Theoretical Example of Decoherence

We now note that the conditions we have thus far considered are highly artificial. What if we consider an interaction with the environment? A simple model for an environmental Hamiltonian might be a set of harmonic oscillators,

$$H_{E}=\sum_j\hbar \omega_j b_j^{\dagger}b_j$$

As developed elsewhere, we can write the interaction Hamiltonian as

$$H_I = \sum_j g_j(a^{\dagger}b_j + ab^{\dagger}_j)$$

A reasonable way to look at this is to see that if the field gains a photon ( $a^{\dagger}$ ) then the single oscillator should loose one $b_j$ and vice versa; the prefactor $g_j$ is a coupling constant that will generally depend on the specifics of the system. The system is now governed by the total Hamiltonian,

$$H=\overbrace{\hbar \omega a^{\dagger}a + \sum_j \hbar \omega_j b_... ...ger}b_j}^{H_0} + \underbrace{\sum_j g_j(a^{\dagger}b_j + ab_j^{\dagger})}_{H_I}$$

Let us call the new density operator corresponding to our environmentally coupled system $\rho_{SE}$ (S: Single original oscillator; E: Environment) where the individual density operators can be retrieved via a trace, i.e.

$$\rho_S = Tr_{E}(\rho_{SE}) = \sum_E\langle E\vert\rho_{SE}\vert E\rangle \$$

$$\rho_E = Tr_{S}(\rho_{SE})=\sum_S\langle S \vert \rho_{SE}\vert S\rangle \ \ \$$

The dynamics of the system evolve as (Liouville equation),

$$i\hbar\frac{d\rho_{SE}}{dt}=[H,\rho_{SE}]$$ (4.1)

We commit a unitary transform to simplify this equation as follows (the so-called interaction picture). Let $U_o=\exp(\frac{i}{\hbar}H_0t)$ , then

$$\rho_{se}=U_o\rho_{SE}U^{\dagger}_o$$

We can then easily find that (since $[H_0, H_0]=0$ , letting $U_oH_IU^{\dagger}_o\equiv H_i$ ),

$$\frac{d\rho_{se}}{dt} = \frac{i}{\hbar}[H_0-U_oHU^{\dagger}_o,\rho_{se}]$$

$$= -\frac{i}{\hbar}[H_i, \rho_{se}]$$ (4.2)

With $n=a^{\dagger}a$ , it is demonstrable that,

$$\exp(\eta n) a \exp(-\eta n) = \exp(-\eta)a$$

$$\exp(\eta n) a^{\dagger} \exp(-\eta n) = \exp(\eta)a^{\dagger}$$ (4.3)

And so we have,

$$H_i = \sum_j \hbar g_j \left(a^{\dagger}b_j \exp(i(\omega-\omega_j)t) + ab_j^{\dagger}\exp(-i(\omega-\omega_j)t)\right)$$ (4.4)

We now follow Orszag and, avoiding some complicated calculations, present a path to the so called master equation[8]. Equation 4.2 is integrated and then reapplied to yield,

$$\frac{d\rho_{se}}{dt}=\frac{1}{i\hbar}[H_i,\rho_{se}(0)]-\frac{1}{\hbar^2}\int^t_0[H_i(t'),[H_i(t'),\rho_{se}(t')]]dt'$$ (4.5)

It is typically not a bad approximation to assume that the environment (bath) is large enough to be unperturbed by the local system, i.e. we assume that (Markovian assumption):

$$\rho_{se}(t)=\rho_{s}(t)\otimes \rho_{e}(0)$$

It follows that

$$Tr_B[H_i(t),\rho_{se}(0)]=0$$

Taking the trace of the Equation 4.5 gives,

$$\frac{d\rho_{s}}{dt}=-i \Delta \omega [a^{\dagger}a, \rho_s(t)] +... ...]+A[a\rho_s(t),a^{\dagger}]+B[a^{\dagger},\rho_s(t)a]+B[a^{\dagger}\rho_s(t),a]$$ (4.6)

Here,

$$\sum_j \rightarrow \int D(\omega_j)d\omega_j$$

$$\Delta \omega = P \int^{\infty}_{0}\frac{g(\omega_j)^2D(\omega_j)}{\omega-\omega_j}d\omega_j$$

$$Tr_e(b_j^{\dagger}b_k\rho_e(0)) = \delta_{jk}\langle n_j\rangle$$

$$Tr_e(b_jb_j\rho_e(0)) =$$ 0

$$A = \pi g(\omega)^2 D(\omega)(1+\langle n(\omega)\rangle)$$

$$B = \pi g(\omega)^2 D(\omega)\langle n(\omega)\rangle$$

Let $\gamma= 2(A-B) = 2\pi g(\omega)^2D(\omega)$ . The master equation becomes

$$\frac{d\rho_s}{dt}=-i \Delta \omega [a^{\dagger}a, \rho_s(t)] -\f... ...}a\rho_s(t)-2a\rho_s(t)a^{\dagger}) - \frac{\gamma}{2}\langle n(\omega)\rangle( )$$ (4.7)

We simplify things even further by considering only the simplest of baths, where $T\approx0$ thus $\langle n(\omega)\rangle \approx 0 \approx \Delta \omega$ , and the master equation is simply

$$\frac{d\rho_s}{dt}=-\frac{\gamma}{2}(\rho_s(t)a^{\dagger}a + a^{\dagger}a\rho_s(t) - 2 a \rho_s(t)a^{\dagger})$$ (4.8)

Now we are ready to examine the evolution of our simple model. We recall that it was $\chi$ which Decoherence need destroy to produce the classical state of $\rho_s$ . We now apply the master equation to this system and see what happens to these off-diagonal terms.

Conventionally, we start with the so-called normally ordered characteristic function,

$$X_N(\eta,t)=Tr(\rho_s(t)\exp(\eta a^{\dagger})\exp(-\eta*a)$$

We take the time derivative of this function, and using the property

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

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

$$[a^{\dagger}a,\exp(\eta a^{\dagger})\exp(-\eta^*a)]=\eta a^{\dagg... ...(\eta a^{\dagger})\exp(-\eta^*a)+\eta^*\exp(\eta a^{\dagger}) \exp(-\eta^* a)a$$

producing [8,9]

$$\frac{\partial X_N}{\partial t} = Tr\left(\frac{d\rho}{dt}\exp(\eta a^{\dagger})\exp(-\eta^*a)\right)$$ (4.9)

$$= -\frac{\gamma}{2}\left(\eta \frac{\partial X_N(\eta,t)}{\partial \eta} + \eta^* \frac{\partial X_N(\eta,t)}{\partial \eta*}\right)$$ (4.10)

$$= \frac{\partial X_N(\eta,t)}{\partial \eta(t)}\frac{\partial \eta(... ...{\partial X_N(\eta,t)}{\partial \eta^*(t)}\frac{\partial \eta^*(t)}{\partial t}$$ (4.11)

The solution to this equation is of the form

$$X_N(\eta,t)=X_N\left(\eta \exp\left(\frac{-\gamma t}{2}\right),0\right)$$

Where $\eta \rightarrow \eta(t) =\eta \exp\left(\frac{-\gamma t}{2}\right)$ . Finding our initial conditions as

$$X_N(\eta,0) = Tr(\rho(0)\exp(\eta a^{\dagger})\exp(-\eta^*a))$$

$$= \sum_{i,j}\left(\langle \alpha_j\vert \exp(\eta a^{\dagger})\exp(-\eta^*a)\vert\alpha_i\rangle\right)$$

$$= \sum_{i,j}\langle \alpha_j \vert \alpha_i \rangle \exp(\eta \alpha_j^* - \eta^*\alpha_i)$$ (4.12)

It then follows that,

$$X_N(\eta,0)=\sum_{i,j}\langle \alpha_j \vert \alpha_i \rangle \exp(\eta \alpha_j^* - \eta^*\alpha_i)\exp(-\frac{\gamma t}{2})$$ (4.13)

Now it is time to note that since

$$\vert\alpha \rangle = \exp(-\vert\alpha\vert^2/2)\sum^{\infty}_{n=0} \frac{\alpha^n}{\sqrt{n!}}\vert n\rangle$$

And thus,

$$\langle \alpha_1 \vert \alpha_2 \rangle = \exp(-(\vert\alpha_1\ve... ...t\alpha_2\vert^2)/2)\sum^{\infty}_{n=0} \frac{\alpha_1^n \alpha_2^n}{\sqrt{n!}}$$

and $\alpha \ \rightarrow \alpha \exp(\gamma t/2)$ , the density matrix becomes,

$$\rho_{i,j} = \langle \alpha_i \vert \alpha _ j \rangle ^{1-\exp(-... ...}\vert\alpha_j \exp(-\gamma t/2)\rangle \langle \alpha_i \exp(-\gamma t/2)\vert$$ (4.14)

When $\gamma t \ll 1$ we can approximate the prefactor on the cross terms by $\exp(-2\vert\alpha\vert^2\gamma t)$ . It is conventional to define $t_c=1/2 \gamma \vert\alpha\vert^2$ so that, along with the condition that $\alpha \rightarrow \alpha \exp(-\gamma t/2)$ , our interference term becomes,

$$\langle q'\vert\rho(t)\vert q'\rangle = I_1^2(q') + I_2^2(q') + 2I_1(q')I_2(q') \cos \theta(t)\exp(-t/t_c)$$ (4.15)

In terms of the density matrix,

$$\rho_s = \int^{\infty}_{-\infty}\left(\begin{array}{cc}I_1^2&I_1I_2\cos \theta(t)\\ I_2I_1 \cos \theta(t)&I_2^2\end{array}\right)dq'$$

$$\overbrace{\longrightarrow}^{\mbox{Decoherence}} \int^{\infty}_{-\infty}\left(\begin{array}{cc}I_1^2&I_1I_2\cos \theta(t)\exp(-t/t_c)\\ I_2I_1 \cos \theta(t)\exp(-t/t_c)&I_2^2\end{array}\right)dq'$$

$$\longrightarrow \int^{\infty}_{-\infty}\left(\begin{array}{cc}I_1^2&0\\ 0&I_2^2\end{array}\right)dq'$$

$$\overbrace{\longrightarrow}^{\mbox{Measurement Problem}} \left(\begin{array}{cc}0&0\\ 0&1\end{array}\right) \ \ \left(\begin{array}{cc}1&0\\ 0&0\end{array}\right)$$

The diagonal states will decay into the ground state as they reach equilibrium with the bath. However, this decay will generally be much slower than the decoherence. Finally we must note that decoherence has brought us into a classical density matrix, but the question still remains about which state we will find the particle in when an actual measurement is taken.

We summarize that decoherence is that part of dampening, caused by a coupling to an environment, which produces a decay (a stronger decay) of off-diagonal elements of a density matrix. Thus, in short time, the local quantum density matrix assumes a classical appearance. Since Decoherence is much quicker than normal dissipation, it has become a major engineering problem in the search for a quantum computer.