## 6.3 Some instructive examples, questions, and remarks

The density matrix corresponding to the state vector

This correspondence is well defined: each

Let’s consider two examples, seeing again how superpositions differ from statistical mixtures.

If Alice prepares a qubit in the superposition state

|\psi\rangle = \alpha|0\rangle + \beta|1\rangle then the corresponding density matrix is the projector|\psi\rangle\langle\psi| = \begin{bmatrix} |\alpha|^2 & \alpha\beta^\star \\\alpha^\star\beta & |\beta|^2 \end{bmatrix}. You are given a qubit and you are told that it was prepared either in state

|0\rangle with probability|\alpha|^2 or in state|1\rangle with probability|\beta|^2 . In this case all you can say is that your qubit is in a mixed state described by the density matrix|\alpha|^2|0\rangle\langle 0| + |\beta|^2|1\rangle\langle 1| = \begin{bmatrix} |\alpha|^2 & 0 \\0 & |\beta|^2 \end{bmatrix}.

The density matrix corresponding to a statistical mixture of states

Suppose you want to distinguish between preparations described by the density matrices in the above two examples.
Assume that you are given sufficiently many qubits, all identically prepared, i.e. either all described by the density matrix

In fact, one of these two measurements is completely useless. Which one, and why?

In general, the diagonal entries of a density matrix describe the probability distributions on the set of basis vectors.
They must add up to one, which is why the trace of any density matrix is one.
The off-diagonal elements, often called **coherences**, signal departure from the classical probability distribution and quantify the degree to which a quantum system can witness interference (we will discuss this in detail later on).
The process in which off-diagonal entries go to zero is called **decoherence**.

- Suppose that your qubit was prepared either in state
\alpha|0\rangle + \beta|1\rangle or in state\alpha|0\rangle - \beta|1\rangle , with equal probability. This means that your qubit is in a mixed state described by the density matrix\frac{1}{2} \begin{bmatrix} |\alpha|^2 & \alpha\beta^\star \\\alpha^\star\beta & |\beta|^2 \end{bmatrix} +\frac{1}{2} \begin{bmatrix} |\alpha|^2 & -\alpha\beta^\star \\-\alpha^\star\beta & |\beta|^2 \end{bmatrix} = \begin{bmatrix} |\alpha|^2 & 0 \\0 & |\beta|^2 \end{bmatrix}.

You cannot tell the difference between the equally weighted mixture of

For

*any*density matrix\rho , the most natural mixture that yields\rho is its**spectral decomposition**:\rho=\sum_i p_i|u_i\rangle\langle u_i| , with eigenvectors|u_i\rangle and eigenvaluesp_i .If the states

|u_1\rangle,\ldots,|u_n\rangle form an orthonormal basis, and each occurs with equal probability1/n , then the resulting density matrix is proportional to the identity:\frac{1}{n}\sum_{i=1}^n |\psi_i\rangle\langle\psi_i| = \frac{1}{n}\mathbf{1}. This is a**maximally mixed**state. For qubits, any pair of orthogonal states taken with equal probabilities gives the maximally mixed state\frac{1}{2}\mathbf{1} .

A state is said to be **maximally mixed** if the outcomes of *any* measurement are completely random.

It is often convenient to write density operators in terms of projectors on states which are *not* normalised, incorporating the probabilities into the length of the state vector:
*not* normalised.
We tend to mark such states with the tilde, e.g.