Reduced States & Partial Trace (Conceptual)

Overview

In multi-qubit systems you often care about only one part of the system: “What happens if I measure qubit A?” or “What statistics do I see locally?”

This page introduces two closely related ideas:

• the reduced state of a subsystem (the best description of a part when you ignore the rest), and
• the partial trace (the mathematical operation that produces that reduced description).

We will keep this conceptual and computation-friendly, without developing full density-matrix formalism.

Intuition

If you have a joint probability table for two classical variables $A$ and $B$, and you only care about $A$, you compute a marginal distribution by summing over $B$.

Quantum subsystems have an analogous idea:

• the full state predicts probabilities for joint outcomes,
• but if you ignore one subsystem, you should still be able to predict probabilities for measurements on the remaining subsystem.

For product states, this is easy: the subsystem really does have its own state vector.

For entangled states, something new happens: the joint state can be perfectly pure and well-defined, but the subsystem can fail to have any single pure state vector that describes all local measurement statistics. The reduced state is the object that captures what you can predict locally.

Formal Description

Marginal probabilities from a two-qubit state

Write a general two-qubit state in the computational basis:

$$|\Psi\rangle = \alpha_{00}|00\rangle + \alpha_{01}|01\rangle + \alpha_{10}|10\rangle + \alpha_{11}|11\rangle.$$

If you measure both qubits in the computational basis, Born’s rule gives:

$$P(ij)=|\alpha_{ij}|^2.$$

If you measure only the first qubit and ignore the second, the probability that the first outcome is 0 is the sum of the two joint probabilities consistent with “first qubit is 0”:

$$P(A=0) = P(00)+P(01) = |\alpha_{00}|^2 + |\alpha_{01}|^2.$$

Similarly,

$$P(A=1) = P(10)+P(11) = |\alpha_{10}|^2 + |\alpha_{11}|^2.$$

This is exactly a quantum version of marginalization for this particular measurement.

What a reduced state means

A reduced state for qubit A is a compact description that lets you compute probabilities for measurements on A without needing to track B explicitly.

• If the joint state is a product state $|a\rangle\otimes|b\rangle$, then the reduced state of A is just $|a\rangle$.
• If the joint state is entangled, the reduced state of A generally cannot be written as a single ket $|a\rangle$.

In that entangled case, the reduced state behaves like a mixture: it produces probabilistic outcomes even for measurements where a pure state could be deterministic.

What “partial trace” does (conceptually)

The partial trace is the operation “discard subsystem B but keep everything needed to predict measurements on A.”

Operationally, you can think of it as:

• “Start from the joint state description.”
• “Average over (sum out) the degrees of freedom of the subsystem you are ignoring.”

The result is the reduced description of the remaining subsystem.

Worked Example

Consider the Bell state

$$|\Phi^+\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle+|11\rangle).$$

Local statistics in the computational basis

The only joint outcomes are 00 and 11, each with probability $1/2$. So for the first qubit,

$$P(A=0)=\tfrac{1}{2},\quad P(A=1)=\tfrac{1}{2}.$$

So qubit A alone looks like a fair random bit in the computational basis.

Local statistics in the $\{|+\rangle,|-\rangle\}$ basis

Define

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

Rewrite $|\Phi^+\rangle$ in this basis using $|0\rangle = \tfrac{1}{\sqrt{2}}(|+\rangle+|-\rangle)$ and $|1\rangle = \tfrac{1}{\sqrt{2}}(|+\rangle-|-\rangle)$.

A standard expansion gives:

$$|\Phi^+\rangle = \tfrac{1}{\sqrt{2}}(|++\rangle + |--\rangle).$$

So if you measure both qubits in the $\{|+\rangle,|-\rangle\}$ basis, you only see $++$ or $--$, each with probability $1/2$.

Therefore, if you measure only qubit A in that basis:

$$P(A=+)=\tfrac{1}{2},\quad P(A=-)=\tfrac{1}{2}.$$

Conclusion: qubit A behaves “maximally uncertain” in both bases, even though the joint state is pure. That is exactly why we need reduced states.

Turtle Tip

Common Pitfalls

Quick Check

What’s Next

Reduced states show that you cannot always describe a subsystem with a single ket. Next we’ll use this “no free information” theme to prove a major limitation: there is no physical process that can perfectly copy an arbitrary unknown quantum state (the no-cloning theorem).