Two-Level Systems and Thermal Baths

Hello! Let's dive into the dynamics of open quantum systems.

In our last lesson, we developed the Lindblad master equation as the fundamental tool for describing how a system's density matrix evolves under environmental influence. We focused on the case of spontaneous emission, where a two-level system decays into a zero-temperature environment. This captured the essence of dissipation and decoherence.

Today, we will address a more general and physically ubiquitous scenario. Our goal is to analyze the dynamics of a two-level system coupled to a thermal bath. We'll extend our framework to include finite temperatures, which is essential for describing almost any realistic quantum system, from qubits in a quantum computer to electronic states in a solid. We will see how the environment can now not only absorb energy from the system but also impart thermal energy to it, leading to a new equilibrium.

1. From a Cold Vacuum to a Hot Bath

Recall that an open quantum system consists of the system of interest (S) coupled to an environment or bath (B).

In the previous lesson, the bath was the electromagnetic vacuum at zero temperature. It was a "cold" environment, capable only of accepting energy from the system, causing the excited state to decay. A thermal bath at a finite temperature $T>0$ is different: it contains thermal excitations (e.g., thermal photons or phonons). This has a profound consequence: the bath can now also give energy to the system, causing it to be excited from a lower energy state to a higher one. This process is called stimulated absorption.

Our task is to incorporate this new physical process into the master equation formalism.

2. The Master Equation in a Thermal Bath

Let's start with the key result. The derivation follows the same steps as in the zero-temperature case (Born-Markov-RWA), but the bath correlations are now calculated for a thermal state instead of the vacuum.

For this, we'll turn to the excellent notes "Quantum Physics of Light-Matter Interactions" which we'll use as our main theoretical guide for this topic.

As you've just read, the master equation for a two-level system with transition frequency ${\omega }_0$ coupled to a thermal bath at temperature $T$ is:

$$\frac{d{\rho }_A}{dt}=-i[H_A,{\rho }_A]+\gamma (\bar{n}+1)\mathcal{D}[{\sigma }^{-}]{\rho }_A+\gamma \bar{n}\mathcal{D}[{\sigma }^{†}]{\rho }_A$$

Let's dissect this. We have two dissipative processes, each with its own Lindblad operator:

1. Emission (Decay): The operator is $L_1={\sigma }^{-}=|g⟩⟨e|$, representing the transition from the excited state $|e⟩$ to the ground state $|g⟩$. The rate is $\gamma (\bar{n}+1)$.

   • The $\gamma \cdot 1$ term represents spontaneous emission, which happens even at $T=0$.
   • The $\gamma \cdot \bar{n}$ term represents stimulated emission, where the presence of thermal excitations in the bath enhances the decay process.

2. Absorption (Excitation): The operator is $L_2={\sigma }^{†}=|e⟩⟨g|$, representing the transition from ground to excited. The rate is $\gamma \bar{n}$.

   • This process is purely thermal. If $T=0$, then $\bar{n}=0$ and this term vanishes, recovering our previous result.

The quantity $\bar{n}\equiv \bar{n}({\omega }_0)$ is the Bose-Einstein distribution:

$$\bar{n}({\omega }_0)=\frac{1}{e^{\hslash {\omega }_0/k_{B}T}-1}$$

It represents the average number of thermal bosons (e.g., photons) in the bath at the system's transition frequency ${\omega }_0$. This is the crucial factor that brings temperature into our quantum dynamics.

3. The Spin-Boson Model: A Canonical Condensed Matter Example

The abstract "two-level system" and "bath" find a concrete home in the spin-boson model, a cornerstone of condensed matter physics. This model describes a single spin-1/2 particle (our TLS) coupled to a bath of harmonic oscillators (the bosons). This is an excellent model for a huge variety of physical systems: a single-molecule magnet, a qubit interacting with phononic vibrations in a crystal, or an atom coupled to the electromagnetic field.

The spin-boson Hamiltonian reads:

$$H=\underset{\text{System }{H}_S}{\underbrace{-\frac{\Delta }{2}{\sigma }_x+\frac{h}{2}{\sigma }_z}}+\underset{\text{Bath }{H}_B}{\underbrace{\sum _n\hslash {\omega }_n{a}_n^{†}{a}_n}}+\underset{\text{Interaction }{H}_{int}}{\underbrace{\frac{1}{2}{\sigma }_z\sum _n{\lambda }_n(a_n+a_n^{†})}}$$

• The system part describes a spin in a magnetic field, where $\Delta$ is a transverse field causing tunneling between spin-up and spin-down, and $h$ is a longitudinal field creating an energy bias.
• The bath is a collection of harmonic oscillators (phonons, photons, etc.).
• The interaction term shows that the spin's $z$-component couples to the displacement of the bath oscillators.

The environment's properties and its coupling to the system are completely captured by the spectral function:

$$J(\omega )=\pi \sum _n{\lambda }_n^2\delta ({\omega }_n-\omega )$$

This function tells us the density of bath modes at frequency $\omega$ weighted by their squared coupling strength. For many systems, it can be approximated by a power law, $J(\omega )\propto {\omega }^s$. The case $s=1$, known as an Ohmic bath, is particularly important as it describes dissipation in many electronic circuits and other condensed matter systems.

4. Dynamics: Relaxation ($T_1$) and Dephasing ($T_2$)

With the master equation established, let's analyze the dynamics it predicts. This is most easily done by writing out the equations of motion for the elements of the density matrix, known as the Bloch equations.

By working through the algebra from our thermal master equation (or adapting the results from the reading by setting $\Omega =0$ and including the absorption term), we get the following equations for the populations and coherences:

Let ${\rho }_{ee}$ be the excited state population and ${\rho }_{eg}$ be the coherence. The rates for upward and downward transitions are $R_{\uparrow }=\gamma \bar{n}$ and $R_{\downarrow }=\gamma (\bar{n}+1)$, respectively.

• Populations (e.g., $⟨{\sigma }_z⟩$):

  $${\dot{\rho }}_{ee}=-R_{\downarrow }{\rho }_{ee}+R_{\uparrow }{\rho }_{gg}=-R_{\downarrow }{\rho }_{ee}+R_{\uparrow }(1-{\rho }_{ee})=-(R_{\downarrow }+R_{\uparrow }){\rho }_{ee}+R_{\uparrow }$$

  This is a first-order ODE describing the relaxation of the population difference. The rate of this relaxation is:

  $$\frac{1}{T_1}=R_{\downarrow }+R_{\uparrow }=\gamma (2\bar{n}+1)$$

  This is the longitudinal relaxation rate, or spin-lattice relaxation rate. $T_1$ is the characteristic time for the system's energy to equilibrate with the bath.

• Coherences ($⟨{\sigma }_x⟩,⟨{\sigma }_y⟩$):

  $${\dot{\rho }}_{eg}=-i{\omega }_0{\rho }_{eg}-\frac{1}{2}(R_{\downarrow }+R_{\uparrow }){\rho }_{eg}=(-i{\omega }_0-\frac{1}{2T_1}){\rho }_{eg}$$

  The solution is ${\rho }_{eg}(t)={\rho }_{eg}(0)e^{-i{\omega }_{0}t}{e}^{-t/T_2}$. The coherence decays at a rate:

  $$\frac{1}{T_2}=\frac{1}{2}(R_{\downarrow }+R_{\uparrow })=\frac{1}{2T_1}$$

  This is the transverse relaxation rate, or dephasing rate. $T_2$ is the characteristic time for the loss of quantum phase information. In this specific model, the relation $T_2=2T_1$ holds. However, if our spin-boson model had interaction via ${\sigma }_z$, it would introduce an additional "pure dephasing" channel ($1/T_{\varphi }$), leading to the more general relation $1/T_2=1/(2T_1)+1/T_{\varphi }$.

Test your understanding!

In the high-temperature limit, $k_{B}T\gg \hslash {\omega }_0$.

1. Show that the Bose-Einstein factor $\bar{n}({\omega }_0)$ can be approximated as $\bar{n}\approx \frac{k_{B}T}{\hslash {\omega }_0}$.
2. Using this approximation, how does the relaxation rate $1/T_1$ depend on temperature?

Show answer

1. For $x=\hslash {\omega }_0/k_{B}T\ll 1$, we can Taylor expand the exponential: $e^x\approx 1+x$.

   $$\bar{n}=\frac{1}{e^{\hslash {\omega }_0/k_{B}T}-1}\approx \frac{1}{(1+\hslash {\omega }_0/k_{B}T)-1}=\frac{k_{B}T}{\hslash {\omega }_0}$$

2. The relaxation rate is $1/T_1=\gamma (2\bar{n}+1)$. Substituting the high-T approximation for $\bar{n}$:

   $$\frac{1}{T_1}\approx \gamma \left(2\frac{k_{B}T}{\hslash {\omega }_0}+1\right)$$

   Since $k_{B}T/\hslash {\omega }_0\gg 1$, the +1 term is negligible. Therefore:

   $$\frac{1}{T_1}\approx \frac{2\gamma k_{B}T}{\hslash {\omega }_0}$$

   In the high-temperature limit, the relaxation rate is directly proportional to the temperature. This makes physical sense: a hotter bath has more thermal energy, leading to more frequent thermal fluctuations that drive the system's relaxation.

5. Steady State and Thermalization

What happens after a long time ($t\to \infty$)? The system will reach a steady state where $\dot{\rho }=0$. For the population equation, this means:

$${\dot{\rho }}_{ee}=-(R_{\downarrow }+R_{\uparrow }){\rho }_{ee}^{ss}+R_{\uparrow }=0$$ $$⟹{\rho }_{ee}^{ss}=\frac{R_{\uparrow }}{R_{\downarrow }+R_{\uparrow }}$$

The ground state population will be ${\rho }_{gg}^{ss}=1-{\rho }_{ee}^{ss}=\frac{R_{\downarrow }}{R_{\downarrow }+R_{\uparrow }}$.

Let's look at the ratio of the populations in the steady state:

$$\frac{{\rho }_{ee}^{ss}}{{\rho }_{gg}^{ss}}=\frac{R_{\uparrow }}{R_{\downarrow }}=\frac{\gamma \bar{n}}{\gamma (\bar{n}+1)}=\frac{\bar{n}}{\bar{n}+1}$$

Substituting the definitions of $\bar{n}$ and $\bar{n}+1$, we find:

$$\frac{{\rho }_{ee}^{ss}}{{\rho }_{gg}^{ss}}=e^{-\hslash {\omega }_0/k_{B}T}$$

This is exactly the Boltzmann distribution for a two-level system! The steady state of the Lindblad master equation for a thermal bath is the thermal Gibbs state of the system, ${\rho }_{ss}=\frac{1}{Z}{e}^{-H_S/k_{B}T}$. This confirms that our model correctly describes the process of a quantum system thermalizing with its environment.

Conclusion

Today, we have taken a significant step towards describing realistic quantum systems by incorporating the effects of finite temperature.

Key Takeaways:

• Coupling to a thermal bath introduces a new process: stimulated absorption, where the system gains energy from the bath.
• The master equation now includes two Lindblad operators: ${\sigma }^{-}$ for emission (rate $\gamma (\bar{n}+1)$) and ${\sigma }^{†}$ for absorption (rate $\gamma \bar{n}$). The rates are governed by the thermal occupation $\bar{n}({\omega }_0)$.
• This dynamics leads to longitudinal relaxation ($T_1$) of populations and transverse relaxation ($T_2$) of coherences, with rates that depend on temperature.
• The spin-boson model is the canonical condensed matter realization of a TLS coupled to a bath of oscillators, characterized by a spectral function $J(\omega )$.
• As $t\to \infty$, the system reaches a thermal steady state described by the Boltzmann distribution, demonstrating that the Lindblad formalism correctly captures quantum thermalization.

Preview of the Next Lesson:
So far in this module, we have focused on how quantum states are represented (density matrix) and how they evolve in open systems (Lindblad equation). The final piece of the puzzle is measurement. In the next lesson, we will introduce Positive-Operator Valued Measures (POVMs) as a generalized framework for quantum measurements, moving beyond the textbook projective measurements to describe the most general types of measurements allowed by quantum mechanics.