frac{-|\alpha|^2}{2}}\sum\frac{\alpha^n}{\sqrt{n!}}\left|n\right\rangle\\
&=e^{\frac{-|\alpha|^2}{2}}\sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}}\phi_n(x)
\end{align*}
\]
\[
\alpha(x)=e^{\frac{-|\alpha|^2}{2}}\sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}}\left(A_ne^{\frac{-x^2}{x_0^2}}H_n(x/x_0)\right)
\]
Matrix representations :
\[
a_{\ell m}=\left\langle \ell\middle|a\middle|m\right\rangle,\qquad a^{\dagger}_{\ell m}=\left\langle \ell\middle|a^{\dagger}\middle|m\right\rangle
\]
\[
x_{\ell m}=\left\langle \ell\middle|x\middle|m\right\rangle,\qquad P_{\ell m}=\left\langle \ell\middle|P\middle|m\right\rangle
\]
\[
A=\vec{J}\cdot\hat{n}
\]
\[
\hat{n}=(\sin\theta\cos\varphi,\ \sin\theta\sin\varphi,\ \cos\theta)
\]
\[
\hat{A}=\sin\theta\cos\varphi\,\hat{J}_x+\sin\theta\sin\varphi\,\hat{J}_y+\cos\theta\,\hat{J}_z
\]
\[
\Delta A=\sqrt{\left\langle A^2\right\rangle-\left\langle A\right\rangle^2}
\]
\[
\begin{align*}
\left\langle A^2\right\rangle=\Big\langle &\sin^2\theta\cos^2\varphi\,J_x^2+\sin^2\theta\sin^2\varphi\,J_y^2+\cos^2\theta\,J_z^2\\
&+\tfrac{\sin^2\theta\sin 2\varphi}{2}J_xJ_y+\tfrac{\sin 2\theta\cos\varphi}{2}J_xJ_z\\
&+\tfrac{\sin 2\varphi}{2}J_yJ_x+\tfrac{\sin 2\theta}{2}\sin\varphi\,J_yJ_z\\
&+\tfrac{\sin 2\theta}{2}\cos\varphi\,J_zJ_x+\tfrac{\sin 2\theta}{2}\sin\varphi\,J_zJ_y\Big\rangle
\end{align*}
\]
(cross terms \(\to 0\))
\[
\left\langle J_xJ_y+J_yJ_x\right\rangle=?\ \longrightarrow\ \left\langle 2J_xJ_y-i\hbar J_z\right\rangle=2\left\langle J_xJ_y\right\rangle-i\hbar m\hbar
\]
We know, \((J_xJ_y-J_yJ_x=i\hbar J_z)\)
\[
=2\left(\tfrac{i\hbar}{2}m\hbar\right)-i\hbar m\hbar=0
\]
\[
\left\langle J_x^2\right\rangle=\left\langle J_y^2\right\rangle=\frac{\left\langle J^2\right\rangle-\left\langle J_z^2\right\rangle}{2}=\frac{j(j+1)\hbar^2-m^2\hbar^2}{2}
\]
\[
\left\langle A^2\right\rangle=\sin^2\theta\left\langle J_x^2\right\rangle+\cos^2\theta\left\langle J_z^2\right\rangle=\sin^2\theta\left\langle J_x^2\right\rangle+\cos^2\theta\,m^2\hbar^2
\]
\[
\begin{align*}
\left\langle J_xJ_y\right\rangle&=\left\langle \left(\frac{J_++J_-}{2}\right)\left(\frac{J_+-J_-}{2i}\right)\right\rangle\\
&=\left\langle \frac{J_+^2-J_+J_-+J_-J_+-J_-^2}{4i}\right\rangle\\
&=\frac{\left\langle [J_-,J_+]\right\rangle}{4i}=\frac{\left\langle \left[(J_x-iJ_y),(J_x+iJ_y)\right]\right\rangle}{4i}\\
&=\left(\frac{\cancel{0}+i^2\hbar J_z+i\,i J_z\hbar}{4i}\right)\\
&=\left\langle \frac{-J_z\hbar}{2i}\right\rangle=\left\langle \frac{i\hbar J_z}{2}\right\rangle\\
&=\frac{i\hbar}{2}m\hbar
\end{align*}
\]
\[
\begin{align*}
\left\langle J_xJ_z\right\rangle&=\left\langle jm\middle|J_xJ_z\middle|jm\right\rangle\\ &=m\hbar\left\langle jm\middle|J_x\middle|jm\right\rangle=0
\end{align*}
\]
\[
\begin{align*}
\left\langle J_zJ_x\right\rangle&=\left\langle jm\middle|J_zJ_x\middle|jm\right\rangle\\ &=m\hbar\left\langle jm\middle|J_x\middle|jm\right\rangle=m\hbar(0)=0
\end{align*}
\]
\[
\left\langle J_zJ_y\right\rangle=\left\langle J_yJ_z\right\rangle=0.
\]
\[
\left\langle A\right\rangle=\sin\theta\cos\varphi\,\overset{0}{\cancel{\left\langle J_x\right\rangle}}+\sin\theta\sin\varphi\,\overset{0}{\cancel{\left\langle J_y\right\rangle}}+\cos\theta\left\langle J_z\right\rangle
\]
\[
=\cos\theta\,m\hbar
\]
\[
\Delta A=\sqrt{\left\langle A^2\right\rangle-\left\langle A\right\rangle^2}
\]
\[
\Delta A=\sqrt{\sin^2\theta\left\langle J_x^2\right\rangle}=\sqrt{\left\langle J_x^2\right\rangle}\sin\theta
\]
\[
\boxed{\Delta A=\left(\sqrt{\frac{j(j+1)-m^2}{2}}\right)\hbar\sin\theta}
\]
This universe is not mine,
I have not created it.
This life itself is not mine,
I am not my creater.
Nobody is creater of anything here.
Every one does is to modify things
for their longivity.
So, Here I am knowing about this universe / Nature or nature of Nature. Training myself from failure to less failure. Making an understanding of World in which I live on.
Just like Artificial machines, I will train my mind to break some unproductive habits, and turn it into an useful machine to understand this nature.
To learn I need to observe, repeat / redo, make conclusions.
\begin{tikzpicture}[scale=0.9,every node/.style={font=\footnotesize}]
\node (a) at (0,0) {Observe};
\node (b) at (2.9,0) {figure out patterns};
\draw[->] (a)--(b);
\node (c) at (2.9,-1.3) {Do it \underline{again \& again \& again}};
\draw[->] (b)--(c);
\node (d1) at (-0.2,-3.2) {make Conclusion $C_1$};
\node (d2) at (3.4,-3.2) {Make Conclusion $C_2$};
\node (d3) at (7.0,-3.2) {Make Conclusion $C_3$};
\node at (8.9,-2.7) {$\cdots$};
\draw[->] (2.4,-1.6)--(0.4,-2.8);
\draw[->] (2.9,-1.6)--(3.2,-2.8);
\draw[->] (3.4,-1.6)--(6.2,-2.8);
\draw[->] (3.6,-1.6)--(8.5,-2.5);
\node at (3.2,-4.0) {Conclusion $\underline{C_n}$ is better than $\underline{C_{n-1}}$.};
\end{tikzpicture}
\[
\frac{d}{dt}\left\langle A\right\rangle=\left\langle \frac{dA}{dt}\right\rangle
\]
where \(\left\langle A\right\rangle=\dfrac{\left\langle \psi(0)\middle|A(t)\middle|\psi(0)\right\rangle}{\left\langle \psi(0)\middle|\psi(0)\right\rangle}\)
or \(\left\langle A\right\rangle=\dfrac{\left\langle \psi_H\middle|A(t)\middle|\psi_H\right\rangle}{\left\langle \psi_H\middle|\psi_H\right\rangle}\)
\[
\frac{d}{dt}\left\langle A\right\rangle=\frac{\left\langle \psi(0)\middle|\frac{dA}{dt}\middle|\psi(0)\right\rangle}{\left\langle \psi(0)\middle|\psi(0)\right\rangle}
\]
\[
\left\{\frac{d}{dt}\left\langle A\right\rangle=\left\langle \frac{dA}{dt}\right\rangle\right\}
\]
From classical mech.
For any \(A=A(q,p)\) \(\Big|\) Heisenberg
\[
\frac{dA}{dt}=\sum_{i=1}^{N}\left(\frac{\partial A}{\partial q}\dot{q}+\frac{\partial A}{\partial p}\dot{p}\right)+\frac{\partial A}{\partial t}
\]
equation of motion.
\[
\frac{dA}{dt}=\{A,H\}+\partial A/\partial t
\]
\[
\frac{d\hat{A}(t)}{dt}=\frac{[A,H]}{i\hbar}+\partial\hat{A}/\partial t
\]
Probability current density associated with probability density :--
\[
\nabla\cdot\vec{J}=-\partial\rho/\partial t
\]
\[
\rho=|\psi|^2,\qquad i\hbar\frac{d}{dt}\left|\psi\right\rangle=H\left|\psi\right\rangle
\]
\[
\hat{H}=\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}+\hat{V}(x)
\]
\[
\left\{\vec{J}=\frac{\hbar}{2mi}\left(\psi^{*}\nabla\psi-\psi\nabla\psi^{*}\right)\right\}
\]
Evolution of state after collapse :--
suppose we have particle inside a (1-D) box.
\[
\psi(x)=\left\langle x\middle|\Psi\right\rangle=\sqrt{\tfrac{2}{L}}\sin\frac{n\pi x}{L}
\]
Any style measurement of position will collapse wavefunction to \(x_0\).
\[
\psi(x=x_0)=\delta(x-x_0)
\]
Now this state will evolve with time. find \(\left\langle x\right\rangle(t)\), \(\left\langle x^2\right\rangle(t)\), \(\left\langle x^3\right\rangle(t)\ldots\)
lets use Heisenberg picture. So state vector remains fixed \(=\delta(x-x_0)\), & \(\hat{x}=\hat{x}(t)\)
To know, how wavefunction changes let us look how \(\left\langle x\right\rangle\), \(\left\langle x^2\right\rangle(t)\), \(\sigma^2\) changes with time.
\[
-\text{(1)}\ \frac{d}{dt}\left\langle x\right\rangle=\left\langle dx/dt\right\rangle=\left\langle \frac{[x,H]}{i\hbar}\right\rangle
\]
\[
=\left\langle \frac{\left[x,\frac{p^2}{2m}+V(x)\right]}{i\hbar}\right\rangle
\]
\[
\frac{d}{dt}\left\langle x\right\rangle=\frac{1}{2mi\hbar}\left\langle 2p\,i\hbar\right\rangle=\frac{\left\langle p\right\rangle}{m}
\]
\[
\left\{\frac{d}{dt}\left\langle x\right\rangle=\frac{\left\langle p\right\rangle}{m}\right\}
\]
\[
\begin{align*}
(2)\ \frac{d}{dt}\left\langle x^2\right\rangle&=\left\langle \frac{dx^2(t)}{dt}\right\rangle=\frac{\left\langle [x^2,H]\right\rangle}{i\hbar}\\
&=\left(\frac{1}{i\hbar}\right)\left\langle \left[x^2,\frac{p^2}{2m}+V(x)\right]\right\rangle\\
&=\frac{1}{2mi\hbar}\left\langle \left[x^2,p^2\right]\right\rangle\\
&=\frac{1}{2mi\hbar}\left\langle x\left[x,p^2\right]+\left[x,p^2\right]x\right\rangle\\
&=\frac{1}{2mi\hbar}\left\langle x(2pi\hbar)+(2pi\hbar)x\right\rangle
\end{align*}
\]
\[
\left\{\frac{d}{dt}\left\langle x^2\right\rangle=\frac{1}{m}\left\langle xp+px\right\rangle\right\}
\]
\(\underset{\text{Not an operator}}{\text{Number}}\)
\[
\begin{align*}
\frac{d^2}{dt^2}\left\langle x^2\right\rangle&=\frac{1}{m}\left\langle \frac{d}{dt}(xp+px)\right\rangle\\
&=\frac{1}{m}\left\langle \frac{\left[xp+px,\frac{p^2}{2m}+V(x)\right]}{i\hbar}+\overset{0}{\cancel{\frac{\partial}{\partial t}\left\langle xp+px\right\rangle}}\right\rangle\\
&=\frac{1}{mi\hbar}\left\langle \left[xp+px,\frac{p^2}{2m}\right]\right\rangle+\frac{\left\langle [xp+px,V]\right\rangle}{mi\hbar}\\
&=\frac{2}{m^2}\left\langle p^2\right\rangle+\frac{\left\langle \left[xp+px,V(x)\right]\right\rangle}{mi\hbar}\\
&=\frac{2}{m^2}\left\langle p^2\right\rangle+\frac{x[p,V(x)]+[p,V(x)]x}{mi\hbar}
\end{align*}
\]
\[
\left\langle \ddot{x^2}\right\rangle=\frac{2}{m^2}\left\langle p^2\right\rangle+\frac{-2}{m}\frac{dV(x)}{dx}
\]
Heisenberg picture of Q.M.
Observables / operators are time dependent, where as state of quantum mechanical system is fixed.
\[
\left\langle A\right\rangle(t)=\frac{\left\langle \psi\middle|A\middle|\psi\right\rangle}{\left\langle \psi\middle|\psi\right\rangle}
\]
\(\downarrow\)
This time dependence came from time dependent \(\left|\psi(t)\right\rangle\) in Schrödinger interpretation of Q.M.
\[
\left\langle A\right\rangle(t)=\frac{\left\langle \psi_s(t)\middle|\hat{A}_s\middle|\psi_s(t)\right\rangle}{\left\langle \psi_s(t)\middle|\psi_s(t)\right\rangle}
\]
\[
\left\{\left|\psi(t)\right\rangle=e^{\frac{-iHt}{\hbar}}\left|\psi(0)\right\rangle\right\}
\]
\[
\left\langle A\right\rangle(t)=\frac{\left\langle \psi(0)\middle|e^{\frac{iHt}{\hbar}}A_se^{\frac{-iHt}{\hbar}}\middle|\psi(0)\right\rangle}{\left\langle \psi(0)\middle|\psi(0)\right\rangle}
\]
\[
\left\{\left\langle A(t)\right\rangle(t)=\frac{\left\langle \psi(0)\middle|A(t)\middle|\psi(0)\right\rangle}{\left\langle \psi(0)\middle|\psi(0)\right\rangle}\right\}
\]
where,
\[
A(t)=A_H(t)=e^{\frac{iH_st}{\hbar}}A_se^{\frac{-iH_st}{\hbar}}
\]
\[
(t=0)\ \text{gives}\ A_H(0)=A_s
\]
So,
\[
A(t)=e^{\frac{iH_st}{\hbar}}A_H(0)e^{\frac{-iH_st}{\hbar}}
\]
\[
H(t)=e^{\frac{iH_st}{\hbar}}H_se^{\frac{-iH_st}{\hbar}}
\]
also, we know \(A_H(0)=A_s\)
i.e. \(H_H(0)=H_s\)
So, \(\left[e^{\frac{iH_st}{\hbar}},H_s\right]=0\)
\[
\Rightarrow H(t)=H_s=H_H(0)
\]
\[
\Rightarrow \left\{A_H(t)=e^{\frac{iHt}{\hbar}}A(0)e^{\frac{-iHt}{\hbar}}\right\}
\]
\[
\begin{align*}
\frac{dA_H(t)}{dt}&=e^{\frac{iHt}{\hbar}}\frac{iH}{\hbar}A(0)e^{\frac{-iHt}{\hbar}}\\
&\quad +e^{\frac{iHt}{\hbar}}A(0)e^{\frac{-iHt}{\hbar}}\left(\frac{-iH}{\hbar}\right)\\
&=\frac{i}{\hbar}HA(t)+A(t)(-iH/\hbar)
\end{align*}
\]
\[
\left\{\frac{dA(t)}{dt}=\frac{[A,H]}{i\hbar}\right\}
\]
Schrödinger Picture :
Each quantum mechanical system is described by a state vector \(\left|\Psi(t)\right\rangle\) which evolves in time by
\[
i\hbar\frac{d}{dt}\left|\Psi(t)\right\rangle=\hat{H}\left|\Psi(t)\right\rangle
\]
\(\underset{\text{sol}^n}{\searrow}\)
where as, observables are time independent, self adjoint, Hermitian operators.
\[
\underline{A+f(t)}\qquad \left|\psi(t)\right\rangle=e^{\frac{-iHt}{\hbar}}\left|\psi(0)\right\rangle
\]
\[
\left\langle A\right\rangle(t)=\frac{\left\langle \psi(t)\middle|A\middle|\psi(t)\right\rangle}{\left\langle \psi\middle|\psi\right\rangle}\quad \left|\psi(t)\right\rangle=e^{\frac{-iH(t-t_0)}{\hbar}}\left|\psi(t_0)\right\rangle
\]
Heisenberg picture :
Each quantum mechanical system is described by a constant / time independent state vector \(\left|\Psi(t_0)\right\rangle\) (an element of seperable hilbert space).
Each physical observable is represented by time dependent self adjoint operator which evolves in time by
\[
\frac{dA_H(t)}{dt}=\frac{[A,H]}{i\hbar}+\frac{\partial A}{\partial t}
\]
whose solution is
\[
A(t)=e^{\frac{iHt}{\hbar}}A(0)e^{\frac{-iHt}{\hbar}}
\]
\[
\left\langle x\middle|\psi_i\right\rangle=\sqrt{\tfrac{2}{L}}\cos\frac{\pi x}{L}
\]
\[
\left\langle x\middle|\psi_f\right\rangle=\sqrt{\tfrac{2}{2L}}\cos\left(\frac{\pi x}{2L}\right)
\]
\[
\left|\psi_i\right\rangle=I\left|\psi_i\right\rangle=\left(\int dx\left|x\right\rangle\left\langle x\right|\right)\left|\psi_i\right\rangle=\int dx\left\langle x\middle|\psi_i\right\rangle\left|x\right\rangle
\]
\[
\left|\psi_f\right\rangle=I\left|\psi_f\right\rangle=\left(\int dx'\left|x'\right\rangle\left\langle x'\right|\right)\left|\psi_f\right\rangle=\int dx'\left\langle x'\middle|\psi_f\right\rangle\left|x'\right\rangle
\]
\[
\begin{align*}
\left\langle \psi_i\middle|\psi_f\right\rangle=\sqrt{P}&=\int dx\left\langle x\middle|\psi_i\right\rangle^{*}\left\langle x\right|\left(\int dx'\underline{\left\langle x'\middle|\psi_f\right\rangle\left|x'\right\rangle}\right)\\
&=\int dx\,\psi_i^{*}(x)\int_{-L}^{L}dx'\left\langle x\middle|x'\right\rangle\psi_f(x')\\
&=\int_{-L/2}^{+L/2}dx\,\psi_i^{*}(x)\int_{-L}^{+L}dx'\,\delta(x-x')\psi_f(x')
\end{align*}
\]
\[
\left\{P=\left|\int_{-L/2}^{+L/2}dx\,\psi_i^{*}(x)\psi_f(x)\right|^2\right\}
\]
\[
\delta(x-x')=\delta(x'-x)
\]
\[
\left\langle x\middle|x'\right\rangle=\left\langle x'\middle|x\right\rangle\ \ \hookrightarrow\text{Real No.}
\]
\[
=\int_{-L/2}^{+L/2}dx\,\psi_i^{*}(x)\underbrace{\left(\int_{-L}^{+L}dx'\,\delta(x'-x)\psi_f(x')\right)}_{\longrightarrow\ \psi_f(x)}
\]
\[
\boxed{\sqrt{P}=\int_{-L/2}^{L/2}dx\,\psi_i^{*}(x)\psi_f(x)}
\]
Some events / phenomena were not explained classically like ex. Colors, color vs temp, rigidity of solid, spectrum of black body radiation, discrete spectrum, stability of atom, temp. dependence of specific heat, photoelectric effect.
Plot :--
\begin{tikzpicture}[scale=0.62]
\draw[->] (0,0)--(5.6,0) node[right]{$\nu$};
\draw[->] (0,0)--(0,5.2);
\draw[dashed] plot[domain=0:1.44,smooth,variable=\x] ({\x},{2.3*\x*\x});
\draw[thick] plot[domain=0.05:5.4,smooth,variable=\x] ({\x},{2.3*\x*\x*\x/(exp(\x)-1)});
\draw[dotted] (2.82,0)--(2.82,3.08);
\node[font=\tiny,align=left] at (3.05,4.85) {ultraviolet\\catastrophy.};
\draw[->,thin] (2.15,4.90)--(1.55,4.72);
\node[font=\tiny] at (2.45,3.82) {(classical)};
\draw[->,thin] (1.92,3.85)--(1.36,3.92);
\node[font=\tiny,align=left] at (4.80,3.35) {observed\\(experimental)};
\draw[->,thin] (4.62,2.90)--(4.40,2.55);
\node[rotate=90,font=\tiny] at (-0.45,2.35) {Spectral density (energy)};
\end{tikzpicture}
(Black body curve)
\[
u_\nu=\frac{8\pi h\nu^3}{c^3}\,\frac{1}{\left(e^{\frac{h\nu}{kT}}-1\right)}
\]
\[
u_\nu=8\pi h\left(\frac{\nu}{c}\right)^{3}\frac{1}{\left(e^{\beta h\nu}-1\right)}
\]
\[
u_\nu=8\pi h\left(\frac{\nu}{c}\right)^{3}\frac{1}{\left(e^{\beta h\nu}-1\right)}
\]
\(u_\nu=\) spectral density \(=\) energy per unit frequency range per unit volume.
\[
\frac{dE}{d\nu}=u_\nu\ \Rightarrow\ dE=u_\nu\,d\nu=u_\lambda\,d\lambda
\]
\[
u_\lambda=8\pi h\,\frac{c}{\lambda^{5}}\left(\frac{1}{e^{\frac{hc}{\lambda kT}}-1}\right)
\]
\[
\frac{du}{d\lambda}=0\ \Rightarrow\ u_\lambda=\max \text{ at }(\lambda_{max}T=\text{const})
\]
\[
\Rightarrow\ \left(\lambda_{max}\propto\frac{1}{T}\right)
\]
\[
h\nu=\phi+K.E_{max}\qquad\text{---(P.E) effect}
\]
Intensity on a given surface :--
\begin{tikzpicture}[scale=0.55]
\fill (0,0) circle (0.07);
\foreach \a in {-40,-20,0,20,40} \draw[->] (0.14,0)--(\a:1.30);
\draw[thick] (-45:2.7) arc (-45:45:2.7);
\draw[line width=1.4pt] (-9:2.7) arc (-9:9:2.7);
\node[right,font=\tiny] at (9:2.85) {$A$};
\draw[dashed] (0,0)--(-58:2.7);
\node[font=\tiny] at (-58:1.55) [below left] {$r$};
\end{tikzpicture}
No. of photons falling on area \(A=\dfrac{N_T\,A}{(4\pi r^2)}\)
\[
\text{Intensity}=\frac{\text{Energy}}{\text{Area}\times t}=\frac{N_p(A)h\nu}{A\cdot t}
\]
\[
I=\frac{(N_T)h\nu}{(4\pi r^2)\,t}\ \Rightarrow\ \boxed{\propto\frac{1}{r^2}}
\]
Radiation pressure :--
(1)
\begin{tikzpicture}[baseline,scale=0.55]
\draw[thick] (1.2,-1.15)--(1.2,1.15);
\foreach \i in {0,...,4} \draw (1.2,-1.0+0.5*\i)--(1.48,-1.18+0.5*\i);
\draw[dashed] (0.05,0)--(1.2,0);
\draw[->] (-0.10,0.95)--(1.15,0.03);
\node[font=\tiny] at (-0.30,1.10) {$I_0$};
\draw (0.75,0) arc (180:143.5:0.45);
\node[font=\tiny] at (0.66,0.24) {$\theta$};
\end{tikzpicture}
\(P=\dfrac{I}{c}\cos^2\theta\)
(2)
\begin{tikzpicture}[baseline,scale=0.55]
\draw[thick] (1.2,-1.15)--(1.2,1.15);
\foreach \i in {0,...,4} \draw (1.2,-1.0+0.5*\i)--(1.48,-1.18+0.5*\i);
\draw[dashed] (0.05,0)--(1.2,0);
\draw[->] (-0.10,0.95)--(1.15,0.03);
\node[font=\tiny] at (-0.30,1.10) {$I_0$};
\draw[->] (1.15,-0.03)--(-0.10,-0.95);
\node[font=\tiny] at (-0.30,-1.10) {$I_R$};
\draw (0.75,0) arc (180:143.5:0.45);
\node[font=\tiny] at (0.66,0.24) {$\theta$};
\draw (0.75,0) arc (180:216.5:0.45);
\node[font=\tiny] at (0.66,-0.24) {$\theta$};
\end{tikzpicture}
\(P=\dfrac{I_0}{c}\cos^2\theta+\dfrac{I_R}{c}\cos^2\theta\)
*Quantization cond\(^n\) for periodic motion*
\[
\text{Area in phase space}=\oint p_i\,dq_i=nh
\]
\((p,q)\) are generalized momenta & coordinates.
A periodic function can be expanded as
\[
f(x)=a_0+\sum_{n=1}^{\infty}a_n\cos nx+\sum_{n=1}^{\infty}b_n\sin nx
\]
A non-periodic function can be expanded as
\[
\begin{align*}
f(x)&=\left\langle x\middle|f\right\rangle=\left\langle x\right|\int dk\,\left|k\right\rangle\left\langle k\middle|f\right\rangle\\
&=\int\left\langle x\middle|k\right\rangle\left\langle k\middle|f\right\rangle\,dk\\
f(x)&=\int_{-\infty}^{\infty}\underbrace{e^{ikx}}_{\text{Basis}}\ \underbrace{f(k)}_{\text{Components}}dk
\end{align*}
\]
Phase velocity :--
velocity of individual wave in a wave packet.
\[
v_p=\frac{\omega}{k}=\frac{\Sigma\omega}{\Sigma k}=\text{speed of fast change}
\]
Group velocity :--
Apparent velocity of group
\[
v_g=\frac{d\omega}{dk}=\frac{\hbar\,d\omega}{\hbar\,dk}=\frac{dE}{dp}
\]
\[
v_g=\frac{\delta\omega}{\delta k}=\frac{|\omega_1-\omega_2|}{|k_1-k_2|}=\text{speed of slow change}
\]
\[
\Delta x\,\Delta p\ge\frac{\hbar}{2}\ ;\quad \Delta E\,\Delta t\ge\frac{\hbar}{2}
\]
\[
\Delta A\,\Delta B\ge\frac{1}{2}\left|\left\langle [A,B]\right\rangle\right|
\]
\begin{tikzpicture}[scale=0.8]
\fill[gray!35] (0,1.42) rectangle (2.6,1.74);
\draw (0,1.58)--(2.6,1.58);
\node[right,font=\tiny] at (2.7,1.58) {$E_2$};
\draw[<->,>=stealth] (-0.55,1.42)--(-0.55,1.74);
\node[left,font=\tiny] at (-0.6,1.58) {$\Delta E$};
\fill[gray!35] (0,0.32) rectangle (2.6,0.58);
\draw (0,0.45)--(2.6,0.45);
\node[right,font=\tiny] at (2.7,0.45) {$E_1$};
\end{tikzpicture}
\(\Big\{\)Actual energy levels of H-atom / any atom
Virial theorem :--
Clausius introduced virial theorem in thermodynamics.
\[
\left\langle K\right\rangle=\frac{\alpha}{2}\left\langle V\right\rangle
\]
\[
\left\langle E\right\rangle=\left\langle T\right\rangle+\left\langle V\right\rangle=\left(\frac{\alpha}{2}+1\right)\left\langle V\right\rangle
\]
LVS \& vectors :--
State of Q.M. system is described by element of some LVS \(\left|\Psi\right\rangle\) which satisfies
\[
\|\Psi\|=\left\langle \Psi\middle|\Psi\right\rangle=1
\]
\[
\rightarrow\ \|\Psi\|+\|\chi\|\ \ge\ \|\Psi+\chi\|
\]
\[
\rightarrow\ \bar a\cdot\bar b\ \le\ |a|\,|b|
\]
\[
\left|\left\langle \Psi\middle|\chi\right\rangle\right|\ \le\ \|\Psi\|\,\|\chi\|
\]
\[
\left\langle \Psi\middle|\phi\right\rangle^{*}=\left\langle \phi\middle|\Psi\right\rangle
\]
Any element of LVS--\(V\) can be expressed as some linear combination of basis vectors.
\[
\left|\Psi\right\rangle=\sum_n C_n\left|n\right\rangle
\]
For continuous basis,
\[
\rightarrow\ \left|\Psi\right\rangle=\int C_x\left|x\right\rangle
\]
abstract state \(\left|\Psi\right\rangle\) \(C_x=\left\langle x\middle|\Psi\right\rangle=\psi(x)\)
In Q.M we use orthonormal set of basis vectors.
\[
\left\langle \phi_i\middle|\phi_j\right\rangle=\delta_{ij}\quad\text{--- orthonormality}
\]
\[
\sum_i\left|\phi_i\right\rangle\left\langle \phi_i\right|=I\quad\text{--- completeness.}
\]
Any operator \(A\) can be expressed as
\[
A=\sum_{n,m}A_{nm}\left|n\right\rangle\left\langle m\right|
\]
\[
\text{(or)}\qquad A=\sum_i\sum_j\left|i\right\rangleA_{ij}\left\langle j\right|
\]
where \(\{\left|i\right\rangle\}\) forms basis for LVS.
\[
A=\begin{pmatrix}a_{11}&a_{12}&a_{13}&\cdots\\ a_{21}&a_{22}&a_{23}&\cdots\\ \vdots&&&\end{pmatrix}
\]
\[
a_{ij}=A_{ij}=\left\langle i\right|\hat A\left|j\right\rangle
\]
\(l_2\) : Space of square summable sequences.
\[
\left|\Psi\right\rangle=(x_1,x_2,x_3,\ldots x_i)
\]
\[
\|\Psi\|\rightarrow\text{finite}\ \Rightarrow\ \left(\sum_{i=1}^{N}|x_i|^2<\infty\right)
\]
\[
\downarrow
\]
\[
\text{True iff}\quad \lim_{i\to\infty}\frac{|x_{i+1}|^2}{|x_i|^2}<1
\]
For \(L_2(a,b)\) to be finite.
\[
\text{for}\quad \int_a^b|f(x)|^2dx<\infty
\]
\[
|f(x)|^2\longrightarrow 0\ \text{faster than}\ |x|\to\infty
\]
\[
|f(x)|^2\longrightarrow 0\ \text{faster than}\ \frac{1}{|x|}\to 0
\]
\[
\text{or,}\quad |f(x)|\longrightarrow 0\ \text{faster than}\ \frac{1}{\sqrt{|x|}}
\]
Any set of functions from \((a,b)\) forms LVS.
\[
\int_a^b f_i\,f_j\,dx=\delta_{ij}
\]
We can redefine normalization conditions to define various special functions.
Generalized normalization condition.
\[
\int_a^b \underset{\downarrow}{dm(x)}\ \underset{\downarrow}{\phi_n^{\dagger}(x)\phi_m(x)}=A_{nm}\delta_{nm}
\]
measureweight factor
Polynomials
Range
\(dm(x)\)
\(A_{nm}\)
Legendre
\((-1,1)\)
\(1\)
\(\frac{2}{2n+1}\)
Hermite
\((-\infty,\infty)\)
\(e^{-x^2}\)
?
Laguerre
\((-\infty,\infty)\)
\(e^{-x}\)
?
Postulates of Q.M
(1) The state of Quantum mechanical system is described by a state vector \(\left|\Psi(t)\right\rangle\) which lives in seperable-hilbert space .
\[
i\hbar\frac{d}{dt}\left|\Psi(t)\right\rangle=\hat H\left|\Psi(t)\right\rangle
\]
\[
\downarrow\ \text{denumerable Basis set (countable }\infty\text{ basis).}
\]
\(\rightarrow\) LVS + limit vector of cauchy sequence = complete LVS
(2) Every physical measurable / observable is represented by Hermitian / self adjoint operators (to ensure the result of measurement (eigen value) is always real.
\[
\{\hat x,\ \hat p,\ \hat H,\ \hat V(x),\ L,\ S,\ J,\ \ldots\}
\]
All are hermitian operators.
Spectrum of an operator is set of all eigen values of an operator. It can be continuous & discrete.
\[
\underset{\text{Bound states}}{\swarrow}\qquad \underset{\text{Scattering states}}{\searrow}
\]
\(\rightarrow\) State of system can be written as linear combination of basis set formed by eigen vectors of observable operator.
\[
\left|\Psi(t)\right\rangle=\sum_n C_n(t)\left|\phi_n\right\rangle
\]
Operators :--
--- \(\left\langle A\right\rangle(t)\) in state \(\left|\Psi(t)\right\rangle\)
\[
=\frac{\left\langle \Psi(t)\right|\hat A\left|\Psi(t)\right\rangle}{\left\langle \Psi(t)\middle|\Psi(t)\right\rangle}
\]
--- Operators in Q.M are self adjoint or Hermitian. \((A^{\dagger}=A)\)
--- Corresponding to every transformation we use unitary operators.
\[
\left|\Psi(t)\right\rangle=\underset{\downarrow}{e^{\frac{-iHt}{\hbar}}}
\left|\Psi(0)\right\rangle
\]
Unitary operator.
Some operators
(1) \(\hat x\left|x_0\right\rangle=x_0\left|x_0\right\rangle\)
(2) \(f(\hat x)\left|x_0\right\rangle=f(x_0)\left|x_0\right\rangle\)
(3) \(\left\langle x\right|\hat p\left|\Psi\right\rangle=-i\hbar\dfrac{\partial}{\partial x}\left\langle x\middle|\Psi\right\rangle\)
(4) Angular momentum operators :--
\[
\vec L=\vec r\times\vec p
\]
\[
r=(x,y,z)\qquad p=(p_x,p_y,p_z)
\]
\[
L_x=y\,p_z-z\,p_y
\]
\[
L_y=z\,p_x-x\,p_z
\]
\[
L_z=x\,p_y-y\,p_x
\]
\[
=i\hbar\begin{pmatrix}0&\sin\varphi&\cot\theta\cos\varphi\\ 0&-\cos\varphi&\cot\theta\sin\varphi\\ 0&0&-1\end{pmatrix}\begin{pmatrix}\partial_r\\ \partial_\theta\\ \partial_\varphi\end{pmatrix}
\]
\[
L_+=L_x+iL_y\qquad L_-=L_x-iL_y
\]
Commutators :
\[
[\hat A,\hat B]=\hat A\hat B-\hat B\hat A
\]
(1) \([x,p]=i\hbar I\)
An operator is said to be hermitian if \(\left\langle A\right\rangle\) is real.
\[
\left\langle A\right\rangle=\left\langle A\right\rangle^{*}
\]
\[
\int\psi^{\dagger}A\psi\,dx=\int\psi(A\psi)^{*}dx\qquad\text{---}(1\sigma)
\]
\[
\text{or}\ \int\psi^{*}A\psi\,d\tau=\int\psi(A\psi)^{*}d\tau\quad\text{---}(3\sigma)
\]
(1) \(p_x\) is hermitian
\[
\left(-i\hbar\frac{\partial}{\partial x}\right)^{\dagger}=-i\hbar\frac{\partial}{\partial x}
\]
\[
i\hbar\left(\frac{\partial}{\partial x}\right)^{\dagger}=-i\hbar\frac{\partial}{\partial x}
\]
\[
\left(\left(\frac{\partial}{\partial x}\right)^{\dagger}=-\frac{\partial}{\partial x}\right)\quad \text{`}\frac{\partial}{\partial x}\text{' is not hermitian}
\]
(2) \((x p_x+p_x x)\) is hermitian
\[
\begin{align*}
(xp_x+p_xx)^{\dagger}&=p_x^{\dagger}x^{\dagger}+x^{\dagger}p_x^{\dagger}\\
&=p_x x+x p_x
\end{align*}
\]
Replace \(x p_x\) from classical to \(\left(\dfrac{xp_x+p_xx}{2}\right)\) in Q.M
\[
\left(i\frac{\partial}{\partial r}\right)\ \text{is Hermitian in 1-D}
\]
But not in 2-D or 3-D
\[
\int\psi^{*}A\psi\,d\tau=\int\psi(A\psi)^{*}d\tau
\]
For \(\hat A\) to be Hermitian.
Ehrenfest theorem :
The expectation value of operators obey classical equations.
\[
\text{classically,}\quad \frac{dA}{dt}=\{A,H\}+\frac{\partial A}{\partial t}
\]
\[
\text{Q.M,}\quad \left\langle \frac{dA}{dt}\right\rangle=\frac{d}{dt}\left\langle A\right\rangle=\left\langle \frac{[A,H]}{i\hbar}\right\rangle+\left\langle \frac{\partial A}{\partial t}\right\rangle
\]
\[
\Rightarrow\ \left\{\frac{d}{dt}\left\langle p_x\right\rangle=\left\langle F_x\right\rangle\ ;\quad \frac{d\left\langle x\right\rangle}{dt}=\frac{\left\langle p_x\right\rangle}{m}\right\}
\]
Unitary Operators :--
(1) space Translation: \(O_T\psi(x)=\psi(x+x_0)\)
\[
O_T=e^{\frac{i\,\vec p\cdot\vec r_0}{\hbar}}\qquad O_T=e^{\frac{i\,p_x x_0}{\hbar}}
\]
(2) Time translation :--
\[
\psi(t)=\underset{(O_t)}{e^{\frac{-iHt}{\hbar}}}
\psi(0)
\]
\[
\boxed{\psi(t+t_0)=e^{\frac{-iHt_0}{\hbar}}\psi(t)}
\]
Rotation about axis \((\hat n)\)
\[
O_R(\theta,\hat n)=e^{\frac{-i\theta(\hat n\cdot\vec J)}{\hbar}}
\]
Schrodinger picture :--
Operators are time independent, but wavefunctions change with time.
\[
\left\{\begin{aligned}&\hat A\to\text{const.}\\ &\psi=\psi(t)\end{aligned}\right\}
\]
\[
\boxed{i\hbar\frac{d}{dt}\left|\Psi(t)\right\rangle=H\left|\Psi(t)\right\rangle}
\]
Heisenberg picture :--
Operators are time dependent but wavefunctions are not.
\[
\psi(t)=\psi(0)\ \text{--- fixed}
\]
\[
\boxed{A_H(t)=e^{\frac{iHt}{\hbar}}A_H(0)\,e^{\frac{-iHt}{\hbar}}}
\]
Angular momentum algebra :
\([J_i,J_j]=i\hbar\,\epsilon_{ijk}J_k\)
\(J^2=\vec J\cdot\vec J=J_1^2+J_2^2+J_3^2\)
\([J^2,J_i]=0\qquad \forall\ i=1,2,3.\)
\([J^2,\vec J\cdot\hat n]=0\) for any \(\hat n(\theta,\varphi)\).
Since \([J^2,J_z]=0\) they forms common set of eigen states.
\[
J^2\left|j,m\right\rangle=j(j+1)\hbar^2\left|j,m\right\rangle
\]
\[
J_z\left|j,m\right\rangle=\hbar m\left|j,m\right\rangle
\]
for \(j\), \(m:-j\) to \(+j\) in steps of 1.
\(J_+=J_x+iJ_y=a^{\dagger}b\)
\(J_-=J_x-iJ_y=a\,b^{\dagger}\)
\(J_+\left|j,m\right\rangle=\sqrt{(j+m+1)(j-m)}\,\left|j,m+1\right\rangle\)
\(J_-\left|j,m\right\rangle=\sqrt{(j+m)(j-m+1)}\,\left|j,m-1\right\rangle\)
\(J_x\left|j,m\right\rangle=\dfrac{(J_++J_-)}{2}\left|j,m\right\rangle\)
\(J_y\left|j,m\right\rangle=\dfrac{(J_+-J_-)}{2i}\left|j,m\right\rangle\)
*\(\left\langle J_i\right\rangle\) in given \(\left|j,m\right\rangle\) state*
(1)
\[
\begin{align*}\left\langle J_x\right\rangle&=\left\langle j,m\right|J_x\left|j,m\right\rangle\\ &=\left\langle j,m\right|\frac{J_++J_-}{2}\left|j,m\right\rangle\\ &=\left\langle j,m\right|\tfrac{1}{2}\big(a\left|j,m+1\right\rangle\\ &+b\left|j,m-1\right\rangle\big)\\ &=0\end{align*}
\]
(2) similarly \(\left\langle J_y\right\rangle\) in \(\left|j,m\right\rangle=0\)
\[
\left\langle J_y\right\rangle=0
\]
(3)
\[
\begin{align*}\left\langle J_z\right\rangle&=\left\langle j,m\right|J_z\left|j,m\right\rangle\\ &=\left\langle j,m\right|m\hbar\left|j,m\right\rangle\\ &=m\hbar\left\langle j,m\middle|j,m\right\rangle\end{align*}
\]
\[
\boxed{\left\langle J_z\right\rangle=m\hbar}
\]
\[
J_+\left|j,m\right\rangle=\hbar\sqrt{(j+m+1)(j-m)}\left|j,m+1\right\rangle
\]
\[
J_-\left|j,m\right\rangle=\hbar\sqrt{(j+m)(j-m+1)}\left|j,m-1\right\rangle
\]
\[
\begin{align*}
[J_+,J_-]&=[J_x+iJ_y,\,J_x-iJ_y]\\
&=0-i(i\hbar J_z)+i(-i\hbar J_z)\\
&=2\hbar J_z
\end{align*}
\]
(4) Find \(\left\langle J_xJ_y\right\rangle\) in \(\left|j,m\right\rangle\) state.
\[
\begin{align*}
\left\langle J_xJ_y\right\rangle&=\left\langle \left(\frac{J_++J_-}{2}\right)\left(\frac{J_+-J_-}{2i}\right)\right\rangle\\
&=\left\langle \frac{(J_+)^2-(J_-)^2+[J_-,J_+]}{4i}\right\rangle\\
&=\frac{\left\langle J_+^2\right\rangle}{4i}-\frac{\left\langle J_-^2\right\rangle}{4i}+\frac{\left\langle [J_-,J_+]\right\rangle}{4i}\\
&=0-0+\left\langle \frac{2\hbar J_z}{4i}\right\rangle\\
&=\frac{\hbar i}{2}\left\langle j,m\right|J_z\left|j,m\right\rangle
\end{align*}
\]
\[
\boxed{\left\langle J_xJ_y\right\rangle=\frac{i\,m\,\hbar^2}{2}}
\]
(5) Find \(\left\langle J_+J_-\right\rangle\)
\[
\begin{align*}
&\left\langle j,m\right|J_+J_-\left|j,m\right\rangle\\
&=\hbar^2\sqrt{(j+m)(j-m+1)(j+m+1)(j-m+1)}\\
&=\hbar^2(j+m)(j-m+1)
\end{align*}
\]
(6) \(\left\langle J_-J_+\right\rangle=\hbar^2(j+m+1)(j-m)\)
(7) \(\left\langle J_xJ_y+J_yJ_x\right\rangle\)
\[
[J_x,J_y]=J_xJ_y-J_yJ_x=i\hbar J_z
\]
\[
\begin{align*}
\left\langle J_xJ_y+J_yJ_x\right\rangle&=\left\langle 2J_xJ_y-i\hbar J_z\right\rangle\\
&=2\left\langle J_xJ_y\right\rangle-i\hbar\left\langle J_z\right\rangle\\
&=2\left(\frac{im\hbar^2}{2}\right)-i\hbar\,m\hbar\\
&=m\hbar^2 i-m\hbar^2 i\\
&=0
\end{align*}
\]
\[
\boxed{\left\langle J_xJ_y+J_yJ_x\right\rangle=0}
\]
(8) \(\left\langle J^2\right\rangle=\left\langle j,m\right|J^2\left|j,m\right\rangle\)
\[
\left\langle J^2\right\rangle=j(j+1)\hbar^2\left\langle j,m\middle|j,m\right\rangle
\]
\[
\left\langle J^2\right\rangle=j(j+1)\hbar^2
\]
For \(j=\frac12\) states are represented by
\[
\left|j,m\right\rangle=\underset{\downarrow}{\left|\tfrac12,\tfrac12\right\rangle}\ \&\ \underset{\downarrow}{\left|\tfrac12,-\tfrac12\right\rangle}
\]
\[
\left|0\right\rangle,\left|\uparrow\right\rangle\left|1\right\rangle\ \text{or}\ \left|\downarrow\right\rangle
\]
\[
\left|0\right\rangle\ \text{or}\ \left|\uparrow\right\rangle=\begin{pmatrix}1\\0\end{pmatrix}\qquad \left|1\right\rangle\ \text{or}\ \left|\downarrow\right\rangle=\begin{pmatrix}0\\1\end{pmatrix}
\]
\[
\bar J_i=\frac{\hbar}{2}\sigma_i\ ;\qquad J^2=J_1^2+J_2^2+J_3^2
\]
\(\sigma_i\) are pauli-matrices.
Pauli matrices \& their properties :--
\[
\sigma_1=\begin{pmatrix}0&1\\1&0\end{pmatrix}\quad \sigma_2=\begin{pmatrix}0&-i\\i&0\end{pmatrix}
\]
\[
\sigma_3=\begin{pmatrix}1&0\\0&-1\end{pmatrix}
\]
\[
|\sigma_i|=-1,\quad \sigma_i^2=I,\quad \sigma_i\sigma_j=\delta_{ij}I+i\epsilon_{ijk}\sigma_k
\]
\[
[\sigma_i,\sigma_j]=2i\,\epsilon_{ijk}\sigma_k
\]
(4)
\[
\begin{align*}
(\bar\sigma\cdot\bar A)(\bar\sigma\cdot\bar B)&=(\sigma_iA_i)(\sigma_jB_j)\\
&=\sigma_i\sigma_j A_iB_j\\
&=(\delta_{ij}I+i\epsilon_{ijk}\sigma_k)A_iB_j\\
&=(\bar A\cdot\bar B)I+i\sigma_k\epsilon_{ijk}A_iB_j\\
&=(A\cdot B)I+i(\sigma_k)(A\times B)_k
\end{align*}
\]
\[
\boxed{(\bar\sigma\cdot\bar A)(\bar\sigma\cdot\bar B)=(\bar A\cdot\bar B)I+i\left(\bar\sigma\cdot(\bar A\times\bar B)\right)}
\]
Operators in matrix Form :--
(1)
\[
\begin{align*}J^2&=J_1^2+J_2^2+J_3^2\\ &=\frac{\hbar^2}{4}(\sigma_1^2+\sigma_2^2+\sigma_3^2)\end{align*}
\]
for \(j=\tfrac12\)
\[
J^2=\frac{3\hbar^2}{4}I
\]
Since \(j=\tfrac12,\tfrac32,\tfrac52\) represents spin angular momentum only,
\[
\text{For } s=\tfrac12\quad S^2=\frac{3\hbar^2}{4}I
\]
\[
\left(\bar S_i=\frac{\hbar}{2}\bar\sigma_i\right)
\]
lets write \((\vec S\cdot\hat n)\) in matrix form.
\[
\begin{align*}
\vec S\cdot\hat n&=\frac{\hbar}{2}\vec\sigma\cdot\big(\sin\theta\cos\varphi\,\hat i+\sin\theta\sin\varphi\,\hat j\\ &+\cos\theta\,\hat k\big)\\
&=\frac{\hbar}{2}\Bigg[\begin{pmatrix}0&1\\1&0\end{pmatrix}\sin\theta\cos\varphi+\begin{pmatrix}0&-i\\i&0\end{pmatrix}\sin\theta\sin\varphi\\ &+\begin{pmatrix}1&0\\0&-1\end{pmatrix}\cos\theta\Bigg]\\
&=\frac{\hbar}{2}\begin{pmatrix}\cos\theta&\sin\theta\cos\varphi-i\sin\theta\sin\varphi\\ \sin\theta\cos\varphi+i\sin\theta\sin\varphi&-\cos\theta\end{pmatrix}
\end{align*}
\]
\[
\vec S\cdot\hat n=\frac{\hbar}{2}\begin{pmatrix}\cos\theta&\sin\theta\,e^{-i\varphi}\\ \sin\theta\,e^{i\varphi}&-\cos\theta\end{pmatrix}
\]
It has eigen values given by
\[
|\vec S\cdot\hat n-\lambda I|=0\qquad \boxed{\lambda=\pm\hbar/2}
\]
\(*\) Eigen vectors of \(\vec S\cdot\hat n\) are
\[
\left|+\tfrac{\hbar}{2}\right\rangle=\begin{pmatrix}\cos(\theta/2)\\ \sin(\theta/2)e^{i\varphi}\end{pmatrix}
\]
\[
\left|-\tfrac{\hbar}{2}\right\rangle=\begin{pmatrix}\sin(\theta/2)\\ -\cos(\theta/2)e^{i\varphi}\end{pmatrix}
\]
\[
\left|+\tfrac{\hbar}{2}\right\rangle=\begin{pmatrix}\cos(\theta/2)\\ \sin(\theta/2)e^{i\varphi}\end{pmatrix}\quad \left|-\tfrac{\hbar}{2}\right\rangle=\begin{pmatrix}\sin(\theta/2)\\ -\cos(\theta/2)e^{i\varphi}\end{pmatrix}
\]
For _\(S_x\) operator_ :
\[
\theta=90^\circ\qquad \varphi=0
\]
\[
\left|+\tfrac{\hbar}{2}\right\rangle=\begin{pmatrix}1/\sqrt2\\ +1/\sqrt2\end{pmatrix}\ ;\quad \left|-\tfrac{\hbar}{2}\right\rangle=\begin{pmatrix}1/\sqrt2\\ -1/\sqrt2\end{pmatrix}
\]
\begin{tikzpicture}[scale=0.62]
\draw[->] (0,0)--(0,1.75) node[above,font=\tiny]{$z$};
\draw[->] (0,0)--(1.85,0) node[right,font=\tiny]{$y$};
\draw[->] (0,0)--(-1.0,-1.0) node[below,font=\tiny]{$x$};
\draw[->] (0,0)--(0.45,1.25) node[above right,font=\tiny]{$\hat n$};
\draw[dashed] (0.45,1.25)--(0.45,-0.50);
\draw[dashed] (0,0)--(0.45,-0.50);
\draw (0,0.75) arc (90:70:0.75);
\node[font=\tiny] at (0.17,0.92) {$\theta$};
\draw (-0.38,-0.38) arc (225:312:0.54);
\node[font=\tiny] at (0.02,-0.66) {$\varphi$};
\end{tikzpicture}
For _\(S_y\) operator_ :--
\[
\theta=90^\circ\qquad \varphi=90^\circ
\]
\[
\left|+\tfrac{\hbar}{2}\right\rangle=\begin{pmatrix}1/\sqrt2\\ 1/\sqrt2\,i\end{pmatrix}\quad \left|-\tfrac{\hbar}{2}\right\rangle=\begin{pmatrix}1/\sqrt2\\ -1/\sqrt2\,i\end{pmatrix}
\]
For _\(S_z\) operator_ :
\[
\theta=0^\circ\qquad \varphi=\varphi
\]
\[
\left|+\tfrac{\hbar}{2}\right\rangle=\begin{pmatrix}1\\0\end{pmatrix}\quad \left|-\tfrac{\hbar}{2}\right\rangle=\begin{pmatrix}0\\ -e^{i\varphi}\end{pmatrix}\ \text{or}\ \begin{pmatrix}0\\1\end{pmatrix}
\]
*Probability of getting \(+\hbar/2\)*
upon measuring \(S_z\) when system is in \(\left|n\uparrow\right\rangle\) state :--
\[
\begin{align*}
\left|n\uparrow\right\rangle=\left|+\tfrac{\hbar}{2}(\hat n)\right\rangle&=\begin{pmatrix}\cos(\theta/2)\\ \sin(\theta/2)e^{i\varphi}\end{pmatrix}\\
&=\cos(\theta/2)\begin{pmatrix}1\\0\end{pmatrix}+\sin\tfrac{\theta}{2}e^{i\varphi}\begin{pmatrix}0\\1\end{pmatrix}
\end{align*}
\]
\[
P\!\left(S_z=+\tfrac{\hbar}{2}\right)=\cos^2(\theta/2)
\]
\[
P\!\left(S_z=-\tfrac{\hbar}{2}\right)=\sin^2(\theta/2)
\]
*Exponential of a matrix (\(2\times2\)) :--*
\[
(\sigma\cdot A)(\sigma\cdot B)=I(A\cdot B)+i\bar\sigma\cdot(\bar A\times\bar B)
\]
\[
(\sigma\cdot\hat n)^2=I
\]
\[
(\sigma\cdot\hat n)^3=\bar\sigma\cdot\hat n\ \Rightarrow\ (\sigma\cdot\hat n)^4=I
\]
\[
e^{i(\bar\sigma\cdot\hat n)\theta}=I+\frac{i(\sigma\cdot\hat n)\theta}{1!}+\frac{i^2(\sigma\cdot\hat n)^2\theta^2}{2!}+\cdots
\]
\[
e^{i(\bar\sigma\cdot\hat n)\theta}=I(\cos\theta)+i(\bar\sigma\cdot\hat n)\sin\theta
\]
\[
e^{i(\bar\sigma\cdot\hat n)\theta}=(\cos\theta)I+i(\bar\sigma\cdot\hat n)\sin\theta
\]
To write matrix representation for \(J_x,J_y,J_z\) (for \(j=j\))
\[
m:-j,\ -j+1,\ \ldots 0,\ \ldots j-1,\ j
\]
Let \(\left|1\right\rangle=\left|j,+j\right\rangle\)
\[
J_i=\begin{pmatrix}\left\langle 1\right|J_i\left|1\right\rangle&\left\langle 1\right|J_i\left|2\right\rangle&\cdots\\ \left\langle 2\right|J_i\left|1\right\rangle&\left\langle 2\right|J_i\left|2\right\rangle&\cdots\\ \vdots&\vdots&\end{pmatrix}
\]
ex. For \(j=1\) \(m=-1,0,1\)
\[
\left|1\right\rangle=\left|1,1\right\rangle\quad \left|2\right\rangle=\left|1,0\right\rangle\quad \left|3\right\rangle=\left|1,-1\right\rangle
\]
\[
J_i=\begin{pmatrix}\left\langle 1\right|J_i\left|1\right\rangle&\left\langle 1\right|J_i\left|2\right\rangle&\left\langle 1\right|J_i\left|3\right\rangle\\ \left\langle 2\right|J_i\left|1\right\rangle&\left\langle 2\right|J_i\left|2\right\rangle&\left\langle 2\right|J_i\left|3\right\rangle\\ \left\langle 3\right|J_i\left|1\right\rangle&\left\langle 3\right|J_i\left|2\right\rangle&\left\langle 3\right|J_i\left|3\right\rangle\end{pmatrix}
\]
\[
i=1,2,3.
\]
Electron in external magnetic field :--
when an \(e^-\) is placed in ext \(\vec B\), its intrinsic magnetic moment \(\vec\mu_e\) interacts with \(\vec B\).
\[
\vec\mu_e=+\frac{g\,q}{2m}\vec S
\]
\[
\vec\mu_e=\frac{2(-|e|)}{2m_e}\frac{\hbar}{2}\bar\sigma
\]
\[
\vec\mu_e=\frac{-|e|\hbar}{2m_e}\bar\sigma
\]
\[
\vec\mu_e=-\mu_B\bar\sigma
\]
\[
\text{Interaction energy}=-\bar m\cdot\vec B=-\vec\mu_e\cdot\vec B
\]
\[
H=-(-\mu_B\bar\sigma)\cdot\vec B
\]
\[
\boxed{\hat H=\mu_B\,\bar\sigma\cdot\vec B}
\]
\(\hat H\) has two eigen values \(\pm\mu_B B\)
\[
\hat H=\mu_B(\bar\sigma\cdot\hat n)B
\]
\[
E=\mu_B B(\pm1)\qquad \boxed{E=\pm\mu_B B}
\]
Any particle with \((J=j)\) in \(\vec B\).
\[
\vec\mu=g\left(\frac{q}{2m}\right)\vec J
\]
\[
H=-\vec\mu\cdot\vec B
\]
\[
\hat H=-\frac{g\,q}{2m}\left(\vec J\cdot\vec B\right)
\]
\[
\text{For}\ \vec B=B\hat n
\]
\[
\hat H=-\frac{g\,qB}{2m}(\vec J\cdot\hat n)
\]
It has eigen values :
\[
E=-\frac{g\,qB}{2m}(m\hbar)\qquad (m:-j\ \text{to}\ +j).
\]
If such particles undergo force due to presence of inhomogeneous \(\vec B\), we can find value of `\(j\)' by noting No. of splits.
\[
F=-\nabla H
\]
\[
F=\frac{q\,g}{2m}(\vec J\cdot\hat n)\,\vec\nabla B
\]
\begin{tikzpicture}[scale=0.55]
\draw (-1.3,1.5)--(1.3,1.5)--(1.3,0.95)--(0.16,0.38)--(-0.16,0.38)--(-1.3,0.95)--cycle;
\node[font=\tiny] at (-0.85,1.15) {N};
\draw (-1.3,-1.5)--(1.3,-1.5)--(1.3,-0.80)--(0.55,-0.80) .. controls (0.2,-0.48) and (-0.2,-0.48) .. (-0.55,-0.80)--(-1.3,-0.80)--cycle;
\node[font=\tiny] at (-0.85,-1.15) {S};
\draw[->] (-0.55,0.62)--(-0.80,-0.78);
\draw[->] (-0.22,0.40)--(-0.30,-0.62);
\draw[->] (0,0.36)--(0,-0.55);
\draw[->] (0.22,0.40)--(0.30,-0.62);
\draw[->] (0.55,0.62)--(0.80,-0.78);
\end{tikzpicture}
\[
\underline{\vec\nabla B}=+\hat z
\]
\(F>0\) for \(m=+j\)
If
\begin{tikzpicture}[scale=0.55]
\draw (-1.3,1.5)--(1.3,1.5)--(1.3,0.95)--(0.16,0.38)--(-0.16,0.38)--(-1.3,0.95)--cycle;
\node[font=\tiny] at (-0.85,1.15) {S};
\draw (-1.3,-1.5)--(1.3,-1.5)--(1.3,-0.80)--(0.55,-0.80) .. controls (0.2,-0.48) and (-0.2,-0.48) .. (-0.55,-0.80)--(-1.3,-0.80)--cycle;
\node[font=\tiny] at (-0.85,-1.15) {N};
\draw[->] (-0.80,-0.78)--(-0.55,0.62);
\draw[->] (-0.30,-0.62)--(-0.22,0.40);
\draw[->] (0,-0.55)--(0,0.36);
\draw[->] (0.30,-0.62)--(0.22,0.40);
\draw[->] (0.80,-0.78)--(0.55,0.62);
\end{tikzpicture}
\(\vec\nabla B<0\) (points in dir\(^n\) of increasing strength).
\(F<0\) for \(m=+j,\ j-1,\ldots\)
\begin{tikzpicture}[scale=0.58]
\draw (-1.1,1.45)--(1.1,1.45)--(1.1,0.90)--(0.14,0.36)--(-0.14,0.36)--(-1.1,0.90)--cycle;
\node[font=\tiny] at (-0.72,1.12) {N};
\draw (-1.1,-1.45)--(1.1,-1.45)--(1.1,-0.78)--(0.5,-0.78) .. controls (0.18,-0.46) and (-0.18,-0.46) .. (-0.5,-0.78)--(-1.1,-0.78)--cycle;
\node[font=\tiny] at (-0.72,-1.12) {S};
\draw[->] (-0.45,0.58)--(-0.68,-0.74);
\draw[->] (0,0.34)--(0,-0.52);
\draw[->] (0.45,0.58)--(0.68,-0.74);
\draw[->] (-2.9,0)--(-1.2,0);
\node[font=\tiny] at (-2.05,0.28) {Beam};
\node[font=\tiny] at (0,-1.75) {$\nabla B>0$};
\draw[->] (1.2,0)--(3.0,1.25) node[right,font=\tiny]{$m=+j$};
\draw[->] (1.2,0)--(3.0,0.62) node[right,font=\tiny]{$m=j-1$};
\draw[->] (1.2,0)--(3.0,0) node[right,font=\tiny]{$m=0$};
\draw[->] (1.2,0)--(3.0,-0.62) node[right,font=\tiny]{$m=-j+1$};
\draw[->] (1.2,0)--(3.0,-1.25) node[right,font=\tiny]{$m=-j$};
\end{tikzpicture}
{ upon seeing the number of splits we can determine the value of `\(j\)' for the particles of Beam. This is known as Stern-Gerlach experiment .}
General case :
\begin{tikzpicture}[scale=0.68]
\draw[->] (-2.4,0)--(-1.0,0);
\node[font=\tiny] at (-1.7,0.32) {$\left|z\uparrow\right\rangle$};
\node[font=\tiny] at (-1.7,-0.32) {$I_0$};
\draw (-1.0,-0.65) rectangle (0.65,0.65);
\node[font=\tiny] at (-0.18,0) {$B\hat n$};
\draw[->] (0.65,0)--(2.1,0.85) node[right,font=\tiny]{$\left|n\uparrow\right\rangle$};
\draw[->] (0.65,0)--(2.1,-0.85) node[right,font=\tiny]{$\left|n\downarrow\right\rangle$};
\node[font=\tiny] at (1.55,0.20) {$I_1$};
\node[font=\tiny] at (1.55,-0.22) {$I_2$};
\end{tikzpicture}
Find the intensities \(I_1\) & \(I_2\).
\[
\left|z\uparrow\right\rangle=a\left|n\uparrow\right\rangle+b\left|n\downarrow\right\rangle
\]
\[
\begin{pmatrix}1\\0\end{pmatrix}=a\begin{pmatrix}\cos\frac{\theta}{2}\\ \sin\frac{\theta}{2}e^{i\varphi}\end{pmatrix}+b\begin{pmatrix}\sin\frac{\theta}{2}\\ -\cos\frac{\theta}{2}e^{i\varphi}\end{pmatrix}
\]
\[
P(n\uparrow)=|a|^2=\cos^2\frac{\theta}{2}
\]
\[
P(n\downarrow)=|b|^2=\left|\left(\sin\tfrac{\theta}{2}\ \ -\cos\tfrac{\theta}{2}e^{i\varphi}\right)\begin{pmatrix}1\\0\end{pmatrix}\right|^2
\]
\[
P(n\downarrow)=\sin^2\frac{\theta}{2}
\]
\[
\left\{\begin{aligned}I_1&=I_0\cos^2\frac{\theta}{2}\\ I_2&=I_0\sin^2\frac{\theta}{2}\end{aligned}\right\}
\]
*Find \(\left\langle J_x^2\right\rangle\), \(\left\langle J_y^2\right\rangle\) \& \(\left\langle J_z^2\right\rangle\)*
(1) \(\left\langle J_x^2\right\rangle=\left\langle j,m\right|J_x^2\left|j,m\right\rangle\)
\[
J_x=\frac{J_++J_-}{2}
\]
\[
J_x^2=\frac{J_+^2+J_-^2+J_+J_-+J_-J_+}{4}
\]
\[
\left\langle J_x^2\right\rangle=\underbrace{\frac{\left\langle J_+^2\right\rangle}{4}}_{0}+\underbrace{\frac{\left\langle J_-^2\right\rangle}{4}}_{0}+\frac{\left\langle J_+J_-+J_-J_+\right\rangle}{4}
\]
\[
\begin{align*}
\left\langle J_+J_-\right\rangle&=\left\langle j,m\right|J_+J_-\left|j,m\right\rangle\\
&=\hbar^2\sqrt{(j+m)(j-m+1)(j+m+1)(j-m+1)}\\
&=\hbar^2(j+m)(j-m+1)
\end{align*}
\]
Similarly, \(\left\langle J_-J_+\right\rangle=\hbar^2(j+m+1)(j-m)\)
\[
\begin{align*}
\left\langle J_x^2\right\rangle&=\frac{\hbar^2}{4}\big[(j+m)(j-m+1)+(j+m+1)(j-m)\big]\\
&=\frac{\hbar^2}{4}\big(j^2-mj+j+mj-m^2+m\\ &+j^2-jm-m^2+j-m+mj\big)\\
&=\frac{\hbar^2}{4}\left(2j^2+2j-2m^2\right)
\end{align*}
\]
\[
\boxed{\left\langle J_x^2\right\rangle=\frac{\hbar^2}{2}\left(j(j+1)-m^2\right)}
\]
\[
\begin{align*}
\left\langle J_z^2\right\rangle&=\left\langle j,m\right|J_z^2\left|j,m\right\rangle\\
&=(m\hbar)^2\left\langle j,m\middle|j,m\right\rangle\\
&=m^2\hbar^2
\end{align*}
\]
\[
\begin{align*}
\left\langle J^2\right\rangle&=\left\langle j,m\right|J^2\left|j,m\right\rangle\\
&=\hbar^2\,j(j+1)
\end{align*}
\]
\[
J^2=J_x^2+J_y^2+J_z^2
\]
\[
\begin{align*}
\left\langle J_y^2\right\rangle&=\left\langle J^2\right\rangle-\left\langle J_x^2\right\rangle-\left\langle J_z^2\right\rangle\\
&=\hbar^2 j(j+1)-\frac{\hbar^2}{2}j(j+1)+\frac{m^2\hbar^2}{2}\\ &-m^2\hbar^2\\
&=\frac{\hbar^2\left(j(j+1)\right)}{2}-\frac{m^2\hbar^2}{2}
\end{align*}
\]
\[
\boxed{\left\langle J_y^2\right\rangle=\frac{\hbar^2}{2}\left(j(j+1)-m^2\right)}
\]
So
\[
\left\langle J_x^2\right\rangle=\left\langle J_y^2\right\rangle=\frac{\hbar^2}{2}\left(j(j+1)-m^2\right)
\]
\[
\left\langle J_z^2\right\rangle=m^2\hbar^2
\]
\[
\left\langle J^2\right\rangle=\hbar^2\,j(j+1)
\]
Addition of Angular momenta and Clebsch Gordan coefficients :--
let we have two particles / or two angular momenta \(\vec J_1\) & \(\vec J_2\) then the resultant angular momentum is given by \(\vec J\)
\[
\vec J=\vec J_1+\vec J_2
\]
states are represented by \(\left|j,m\right\rangle\)
\[
J^2\left|j,m\right\rangle=\hbar^2 j(j+1)\left|j,m\right\rangle
\]
\[
J_z\left|j,m\right\rangle=m\hbar\left|j,m\right\rangle
\]
Here `\(j\)' represents total angular momentum quantum number.
\[
m:-j\ \text{to}\ +j\ \text{in steps of 1}
\]
states \(\left|j,m\right\rangle\) are total angular momentum states.
\[
j:\ |j_1-j_2|\ \text{to}\ j_1+j_2
\]
\(j_1\) & \(j_2\) are quantum numbers of \(J_1\) & \(J_2\) respectively.
\(\rightarrow (J_1^2,J_2^2,J_{1z},J_{2z})\) commute with each other so they have common eigen states labelled by
\[
\left|j_1 j_2,\ m_1 m_2\right\rangle
\]
\[
\text{(or)}\qquad \left|j_1,m_1\right\rangle\otimes\left|j_2,m_2\right\rangle
\]
\[
\text{(or)}\qquad \left|m_1 m_2\right\rangle
\]
\(\rightarrow (J_1^2,J_2^2,J^2,J_z)\) also commute with each other so, common eigen states are represented by
\[
\left|j_1 j_2;\,j\,m\right\rangle\quad\text{or}\quad\left|j,m\right\rangle.
\]
We can express any state \(\left|j,m\right\rangle\) in basis of \(\{\left|m_1m_2\right\rangle\}\)
\[
\left|j,m\right\rangle=\sum_{m_1,m_2}C_{m_1m_2}\left|m_1m_2\right\rangle
\]
\(C_{m_1m_2}\) are called as CB coefficients.
\[
C_{m_1m_2}=\left\langle m_1m_2\middle|j,m\right\rangle.
\]
To find CB coefficients :--
look for states \(m=m_1+m_2\)
\[
\left|j,m\right\rangle=\sum_{m_1,m_2}C_{m_1m_2}\left|m_1,m_2\right\rangle
\]
\[
\underset{m}{\downarrow}\underset{(m_1+m_2)}{\downarrow}
\]
ex \(j_1=\tfrac12\), \(j_2=\tfrac12\), \(j=|j_1-j_2|\) to \(j_1+j_2\)
\[
\left|\tfrac{-1}{2}\tfrac{-1}{2}\right\rangle,\ \left|\tfrac{-1}{2}\tfrac{1}{2}\right\rangle,\ \left|\tfrac{1}{2}\tfrac{-1}{2}\right\rangle,\ \left|\tfrac12\tfrac12\right\rangle
\]
we can see that
\[
\underset{\substack{\downarrow\ \ \downarrow\\ j\ \ \ m}}{\left|1,1\right\rangle}=\underset{\substack{\downarrow\ \ \downarrow\\ m_1\ m_2}}{\left|\tfrac12\tfrac12\right\rangle}\qquad \begin{aligned}&(1=\tfrac12+\tfrac12)\\ &(m=m_1+m_2)\end{aligned}
\]
We can apply \(J_-\) to \(\left|1,1\right\rangle\) to get \(\left|1,0\right\rangle\).
\[
J_-\left|1,1\right\rangle=(J_{1-}+J_{2-})\left|\tfrac12\tfrac12\right\rangle
\]
also
\[
\boxed{\left|1,-1\right\rangle=\left|\tfrac{-1}{2}\tfrac{-1}{2}\right\rangle}
\]