• #Lecture 14
  • #spread of a wave packet :-
• #Lecture 15
  • #scattering from a potential barrier (1-D) :-
  • #Particle in a constant Force field
  • #charge particle in constant B. (megnetic field) :-

overview | << prev | next >>

[The scan page that opens this set begins with the tail of the previous lecture -- "if \(H = H(t)\) non-autonomous quantum system, Heisenberg eq\(^{\text{n}}\) takes the form \(\frac{dA_H(t)}{dt} = \frac{[A_H(t), H_H(t)]}{i\hbar} + \frac{\partial A_H(t)}{\partial t}\), Most general Heisenberg equation of motion." -- that material belongs to the Lec 12-13 page. This page begins at the "lec 14" heading in the scans.]

letus go back a little bit to shroging shrödinger eq\(^{\text{n}}\) of a particle in 3-D. and show associated with probability density \(|\psi|^2\), there is a probability current.

from classical,

\[ \frac{\partial \rho}{\partial t} + \nabla\cdot\vec{J} = 0 \]

\(\rho = \) density

\(J = \) current density

shrödinger eq\(^{\text{n}}\) of motion in position basis.

\[ \psi^*\left[i\hbar\frac{\partial \psi(r,t)}{\partial t}\Big\rangle = H|\psi(r,t)\rangle = \frac{-\hbar^2}{2m}\nabla^2\psi(r,t)\rangle + V(r).|\psi(r,t)\rangle\right] - \text{\textcircled{1}} \]

complex conjugate eq\(^{\text{n}}\).

\[ \psi\left(-i\hbar\frac{\partial \psi^*(r,t)}{\partial t} = \frac{-\hbar^2}{2m}\nabla^2\psi^*(r,t) + V(r)\,\psi^*(r,t)\right] - \text{\textcircled{2}} \] \[ \text{\textcircled{1}} * \psi^* - \text{\textcircled{2}}*(\psi) \Rightarrow \] \[ i\hbar\frac{\partial}{\partial t}(\psi^*\psi) = \frac{-\hbar^2}{2m}\left[\psi^*\nabla^2\psi - \psi\nabla^2\psi^*\right] + \psi^* V(r)\psi - \psi V(r)\psi^* \]

\(\left(\psi^* V(r)\psi = \psi V(r)\psi^* \right\}\) How (?)

\[ i\hbar\frac{\partial}{\partial t}|\psi|^2 = \frac{-\hbar^2}{2m}\left(\psi^*\nabla^2\psi - \psi\nabla^2\psi^*\right) \] \[ \boxed{i\hbar\frac{\partial}{\partial t}|\psi|^2 = \frac{-\hbar^2}{2m}\nabla\cdot\left(\psi^*\nabla\psi - \psi\nabla\psi^*\right)} \]

\(\left\{\nabla\cdot(\psi^*\nabla\psi - \psi\nabla\psi^*)\right.\)

\(\Rightarrow \nabla\psi^*\cdot\nabla\psi + \psi^*\nabla^2\psi - \nabla\psi\cdot\nabla\psi^* - \psi\nabla^2\psi^*\)

\[ \frac{\partial}{\partial t}|\psi|^2 = \frac{-\hbar^2}{(i\hbar)2m}\nabla\cdot(\vec{J}) \] \[ \frac{\partial}{\partial t}|\psi|^2 + \frac{\hbar}{2mi}\nabla\cdot\vec{J} = 0 \] \[ \rho = |\psi|^2 \] \[ J = \left(\psi^*\nabla\psi - \psi\nabla\psi^*\right)\frac{\hbar}{2mi} = \text{probability current density.} \]

we have a particle (free or in a potential), its propagating in a direction, If we prepare an initial state of this particle such that it is localized around some point \(x_0\), (so a wave packet peaked at \(x_0\)) and let go. This particle propogate under shrödinger eq\(^{\text{n}}\), let us ask, "what happens to shape of this wave packet ?" as time goes along.

Let us assume we have a state vector \(|\Psi(t)\rangle\)

What happens to mean value of position \(\langle x\rangle\). ?

or mean squared value of the position \(\langle x^2\rangle\). and so on.

Now we know that if we have a gaussian wave packet. the peak position is actual value of the mean, if it spreads out or it contracts, we know that wave packet is either being dispersed or getting tighter.

let us ask what is \(\langle x^2\rangle(t)\)?

\[ H = \frac{p^2}{2m} + V(x) \] \[ \langle x^2\rangle(t) = \langle \Psi(t)|x^2|\Psi(t)\rangle \]

This can be calculated either from shrödinger picture or Heisenberg picture.

\[ \frac{d}{dt}\langle x^2\rangle = \left(\frac{d}{dt}\langle \Psi(t)|\right)x^2|\Psi(t)\rangle + \langle \Psi(t)|x^2\left(\frac{d}{dt}|\Psi(t)\rangle\right) \]

using shrödingers eq\(^{\text{n}}\) of motion

\[ i\hbar\frac{d}{dt}|\Psi\rangle = H|\Psi\rangle \] \[ -i\hbar\frac{d}{dt}\langle \Psi| = \langle\Psi|H^{\dagger} = \langle\Psi|H \]

\(\{\) \(H\) is hermitian

so,

\[ \frac{d}{dt}\langle x^2\rangle = \left\langle \Psi(t)\left|\frac{Hx^2}{-i\hbar}\right|\Psi(t)\right\rangle + \left\langle \Psi(t)\left|\frac{x^2H}{i\hbar}\right|\Psi(t)\right\rangle \] \[ = \left\langle \Psi(t)\left|\frac{x^2H - Hx^2}{i\hbar}\right|\Psi(t)\right\rangle \] \[ = \left\langle \Psi(t)\left|\frac{[x^2,H]}{i\hbar}\right|\Psi(t)\right\rangle \ ---- \ \text{\textcircled{1}} \]

[\(\left\langle\frac{dx^2}{dt}\right\rangle\)]

Let us find

\[ [x^2,H] = \left[x^2, \frac{p^2}{2m} + V(r)\right] = \left[x^2,\frac{p^2}{2m}\right] + \left[x^2, V(r)\right]^{\nearrow 0} \] \[ = \left[x^2,\frac{p^2}{2m}\right] \] \[ = \frac{1}{2m}\left[x^2, p^2\right] \] \[ = \frac{1}{2m}\left[x\cdot x, p^2\right] = \frac{1}{2m}\left[x[x,p^2] + [x,p^2]x\right] \] \[ = \frac{1}{2m}\left(x\big(p[x,p] + [x,p]p\big) + \big(p[x,p] + [x,p]p\big)x\right) \] \[ = \frac{1}{2m}\left(xp(i\hbar I) + x(i\hbar)Ip + i\hbar\,px + i\hbar\,px\right) \] \[ = \frac{2i\hbar}{2m}\left(xp + px\right) = \frac{i\hbar}{m}(xp+px) \]

eq\(^{\text{n}}\) \(\text{\textcircled{1}}\) becomes.

\[ \frac{d}{dt}\langle x^2\rangle = \left\langle \Psi(t)\left|\frac{i\hbar(xp+px)}{i\hbar\,m}\right|\Psi(t)\right\rangle \] \[ \frac{d}{dt}\langle x^2\rangle = \frac{1}{m}\langle xp+px\rangle \]

since \(x\), \(x^2\) are physical observables so is \(xp\), \(px\) and \((xp+px)\).

"So we can also find \(\frac{d}{dt}\langle xp+px\rangle\) and again we will have new physical measurables, and the series of generation of such new physical operators do not truncate, and end up having \(\infty -\) of such operators."

\[ \frac{1}{m}\langle xp+px\rangle \equiv \frac{1}{m}\langle \Psi(t)|xp+px|\Psi(t)\rangle \qquad \text{(in position basis)} \] \[ = \frac{-i\hbar}{m} \] \[ = \frac{1}{m}\int_{-\infty}^{\infty} dx\ \psi^*(x,t)\cdot\left(\frac{\partial}{\partial x}\big(x\,\psi(x,t)\big)\right)\left(\frac{-i\hbar}{m}\right) \]

[the two lines above are struck out in the notebook]

\[ = \frac{-i\hbar}{m}\int_{-\infty}^{\infty} dx\ \psi^*(x,t)\left\{x\frac{\partial}{\partial x}\psi(x,t) + \frac{\partial}{\partial x}\big(x\,\psi(x,t)\big)\right\} \] \[ = \frac{\hbar}{mi}\left(\int_{-\infty}^{\infty} dx\cdot x\left(\psi^*\frac{\partial\psi}{\partial x} - \frac{\partial\psi^*}{\partial x}\psi\right)\right) \]

we have

\[ \frac{d}{dt}\langle x^2\rangle = \frac{1}{m}\langle xp+px\rangle \] \[ \frac{d^2}{dt^2}\langle x^2\rangle = \frac{1}{m}\left\langle \frac{[xp+px, H]}{i\hbar}\right\rangle \]

\(\left\{\frac{dA}{dt} = \left\langle\frac{[A,H]}{i\hbar}\right\rangle\right.\)

\(\left\{\frac{dA}{dt} = \frac{(A,H)}{i\hbar}\right.\)

for free particle \(V(x) = 0\)

\[ \frac{d^2}{dt^2}\langle x^2\rangle = \frac{1}{mi\hbar}\left\langle [xp+px, H]\right\rangle \] \[ = \frac{1}{mi\hbar}\left\langle [xp,H] + [px,H]\right\rangle \]

since \(\hat{H} = \frac{p^2}{2m}\)

\[ = \frac{1}{2m^2 i\hbar}\left\langle [xp, p^2] + [px, p^2]\right\rangle \] \[ = \frac{1}{2m^2 i\hbar}\left\langle [x,p^2]p + x[p,p^2]^{\nearrow 0} + [p,p^2]^{\nearrow 0}x + p[x,p^2]\right\rangle \] \[ = \frac{1}{2im^2\hbar}\left\langle \big([x,p]p + p[x,p]\big)p + p\big(p[x,p] + [x,p]p\big)\right\rangle \] \[ = \frac{1}{2im^2\hbar}\left\langle 2i\hbar p^2 + 2i\hbar p^2\right\rangle \] \[ = \frac{2}{m^2}\langle p^2\rangle \]

\(\left\langle \frac{p^2}{2m}\right\rangle\) is just the energy

\(\Rightarrow\) \(\langle p^2\rangle\) is independent of time.

\[ \frac{d}{dt}\langle x^2\rangle = \frac{2}{m^2}\langle p^2\rangle\, t + \frac{d}{dt}\langle x^2\rangle\bigg|_{t=0} \]

[integrating]

\[ \langle x^2\rangle(t) = ? \]

integrating once again

\[ \langle x^2\rangle(t) = \frac{2}{m^2}\langle p^2\rangle t^2 + \left(\frac{d}{dt}\langle x^2\rangle\bigg|_{t=0}\right)t + \langle x^2\rangle\bigg|_{t=0} \]

lets find :-

\[ \frac{d}{dt}\langle x\rangle = \frac{1}{i\hbar}\left\langle [x,H]\right\rangle \] \[ \frac{d}{dt}\langle x\rangle = \frac{1}{i\hbar}\left\langle \left[x, \frac{p^2}{2m}+V(x)\right]\right\rangle = \frac{1}{i\hbar}\left\langle \left[x,\frac{p^2}{2m}\right] + [x,V(x)]^{\nearrow 0}\right\rangle \] \[ \frac{d}{dt}\langle x\rangle = \frac{1}{2im\hbar}\left\langle [x,p^2]\right\rangle \] \[ = \frac{1}{2m i \hbar}\left\langle p[x,p] + [x,p]p\right\rangle \] \[ = \frac{1}{2im\hbar}\left\langle 2i\hbar p\right\rangle = \frac{\langle p\rangle}{m} \] \[ \frac{d}{dt}\langle x\rangle = \frac{\langle p\rangle}{m} \]

integrating once. (For free particle)

[\(\langle p\rangle\) is time independent and we can integrate directly ).]

\[ \langle x\rangle(t) = \frac{\langle p\rangle}{m}t + \langle x\rangle(t=0) \]

This eq\(^{\text{n}}\) resembles with classical picture of \(x = ut + x_0\)

So quantum mechanical equation of motion expectation values obey classical equation of motion. \(\}\) ernfest theorem

This theorem breaks down when chaose occurs in classical dynamics.

(How uncertainty varies in time) ?

At \(t=0\) :-

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw (-3.6,0) -- (3.4,0);
  \draw (-1.6,-0.5) -- (-1.6,2.6);
  \draw[thick] (-3.4,0.12) .. controls (-2.2,0.35) and (-1.4,2.2) .. (-0.2,2.25)
      .. controls (1.0,2.2) and (1.9,0.35) .. (3.2,0.12);
  \draw[dashed] (-0.2,0) -- (-0.2,2.25);
  \node[below] at (-0.2,-0.05) {$x_0$};
  \draw[-{Stealth}] (-2.0,1.25) -- (-1.15,1.25);
  \draw[-{Stealth}] (1.5,1.25) -- (0.7,1.25);
  \node[right] at (2.2,1.6) {$\Delta x = \langle (x-\langle x\rangle)^2\rangle^{1/2}$};
  \node[right] at (2.2,-0.7) {(initially)};
  \draw[-{Stealth}] (1.6,-1.15) -- (2.9,-1.15);
  \node[right] at (3.0,-1.15) {forward momentum};
  \node[left] at (-3.7,0.6) {initial wave packet.};
\end{tikzpicture}

\[ (\Delta x)_{(t)} = \langle x^2\rangle(t) - \big(\langle x\rangle(t)\big)^2 \] \[ = \left[\frac{2}{m^2}\langle p^2\rangle t^2 + \frac{d\langle x^2\rangle}{dt}\bigg|_{t=0} + \langle x^2\rangle(0)\right]^2 - \left(\frac{\langle p\rangle t}{m} + \langle x\rangle(0)\right)^2 \] \[ \geq 0 \]

so, \(\Delta x\) will increase with time. \(\Delta x(t)\) is increasing fun.

There is dispercion, the wave packet is actually increase with time.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw (-3.4,0) -- (3.4,0) node[right] {$x$};
  \draw (0,-1.2) -- (0,2.4) node[above] {$V(x)$};
  \draw[dotted] (-3.0,1.7) -- (2.6,1.7);
  \node[right] at (2.7,1.7) {$V_0$};
  \draw[thick] (-3.0,0.08) .. controls (-1.6,0.15) and (-0.9,1.6) .. (0.4,1.68)
      .. controls (1.4,1.72) and (2.0,1.7) .. (2.6,1.7);
  \draw[-{Stealth}] (-3.3,0.9) -- (-2.9,0.9);
  \node[draw,circle,inner sep=1pt] at (-2.4,0.9) {I};
  \node[draw,circle,inner sep=1pt] at (-1.0,0.55) {II};
  \node at (2.6,0.9) {$(\mathrm{T.E} = E)$};
\end{tikzpicture}

\((E < V_0)\)

classically any particle with energy \(< V_0\), will climb up the hill and roll back.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw (-1.6,0) -- (1.4,0) node[right] {$x$};
  \draw (0,-0.5) -- (0,1.5) node[above] {$V$};
  \draw[thick] (-1.5,0.15) -- (0.6,0.95);
  \draw[-{Stealth}] (-1.4,0.5) -- (-0.4,0.5);
  \draw[-{Stealth}] (-0.4,0.32) -- (-1.4,0.32);
\end{tikzpicture}

Quantum mechanically, if we shoot a particle in \(x\)-direction, If total energy \(< V_0\), then particle comes along and portion of state vector / wave function \(\psi(r,t)\), actually exists in the \(\text{\textcircled{II}}^{\text{rd}}\) region but dies down exponentially. and there is also a reflected wave fun. But there would be exactly total internal reflection, The incident wave would be reflected back, & there is no transmission as such.

\((E > V_0)\)

classically if a particle is having T.E \(> V_0\), the particle climbs up the hill and moves in \(+x\) direction with reduced velocity. \((K = T-V)\) and contineus to move upto \(+\infty\).

Quantum mechanically something interesting happens, A particle with T.E \(> V_0\), a portion of its wave is transmitted and a portion of its incident wave is reflected back. And the amount which is transmitted or reflected will tell us, transmisson coefficient and reflection co-efficient.

(letus solve first for \(E > V_0\))

Its incident wave fun. (stationary) \(\sim e^{ikx}\)

[approximation because of some potential]

where,

\[ \frac{2mE}{\hbar^2} = k^2 \]

(This state is not normalizable, its extends from \(-\infty\) to \(+\infty\))

Transmitted wave fun. \(\sim A e^{ik'x}\)

where,

\[ k'^2 = \frac{2m(E-v)}{\hbar^2} \]

Reflected wave function : \(\sim B e^{-ikx}\)

\[ \text{as } x\to(-\infty) \qquad \left(\psi(x) = e^{ikx} + Be^{-ikx}\right) \] \[ x\to(+\infty) \qquad \left(\psi(x) = A e^{ik'x}\right) \] \[ \text{Probability current} = \frac{\hbar}{2mi}\left(\psi^*\frac{\partial\psi}{\partial x} - \psi\frac{\partial\psi^*}{\partial x}\right) \] \[ \text{for } \psi = e^{ikx}, \quad \text{current} = \frac{\hbar k}{m} \]

so let us define :-

\((T)\) Transmission coefficient \(= \dfrac{\text{Transmitted Probability current}}{\text{Incident Probability current}}\)

\[ = \frac{\frac{\hbar k'}{m}|A|^2}{\left(\frac{\hbar k}{m}\right)} = \frac{k'}{k}|A|^2 . \]

\((R)\) Reflection coefficient \(= \dfrac{\text{Reflected Probability current}}{\text{Incident probability current}}\)

\[ = |B|^2\frac{k}{k} = |B|^2 \]

Now Probability must be conserved, If initially our wave fun. is normalized, then Probability sum of reflected and transmitted should also be equal to 1.

\[ \Rightarrow \quad T+R = 1 \] \[ \left(\frac{k'}{k}|A|^2 + |B|^2 = 1\right) \]

To find actual value of \(|A|^2\), \(|B|^2\) one should solve shrödinger equation, with that specific form of potential, then find out what happens asymptotically.

"Let us look into possibility of absorption. Absorption only happens if a reaction is causing particles to disappear.

Suppose we incident a beam of neutrons, and Nucleus absorbs some of it, then the total flux of incident neutrons decreases / disappear, then we indeed have real absorption coefficient and part of it is scattered. This is taken into account by making the potential complex (optical potential method).

where imaginary part of \(V(x)\) is related with absorption coefficient".

Let's take a simple potential like

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw[->] (-3.0,0) -- (3.0,0) node[right] {$x$};
  \draw[->] (0,-1.0) -- (0,2.2) node[above] {$V(x)$};
  \draw[very thick] (-2.8,0) -- (0,0);
  \draw[very thick] (0,0) -- (0,1.5);
  \draw[very thick] (0,1.5) -- (2.6,1.5);
  \node[right] at (2.65,1.5) {$V_0$};
  \node[draw,circle,inner sep=1pt] at (-1.6,0.65) {I};
  \node[draw,circle,inner sep=1pt] at (0.9,1.9) {II};
  \node[below left] at (0,0) {$0$};
\end{tikzpicture}

In \(\text{\textcircled{I}}\)

\[ \psi(x) = e^{ikx} + Be^{-ikx} \]

In \(\text{\textcircled{II}}\)

\[ \psi(x) = A e^{ik'x} \]

where,

\[ \left\{ \begin{aligned} k^2 &= \frac{2mE}{\hbar^2}\\[4pt] k'^2 &= \frac{2m(E-V_0)}{\hbar^2} \end{aligned} \right. \]

Since \((T+R=1)\)

\[ \frac{k'}{k}|A|^2 + |B|^2 = 1 \]

But to find \(|A|\), \(|B|\) we have to solve for shrödingers eq\(^{\text{n}}\) and apply some boundary cond\(^{\text{n}}\).

\[ \psi(0^-) = \psi(0^+) \qquad \text{(wave fun is contineous)} \] \[ 1+B = A \]

we know that for any stationary state / eigen state of \(H\).

\[ \frac{-\hbar^2}{2m}\psi''(x) + V(x)\,\psi(x) = E\,\psi(x) \]

\(\uparrow\) finite discountinuity ; \(\uparrow\) finite discountinuity ; \(\uparrow\) contineous

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw (-1.6,0) -- (1.8,0);
  \draw (0,-0.6) -- (0,1.4) node[above] {$\psi'(x)$};
  \draw[thick] (-1.4,0.35) -- (0,0.35);
  \draw[thick] (0,0.9) -- (1.6,0.9);
\end{tikzpicture}

slopes will be same \(\psi'(0^-) = \psi'(0^+)\)

such that \(\psi''(0^-) \neq \psi''(0^+)\).

(finite discontinuity).

So, using \(\psi'(0^-) = \psi'(0^+)\)

\[ ik\left(e^{ikx} - Be^{-ikx}\right)\bigg|_{x=0} = (ik')A\,e^{ik'x} \] \[ k(1-B) = k'A \] \[ k(1-B) = k'(1+B) \] \[ \frac{k}{k'} = \frac{1+B}{1-B} \quad \Rightarrow \quad B(k'+k) = (k-k') \] \[ \left(B = \frac{k-k'}{k'+k}\right) \qquad A = 1+B = \frac{2k}{k'+k} \]

checking \((T+R=1)\)

\[ \left(\frac{k'}{k}|A|^2 + |B|^2 = ?\right. \] \[ = \frac{k'}{k}\frac{4k^2}{(k'+k)^2} + \frac{(k'-k)^2}{(k'+k)^2} \] \[ = \frac{4kk' + k'^2 + k^2 - 2kk'}{(k'+k)^2} = \frac{(k'+k)^2}{(k'+k)^2} = 1. \]

What happens if \((E < V_0)\).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw (-3.0,0) -- (3.0,0);
  \draw (0,-0.8) -- (0,2.0);
  \draw[very thick] (-2.8,0) -- (0,0) -- (0,1.2) -- (2.6,1.2);
  \node[right] at (2.65,1.2) {$V_0$};
  \draw[dashed] (-2.6,0.65) -- (0,0.65);
  \node[left] at (-2.65,0.65) {$E$};
  \node[draw,circle,inner sep=1pt] at (-1.5,1.15) {I};
  \node[draw,circle,inner sep=1pt] at (0.9,1.7) {II};
\end{tikzpicture}

since

\[ k'^2 = \frac{2m(E-V_0)}{\hbar^2} \] \[ k' = \sqrt{\frac{2m(E-V)}{\hbar^2}} \]

for \(E<V_0\), \(k'\) becomes imaginary

\[ \psi(x) \text{ in } \text{\textcircled{II}} \text{ region } = A e^{ik'x} \ \nearrow \ A e^{-|k'|x} \]

So wave dies down exponentially

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw (-3.2,0) -- (3.2,0) node[right] {$x$};
  \draw (0,-1.0) -- (0,2.2) node[above] {$V(x)$};
  \draw[very thick] (-3.0,0) -- (0,0) -- (0,1.1) -- (2.8,1.1);
  \node[right] at (2.85,1.1) {$V_0$};
  \draw[dashed,thick] (-2.9,0.35) sin (-2.6,0.75) cos (-2.3,0.35) sin (-2.0,-0.05) cos (-1.7,0.35)
       sin (-1.4,0.85) cos (-1.1,0.35) sin (-0.8,-0.15) cos (-0.5,0.35) sin (-0.2,0.95) cos (0,0.6);
  \draw[dashed,thick] (0,0.6) .. controls (0.5,0.3) and (1.0,0.12) .. (1.8,0.05);
\end{tikzpicture}

So although Transmisson coefficient is zero (in this case) [. as \(x\to\infty\) these is no wave function) (no flux of particles). However there is a penetration in this barrier, and it dies down exponentially fast (with an exponent \(\sqrt{\frac{2m(V_0-E)}{\hbar^2}}\).)

"Since \(\psi_I(x)\) is not normalized for infinite domain, but we can put whole thing in a box, so that \(e^{ikx}\) is member of \(L_2\) (normalizable). so that \(T+R=1\)."

But if we put the whole thing in huge box \([-L\) to \(+L]\), then momentum gets quantized (because it gotta satisfy boundary conditions), called box normalization, let's extend the box to infinity then the levels become essentially contineous."

one more rigerous way of doing it (because we can't find sharp momentum eigen state. i.e. plane wave is a fiction), so let's find a wave packet localized about some wave number \((k_0)\) maybe a gaussian wave function, we shoot that and it comes back and we can do same thing for superposition of waves.

"Also in case of total internal reflection there is a wave which penetrates another medium, but it also dies down exponentially"

What happens if we cut the barrier at \(x=a\).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw (-3.4,0) -- (3.4,0) node[right] {$x$};
  \draw[->] (-1.0,-0.2) -- (-1.0,2.2) node[above] {$V(x)$};
  \draw[very thick] (-1.0,0) -- (-1.0,1.3) -- (0.4,1.3) -- (0.4,0);
  \node[below] at (0.1,-0.05) {$a$};
  \node[right] at (0.55,1.35) {$V_0$};
  \node[right] at (1.2,1.35) {$Ce^{ikx}$};
  \node[left] at (-1.4,1.5) {$e^{ikx} + Be^{-ikx}$};
  \draw[-{Stealth}] (-0.7,-0.15) .. controls (-0.7,-0.6) .. (-1.0,-0.95);
  \node[right] at (-1.05,-1.15) {$\left(De^{ik'x} + Ee^{-ik'x}\right)$};
\end{tikzpicture}

[equate wave fun. and its derivative \(x = 0, a\).]

\[ \left\{ \begin{aligned} \psi(0^-) &= \psi(0^+)\\ \psi'(0^-) &= \psi'(0^+) \end{aligned} \right. \] \[ \left\{ \begin{aligned} \psi(a^-) &= \psi(a^+)\\ \psi'(a^-) &= \psi'(a^+) \end{aligned} \right. \]

what happens if we decrease width and increase height to \(\infty\) such that product remains const. (we get a delta fun.)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw[->] (-2.6,0) -- (2.6,0) node[right] {$x$};
  \draw[->] (0,-0.3) -- (0,2.2) node[above] {$V(x)$};
  \draw[very thick] (0,0) -- (0,1.9);
  \node[right] at (0.8,1.2) {$V(x) = +\lambda\,\delta(x)$};
\end{tikzpicture}

\[ \frac{-\hbar^2}{2m}\frac{d^2\phi_E(x)}{dx^2} + V(x)\,\phi_E(x) = E\,\phi_E(x) \]

\(\uparrow\) (\(\infty\) discontinuity) ; \(\uparrow\) (\(\infty\) discontinuity) ; \(\uparrow\) contineous

If \(\phi''\) has \(\infty\)-discontinuity, this means \(\phi'(x)\) has finite jum, which implies \(\phi(x)\) has cusp at origin.

so only boundary cond\(^{\text{n}}\). left is.

\[ \phi(0^-) = \phi(0^+) \] \[ \int_{-\epsilon}^{+\epsilon}\left(\frac{-\hbar^2}{2m}\phi''(x) + V(x)\,\phi(x)\right)dx = \int_{-\epsilon}^{+\epsilon} E\,\phi(x)\,dx \]

Now discontinuity in \(\phi'(x)\) will be related to \(\delta(x)\) function.

(It is given as an exercise) [solve it]

particle will get through the barrier in \(\delta(x)\) potential.

\[ \left(\psi_T \sim e^{ikx} \ \text{plane wave}\right) \] \[ V(x) = -Fx \qquad (F = \text{const}) \]

so that

\[ \text{Force} = -\frac{dV}{dx} = F \ \text{(constant force)}. \]

(i.e. could be electic force \(qE\) if \(E\)=const).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw[->] (-2.8,0) -- (2.8,0) node[right] {$x$};
  \draw[->] (0,-2.0) -- (0,2.2) node[above] {$V(x)$};
  \draw[thick] (-2.4,1.8) -- (2.0,-1.5);
\end{tikzpicture}

Every Energy is allowed \((-\infty < E < +\infty)\)

classically any particle shot in \((-x)\) will climb up the hill, loose the K.E and will roll back accelarated upto \(+\infty\).

(Quantum mechanically what would we expect ?)

\[ -\infty < E < \infty \]

and No bound states (no confineing potential)

so, solution would not be normalizable.

so what kind of wave would it be ?

let \(\phi_E\) be stationary state of \(H\). satisfying

\[ \frac{-\hbar^2}{2m}\frac{d^2\phi_E(x)}{dx^2} + (-Fx)|\phi_E(x)\rangle = E|\phi_E(x)\rangle \] \[ \phi_E''(x) + \frac{2m}{\hbar^2}(E+Fx)\,\phi_E(x) = 0 \]

Let

\[ \xi = \frac{2m}{\hbar^2}(E+Fx), \qquad x = \left(\frac{\hbar^2}{2m}\frac{\xi}{F} - \frac{E}{F}\right) \]

[side working in the margin:]

\[ \frac{d}{dx}\left(\frac{d\phi_E'(\xi)}{d\xi}\cdot\frac{d\xi}{dx}\right) + \xi\,\phi_E = 0 \] \[ \frac{d}{d\xi}\left(\frac{d\phi'(\xi)}{d\xi}\cdot\frac{2mF}{\hbar^2}\right)\frac{d\xi}{dx} + \xi\phi(\xi) = 0 \] \[ \phi''(\xi) + \frac{\hbar^4}{(2mF)^2}\,\xi\,\phi(\xi) = 0 \] \[ \phi_E''(\xi) + c\,\xi\,\phi_E(\xi) = 0 \]

(Airy's equation)

\[ \left(c = \frac{\hbar^4}{(2mF)^2}\right) \ \text{which is independent of } E \ \big) \] \[ \phi_E(\xi) = (\text{const.})\,Ai(-\xi) \]

where

\[ Ai(z) = \frac{1}{\sqrt{\pi}}\int_0^{\infty} du\cdot\cos\left(\frac{u^3}{2} + u\xi\right) \]

(since \(c\) is independent of \(E\), every energy eigenstate has same \(\phi(x)\) \(\}\)

Let's work in momentum basis \(\rangle\) (This problem is easier in \(|p\rangle\)).

\[ H\hat{\phi}_E(p) = E\,\tilde{\phi}_E(p) \qquad - \text{\textcircled{1}} \] \[ H = \frac{p^2}{2m} - F\hat{x} = \frac{p^2}{2m} - F\left(i\hbar\frac{d}{dp}\right) \]

so above eq\(^{\text{n}}\) \(\text{\textcircled{1}}\) becomes

\[ \left(\frac{p^2}{2m} - i\hbar F\frac{d}{dp}\right)\tilde{\phi}_E(p) = E\,\tilde{\phi}_E(p) \] \[ \frac{d\tilde{\phi}_E(p)}{dp} = \frac{i\left(E - \frac{p^2}{2m}\right)\tilde{\phi}_E(p)}{\hbar F} \] \[ \int \frac{d\tilde{\phi}_E(p)}{\tilde{\phi}_E(p)} = \frac{1}{\hbar F}\int i\left(E - \frac{p^2}{2m}\right)dp \] \[ \ln\left(\tilde{\phi}_E(p)\right) = \frac{i}{\hbar F}\left(Ep - \frac{p^3}{6m}\right) + c \] \[ \boxed{\tilde{\phi}_E(p) = A\,e^{\frac{i}{\hbar F}\left(Ep - \frac{p^3}{6m}\right)}} \]

normalization cond\(^{\text{n}}\) :-

\[ \int_{-\infty}^{\infty} dp\ \tilde{\phi}_E^*(p)\,\tilde{\phi}_{E'}(p) = \delta(E-E') \]

[normalization cond\(^{\text{n}}\) for the wavefun. which are not normalizable in position space.]

[orthonormality cond\(^{\text{n}}\)]

[Energy normalization]

\[ \text{as } \xi\to-\infty \qquad \phi \longrightarrow e^{-|\xi|^{3/2}} \] \[ \xi\to+\infty \qquad \phi \longrightarrow \xi^{-1/4}\sin\left([?]\right) \]

classically :-

\[ L(q,\dot{q}) = \frac{1}{2}m\vec{V}^2 + e(\vec{A}\cdot\vec{V}) - e\phi \]

\(\vec{A} = \vec{A}(r,t)\)

\(\phi = \phi(r,t)\)

\[ \vec{P} = \frac{\partial L}{\partial \vec{V}} = m\vec{V} + e\vec{A} \] \[ H \text{ (free particles) } = \frac{p^2}{2m} + e\phi \] \[ H(B) = \frac{(\vec{p}-e\vec{A})^2}{2m} + e\phi \]

since we only have const \(B\).

\[ (\phi = 0) \] \[ H = \frac{(\vec{p}-e\vec{A})^2}{2m} \]

\(\left\{[p, A(r)] \neq 0 \right.\) Because \([P,r] \neq 0\)

lets first find \([p,A]\) =?

\[ \vec{p}\cdot\vec{A}(r) - \vec{A}(r)\cdot\vec{p} \] \[ = -[A,p] = -i\hbar\,\nabla\cdot\vec{A}(r) \]

\(\left\{\text{if } [x, P_x] = i\hbar\,\mathbb{1}\right.\)

then what is \([f(x), P_x]\)

\(= i\hbar f'(x)\) (How?)

\[ f(x)\left(-i\hbar\frac{\partial}{\partial x}\phi\right) + \left(i\hbar\frac{\partial}{\partial x}\right)f(x)\cdot\phi(x) \] \[ = [f(x), P_x]\,\phi(x) \] \[ = -i\hbar f(x)\frac{\partial\phi}{\partial x} + \phi(x)\,i\hbar\frac{\partial f}{\partial x} + f(i\hbar)\frac{\partial\phi}{\partial x} \] \[ = (i\hbar)\,f'(x)\cdot\phi(x) \] \[ \boxed{[f(x), p] = i\hbar f'(x)} \]

"we can always work in gauge where \(\nabla\cdot\vec{A} = 0\) (coulamb gauge)"

We know that

\[ \nabla\cdot\vec{B} = 0 \quad \Rightarrow \quad \vec{B} = \nabla\times\vec{A}(r) \]

But \(\vec{B}\) must be const, so what is \(\vec{A}\) in terms of \(B\).

\[ \vec{A}(\vec{r}) = \left(\tfrac{1}{2}\right)\cdot\vec{B}\times\vec{r} \qquad ----\ (\text{How}). \]

Is \(\nabla\cdot\vec{A} = 0\)

\[ \nabla\cdot\vec{A} = \partial_i A_i \] \[ = \frac{1}{2}\partial_i\,\epsilon_{ijk}x_k = 0 \]

so \(\nabla\cdot\vec{A} = 0\) for const \(\vec{B}\).

\[ \Rightarrow \quad [p, A(r)] = 0 \]

\(\vec{p}\) & \(\vec{A}\) commute.

But \(A\) is not fixed, we can choose any \(A'\)

such that

\[ A' = A + \nabla\chi(\vec{r}) \]

so that

\[ (\nabla\times A' = \nabla\times A) = B. \]

By doing so, \(\hat{H}\) will also change, but it will be a canonical transformation.

\[ H = \frac{(p-eA)^2}{2m} \] \[ H' = \frac{(p-eA')^2}{2m} \]

provided shrödinger equation does not change.

\[ H\phi = E\phi \qquad \text{(should not change)} \]

"\(|\)lec 16\(\rangle\) is written after \(|\)lec 17\(\rangle\)"

[The notebook's "lec 17" heading begins here, part-way down scan page qm6-05; the material that follows it is transcribed on the Lec 16-17 page.]