# What information is contained in the quantum spectral density?

The spectral density, or spectral function, describes the coupling between a small quantum system that is coupled to a larger environment. In many cases, this environment can be modelled effectively as a system of free bosonic or fermionic modes, with Hamiltonian (working in units with $\hbar = 1$) $$ H_B = \sum_k \omega_k b_k^{\dagger}b_k. $$ The mode operators satisfy $[b_k,b_l^{\dagger}] = \delta_{k,l}$ or $\{b_k,b_l^{\dagger}\} = \delta_{k,l}$ for bosonic and fermionic modes respectively.

The small quantum system (hereafter referred to simply as the *system*) is described by an autonomous Hamiltonian $H_A$, which we leave unspecified. In many cases, the system-environment coupling takes the form $H_{AB} = AB,$ where the operator $A$ acts on the system only, while the environment noise operator is *linear*:
$$ B = \sum_k \left( g_k b_k + g_k^{\ast} b_k^{\dagger}\right ).$$
This type of Hamiltonian (or its close relatives) can successfully model atoms coupled to the radiation field, electron-phonon coupling in solids, the coupling between a mesoscopic quantum conductor and macroscopic electrical leads or a superconducting qubit coupled to its electromagnetic environment, to name just a handful of examples.

In this setting, one finds in practice that all the effects of the environment are encapsulated by a single quantity, the **spectral density**, defined as
$$ J(\omega) = 2\pi\sum_k \lvert g_k\rvert^2 \delta(\omega-\omega_k).$$
This is the coupling strength weighted by the density of states of the environment. It describes how easy it is to exchange a quantum of energy $\omega$ with the environment.

In the simplest cases, the dissipation can be modelled by a Markovian master equation (Lindblad equation). I am not going to give tedious details of the derivation here, for more information see Breuer & Petruccione. In the master equation description, the effect of the environment is to cause incoherent transitions between energy eigenstates of $H_A$. The transition rate for the process that *increases* the energy of the system by an amount $\epsilon$ (which may be positive or negative) is found to be
\begin{align*}
\gamma(\epsilon) & = \int_{-\infty}^{\infty}\mathrm{d}t\; e^{-\mathrm{i}\epsilon t} \langle B(t) B(0)\rangle, \\
& = \int_0^{\infty}\mathrm{d}\omega\; J(\omega)\left [ n(\omega) \delta(\omega-\epsilon) + (1\pm n(\omega)) \delta(\omega+\epsilon) \right ],
\end{align*}
where, in the factor $(1\pm n(\omega)),$ the plus sign is for bosons and the minus sign for fermions. Here, $\langle \bullet\rangle$ indicates a thermal average over the environment variables, $B(t)$ indicates the Heisenberg picture evolution under $H_B$, while $n(\omega)$ indicates the thermal occupation number of a mode with energy $\omega$. Of course, one finds that $n(\omega) = (e^{\beta (\omega-\mu)} \mp 1)^{-1}$ is the Bose-Einstein (Fermi-Dirac) distribution at temperature $T = 1/k_B\beta$ and chemical potential $\mu$.

Now the similarity with the OP's expression should be clear (one can rewrite the $\coth$ terms into the form I have given). The dissipation is determined completely by the power spectrum of the quantum noise operator $B$. In evaluating this expression, one finds two terms, both proportional to the spectral density. The first term describes the probability of absorption of energy from the environment, which is only possible if a mode of energy $\epsilon$ is occupied. The second term describes emission of energy into the environment, which comes with an additional $+1$ due to quantum non-commutativity. This $+1$ allows spontaneous emission even when the environment mode is empty (this may be colourfully attributed to quantum zero-point fluctuations). When the environment mode is occupied, we have either enhanced (stimulated) emission due to bosonic bunching, or suppressed emission due to the Pauli exclusion principle.

$\newcommand{\bra}[1]{\langle #1 |}$ $\newcommand{\ket}[1]{| #1 \rangle}$ $\newcommand{\braket}[2]{\langle #1 | #2 \rangle}$ $\newcommand{\bbraket}[3]{\langle #1 | #2 | #3 \rangle}$ Although the question asks specifically about a harmonic oscillator, we can understand the meaning of the spectral density by considering a somewhat more general problem. Consider a quantum system $S$ coupled to an environment $E$. Focus attention on a single pair of states of $S$ which we denote $\ket{0}$ and $\ket{1}$. Denote the energy difference between these levels as $E_{10} = E_1 - E_0$ and define the transition frequency by

$$\omega_{10} \equiv E_{10}/\hbar \, .$$

With this notation, the Hamiltonian of $S$ is just

$$H_S = - \frac{\hbar \omega_{10}}{2} \sigma_z \, .$$

Suppose $E$ couples to the $S$ via an interaction Hamiltonian given by

$$ V = g (F \otimes \sigma_x)$$

where here $F$ is an operator acting on $E$ and $\sigma_x$ is the Pauli $x$ operator on the two-level subspace of $S$ consisting of $\ket{0}$ and $\ket{1}$. We use perturbation theory to compute the affect of this interaction on $S$. There are several ways to think about perturbation theory. I prefer to use the rotating frame in which the system and environment self Hamiltonians go to zero and the interaction Hamiltonian becomes

$$V(t) = g ( F(t) \otimes (\cos(\omega_{10} t )\sigma_x + \sin(\omega_{10} t) \sigma_y ) )$$

In order to compute the evolution of $S$ we use first order perturbation theory. We start with Schrodinger's equation

$$ i \hbar (d/dt)\ket{\Psi(t)} = V(t) \ket{\Psi(t)}$$

which is formally solved by

$$ \ket{\Psi(t)} = \ket{\Psi(0)} - \frac{i}{\hbar} \int_0^t dt' \, V(t') \ket{\Psi(t')} \, . $$

Iterating this equation once gives

$$ \ket{\Psi(t)} = \ket{\Psi(0)} - \frac{i}{\hbar} \int_0^t dt' \, V(t') \ket{\Psi(0)} \, . $$

Suppose $S$ starts in the lower state $\ket{0}$ and $E$ starts in some particular state $\ket{n}$. In other words, $\ket{\Psi(0)} = \ket{n} \otimes \ket{0}$. Let us compute the amplitude for $S$ to make a transition to $\ket{1}$: \begin{align} \braket{1}{\Psi(t)} = &= - \frac{i}{\hbar} \int_0^t \bbraket{1}{\tilde{V}(t')}{\Psi(0)} \, dt' \\ &= \frac{-ig}{\hbar} \int_0^t F(t')\ket{n} \otimes \bbraket{1}{e^{i \omega_{10} t'} \sigma_+ + e^{-i \omega_{10} t'} \sigma_-}{0} \, dt' \\ &= \frac{-ig}{\hbar} \int_0^t F(t')\ket{n} e^{i \omega_{10} t'} \, dt' \, . \qquad (1) \end{align}

Note that the expression on the left hand side is a state vector for the environment.
This is because we have only taken the inner product with respect to a final state of $S$, but have not specified a final state of $E$.
Note also that our initial state $\ket{\Psi(0)}$ had the environment in a specific state $\ket{n}$.
Let's consider a more interesting case where the environment is initially in a mixed state $\rho_E = \sum_n \rho_{nn} \ket{n}\bra{n}$.
To accommodate this into our analysis we form the density matrix for the environment state given in equation (1) and sum it over the initial states of the environment:
\begin{equation}
\rho(t) = \left(\frac{g}{\hbar}\right)^2 \sum_n \rho_{nn} \int_0^t \int_0^t \, dt' \, dt'' \, F(t')\ket{n}\bra{n}F(t'') e^{i \omega_{10} (t'-t'')} \, .
\end{equation}
Now, to finally get a probability $p_1(t)$ for the system to be in $\ket{1}$ *averaged over all final states of the environment*, we trace over the environmental states:
\begin{align}
p_1(t)
&= \left(\frac{g}{\hbar}\right)^2 \sum_{nm} \int_0^t \int_0^t \, dt' \, dt'' \, \rho_{nn} \bbraket{m}{F(t')}{n} \bbraket{n}{F(t'')}{m} e^{i \omega_{10} (t' - t'')} \\
&= \left(\frac{g}{\hbar}\right)^2 \sum_{nm} \int_0^t \int_0^t \, dt' \, dt'' \, \rho_{nn} \bbraket{n}{F(t'')}{m} \bbraket{m}{F(t')}{n} e^{i \omega_{10} (t' - t'')} \\
&= \left(\frac{g}{\hbar}\right)^2 \sum_n \int_0^t \int_0^t \, dt' \, dt'' \, \rho_{nn} \bbraket{n}{F(t'') F(t')}{n} e^{i \omega_{10} (t' - t'')} \\
&= \left(\frac{g}{\hbar}\right)^2 \int_0^t \int_0^t \, dt' \, dt'' \, \langle F(t'') F(t') \rangle e^{i \omega_{10} (t' - t'')} \, .
\end{align}
Changing variables to $\tau \equiv t'' - t'$ and assuming the average of $F$ is stationary gives
\begin{equation}
p_1(t)
= \left(\frac{g}{\hbar}\right)^2 \int_0^t \int_{-t'}^{t-t'} \, dt' \, d\tau \, \langle F(\tau) F(0) \rangle e^{-i \omega_{10} \tau} \, .
\end{equation}
If the correlation function of $F$ is very short then we can extend the limits of integration of $\tau$ to positive and negative infinity.
Doing this and defining the spectral density as
\begin{equation}
S_{FF}(\Omega) \equiv \int_{-\infty}^\infty \, dt \, \langle F(t) F(0) \rangle e^{i \Omega t}
\end{equation}
we find
\begin{equation}
p_1(t) = \left( \frac{g}{\hbar} \right)^2 \int_0^t S_{FF}(-\omega_{10})
\end{equation}
which we can interpret as an upward transition rate
\begin{equation}
\Gamma_{\uparrow} = \frac{dp_1}{dt} = \left(\frac{g}{\hbar}\right)^2 S_{FF}(-\Omega) \, .
\end{equation}
Similar arguments can be used to show
\begin{equation}
\Gamma_\downarrow = \left(\frac{g}{\hbar}\right)^2 S_{FF}(\Omega) \, .
\end{equation}

Therefore, we have shown that when a system is connected to a quantum environment via an operator $F$, the quantum spectral density $S_{FF}$ characterizes the upward and downward transition rates induced on the system by the environment. We note that positive frequencies in the spectral density give downward transitions in the system (absorption by the environment) while negative frequencies give upward transitions (emission by the environment).

**Bonus:**

For those readers interested in quantum electronics, the quantum spectral density of current for an arbitrary linear element with admittance $Y(\omega)$ which is in thermal equilibrium is

$$S_{II}(\omega) = \hbar \omega \Re Y(\omega) \left[ \coth \left( \frac{\beta \hbar \omega}{2}\right) +1 \right]$$ where $\beta \equiv 1 / k_b T$.