Quantum decoherence I: coherent states

In order to understand how thermal effects (among other things) can cause decoherence of quantum states, we first need to understand what a quantum coherent state is.
Coherent states are packets that behave very nearly like classical particles: they are nondispersive packets whose centers move according to the classical EOMs. Define first, then justify this statement.

$a|z\rangle=z|z\rangle, \quad |z\rangle=\sum_n c_n|n\rangle, \qquad z|z\rangle=z\sum_n c_n |n\rangle=\sum_n c_n\sqrt{n}|n-1\rangle$

so $c_n=N \, z^n/\sqrt{n!}$ and

$1=\langle z|z\rangle=|N|^2\sum_{n,m}{z^n (z^*)^m\over \sqrt{n!m!}}\langle m|n\rangle=|N|^2\sum_{n}{|z|^n\over n!}=|N|^2 e^{|z|^2}$

so $N=e^{-|z|^2/2}$

$\implies \qquad |z\rangle=e^{-|z|^2/2}\sum_n {z^n\over \sqrt{n!}}|n\rangle$

Pass to Heisenberg picture

$H=\hbar\omega a^\dagger a, \qquad |\psi(t)\rangle=e^{-i\omega t \, a^\dagger a}|\psi\rangle$

$e^{b a^\dagger a}|n\rangle =e^{b n}|n\rangle, \qquad e^{b a^\dagger a}e^{-|z|^2/2}\sum_n {z^n\over \sqrt{n!}}|n\rangle$

$=e^{-|z|^2/2}\sum_n {z^ne^{bn}\over \sqrt{n!}}|n\rangle=|ze^b\rangle$

For a single coherent state $|z\rangle$

$\langle x|z\rangle =\sum_{n=0}^\infty {z^n \over n!} \langle x|(a^\dagger)^n|0\rangle$

$=e^{-|z|^2/2}\sum_n {(z e^{-i\omega t})^n\over \sqrt{n!}}\langle x|n\rangle$

we know that

$\langle x|n\rangle=\sqrt{{1\over 2^n n!}\sqrt{{m\omega \over \hbar\pi}}} \, H_n(q) \, e^{-{1\over 2} q^2},$

$q=\sqrt{m\omega/\hbar}\, x, \quad H_n(q)= (-1)^n e^{q^2} \big({d\over dq}\big)^n e^{-q^2}$

and the summation in the definition of the coherent state can be performed

$\langle x|z(t)\rangle=\big( {m\omega\over \pi\hbar}\big)^{{1\over 4}} e^{-|z|^2/2}\sum_{n=0}^\infty (-{ze^{-i\omega t}\over \sqrt{2}})^n {1\over n!} e^{-{1\over 2} q^2} e^{q^2} \big({d\over dq}\big)^n e^{-q^2}$

$=\big( {m\omega\over \pi\hbar}\big)^{{1\over 4}} e^{{1\over 2} q^2} e^{-|z|^2/2}\sum_{n=0}^\infty {(-{z e^{-i\omega t}\over \sqrt{2}})^n \over n!} \big({d\over dq}\big)^n e^{-q^2}$

Use the Taylor theorem

$\sum_{n=0}^\infty {a^n\over n!} \big({d\over dq}\big)^n f(q)=f(q+a)$

to get

$\langle x|\psi(t)\rangle=\big( {m\omega\over \pi\hbar}\big)^{{1\over 4}} e^{-|z|^2/2} e^{{1\over 2} q^2} e^{-(q-{z e^{-i\omega t}\over \sqrt{2}})^2}$

$=\big( {m\omega\over \pi\hbar}\big)^{{1\over 4}}e^{-{q^2\over 2}+\sqrt{2}\xi q-{(|\xi|^2+\xi^2)\over 2}}, \qquad \xi=z e^{-i\omega t}$

$|\langle x|\psi(t)\rangle|^2=\big( {m\omega\over \pi\hbar}\big)^{{1\over 2}} e^{-|z|^2} e^{ q^2} e^{-(q-{z e^{-i\omega t}\over \sqrt{2}})^2-(q-{z^* e^{i\omega t}\over \sqrt{2}})^2}$

$=\big( {m\omega\over \pi\hbar}\big)^{{1\over 2}} e^{-(q-({ze^{-i\omega t}+z^*e^{i\omega t}\over \sqrt{2}}))^2}=\big( {m\omega\over \pi\hbar}\big)^{{1\over 2}} e^{-(q-({\xi+\xi^*\over \sqrt{2}}))^2}$

If $z$ is real

$|\langle x|\psi(t)\rangle|^2=\big( {m\omega\over \pi\hbar}\big)^{{1\over 2}} e^{-(q-\sqrt{2}z \cos(\omega t))^2}, \quad q=\sqrt{{m\omega\over \hbar}}x$

Inspection shows that the center of the wavepacket moves according to the classical equations of motion, and that the width of the packet does not change in time. There is no dispersion.
Some additional math fun, for later:

$\mbox{if}\quad \langle x|z\rangle=e^{A}, \qquad |\langle x|z\rangle|^2=e^{A+A^*}$

$\langle x|z\rangle=\sqrt{|\langle x|z\rangle|^2}e^{i\phi}=e^{(A+A^*)/2+i\phi}, \qquad \phi={A-A^*\over 2i}$

For us

$A=-{q^2\over 2}+\sqrt{2}\xi q-{(|\xi|^2+\xi^2)\over 2},$

$\phi=\sqrt{2}q{(\xi-\xi^*)\over 2i}-{(\xi^2-(\xi^*)^2)\over 4i}=-\sqrt{2}q\sin(\omega t)+{z^2\over 2}\sin(2\omega t), \quad \mbox{real}\quad z$

If we call $z=r e^{i\phi}$ then

$|\langle x|\psi(t)\rangle|^2=\big( {m\omega\over \pi\hbar}\big)^{{1\over 2}} \, \, e^{-(q-\sqrt{2} r \cos(\omega t-\phi))^2}$

There is closure for coherent states

$\int {dz dz^*\over 2\pi i} e^{-zz^*}\, |z\rangle \langle z|=1$

but they are not orthogonal

$\langle z_1|z_2\rangle=e^{-(|z_1|^2+|z_2|^2)/2}\sum_{n,m}{(z_1^*)^n\over \sqrt{n!}}{(z_2)^m\over \sqrt{m!}}\langle n|m\rangle$

$=e^{-(|z_1|^2+|z_2|^2)/2}e^{z_1^*z_2}$

Matrix elements
Write the coherent state as a normal-ordered form acting on the vacuum

$|z\rangle=e^{-zz^*/2} \sum_n {z^n\over \sqrt{n!}}|n\rangle=e^{-zz^*/2} e^{za^\dagger} e^{-z^*a}|0\rangle$

Recall that commutators obey the derivation rule, and since $1=[a,a^\dagger]$ is a $c$ -number

$[a,f(a^\dagger)]={\partial f\over \partial a^\dagger}[a,a^\dagger]={\partial f\over \partial a^\dagger}$

$a^\dagger,g(a)]={\partial g\over \partial a}[a^\dagger,a]=-{\partial g\over \partial a}$

$[a,f(a^\dagger)g(a)]=[a,f(a^\dagger)] g(a)+0={\partial f\over \partial a^\dagger}\, g(a)$

$[a^\dagger,f(a^\dagger)g(a)]=f(a^\dagger)[a^\dagger,g(a)]=-f(a^\dagger){\partial g\over \partial a}$

$[a^\dagger a, f(a^\dagger) g(a)]=a^\dagger [a,f(a^\dagger)g(a)]+[a^\dagger,f(a^\dagger)g(a)]a$

$=a^\dagger{\partial f\over \partial a^\dagger}\, g(a)-f(a^\dagger){\partial g\over \partial a}a$

Let $f=e^{za^\dagger}$ and $g=e^{-z^* a}$ ;

$[a^\dagger a,e^{za^\dagger}e^{-z^* a}]=za^\dagger \, e^{za^\dagger}e^{-z^* a}-ze^{za^\dagger}e^{-z^* a}a$

$=\Big(z{\partial\over \partial z}+z^*{\partial \over \partial z^*}\Big)e^{za^\dagger}e^{-z^* a}$

Superposition and density matrix
Next we consider a superposition of such states and build a density matrix, which we will ultimately reduce. Let

$|\psi\rangle=a_1|z_1\rangle+a_2|z_2\rangle$

so in the Heisenberg picture the state and its projector/density matrix is

$|\psi(t)\rangle=a_1|z_1e^{-i\omega t}\rangle+a_2|z_2e^{-i\omega t}\rangle, \quad \rho(t)$

$=|\psi(t)\rangle\langle\psi(t)|=\sum_{j,k} a_ja^*_k|z_je^{-i\omega t}\rangle\langle z_ke^{-i\omega t}|$

$+a^*_1a_2\langle x|\xi_1\rangle^*\langle x|\xi_2\rangle$

The cross term is the interference term that a process of decoherence would need to kill off. We can think of the first two terms and the accompanying probabilities $|a_i|^2$ as representing a classical limit because the packets that correspond to these states are as close as we can get to classically behaved objects. The interference term is

$a_1a^*_2\langle x|\xi_1\rangle\langle x|\xi_2\rangle^*+a^*_1a_2\langle x|\xi_1\rangle^*\langle x|\xi_2\rangle$

$=2 \, Re \, \Big(a_1a^*_2\langle x|\xi_1\rangle\langle x|\xi_2\rangle^*\Big)$

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Quantum decoherence I: coherent states

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