# 4.1: Mixed States

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First, we recall some properties of the trace:

• $$\operatorname{Tr}(a A)=a \operatorname{Tr}(A),$$
• $$\operatorname{Tr}(A+B)=\operatorname{Tr}(A)+\operatorname{Tr}(B).$$

Also remember that we can write the expectation value of $$A$$ as

$\langle A\rangle=\operatorname{Tr}(|\psi\rangle\langle\psi| A),\tag{4.1}$

where $$|\psi\rangle$$ is the state of the system. It tells us everything there is to know about the system. But what if we don’t know everything?

As an example, consider that Alice prepares a qubit in the state $$|0\rangle$$ or in the state $$|+\rangle=(|0\rangle+|1\rangle) / \sqrt{2}$$ depending on the outcome of a balanced (50:50) coin toss. How does Bob describe the state before any measurement? First, we cannot say that the state is $$\frac{1}{2}|0\rangle+\frac{1}{2}|+\rangle$$, because this is not normalized!

The key to the solution is to observe that the expectation values must behave correctly. The expectation value $$\langle A\rangle$$ is the average of the eigenvalues of $$A$$ for a given state. If the state is itself a statistical mixture (as in the example above), then the expectation values must also be averaged. So for the example, we require that for any A

\begin{aligned} \langle A\rangle &=\frac{1}{2}\langle A\rangle_{0}+\frac{1}{2}\langle A\rangle_{+}=\frac{1}{2} \operatorname{Tr}(|0\rangle\langle 0| A)+\frac{1}{2} \operatorname{Tr}(|+\rangle\langle+| A) \\ &=\operatorname{Tr}\left[\left(\frac{1}{2}|0\rangle\langle 0|+\frac{1}{2}|+\rangle\langle+|\right) A\right] \\ & \equiv \operatorname{Tr}(\rho A), \end{aligned}\tag{4.2}

where we defined

$\rho=\frac{1}{2}|0\rangle\langle 0|+\frac{1}{2}|+\rangle\langle+|.\tag{4.3}$

The statistical mixture is therefore properly described by an operator, rather than a simple vector. We can generalize this as

$\rho=\sum_{k} p_{k}\left|\psi_{k}\right\rangle\left\langle\psi_{k}\right|,\tag{4.4}$

where the $$p_{k}$$ are probabilities that sum up to one $$\left(\sum_{k} p_{k}=1\right)$$ and the $$\left|\psi_{k}\right\rangle$$ are normalized states (not necessarily complete or orthogonal). Since $$\rho$$ acts as a weight, or a density, in the expectation value, we call it the density operator. We can diagonalize $$\rho$$ to find the spectral decomposition

$\rho=\sum_{j} \lambda_{j}\left|\lambda_{j}\right\rangle\left\langle\lambda_{j}\right|,\tag{4.5}$

where $$\left\{\left|\lambda_{j}\right\rangle\right\}_{j}$$ forms a complete orthonormal basis, $$0 \leq \lambda_{j} \leq 1$$, and $$\sum_{j} \lambda_{j}=1$$. We can also show that $$\rho$$ is a positive operator:

\begin{aligned} \langle\psi|\rho| \psi\rangle &=\sum_{j k} c_{j}^{*} c_{k}\left\langle\lambda_{j}|\rho| \lambda_{k}\right\rangle=\sum_{j k} c_{j}^{*} c_{k}\left\langle\lambda_{j}\left|\sum_{l} \lambda_{l}\right| \lambda_{l}\right\rangle\left\langle\lambda_{l} \mid \lambda_{k}\right\rangle \\ &=\sum_{j k l} c_{j}^{*} c_{k} \lambda_{l}\left\langle\lambda_{j} \mid \lambda_{l}\right\rangle\left\langle\lambda_{l} \mid \lambda_{k}\right\rangle=\sum_{j k l} c_{j}^{*} c_{k} \lambda_{l} \delta_{j l} \delta_{l k} \\ &=\sum_{l} \lambda_{l}\left|c_{l}\right|^{2} \\ & \geq 0 \end{aligned}\tag{4.6}

In general, an operator $$\rho$$ is a valid density operator if and only if it has the following three properties:

1. $$\rho^{\dagger}=\rho,$$
2. $$\operatorname{Tr}(\rho)=1,$$
3. $$\rho \geq 0$$.

The density operator is a generalization of the state of a quantum system when we have incomplete information. In the special case where one of the $$p_{j}=1$$ and the others are zero, the density operator becomes the projector $$\left|\psi_{j}\right\rangle\left\langle\psi_{j}\right|$$. In other words, it is completely determined by the state vector $$\left|\psi_{j}\right\rangle$$. We call these pure states. The statistical mixture of pure states giving rise to the density operator is called a mixed state.

The unitary evolution of the density operator can be derived directly from the Schrödinger equation $$i \hbar \partial_{t}|\psi\rangle=H|\psi\rangle$$:

\begin{aligned} i \hbar \frac{d \rho}{d t} &=i \hbar \frac{d}{d t} \sum_{j} p_{j}\left|\psi_{j}\right\rangle\left\langle\psi_{j}\right| \\ &=i \hbar \sum_{j}\left\{\frac{d p_{j}}{d t}\left|\psi_{j}\right\rangle\left\langle\psi_{j}\right|+p_{j}\left[\left(\frac{d}{d t}\left|\psi_{j}\right\rangle\right)\left\langle\psi_{j}|+| \psi_{j}\right\rangle\left(\frac{d}{d t}\left\langle\psi_{j}\right|\right)\right]\right\} \\ &=i \hbar \frac{\partial \rho}{\partial t}+H \rho-\rho H \\ &=[H, \rho]+i \hbar \frac{\partial \rho}{\partial t} \end{aligned}\tag{4.7}

This agrees with the Heisenberg equation for operators, and it is sometimes known as the Von Neumann equation. In most problems the probabilities $$p_{j}$$ have no explicit time-dependence, and $$\partial_{t} \rho=0$$.

This page titled 4.1: Mixed States is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pieter Kok via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.