• #PDF
• #Recall ---
• #Pictures in QM ---
• #Heisenberg picture ---
• #The forced oscillator ---
• [[#in' and out' operators ---]]
• #Interaction Picture ---

Prof. Sachindeo Vaidya (CHEP, IISc) | PDF

Previous: Lecture 13

Next: Lecture 15

Quantised EM field interacting with

(Intro to infrared problem in QED)
\(\hookrightarrow\) topic of current research.

Warmup : SHO with a forcing function (1-d) --- To build necessary technology

1. Time evolution in QM is given by \[ \left|\psi(t)\right\rangle \;=\; T\left(t,t_0\right)\left|\psi(t_0)\right\rangle \hspace{2em}\text{---(1)} \] \[ TT^{\dagger} \;=\; T^{\dagger}T \;=\; I \qquad\text{(Probability conservation)} \] \[ T\left(t,t_0\right) \;=\; T\left(t,t_1\right)T\left(t_1,t_0\right) \] \[ T\left(t,t_0\right)^{-1} \;=\; T\left(t_0,t\right) \]

Also eqn (1) requires

\[ \left.\left|\psi(t)\right\rangle\right|_{t=t_0} \;=\; \left|\psi(t_0)\right\rangle \qquad\text{for } t = t_0 \] \[ \text{i.e.}\qquad T\left(t_0,t_0\right) \;=\; I \] \[ T\left(t+\epsilon, t\right) \;=\; 1 - \frac{i}{\hbar}\epsilon H(t) \hspace{2em}\text{--- defn of Hamiltonian} \]

\(H(t)\) is defined as leading operator for time evolution operator (when expanded as first order
in \(\epsilon\)). (\(\epsilon\ \downarrow\) time step)

\[ i\hbar\frac{d}{dt}T\left(t,t_0\right) \;=\; H(t)\,T\left(t,t_0\right) \]

for \(H\) to be time independent ---

\[ T\left(t,t_0\right) \;=\; e^{-iH(t-t_0)/\hbar} \]

1. Schrödinger picture --- observables (operators) are fixed & wavefun / state evolve
2. Heisenberg picture --- wave fun are fixed but observables evolve with time
3. Interacting picture --- (Dirac picture) --- Both states & observables evolve via convenient

operators \(T_0\) or \(H_0\).

(2) \underline{define} \(\quad U(t) \;=\; T(0,t)\)

This operator doesn't change state ---

\[ \begin{aligned} \text{i.e. } U(t)\left|\psi(t)\right\rangle &= U(t)\,T(t,0)\left|\psi(0)\right\rangle\\ &= T(0,t)\,T(t,0)\left|\psi(0)\right\rangle\\ &= I\left|\psi(0)\right\rangle \end{aligned} \]

define

\[ \bar{L}(t) \;=\; U(t)\,L\,U^{\dagger}(t) \] \[ i\hbar\frac{d}{dt}\bar{L} \;=\; \left[\bar{L},\bar{H}\right] + i\hbar\frac{\partial\bar{L}}{\partial t} \]

(3) For dirac picture ---

Choose \(U(t)\) a soln of \( i\hbar\dfrac{\partial U}{\partial t} = -U H_0\)

\[ \bar{L} \;=\; U(t)\,L\,U^{\dagger}(t) \] \[ i\hbar\frac{d}{dt}\bar{L} \;=\; \left[\bar{L},\bar{H}_0\right] + i\hbar\frac{\partial\bar{L}}{\partial t} \]

where as states change

\[ i\hbar\frac{d}{dt}\left|\bar{\psi}(t)\right\rangle \;=\; U(t)\left(H-H_0\right)U^{\dagger}(t) \left|\bar{\psi}(t)\right\rangle \]

where,

\[ \left|\bar{\psi}(t)\right\rangle \;=\; U(t)\left|\psi(t)\right\rangle \] \[ H - H_0 \;=\; V \;=\; \text{interaction} \] \[ i\hbar\frac{d}{dt}\left|\bar{\psi}(t)\right\rangle \;=\; \bar{V}(t)\left|\bar{\psi}(t)\right\rangle \qquad \left(\bar V = U V U^{\dagger}\right) \]

----------------- \(\times\) -----------------

For

\[ H \;=\; \frac{p^2}{2m} + \frac{1}{2}m\omega^2 q^2 - q\,Q(t) - p\,P(t) \] \[ H \;=\; \hbar\omega\left(a^{\dagger}a + \frac{1}{2}\right) + f(t)\,a + f^{*}(t)\,a^{\dagger} \] \[ f(t) \;=\; -\sqrt{\frac{\hbar}{2m\omega}}\,Q(t) + i\sqrt{\frac{\hbar m\omega}{2}}\,P(t) \]

We are interested in

\[ \begin{aligned} f &\neq 0 &\qquad& T_1 < t < T_2\\ f &= 0 &\qquad& t < T_1\ \ \&\ \ t > T_2 \end{aligned} \]

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\[ H \;=\; \hbar\omega\left(\bar{a}^{\dagger}(t)\,\bar{a}(t) + \frac{1}{2}\right) + f(t)\,\bar{a}(t) + f^{*}(t)\,\bar{a}^{\dagger}(t) \] \[ i\hbar\frac{d\bar{a}}{dt} \;=\; \left[\bar{a},\bar{H}\right] \;=\; \left[\bar{a},H\right] \qquad \left(\bar{H} = H\right)\ \left\{\text{in Heisenberg picture } \hat{H}_H = \hat{H}_S\right\} \] \[ i\hbar\frac{d\bar{a}}{dt} \;=\; \hbar\omega\,\bar{a} + f^{*}(t) \hspace{2em}\text{---(2)} \]

\((t_0 = 0)\)

\[ \bar a(t) \;=\; \bar{a}\,e^{-\frac{i\omega}{\hbar}(t)} - \frac{i}{\hbar}\int_0^{t} e^{-i\omega(t-t')}f^{*}(t')\,dt' \]

we can solve eqn (2) using greens fun.

\[ \left(\frac{d}{dt} + i\omega\right)G\left(t,t'\right) \;=\; \delta\left(t-t'\right) \] \[ \bar{a}(t) \;=\; -\frac{i}{\hbar}\int_{-\infty}^{\infty}G\left(t,t'\right)f^{*}(t')\,dt' \]

Retarded green's fun. ---

\[ G_R\left(t-t'\right) \;=\; \theta\left(t-t'\right)e^{-\frac{i\omega}{\hbar}(t-t')} \] \[ \theta(x) = 0\ ;\ x<0\ ,\qquad = 1\ ;\ x>0 \]

Advanced greens fun ---

\[ G_A\left(t-t'\right) \;=\; -\theta\left(t'-t\right)e^{-i\omega(t-t')} \;=\; -G_R\left(t-t'\right) \]

So,

\[ \bar{a}(t) \;=\; -\frac{i}{\hbar}\int_{-\infty}^{\infty}G_R\left(t-t'\right)f^{*}(t')\,dt' \qquad\text{vanishes for } t < T_1 \] \[ \phantom{\bar a(t)}\;=\; -\frac{i}{\hbar}\int_{-\infty}^{\infty}G_A\,f^{*} \qquad\text{vanishes if } t > T_2 \]

let \(\bar{a}_{\text{in}}\) be soln of \(\dfrac{d\bar{a}}{dt} + i\omega\bar{a} = 0\)
which coincide with soln using \(G_R\) \(\left(G_A\right)\) resp.

\[ \bar{a}(t) \;=\; \bar{a}_{\text{in}}(t) - \frac{i}{\hbar}\int_{-\infty}^{\infty} G_R\left(t-t'\right)f^{*}(t')\,dt' \qquad\left(\bar{a}_{\text{in}} \to \bar{a}\,e^{-i\omega t}\right) \] \[ \Rightarrow\qquad \bar{a}_{\text{out}}(t) \;=\; -\frac{i}{\hbar}\int_{-\infty}^{\infty} G_A(\ \ )\,f^{*}(\ \ )\,dt' \] \[ \bar{a}_{\text{out}}(t) \;=\; \bar{a}_{\text{in}}(t) \;=\; -\frac{i}{\hbar} \int_{-\infty}^{\infty}\left(G_R - G_A\right)f^{*} \] \[ \bar{a}_{\text{out}}(t) \;=\; \bar{a}_{\text{in}}(t) - \frac{i}{\hbar}\int_{-\infty}^{\infty} dt'\ e^{-i\omega(t-t')}f^{*}(t') \]

\(\left(\bar{a}_{\text{out}}\,e^{-i\omega t}\ ,\quad \bar{a}_{\text{in}}\,e^{-i\omega t}\right)\)

\[ \bar{a}_{\text{out}} \;=\; \bar{a}_{\text{in}} - \frac{i}{\hbar}\int_{-\infty}^{\infty} dt'\ e^{i\omega t'}f^{*}(t) \]

\(\underbrace{\hspace{5cm}}\)
just a complex number \(\tilde{g}(\omega)\).

Compute \(\left|\left\langle n_{\text{out}}\,\middle|\,0_{\text{in}}\right\rangle\right|^2\)

Take \(H_0 = \hbar\omega\left(a^{\dagger}a + \frac{1}{2}\right)\) & \(V = f(t)a + f^{*}(t)a^{\dagger}\)

Then,

\[ \bar{V}(t) \;=\; e^{i\hbar\omega\left(a^{\dagger}a+\frac{1}{2}\right)t}\ V\ e^{-i\hbar\omega\left(a^{\dagger}a+\frac{1}{2}\right)t} \]

use BCH formula to find

\[ \left\{\begin{aligned} &e^{i\omega\hbar a^{\dagger}a}\ a\ e^{-i\omega\hbar a^{\dagger}a}\\ &e^{-i\omega\hbar a^{\dagger}a}\ a^{\dagger}\ e^{-i\omega\hbar a^{\dagger}a} \end{aligned}\right. \] \[ =\; f(t)\,a\,e^{-i\omega t} + f^{*}a^{\dagger}e^{i\omega t} \]

Time evolution \(\bar{T} = U T U^{\dagger}\)

where \(\bar{T}\) obeys :

\[ i\hbar\frac{d}{dt}\bar{T}(t) \;=\; \bar{V}(t)\,\bar{T}(t) \]

Since \(\bar{V} = \bar{V}(t)\) time dependent, we have \(\bar{T}\) as time ordered exponential ---

\[ \bar{T}\left(t,t_1\right) \;=\; \mathcal{T}\left\{\exp\left(-\frac{i}{\hbar}\int_{t_1}^{t} \bar{V}(t')\,dt'\right)\right\} \qquad\left(\mathcal{T}: \text{Time ordering}\right) \] \[ =\; 1 + \sum_{n=1}^{\infty}\frac{1}{n!}\int_{t_1}^{t}\cdots\int_{t_1}^{t}dt_1'\ldots dt_n'\ \mathcal{T}\left(\bar{V}(t_1')\,\bar{V}(t_2')\ldots\bar{V}(t_n')\right) \]

For one \(V(t)\) which is linear in \(a\) & \(a^{\dagger}\) we can exactly solve for
\(\bar{T}\left(t,t_1\right)\)

\[ \bar{T}\left(t,t_1\right) \;=\; \lim_{N\to\infty}\ e^{V_N}e^{V_{N-1}}\cdots e^{V_2}e^{V_1} \]

$\left(\ \cdots\ \underset{T\left(t_1+2\epsilon,\,t_1+\epsilon\right)}{\downarrow}\ \
\underset{T\left(t_1+\epsilon,\,t_1\right)}{\downarrow}\ \right)$

Where

\[ V_k \;=\; -\frac{i}{\hbar}\int_{t_1 + (k-1)\epsilon}^{t_1 + k\epsilon}\bar{V}(t')\,dt' \qquad\left(t - t_1 = N\epsilon\right) \]

We have

\[ \bar{T}\left(t,t_1\right) \;=\; \cdots\ e^{V_k}e^{V_{k-1}}\cdots \]

using BCH formula---

\[ e^{A}e^{B} \;=\; e^{A + B + \frac{1}{2}[A,B]} \qquad\text{if}\quad \left[A,[A,B]\right] = \left[B,[A,B]\right] = 0 \]

We combine \(e^{V_k}e^{V_{k-1}}\) terms using above

\[ \left[\bar{V}(t'),\bar{V}(t'')\right] \;=\; f(t')f^{*}(t'')e^{-i\omega(t'-t'')} - f^{*}(t')f(t'')e^{i\omega(t'-t'')} \] \[ \bar T\left(t,t_1\right) \;=\; \exp\left(-\frac{i}{\hbar}\int_{t_1}^{t}\bar{V}(t')\,dt' - \frac{1}{2\hbar^2}\int_{t_1}^{t}\int_{t_1}^{t}dt'dt'' \left[\bar{V}(t'),\bar{V}(t'')\right]\right) \] \[ =\; e^{i\beta\left(t,t_1\right)}\ e^{-\zeta^{*}\left(t,t_1\right)\bar{a} + \zeta\left(t,t_1\right)\bar{a}^{\dagger}} \] \[ \zeta\left(t,t_1\right) \;=\; -\frac{i}{\hbar}\int_{t_1}^{t}e^{i\omega t'}f^{*}(t)\,dt' \] \[ g(\omega) \;=\; -i\hbar\,\zeta^{*}\left(T_1,T_2\right) \;=\; -i\hbar\,\zeta^{*}(-\infty,\infty) \]

Defining scattering operator \(S = \bar{T}(-\infty,\infty)\)

Then

\[ S \;=\; e^{i\beta}\exp\left(-\frac{i}{\hbar}g(\omega)a - \frac{i}{\hbar}g^{*}a^{\dagger}\right) \] \[ \left|\left\langle n\left|S\right|0\right\rangle\right|^2 \;=\; \text{Transition probability for }\left|0\right\rangle \longrightarrow \left|n\right\rangle \] \[ =\; \frac{1}{n!}\left|\frac{g(\omega)}{\hbar}\right|^{2n}e^{-\left|g(\omega)\right|^2/\hbar^2} \] \[ =\; \frac{\left(\langle n\rangle\right)^n}{n!}\,e^{-\langle n\rangle} \qquad\text{(poisson distribution)} \]

If we don't turn off \(f(t)\) fast enough the fourier transform \(g(\omega)\) does not exist !

\(\hookrightarrow\) \underline{IR divergence !}

----------------- \(\times\) -----------------

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\node at (0,0.7) {$\left|1\right\rangle_{\text{in}}$};
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\node at (4.0,2.4) {$\left|3\right\rangle_{\text{out}}$};
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So \(\left|0\right\rangle_{\text{in}} \neq \left|0\right\rangle_{\text{out}}\)

But physically speaking \(\left|0\right\rangle_{\text{in}}\) seems same !