Density Matrices and the von Neumann Equation

Hello! Welcome to the fifth module of our course, "Open Quantum Systems and Measurement." In all our discussions so far, including symmetries, angular momentum, and scattering, we've implicitly assumed that we are dealing with isolated systems. For such systems, if we have the maximum possible information, their state can be completely described by a state vector $|\psi ⟩$, which we call a pure state.

However, in many realistic scenarios, especially in condensed matter physics, systems are rarely isolated. They might be in thermal contact with a heat bath, or we might only be interested in a small part of a larger, more complex system. In these cases, we have incomplete information, and the system is in a mixed state. The wavefunction formalism is insufficient for this.

This lesson introduces the density matrix (or density operator), $\hat{\rho }$, a powerful mathematical tool that provides a unified framework to describe both pure and mixed states. It's the cornerstone for understanding decoherence, entanglement, and quantum statistical mechanics.

Our goal today is to learn how to construct density matrices for pure and mixed quantum states and derive the von Neumann equation, which governs their time evolution.

1. From Complete to Incomplete Knowledge: Pure vs. Mixed States

Let's begin by clearly distinguishing between the two types of states we need to describe. A pure state is what you're familiar with: a system described by a single ket $|\psi ⟩$. The uncertainty in a pure state is fundamentally quantum mechanical; for instance, even in a state $|\psi ⟩=\frac{1}{\sqrt{2}}(|\uparrow ⟩+|\downarrow ⟩)$, where the state is perfectly known, a measurement of $S_z$ will yield $\hslash /2$ or $-\hslash /2$ with 50% probability each.

A mixed state introduces a second layer of uncertainty—a classical one, born from our ignorance. Imagine a machine that prepares spin-1/2 particles. 50% of the time it produces a particle in the state $|\uparrow ⟩$, and 50% of the time it produces one in $|\downarrow ⟩$. If you pick a particle at random, you don't know if it's $|\uparrow ⟩$ or $|\downarrow ⟩$. This statistical ensemble is a mixed state. It is fundamentally different from the pure superposition state $|\psi ⟩$ mentioned above.

To build a solid intuition for this crucial distinction, please watch the first part of the following video.

2. The Density Operator

To handle both pure and mixed states within a single framework, we introduce the density operator, $\hat{\rho }$.

For Pure States

For a pure state described by the ket $|\psi ⟩$, the density operator is defined as the outer product:

$$\hat{\rho }=|\psi ⟩⟨\psi |$$

This might seem like an unnecessary complication, but it's the bridge to the more general case. All physical predictions can be reformulated in terms of $\hat{\rho }$. For example, the expectation value of an observable $\hat{A}$ is:

$$⟨\hat{A}⟩=⟨\psi |\hat{A}|\psi ⟩=\sum _n⟨\psi |n⟩⟨n|\hat{A}|\psi ⟩=\sum _n⟨n|\hat{A}|\psi ⟩⟨\psi |n⟩=\text{Tr}(\hat{A}|\psi ⟩⟨\psi |)=\text{Tr}(\hat{A}\hat{\rho })$$

(Here, we inserted a complete set of states $|n⟩$ and used the cyclic property of the trace, $\text{Tr}(XYZ)=\text{Tr}(ZXY)$).

Pure state density operators have two key properties:

1. Unit Trace (Normalization): $\text{Tr}(\hat{\rho })=\text{Tr}(|\psi ⟩⟨\psi |)=⟨\psi |\psi ⟩=1$.
2. Idempotency (It's a projector): ${\hat{\rho }}^2=(|\psi ⟩⟨\psi |)(|\psi ⟩⟨\psi |)=|\psi ⟩(⟨\psi |\psi ⟩)⟨\psi |=|\psi ⟩⟨\psi |=\hat{\rho }$.

From idempotency, it follows that for a pure state, $\text{Tr}({\hat{\rho }}^2)=\text{Tr}(\hat{\rho })=1$. This will be our litmus test.

For Mixed States

Now for the crucial step. How do we define $\hat{\rho }$ for a statistical ensemble where the system is in state $|{\psi }_i⟩$ with classical probability $p_i$? The answer is to take a weighted average of the pure-state density operators:

$$\hat{\rho }=\sum _i{p}_i|{\psi }_i⟩⟨{\psi }_i|=\sum _i{p}_i{\hat{\rho }}_i$$

where ${\sum }_i{p}_i=1$. This definition is natural because it correctly reproduces the average expectation value:

$$⟨\hat{A}{⟩}_{\text{ensemble}}=\sum _i{p}_i⟨\hat{A}{⟩}_i=\sum _i{p}_i\text{Tr}({\hat{\rho }}_i\hat{A})=\text{Tr}\left(\left(\sum _i{p}_i{\hat{\rho }}_i\right)\hat{A}\right)=\text{Tr}(\hat{\rho }\hat{A})$$

The formula $⟨\hat{A}⟩=\text{Tr}(\hat{\rho }\hat{A})$ is universal! The following segment of the video you started derives this definition in a very elegant way by considering measurement probabilities.

The properties of a mixed state density operator are:

1. Unit Trace: $\text{Tr}(\hat{\rho })=\text{Tr}({\sum }_i{p}_i{\hat{\rho }}_i)={\sum }_i{p}_i\text{Tr}({\hat{\rho }}_i)={\sum }_i{p}_i=1$. It's still normalized.
2. Idempotency: It is not idempotent. In general, ${\hat{\rho }}^2\ne \hat{\rho }$. This leads to the crucial test: for a mixed state, $\text{Tr}({\hat{\rho }}^2)<1$.

The Criterion: Purity

This gives us a simple, powerful way to determine if a state described by $\hat{\rho }$ is pure or mixed: calculate the purity, $\gamma =\text{Tr}({\hat{\rho }}^2)$.

• If $\gamma =1$, the state is pure.
• If $\gamma <1$, the state is mixed.

The final segment of the "Professor M does Science" video provides an excellent walkthrough of this result.

A Subtle Point: Non-Unique Ensembles

A fascinating and non-intuitive feature of the density matrix is that different statistical mixtures (different ensembles) can produce the exact same density matrix. This means they are experimentally indistinguishable.

3. Example: The Spin-1/2 System

Let's make this concrete with the most fundamental two-level system. We will work in the $S_z$ basis, where $|\uparrow ⟩=\left(\begin{array}{c}1\\ 0\end{array}\right)$ and $|\downarrow ⟩=\left(\begin{array}{c}0\\ 1\end{array}\right)$.

The following reading provides clear examples of constructing density matrices for both pure and mixed spin-1/2 states.

Let's recap the two key cases you just read about:

1. Pure State (polarized along +x): $|{\psi }_x⟩=\frac{1}{\sqrt{2}}(|\uparrow ⟩+|\downarrow ⟩)$
   The density matrix is ${\hat{\rho }}_{\text{pure}}=|{\psi }_x⟩⟨{\psi }_x|=\frac{1}{2}(|\uparrow ⟩⟨\uparrow |+|\uparrow ⟩⟨\downarrow |+|\downarrow ⟩⟨\uparrow |+|\downarrow ⟩⟨\downarrow |)$.
   In matrix form:

   $${\rho }_{\text{pure}}=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$$

   The non-zero off-diagonal elements, known as coherences, encode the specific phase relationship between the $|\uparrow ⟩$ and $|\downarrow ⟩$ components. The purity is $\text{Tr}({\rho }_{\text{pure}}^2)=1$.

2. Mixed State (unpolarized): An ensemble with 50% $|\uparrow ⟩$ and 50% $|\downarrow ⟩$.
   The density matrix is ${\hat{\rho }}_{\text{mixed}}=\frac{1}{2}|\uparrow ⟩⟨\uparrow |+\frac{1}{2}|\downarrow ⟩⟨\downarrow |$.
   In matrix form:

   $${\rho }_{\text{mixed}}=\frac{1}{2}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=\frac{1}{2}\hat{I}$$

   Here, the coherences are zero. This represents a complete lack of phase information. The purity is $\text{Tr}({\rho }_{\text{mixed}}^2)=\text{Tr}(\frac{1}{4}\hat{I})=\frac{1}{2}<1$.

Test your understanding!

Calculate the expectation value of the spin operator ${\hat{S}}_x=\frac{\hslash }{2}{\sigma }_x=\frac{\hslash }{2}\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ for both ${\rho }_{\text{pure}}$ and ${\rho }_{\text{mixed}}$ above. What does the result tell you about the importance of the off-diagonal elements?

Show answer

For the pure state:

$$⟨{\hat{S}}_x⟩=\text{Tr}({\hat{S}}_x{\rho }_{\text{pure}})=\text{Tr}\left(\frac{\hslash }{2}\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\frac{1}{2}\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\right)=\text{Tr}\left(\frac{\hslash }{4}\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\right)=\frac{\hslash }{4}(1+1)=\frac{\hslash }{2}$$

This makes sense, as the state is polarized along the x-axis.

For the mixed state:

$$⟨{\hat{S}}_x⟩=\text{Tr}({\hat{S}}_x{\rho }_{\text{mixed}})=\text{Tr}\left(\frac{\hslash }{2}\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\frac{1}{2}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\right)=\text{Tr}\left(\frac{\hslash }{4}\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\right)=\frac{\hslash }{4}(0+0)=0$$

This also makes sense, as the unpolarized mixture has no preferred direction. The calculation shows that the expectation value of an operator that probes off-diagonal elements (like ${\hat{S}}_x$) is directly sensitive to the coherences. The off-diagonal elements are not just mathematical artifacts; they have direct physical consequences.

4. Time Evolution: The von Neumann Equation

The final piece of our formalism is to determine how the density matrix evolves in time for a closed system governed by a Hamiltonian $\hat{H}$. The result is a beautiful and compact equation of motion.

The derivation starts from the Schrödinger equation for each pure state in an ensemble and applies the product rule for differentiation.

Let's walk through the steps shown in the image. For any state $|{\psi }_j(t)⟩$ in the ensemble, its evolution is given by the Schrödinger equation:

$$i\hslash \frac{d}{dt}|{\psi }_j(t)⟩=\hat{H}|{\psi }_j(t)⟩⟹-i\hslash \frac{d}{dt}⟨{\psi }_j(t)|=⟨{\psi }_j(t)|\hat{H}$$

Now we take the time derivative of the density matrix $\hat{\rho }(t)={\sum }_j{p}_j|{\psi }_j(t)⟩⟨{\psi }_j(t)|$. Since the probabilities $p_j$ are time-independent for a closed system, we have:

$$i\hslash \frac{d\hat{\rho }}{dt}=\sum _j{p}_j\left(\left(i\hslash \frac{d|{\psi }_j⟩}{dt}\right)⟨{\psi }_j|+|{\psi }_j⟩\left(i\hslash \frac{d⟨{\psi }_j|}{dt}\right)\right)$$

Substituting the Schrödinger equation and its conjugate:

$$i\hslash \frac{d\hat{\rho }}{dt}=\sum _j{p}_j\left((\hat{H}|{\psi }_j⟩)⟨{\psi }_j|-|{\psi }_j⟩(⟨{\psi }_j|\hat{H})\right)$$ $$=\hat{H}\left(\sum _j{p}_j|{\psi }_j⟩⟨{\psi }_j|\right)-\left(\sum _j{p}_j|{\psi }_j⟩⟨{\psi }_j|\right)\hat{H}$$

This gives us the celebrated von Neumann equation:

$$i\hslash \frac{\partial \hat{\rho }}{\partial t}=[\hat{H},\hat{\rho }]$$

This is the quantum mechanical analogue of the classical Liouville equation. Notice its structural similarity to the Heisenberg equation of motion for an operator $\hat{A}(t)$, but with a crucial sign difference: $i\hslash \frac{d{\hat{A}}_H}{dt}=[{\hat{A}}_H,\hat{H}]$. The derivation is also presented concisely in the context of linear response theory (a topic we will touch upon later) in the KIT lecture notes (LINK, Section 1.1).

Conclusion

In this lesson, we have moved beyond the limitations of the wavefunction and introduced the density matrix as a universal tool for describing quantum systems.

Key Takeaways:

• Quantum systems can be in pure states (maximum information, described by a ket $|\psi ⟩$) or mixed states (statistical ensembles with classical uncertainty).
• The density operator, $\hat{\rho }$, provides a unified description. For pure states, $\hat{\rho }=|\psi ⟩⟨\psi |$. For mixed states, $\hat{\rho }={\sum }_i{p}_i|{\psi }_i⟩⟨{\psi }_i|$.
• Expectation values for any state are given by the universal formula $⟨\hat{A}⟩=\text{Tr}(\hat{\rho }\hat{A})$.
• The purity, $\gamma =\text{Tr}({\hat{\rho }}^2)$, serves as a criterion to distinguish states: $\gamma =1$ for pure states and $\gamma <1$ for mixed states.
• The time evolution of the density matrix for a closed system is governed by the von Neumann equation: $i\hslash \frac{\partial \hat{\rho }}{\partial t}=[\hat{H},\hat{\rho }]$.

Preview of the Next Lesson:
We now have the necessary machinery to tackle one of the most important applications of the density matrix: describing subsystems. In the next lesson, we will learn how to take a large, entangled quantum system and "trace out" parts of it to find the reduced density matrix of the subsystem we are interested in. This will allow us to formalize the concept of entanglement and quantify it using measures like purity and von Neumann entropy, topics that are essential in condensed matter and quantum information theory.