Consider a harmonic oscillator system with Hamiltonian

$$\hat{H} = \frac{1}{2} A \hat{u}^2 + \frac{1}{2} B \hat{v}^2 \qquad [\hat{u}, \hat{v}]=i \gamma $$

where $A$, $B$, and $\gamma$ are all real. This system has resonance frequency $\omega_0 = \gamma \sqrt{A B}$. Suppose we are at nonzero temperature with $\beta \equiv 1/k_b T$. Denote the zero point fluctuations in $\hat{u}$ as

$$u_{\text{zpf}}^2 \equiv \langle 0 | \hat{u}^2 | 0 \rangle = \frac{\gamma}{2} \left( \frac{B}{A} \right)^{1/2}\,.$$

Using the usual Heisenberg time dependence $a(t) = a(0)e^{-i \omega_0 t}$ and the Planck distribution

$$\langle \hat{n} \rangle = \frac{1}{\exp \left( \beta \hbar \omega_0 \right) - 1}$$

one computes the correlation function for $\hat{u}$ as

$$ \langle \hat{u}(t) \hat{u}(0)\rangle = u_{\text{zpf}}^2 \left[ \coth(\hbar \omega_0 \beta / 2)\cos(\omega_0 t) - i \sin(\omega_0 t) \right] \, . $$

One can then define a spectral density as the Fourier transform of the time correlation function

$$ \begin{align} S_{uu}(\omega) &\equiv \int dt \langle \hat{u}(t)\hat{u}(0)\rangle e^{i \omega t} \\ &= \frac{u_{\text{zpf}}^2}{2} \left\{ \delta(\omega + \omega_0) \left[ \coth(\hbar \omega_0 \beta / 2) - 1 \right] - \delta(\omega - \omega_0) \left[ \coth(-\hbar \omega_0 \beta / 2) - 1 \right] \right\} \, . \end{align} $$

What does $S_{uu}(\omega)$ mean? In other words, what information about the oscillator does $S_{uu}(\omega)$ tell me?

Some notes:

I understand the spectral density of a random process in classical physics. In the simplest sense it is the amount of power per unit frequency range in the process. However, the quantum version is somewhat different. Unlike the usual classical spectral density it is different at positive and negative frequencies (because the quantum correlation function is complex). I have read that the quantum spectral density is related to emission and absorption rates into and out of a thermal bath. In particular, the negative frequency part of the spectral density supposedly corresponds to emission of a quantum of energy, whereas the positive frequency part corresponds to absorption. However, I have never seen a proof of this idea or an example problem in which it can be seen that those precesses are described by the spectral density. A good answer to this question could focus on that relationship, possibly showing the connection between the two delta functions in the $S_{uu}(\omega)$ computed in this example and their corresponding emission and absorption processes.


Michel Devoret's Les Houches notes on quantum fluctuations in electrical systems

Paper by J. Martinis which partially rehashes the Devoret notes and uses the spectral density to compute decay rates, but doesn't explain why that works

Fairly comprehensive notes by Ingold

Similar question with answer unsatisfactory for my question

will add more as they are found and/or suggested

  • $\begingroup$ The term "spectral density" is usually applied to describe the properties of the coupling between a simple quantum system and an infinite reservoir. In that case the Fourier transform of auto-correlation functions of certain bath variables indeed contains information about emission and absorption of energy. That is obviously not the case here, since you are describing an isolated system: there is no bath. However, if you consider the oscillator as belonging to the reservoir then what you are asking will start to make sense to me. Does that sound like what you are asking for? $\endgroup$ Jan 5, 2015 at 0:35
  • $\begingroup$ @MarkMitchison: That is indeed what I have in mind. I didn't say so in the question (although I did note that the oscillator might be in a thermal state) because I didn't think that having a thermal state was necessary for the spectral density to be useful. I actually tried doing a from-first-principles calculation of the transition rates for a two level system coupled to a thermal reservoir, but the answer didn't make sense and I couldn't see a way to work the spectral density in. If you know about these topics I look forward to hearing more :) $\endgroup$
    – DanielSank
    Jan 5, 2015 at 1:31
  • $\begingroup$ You are right: the spectral density (as normally defined) is independent of the state. I will try to write up an answer later with some more detail, using the standard example of a qubit under thermal dissipation. $\endgroup$ Jan 5, 2015 at 13:01

3 Answers 3


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.

  • $\begingroup$ how come the + and - are related to spin statistics? $\endgroup$
    – Phoenix87
    Jan 5, 2015 at 15:43
  • $\begingroup$ @Phoenix87 I'm not sure I understand your question. I gave a physical explanation for the $\pm$ sign at the end of the last paragraph: it represents the effect of the Bose or Fermi statistics on the emission probabilities. Mathematically, the different signs arise due to whether the bath mode operators satisfy canonical anti-commutation or commutation relations. Does that answer your question? $\endgroup$ Jan 5, 2015 at 15:56
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    $\begingroup$ @Phoenix87 Ah, I understand the confusion. The $\pm$ sign relevant for statistics appears in the factor $(1\pm n(\omega))$. There are also plus and minus signs appearing in the delta functions: these correspond to emission and absorption of energy (i.e. whether $\epsilon$ is positive or negative) and are the same for bosons and fermions. I have edited to try and clarify this point: please read and let me know if it makes sense. $\endgroup$ Jan 5, 2015 at 16:06
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    $\begingroup$ @MarkMitchison: I didn't realize it was that complicated. Shouldn't it be possible to find the decay rate using the same kind of arguments used to develop Fermi's Golden rule? Write down the equation of motion for the density matrix, expand to first order, trace out the bath, and presto! (But it didn't work for me) $\endgroup$
    – DanielSank
    Jan 5, 2015 at 17:40
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    $\begingroup$ This is a great answer. The one thing which would improve it is to explicitly show why the transition rate is given by the integral shown in the answer. I worked this out today on the blackboard and I think it's simple enough to add here so I may do it in the future. $\endgroup$
    – DanielSank
    Jan 6, 2015 at 7:15

$\newcommand{\bra}[1]{\langle #1 |}$ $\newcommand{\ket}[1]{| #1 \rangle}$ $\newcommand{\braket}[2]{\langle #1 | #2 \rangle}$ $\newcommand{\bbraket}[3]{\langle #1 | #2 | #3 \rangle}$ 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).


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 $\beta \equiv 1 / k_b T$.

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    $\begingroup$ Nicely done. A brief comment for any future readers: taking the $\tau$ integral limits to $\infty$ is called a Markovian approximation, it is valid when the spectral density varies slowly with frequency and over a large range (bandwidth). Also, this definition of the quantum spectral density is related to the one in my answer (which is standard in quantum optics) by $$J(\omega) = \left (\frac{g}{\hbar}\right)^2 \left[ S_{FF}(\omega) \mp S_{FF}(-\omega)\right], $$ where the $-$ ($+$) sign is for bosons (fermions), assuming thermal equilibrium for the bath. $\endgroup$ Jan 29, 2015 at 11:31

I think that, as any other Fourier transform, this is telling you which natural frequencies occur in the correlation function, and with which "weight". In the case of a harmonic oscillator you have only one characteristic frequency, namely $\omega_0^2$ (I believe the sign is just the possibility of oscillating with $\omega_0$ or in the opposite orientation $-\omega_0$), therefore the only frequency component you get are the ones at $\pm\omega_0$. My guess is that perhaps you can describe the effects of anharmonic terms by replacing the deltas with some other frequency density distributions, i.e. some functions in $\omega$ from, e.g. the Schwartz space on $\mathbb R$.

  • $\begingroup$ It's not clear what you mean by "opposite orientation". $\endgroup$
    – DanielSank
    Dec 31, 2014 at 0:23
  • $\begingroup$ Assume $\omega_0>0$ and that this corresponds to an anticlockwise rotation. Then $-\omega_0$ is a clockwise rotation. $\endgroup$
    – Phoenix87
    Dec 31, 2014 at 0:27
  • $\begingroup$ "Rotation" of what? This is a 1D harmonic oscillator. I'm fairly certain the frequency sign has something to do with emission versus absorption. In that case the "rotation" can be thought of as the trajectory of the oscillator in phase space relative to another energy scale. $\endgroup$
    – DanielSank
    Dec 31, 2014 at 0:31
  • $\begingroup$ Oh that's right! Perhaps the correct interpretation is emission/absorption of a quantum of $\omega_0$? $\endgroup$
    – Phoenix87
    Dec 31, 2014 at 0:37
  • $\begingroup$ I'm pretty sure that's exactly it. I should have made that clear in the question. What I would like is an example showing that this is the case. $\endgroup$
    – DanielSank
    Dec 31, 2014 at 0:43

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