6.3 Some instructive examples, questions, and remarks

The density matrix corresponding to the state vector |\psi\rangle is the rank-one projector |\psi\rangle\langle\psi|.

This correspondence is well defined: each |\psi\rangle gives rise to a distinct density matrix, and the fact that we ignore global phases for state vectors doesn’t introduce any ambiguity for the density matrices, since |\psi\rangle and e^{i\phi}|\psi\rangle give the same density matrix.

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

  1. 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}.

  2. 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 |\psi_1\rangle,\ldots,|\psi_n\rangle with probability distribution p_1,\ldots,p_n is the convex combination \sum_i p_i|\psi_i\rangle\langle\psi_i|.

The resulting density matrix is always diagonal.

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 \left[\begin{smallmatrix}|\alpha|^2&\alpha\beta^\star\\\alpha^\star\beta&|\beta|^2\end{smallmatrix}\right], or all described by the density matrix \left[\begin{smallmatrix}|\alpha|^2&0\\0&|\beta|^2\end{smallmatrix}\right]. 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\} where |\psi^\perp\rangle is orthonormal to |\psi\rangle?

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. \begin{bmatrix} |\alpha|^2 & \alpha\beta^\star \\\alpha^\star\beta & |\beta|^2 \end{bmatrix} \longmapsto \begin{bmatrix} |\alpha|^2 & \varepsilon \\\varepsilon^\star & |\beta|^2 \end{bmatrix} \longmapsto \begin{bmatrix} |\alpha|^2 & 0 \\0 & |\beta|^2 \end{bmatrix} For \varepsilon = \alpha\beta^\star we have a pure quantum state (“full interference capability”) and for \varepsilon=0 we have a classical probability distribution over the standard basis (“no interference capability”).

  1. 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 \alpha|0\rangle\pm\beta|1\rangle and a mixture of |0\rangle and |1\rangle with (respective) probabilities |\alpha|^2 and |\beta|^2.

  1. 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 eigenvalues p_i.

  2. If the states |u_1\rangle,\ldots,|u_n\rangle form an orthonormal basis, and each occurs with equal probability 1/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: \rho = \sum_i|\tilde\psi_i\rangle\langle\tilde\psi_i| where |\tilde\psi_i\rangle = \sqrt{p_i}|\psi_i\rangle, i.e. p_i=\langle\tilde\psi_i|\tilde\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. |\tilde\psi\rangle, but you may have your own way to remember.