Construction of the Hamiltonian and Density Operators

We assume the system can be modeled with the Hamiltonian,

$$H_{s}=\hbar \omega a^{\dagger}a, \ \ \ [a,a^{\dagger}]=1$$ (2.1)

Separately we assume that there exists a bath which can be modeled with the Hamiltonian,

$$H_{b}=\sum_i \hbar \omega_i b_i^{\dagger}b_i, \ \ \ [b_i,b_k^{\dagger}]=\delta_{ik}$$ (2.2)

Finally, as stated before, the interaction of the bath and system is taken in the form of

$$\sum_i g_i\left(a^{\dagger}b_i + ab_i^{\dagger}\right), \ \ \ g_i \in \Re$$ (2.3)

The total Hamiltonian is,

$$H=\overbrace{\hbar \omega a^{\dagger}a + \sum_j \hbar \omega_j b_... ...ger}b_j}^{H_0} + \underbrace{\sum_i g_i(a^{\dagger}b_i + ab_i^{\dagger})}_{H_I}$$ (2.4)

We assume uncertainty in the preparation of states, so we switch to the density operator regime. We call the density operator corresponding to our bath coupled system $\rho_{SB}$ (S: Single original oscillator; B: Environment) where the individual density operators can be retrieved via a trace, i.e.

$$\rho_S = Tr_{B}(\rho_{SB}) = \sum_B\langle B\vert\rho_{SB}\vert B\rangle \$$

$$\rho_B = Tr_{S}(\rho_{SB})=\sum_S\langle S \vert \rho_{SB}\vert S\rangle \ \ \$$

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

$$i\hbar\frac{d\rho_{SE}}{dt}=[H,\rho_{SE}]$$ (2.5)

We commit a unitary transform to simplify this equation as follows (the so-called interaction picture). It can be shown that,

$$\exp(\alpha A)B\exp(-\alpha A)= B + \alpha[A,B] + \frac{\alpha^2}{2!}[A,[A,B]] + \cdots$$ (2.6)

Furthermore, from the premises of Quantum Mechanics, it is legitimate to commit transforms of the type,

$$G'=\Lambda G \Lambda^{\dagger}, \ \ \ \Lambda \in \$$ (2.7)

These so called similarity transformations preserve rank, determinant, trace, and eigenvalues. We let,

$$\rho_{se} = \exp\left(\frac{i}{\hbar}H_o t\right)\rho_{SE}\exp\left(-\frac{i}{\hbar}H_o t\right)$$ (2.8)

We take the time derivative and find,

$$\frac{d\rho_{se}}{dt}=\frac{i}{\hbar}[H_o,\rho_{se}]-\frac{i}{\hb... ...(\frac{i}{\hbar}H_o t\right)[H,\rho_{SE}]\exp\left(-\frac{i}{\hbar}H_o t\right)$$ (2.9)

From equation 2.6, since

$$\Lambda G H \Lambda^{\dagger} = \Lambda G \Lambda^{\dagger} \Lambda H \Lambda ^{\dagger}$$

and $[H_o,H_o] = 0$,

$$\exp\left(\frac{i}{\hbar}H_o t\right)H_o\exp\left(-\frac{i}{\hbar}H_o t\right)=H_o$$

$$\frac{d\rho_{se}}{dt}=-\frac{i}{\hbar}[H_i, \rho_{se}]$$ (2.10)

where $H_i$ is the transform of $H_I$ and can be calculated as follows. We know that,

$$[a^{\dagger}a,a^{\dagger}] = a^{\dagger}aa^{\dagger}-a^{\dagger}a^{\dagger}a=[a,a^{\dagger}]a^{\dagger}=a^{\dagger}$$

$$[a^{\dagger}a,a] = a^{\dagger}aa-aa^{\dagger}a=[a^{\dagger},a]a=-a$$

and whereas $H_o = \hbar \omega a^{\dagger}a$, equation 2.6 gives

$$\exp\left(\alpha a^{\dagger}a\right)a^{\dagger}\exp(-\alpha aa^{\dagger}) = \exp(\alpha)a^{\dagger},$$

$$\exp\left(\alpha a^{\dagger}a\right)a\exp(-\alpha aa^{\dagger}) = \exp(-\alpha)a, \ \ \ \alpha \in \Im$$

$H_i$ can be calculated in this fashion for both the $aa^{\dagger}$ and the multiple set $b_ib_i^{\dagger}$, where it is thus now trivial to find that,

$$H_i(t) = \sum_i \hbar g_i \left(a^{\dagger}b_j\exp(\left(i(\omega-\omega_i)t\right) + ab_i^{\dagger}\exp \left(-i(\omega-\omega_i)t\right)\right)$$ (2.11)

If we now define

$$G(t)\equiv \sum_i g_ib_i\exp\left(i(\omega-\omega_j)t\right)$$

we can simplify to

$$H_i(t) = \hbar\left(G(t)a^{\dagger} + G^{\dagger}(t)a\right)$$ (2.12)

Since the bath is in equilibrium, we can construct $\rho_e(0)$ easily. Let $Z_i=\exp \left(-\hbar \omega_i/kT\right)$. Then the probability that one mode (i) of the field is excited with n photons is

$$P_{i,n}=\frac{Z_i^n}{\sum_n Z_i^n}=(1-Z_i)Z_i^n=\left(1-\exp\left... ...{-\hbar \omega_i}{kT}\right)\right)\exp\left(\frac{-n\hbar \omega_i}{kT}\right)$$ (2.13)

The density matrix for this mode is then

$$\rho_{e,i} = \left(1-\exp\left(\frac{-\hbar \omega_i}{kT}\right)\right)\sum_n\exp\left(\frac{-n\hbar \omega_i}{kT}\right)\vert n\rangle \langle n \vert$$ (2.14)

$$= \sum_n\left(1-\exp\left(\frac{-\hbar \omega_i}{kT}\right)\right)\... ...\frac{-b_i^{\dagger}b_i\hbar \omega_i}{kT}\right)\vert n\rangle \langle n \vert$$

To simplify this further, we imagine a three states system in which we can represent $b^{\dagger}b\vert n\rangle = n\vert n\rangle$ as, for example letting $n=2$,

$$\left(\begin{array}{ccc}0&0&0\\ 0&1&0\\ 0&0&2\end{array}\right)\l... ...}0\\ 0\\ 1\end{array}\right)=2\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)$$

The above sum could then be written,

$$0\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\left[\begin{ar... ...y}{c}0\\ 0\\ 1\end{array}\right)\left[\begin{array}{ccc}0&0&1\end{array}\right]$$

We recover the matrix, and so we can write just as well,

$$\rho_{e,i}=\left(1-\exp\left(\frac{-\hbar \omega_i}{kT}\right)\right)\exp\left(\frac{-b_i^{\dagger}b_i\hbar \omega_i}{kT}\right)$$ (2.15)

By the rules of probability, the density operator for the entire bath becomes,

$$\rho_{e}=\prod_i \left(1-\exp\left(\frac{-\hbar \omega_i}{kT}\right)\right)\exp\left(\frac{-b_i^{\dagger}b_i\hbar \omega_i}{kT}\right)$$ (2.16)

By our assumptions, this is the state of the bath for all time, and so we will not write this as a function of time. The form $\rho_s(t)$ will take will be much more complicated, and it is the purpose of this report to derive the differential equation defining its time evolution.