Mixed States (Conceptual Introduction)

Overview

So far, we described systems using pure states like $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$. Pure states are powerful, but they are not always enough.

This page introduces mixed states conceptually: situations where the best description is not a single ket, but a probabilistic mixture of different possible pure states.

The main goal is to remove a common confusion: uncertainty (mixture) is not the same thing as superposition.

Intuition

There are two different ways a system can look “uncertain,” and they have different meanings.

1) Classical uncertainty about preparation (mixture)

Imagine a lab procedure:

• Flip a classical coin.
• If heads, prepare $|0\rangle$.
• If tails, prepare $|1\rangle$.

After this procedure, the qubit is definitely in either $|0\rangle$ or $|1\rangle$. The uncertainty is in your information: you do not know which branch happened, because you did not look at the coin.

That is a mixed-state situation.

2) Quantum superposition (pure state)

Now compare with preparing

$$|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle+|1\rangle).$$

Here there is no hidden coin and no “which branch.” There is a single pure state with well-defined amplitudes.

Both procedures can produce the same probabilities for some measurements (for example, measuring 0/1), but they are still physically different descriptions because they behave differently under other measurements.

This is the key intuition:

• A mixture is “one of several states, chosen by classical randomness.”
• A superposition is “one state with multiple amplitude components.”

Formal Description

We will describe mixtures using an “ensemble” viewpoint.

An ensemble is a list of possible pure states together with classical probabilities:

• $|\psi_1\rangle$ with probability $p_1$
• $|\psi_2\rangle$ with probability $p_2$
• …

where:

• each $p_k \ge 0$
• and $\sum_k p_k = 1$.

If you perform a measurement, the overall outcome probabilities are computed by averaging over the classical uncertainty:

1. First compute probabilities within each branch using Born’s rule.
2. Then average those probabilities using the classical weights $p_k$.

In words: “probability of an outcome = expected value of the branch-probability.”

This is the correct rule as long as the randomness is classical and you truly do not condition on which branch occurred.

Worked Example

Consider two different preparations.

Preparation A: classical mixture

• With probability $1/2$, prepare $|0\rangle$.
• With probability $1/2$, prepare $|1\rangle$.

If you measure in the computational basis:

• In the $|0\rangle$ branch, you get outcome 0 with probability 1.
• In the $|1\rangle$ branch, you get outcome 0 with probability 0.

So overall:

$$P(0)=\tfrac{1}{2}\cdot 1 + \tfrac{1}{2}\cdot 0 = \tfrac{1}{2}.$$

Similarly $P(1)=1/2$.

Preparation B: pure superposition

Prepare $|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$.

Measuring in the computational basis also gives $P(0)=P(1)=1/2$.

So far, the two preparations look identical.

Now measure in the $\{|+\rangle,|-\rangle\}$ basis.

• In Preparation B, if the state is $|+\rangle$, you get “$+$” with probability 1.
• In Preparation A, the state is sometimes $|0\rangle$ and sometimes $|1\rangle$. Each of those gives “$+$” with probability $1/2$ in that basis.

So for Preparation A:

$$P(+)=\tfrac{1}{2}\cdot\tfrac{1}{2} + \tfrac{1}{2}\cdot\tfrac{1}{2} = \tfrac{1}{2},$$

while for Preparation B:

$$P(+)=1.$$

Conclusion: mixture and superposition can agree on one measurement but disagree on another. That is why mixed states are a genuinely new idea.

Turtle Tip

Common Pitfalls

Quick Check

What’s Next

We described mixtures using a list of states and classical probabilities. That list is not unique: different ensembles can describe the same mixed situation. Next we introduce a single mathematical object—the density matrix—that represents mixed states cleanly and avoids keeping track of arbitrary ensemble lists.