## 6.3 A few instructive examples, and some less instructive remarks

The density matrix corresponding to the state vector

|\psi\rangle is the rank one projector|\psi\rangle\langle\psi| . Observe that there is no phase ambiguity, since|\psi\rangle\mapsto e^{i\phi}|\psi\rangle leaves the density matrix unchanged, and each|\psi\rangle gives rise to a distinct density matrix.If Alice prepares a qubit in the 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}. Diagonal density matrices correspond to classical probability distributions on the set of basis vectors.Suppose you want to distinguish between preparations described by the density matrices in examples 2 and 3. Assume that you are given sufficiently many identically prepared qubits described either by the density matrix in example 2 or by the density matrix in example 3. Which of the two measurements would you choose: the measurement in the standard basis

\{|0\rangle,|1\rangle\} , or the measurement in the basis\{|\psi\rangle,|\psi_\perp\rangle\} ? One of the 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 interfere (we will discuss this in detail later on). The process in which off-diagonal entries (the parameter\epsilon in the matrices below) go to zero is called**decoherence**.\begin{bmatrix} |\alpha|^2 & \alpha\beta^\star \\\alpha^\star\beta & |\beta|^2 \end{bmatrix} \mapsto \begin{bmatrix} |\alpha|^2 & \epsilon \\\epsilon^\star & |\beta|^2 \end{bmatrix} \mapsto \begin{bmatrix} |\alpha|^2 & 0 \\0 & |\beta|^2 \end{bmatrix} For\epsilon = \alpha\beta^\star we have a pure quantum state (“full interference capability”) and for\epsilon=0 we have a classical probability distribution over the standard basis (“no interference capability”).Suppose it is equally likely that your qubit was prepared either in state

\alpha|0\rangle + \beta|1\rangle or in state\alpha|0\rangle - \beta|1\rangle . This means that your qubit is in a mixed state described by the density matrix\frac12 \begin{bmatrix} |\alpha|^2 & \alpha\beta^\star \\\alpha^\star\beta & |\beta|^2 \end{bmatrix} + \frac12 \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\alpha|0\rangle\pm\beta|1\rangle and a mixture of|0\rangle and|1\rangle with (respective) probabilities|\alpha|^2 and|\beta|^2 .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_m\rangle form an orthonormal basis, and each occurs with equal probability1/m , then the resulting density matrix is proportional to the identity:\frac{1}{m}\sum_{i=1}^m |\psi_i\rangle\langle\psi_i| = \frac{1}{m}\mathbf{1}. This is called the**maximally mixed state**. For qubits, any pair of orthogonal states taken with equal probabilities gives the maximally mixed state\frac12\mathbf{1} . In maximally mixed states, 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:

\rho = \sum_i|\widetilde\psi_i\rangle\langle\widetilde\psi_i| where|\widetilde\psi_i\rangle = \sqrt{p_i}|\psi_i\rangle , i.e.p_i=\langle\widetilde\psi_i|\widetilde\psi_i\rangle . This form is more compact, but you have to remember that the state vectors are*not*normalised. We tend to mark such states with the tilde, e.g.|\widetilde\psi\rangle , but you may have your own way to remember.