Given that a qubit in equal superposition of $|0\rangle$ and $|1\rangle$ is represented by following wave function

\begin{equation} \Psi = \frac{1}{\sqrt 2}(|0\rangle + |1\rangle) \end{equation}

and associated density matrix

\begin{equation} \rho_1 = |\Psi\rangle \langle\Psi| = \frac{1}{2} \begin{pmatrix} 1 && 1 \\ 1 && 1 \end{pmatrix} \end{equation}

Further a qubit in this state is considered to be in a pure state (even though its in a superposition) since the density matrix can be cleanly factored into a product.

Fact $F$ : When measured this qubit will turn out to be 0 or 1 with 50% probability each.

But there exists another density matrix consistent with the same fact $F$ as follows:

\begin{equation} \rho_2 = \frac{1}{2} (|0\rangle \langle0|) + \frac{1}{2} (|1\rangle \langle1|) = \frac{1}{2} \begin{pmatrix} 1 && 0 \\ 0 && 1 \end{pmatrix} \end{equation}

and this matrix corresponds to a mixed state. Clearly $\rho_1 \neq \rho_2$ but they have same physical characteristics which is the fact $F$.

My Question: What is a physical fact or property that can be used to distinguish $\rho_1$ from $\rho_2$?


The mixed state is invariant under unitary operations; no choice of $U$ that you might apply to $\rho_2$ will change its output statistics from 50/50 distribution of "0" versus "1". You can see this since $U \rho_2 U^\dagger = \rho_2 U U^\dagger = \rho_2$ since $\rho_2$ is proportional to the identity and therefore commutes with any operator.

Meanwhile, you can apply the Hadamard gate $H$ to $\rho_1$ to change its statistics since $H\rho_1 H = |0\rangle \langle 0|$, resulting in measurement of "0" 100% of the time. So you can use certainly use unitary operations to distinguish $\rho_1$ from $\rho_2$.

The more general idea at play here is that $\rho_2$ (and mixed states in general) describes a classical probability distribution, not a quantum state with probability amplitudes. The whole idea of quantum computing is that we can manipulate probability amplitudes and take advantage of superposition and interference effects (among others) to manipulate the state into doing something useful for us. Classical probability distributions don't have these interesting effects so there's a lot less we can use them for algorithmically.

This doesn't mean classical distributions or mixed states are uninteresting for quantum computing. For example, classically random distributions are an essential part of many quantum communication protocols.


Not the answer you're looking for? Browse other questions tagged or ask your own question.