Density Matrices (Introductory)

Overview

Mixed states arise when a system is prepared with classical randomness or when you focus on a subsystem of a larger quantum system.

To reason about mixed states efficiently, we need a representation that:

• predicts measurement probabilities correctly,
• does not depend on a particular “ensemble story,” and
• includes pure states as a special case.

That representation is the density matrix (also called the density operator).

Intuition

A ket $|\psi\rangle$ is a compact way to predict measurement outcomes when the system is in a single pure state.

A mixed situation is more like: “with probability $p_1$ the state is $|\psi_1\rangle$, with probability $p_2$ it is $|\psi_2\rangle$, …”.

If you try to carry that list around, you run into a problem: there can be multiple different lists that produce the same observable behavior.

A density matrix is the object that keeps exactly what measurements can reveal, and forgets the non-unique bookkeeping details.

Formal Description

Pure states as matrices

Given a pure state $|\psi\rangle$, define its density matrix as

$$\rho = |\psi\rangle\langle\psi|.$$

Here:

• $|\psi\rangle$ is a column vector (a ket).
• $\langle\psi|$ is its conjugate transpose (a bra).
• The product $|\psi\rangle\langle\psi|$ is a matrix (an operator).

Example: if $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$, then in the computational basis the column vector is

$$|\psi\rangle = \begin{pmatrix}\alpha\\\beta\end{pmatrix},$$

and the density matrix is

$$\rho = \begin{pmatrix}\alpha\\\beta\end{pmatrix}\begin{pmatrix}\alpha^* & \beta^*\end{pmatrix} = \begin{pmatrix} |\alpha|^2 & \alpha\beta^*\\ \alpha^*\beta & |\beta|^2 \end{pmatrix}.$$

The $*$ means complex conjugate.

Mixed states as averages of pure-state matrices

If the system is in $|\psi_k\rangle$ with classical probability $p_k$, define

$$\rho = \sum_k p_k\,|\psi_k\rangle\langle\psi_k|.$$

This single matrix $\rho$ represents the mixed state.

Two core properties (kept conceptual)

A valid density matrix has two crucial properties.

1. Trace equals 1:

$$\mathrm{Tr}(\rho)=1.$$

The trace $\mathrm{Tr}(\cdot)$ means “sum of the diagonal entries.” Trace $=1$ is the matrix version of “total probability is 1.”

2. Positivity (conceptual):

• $\rho$ never predicts negative probabilities.
• Formally, it means all measurement probabilities computed from $\rho$ are $\ge 0$.

You do not need advanced operator algebra here; the takeaway is: density matrices are constrained so that they behave like probability-bearing objects.

Worked Example

Consider the 50/50 mixture:

• with probability $1/2$, the state is $|0\rangle$,
• with probability $1/2$, the state is $|1\rangle$.

Compute its density matrix:

$$\rho = \tfrac{1}{2}|0\rangle\langle 0| + \tfrac{1}{2}|1\rangle\langle 1|.$$

In the computational basis,

$$|0\rangle\langle 0| = \begin{pmatrix}1&0\\0&0\end{pmatrix},\quad |1\rangle\langle 1| = \begin{pmatrix}0&0\\0&1\end{pmatrix}.$$

So

$$\rho = \tfrac{1}{2}\begin{pmatrix}1&0\\0&0\end{pmatrix} + \tfrac{1}{2}\begin{pmatrix}0&0\\0&1\end{pmatrix} = \begin{pmatrix}1/2&0\\0&1/2\end{pmatrix}.$$

This matrix encodes “a fair coin over $|0\rangle$ and $|1\rangle$.”

Turtle Tip

Common Pitfalls

Quick Check

What’s Next

Now that we have density matrices, we can restate Born’s rule in a way that works uniformly for pure states, mixtures, and subsystems. Next: measurement using $\rho$, and why it simplifies “ignore part of the system” calculations.