Chapter 6 Density matrices

About density matrices, and how they help to solve the problem introduced by entangled states, as well as how they let us talk about mixtures and subsystems. Also a first look at the partial trace.

We cannot always assign a definite state vector to a quantum system. It may be that the system is part of a composite system that is in an entangled state, or it may be that our knowledge of the preparation of a particular system is insufficient to determine its state (for example, someone may prepare a particle in one of the states |\psi_1\rangle, |\psi_2\rangle, \ldots, |\psi_n\rangle, with (respective) probabilities p_1, p_2, \ldots, p_n). Nevertheless, in either case we are able to make statistical predictions about the outcomes of measurements performed on the system using a more general description of quantum states.

We have already mentioned that the existence of entangled states begs an obvious question: if we cannot attribute a state vectors to an individual quantum system then how shall we describe its quantum state? In this chapter we will introduce an alternate description of quantum states that can be applied both to a composite system and to any of its subsystems. Our new mathematical tool is called a density operator.83 We will start with the density operator as a description of the mixture of quantum states, and will then discuss the partial trace, which is a unique operation that takes care of the reduction of a density operator of a composite system to density operators of its components.

6.1 Definitions

If you are an impatient mathematically minded person, who feels more comfortable when things are properly defined right from the beginning, here is your definition:84

A density operator \rho on a finite dimensional Hilbert space \mathcal{H} is any non-negative self-adjoint operator with trace equal to one.

It follows that any such \rho can always be diagonalised, that the eigenvalues are all real and non-negative, and that the eigenvalues sum to one. Moreover, given two density operators \rho_1 and \rho_2, we can always construct another density operator as a convex sum of the two: \rho = p_1\rho_1 + p_2\rho_2 \qquad\text{where}\quad p_1, p_2 \geqslant 0 \text{ and } p_1+p_2 = 1. You should check that \rho has all the defining properties of a density matrix, i.e. that it is self-adjoint, non-negative, and that its trace is one. This means that density operators form a convex set.85

An important example of a density operator is a rank one projector.86 Any quantum state that can be described by the state vector |\psi\rangle, called a pure state, can be also described by the density operator \rho=|\psi\rangle\langle\psi|. Pure states are the extremal points in the convex set of density operators: they cannot be expressed as a convex sum of other elements in the set. In contrast, all other states, called mixed states, can be always written as the convex sum of pure states: \sum_i p_i |\psi_i\rangle\langle\psi_i| (p_i\geqslant 0 and \sum_i p_i=1). Now that we have cleared the mathematical essentials, we will turn to physical applications.

6.2 Statistical mixtures

Let us start with probability distributions over state vectors. Suppose Alice prepares a quantum system and hands it over to Bob who subsequently measures observable M. If Alice’s preparation is described by a state vector |\psi\rangle, then, quantum theory declares, the average value of any observable M is given by \langle\psi|M|\psi\rangle, which can be also written as87 \langle M\rangle = \operatorname{tr}M|\psi\rangle\langle\psi|. This way of expressing the average value makes a clear separation between the contributions from the state preparation and from the choice of the measurement. We have two operators under the trace: one of them, |\psi\rangle\langle\psi|, describes the state preparation, and the other one, M, the measurement. Now, suppose Alice prepares the quantum system in one of the states |\psi_1\rangle,\ldots,|\psi_m\rangle, choosing state |\psi_i\rangle with probability p_i, and hands the system to Bob without telling him which state was chosen. The possible states |\psi_i\rangle are normalised but need not be orthogonal. We call this situation a mixture of the states |\psi_i\rangle, or a mixed state for short.

Remember, a mixture of states is very different from a superposition of states: a superposition always yields a definite state vector, whereas a mixture does not, and so must be described by a density operator.

Bob knows the ensemble of states |\psi_1\rangle,\ldots,|\psi_m\rangle and the corresponding probability distribution p_1,\ldots,p_m, and can hence calculate \langle M\rangle as88 \begin{aligned} \langle M\rangle &= \sum_i p_i\left( \operatorname{tr}M|\psi_i\rangle\langle\psi_i| \right) \\&= \operatorname{tr}M \underbrace{\left( \sum_i p_i|\psi_i\rangle\langle\psi_i| \right)}_{\rho} \\&=\operatorname{tr}M\rho. \end{aligned} Again, we have two operators under the trace: \rho=\sum_i p_i|\psi_i\rangle\langle\psi_i|, which pertains to the state preparation, and M, which describes the measurement. We shall call the operator \rho = \sum_i p_i |\psi_i\rangle\langle\psi_i| the density operator, since it has all the defining properties of the density operator (the convex sum of rank one projectors). It depends on the constituent states |\psi_i\rangle and their probabilities, and it describes our ignorance about the state preparation.

Once we have \rho we can make statistical predictions: for any observable M we have \langle M\rangle = \operatorname{tr}M\rho. We see that the exact composition of the mixture does not enter this formula: for computing the statistics associated with any observable property of a system, all that matters is the density operator itself, and not its decomposition into the mixture of states. This is important because any given density operator, with the remarkable exception of a pure state, can arise from many different mixtures of pure states. Consider, for example, the following three scenarios:

  1. Alice flips a fair coin. If the result is \texttt{Heads} then she prepares the qubit in the state |0\rangle, and if the result is \texttt{Tails} then she prepares the qubit in the state |1\rangle. She gives Bob the qubit without revealing the result of the coin-flip. Bob’s knowledge of the qubit is described by the density matrix \frac12|0\rangle\langle 0| + \frac12|1\rangle\langle 1| = \begin{bmatrix} \frac12 & 0 \\0 & \frac12 \end{bmatrix}.

  2. Suppose Alice flips a fair coin, as before, but now if the result is \texttt{Heads} then she prepares the qubit in the state |\bar{0}\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle), and if the result is \texttt{Tails} then she prepares the qubit in the state |\bar{1}\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle). Bob’s knowledge of the qubit is now described by the density matrix \begin{aligned} \frac12|\bar{0}\rangle\langle\bar{0}| + \frac12|\bar{1}\rangle\langle\bar{1}| &= \frac12 \begin{bmatrix} \frac12 & \frac12 \\\frac12 & \frac12 \end{bmatrix} + \frac12 \begin{bmatrix} \frac12 & -\frac12 \\-\frac12 & \frac12 \end{bmatrix} \\&= \begin{bmatrix} \frac12 & 0 \\0 & \frac12 \end{bmatrix}. \end{aligned}

  3. Suppose Alice picks up any pair of orthogonal states of a qubit and then flips the coin to chose one of them. Any two orthonormal states of a qubit, |u_1\rangle, |u_2\rangle, form a complete basis, so the mixture \frac12|u_1\rangle\langle u_1|+\frac12|u_2\rangle\langle u_2| gives \frac12\mathbf{1}.

As you can see, these three different preparations yield precisely the same density matrix and are hence statistically indistinguishable. In general, two different mixtures can be distinguished (in a statistical sense) if and only if they yield different density matrices. In fact, the optimal way of distinguishing quantum states with different density operators is still an active area of research.

6.3 A few instructive examples, and some less instructive remarks

  1. 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.

  2. 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}.

  3. 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.

  4. 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?

  5. 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”).

  6. 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.

  7. 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.

  8. If the states |u_1\rangle,\ldots,|u_m\rangle form an orthonormal basis, and each occurs with equal probability 1/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.

  9. 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.

6.4 The Bloch ball

We have already talked in some depth about the Bloch sphere in Chapter 2 and Chapter 3, but now that we are considering density operators (which are strictly more general than state vectors), we are actually interested in the Bloch ball, i.e. not just the sphere of vectors of magnitude 1, but instead the ball of vectors of magnitude less than or equal to 1.

The most general Hermitian (2\times 2) matrix has four real parameters and can be expanded in the basis composed of the identity and the three Pauli matrices: \{\mathbf{1}, \sigma_x, \sigma_y, \sigma_z\}. Since the Pauli matrices are traceless, the coefficient of \mathbf{1} in the expansion of a density matrix \rho must be \frac12, so that \operatorname{tr}\rho=1. Thus \rho may be expressed as89 \begin{aligned} \rho &= \frac12\left( \mathbf{1}+\vec{s}\cdot\vec{\sigma} \right) \\&= \frac12\left( \mathbf{1}+ s_x\sigma_x + s_y\sigma_y + s_z\sigma_z \right) \\&= \frac12 \begin{bmatrix} 1 + s_z & s_x - is_y \\s_x + is_y & 1 - s_z \end{bmatrix}. \end{aligned} The vector \vec{s} is called the Bloch vector for the density operator \rho. Any real Bloch vector \vec{s} defines a trace one Hermitian operator \rho, but in order for \rho to be a density operator it must also be non-negative. Which Bloch vectors yield legitimate density operators?

Let us compute the eigenvalues of \rho. The sum of the two eigenvalues of \rho is, of course, equal to one (\operatorname{tr}\rho=1) and the product is equal to the determinant of \rho, which can be computed from the matrix form above: \det\rho = \frac{1}{4}(1-s^2) = \frac12(1+s)\frac12(1-s) where s=|\vec{s}|. It follows that the two eigenvalues of \rho are \frac12(1\pm s). They have to be non-negative, and so s, the length of the Bloch vector, cannot exceed one.90 We can now visualise the convex set of (2\times 2) density matrices as a unit ball in three-dimensional Euclidean space: the extremal points, which represent pure states, are the points on the boundary (s=1), i.e. the surface of the ball; the maximally mixed state \mathbf{1}/2 corresponds to s=0, i.e. the centre of the ball. In general, the length of the Bloch vector s can be thought of as a “purity” of a state.

6.5 Subsystems of entangled systems

We have already trumpeted that one of the most important features of the density operator formalism is its ability to describe the quantum state of a subsystem of a composite system. Let me now show you how it works.

Given a quantum state of the composite system \mathcal{AB}, described by some density operator \rho^{\mathcal{AB}}, we obtain reduced density operators \rho^{\mathcal{A}} and \rho^{\mathcal{B}} of subsystems \mathcal{A} and \mathcal{B}, respectively, by the partial trace: \begin{aligned} \rho^{\mathcal{AB}} &\longmapsto \underbrace{\rho^\mathcal{A}=\operatorname{tr}_{\mathcal{B}}\rho^{\mathcal{AB}}}_{\mathrm{partial\,trace\,over}\,\mathcal{B}}\qquad \\\rho^{\mathcal{AB}} &\longmapsto \underbrace{\rho^\mathcal{B}=\operatorname{tr}_{\mathcal{A}}\rho^{\mathcal{AB}}}_{\mathrm{partial\,trace\,over}\,\mathcal{A}} \end{aligned} We define the partial trace over \mathcal{B}, or \mathcal{A}, first on a tensor product of two operators A\otimes B as \begin{aligned} \operatorname{tr}_{\mathcal{B}} (A\otimes B) &= A(\operatorname{tr}B) \\\operatorname{tr}_{\mathcal{A}} (A\otimes B) &= (\operatorname{tr}A) B, \end{aligned} and then extend to any operator on \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}} by linearity.

Here is a simple example. Suppose a composite system \mathcal{AB} is in a pure entangled state, which we can always write as |\psi_{\mathcal{AB}}\rangle = \sum_{i} c_{i} |a_i\rangle\otimes|b_i\rangle, where |a_i\rangle and |b_j\rangle are two orthonormal bases (e.g. the Schmidt bases), and where \sum_i |c_i|^2 = 1 (due to the normalisation). The corresponding density operator of the composite system is the projector \rho^{\mathcal{AB}}= |\psi_{\mathcal{AB}}\rangle\langle\psi_{\mathcal{AB}}|, which we can write as \rho^{\mathcal{AB}} = |\psi_{\mathcal{AB}}\rangle\langle\psi_{\mathcal{AB}}| = \sum_{ij} c_i c^\star_j |a_i\rangle\langle a_j| \otimes |b_i\rangle\langle b_j| Let us compute the reduced density operator \rho^{\mathcal{A}} by taking the partial trace over \mathcal{B}: \begin{aligned} \rho^\mathcal{A} &= \operatorname{tr}_{\mathcal{B}}\rho^{\mathcal{AB}} \\&= \operatorname{tr}_{\mathcal{B}} |\psi_{\mathcal{AB}}\rangle\langle\psi_{\mathcal{AB}}| \\&= \operatorname{tr}_{\mathcal{B}} \sum_{ij} c_i c^\star_j |a_i\rangle\langle a_j| \otimes |b_i\rangle\langle b_j| \\&= \sum_{ij} c_i c^\star_j |a_i\rangle\langle a_j|(\operatorname{tr}|b_i\rangle\langle b_j|) \\&= \sum_{ij} c_i c^\star_j |a_i\rangle\langle a_j| \underbrace{\langle b_i|b_j\rangle}_{\delta_{ij}} \\& = \sum_{i} |c_i|^2 |a_i\rangle\langle a_i| \end{aligned} where we have used the fact that \operatorname{tr}|b_i\rangle\langle b_j| = \langle b_j|b_i\rangle=\delta_{ij}. In the |a_i\rangle basis, the reduced density matrix \rho^{\mathcal{A}} is diagonal, with entries p_i=|c_i|^2. We can also take the partial trace over \mathcal{A} and obtain \rho^\mathcal{B} = \sum_{i} |c_i|^2 |b_i\rangle\langle b_i|. In particular, for the maximally entangled states in the (d\times d)-dimensional Hilbert spaces \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}}, |\psi_{\mathcal{AB}}\rangle = \frac{1}{\sqrt{d}} \sum_{i}^d |a_i\rangle|b_i\rangle, and the reduced density operators, \rho^\mathcal{A} and \rho^\mathcal{B}, are the maximally mixed states: \rho^\mathcal{A}=\rho^\mathcal{B}=\frac{1}{d}\mathbf{1}. It follows that the quantum states of individual qubits in any of the Bell states are maximally mixed: their density matrix is \frac12\mathbf{1}. A state such as \frac{1}{\sqrt 2} \left( |00\rangle + |11\rangle \right) guarantees perfect correlations when each qubit is measured in the standard basis: the two equally likely outcomes are (0 and 0) or (1 and 1), but any single qubit outcome, be it 0 or 1 or anything else, is completely random.

6.6 Partial trace, revisited

If you are given a matrix you calculate the trace by summing its diagonal entries. How about the partial trace? Suppose someone writes down for you a density matrix of two qubits in the standard basis, \{|00\rangle, |01\rangle, |10\rangle, |11\rangle\}, and asks you to find the reduced density matrices of the individual qubits. The tensor product structure of this (4\times 4) matrix means that it is has a block form: \rho^{\mathcal{AB}} = \left[ \begin{array}{c|c} P & Q \\\hline R & S \end{array} \right] where P,Q,R,S are (2\times 2) sized sub-matrices. The two partial traces can then be evaluated as91 \begin{aligned} \rho^\mathcal{A} &= \operatorname{tr}_{B}\rho^{\mathcal{AB}} = \left[ \begin{array}{c|c} \operatorname{tr}P & \operatorname{tr}Q \\\hline \operatorname{tr}R & \operatorname{tr}S \end{array} \right] \\\rho^\mathcal{B} &= \operatorname{tr}_{A}\rho^{\mathcal{AB}} = P+S. \end{aligned} The same holds for a general \rho^{\mathcal{AB}} on any \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}} with corresponding block form ((m\times m) blocks of (n\times n)-sized sub-matrices, where m and n are the dimensions of \mathcal{H}_{\mathcal{A}} and \mathcal{H}_{\mathcal{B}}, respectively).

6.7 Mixtures and subsystems

We have used the density operators to describe two distinct situations: the statistical properties of the mixtures of states, and the statistical properties of subsystems of composite systems. In order to see the relationship between the two, consider a joint state of a bipartite system \mathcal{AB}, written in a product basis in \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}} as \begin{aligned} |\psi_{\mathcal{AB}}\rangle &= \sum_{ij} c_{ij}|a_i\rangle\otimes|b_j\rangle \\&= \sum_{j=1} |\widetilde\psi_j\rangle|b_j\rangle \\&= \sum_{j=1} \sqrt{p_j}|\psi_j\rangle|b_j\rangle \end{aligned} \tag{9.7.1} where |\widetilde\psi_j\rangle = \sum_i c_{ij}|a_i\rangle = \sqrt{p_j}|\psi_j\rangle, and the vectors |\psi_j\rangle are the normalised versions of the |\widetilde\psi_j\rangle. Note that p_j=\langle\widetilde\psi_j|\widetilde\psi_j\rangle.

Then the partial trace over \mathcal{B} gives the reduced density operator of subsystem \mathcal{A}: \begin{aligned} \rho^{\mathcal{A}} &=\operatorname{tr}_{\mathcal{B}} \sum_{ij} |\widetilde\psi_i\rangle\langle\widetilde\psi_j| \otimes |b_i\rangle\langle b_j| \\&= \sum_{ij} |\widetilde\psi_i\rangle\langle\widetilde\psi_j| (\operatorname{tr}|b_i\rangle\langle b_j|) \\&= \sum_{ij} |\widetilde\psi_i\rangle\langle\widetilde\psi_j| \langle b_j|b_i\rangle \\&= \sum_{i} |\widetilde\psi_i\rangle\langle\widetilde\psi_i| = \sum_{i} p_i |\psi_i\rangle\langle\psi_i|. \end{aligned}

Now, let us see how \rho^{\mathcal{A}} can be understood in terms of mixtures. Let us place subsystems \mathcal{A} and \mathcal{B} in separate labs, run by Alice and Bob, respectively. When Bob measures part \mathcal{B} in the |b_j\rangle basis and obtains result k, which happens with the probability p_k, he prepares subsystem \mathcal{A} in the state |\psi_k\rangle: \sum_{i=1} \sqrt{p_j}|\psi_i\rangle|b_i\rangle \overset{\mathrm{outcome}\,k}{\longmapsto} |\psi_k\rangle|b_k\rangle. Bob does not communicate the outcome of his measurement. Thus, from Alice’s perspective, Bob prepares a mixture of |\psi_1\rangle,\ldots,|\psi_m\rangle, with probabilities p_1,\ldots,p_m, which means that Alice, who knows the joint state but not the outcomes of Bob’s measurement, may associate density matrix \rho^\mathcal{A}=\sum_i p_i|\psi_i\rangle\langle\psi_i| with her subsystem \mathcal{A}. This is the same \rho^{\mathcal{A}} which we obtained by the partial trace.

But suppose Bob chooses to measure his subsystem in some other basis. Will it have any impact on Alice’s statistical predictions? Measurement in the new basis will result in a different mixture, but Alice’s density operator will not change. Suppose Bob chooses basis |d_i\rangle for his measurement. Any two orthonormal bases are connected by some unitary transformation, and so we can write |b_i\rangle = U|d_i\rangle for some unitary U. In terms of components, |b_i\rangle = \sum_j U_{ij}|d_j\rangle. The joint state can now be expressed as \begin{aligned} |\psi_{\mathcal{AB}}\rangle &= \sum_{i} |\widetilde\psi_i\rangle|b_i\rangle \\&= \sum_{i} |\widetilde\psi_i\rangle \left( \sum_j U_{ij}|d_j\rangle \right) \\&= \sum_j \underbrace{\left( \sum_i U_{ij}|\widetilde\psi_i\rangle \right)}_{|\widetilde\phi_j\rangle}|d_j\rangle \\&= \sum_j|\widetilde\phi_j\rangle|d_j\rangle. \end{aligned}

If Bob measures in the |d_i\rangle basis then he generates a new mixture of states |\phi_1\rangle,\ldots|\phi_m\rangle, which are the normalised versions of |\widetilde\phi_1\rangle,\ldots|\widetilde\phi_m\rangle, with each |\phi_k\rangle occurring with probability p_k=\langle\widetilde\phi_k|\widetilde\phi_k\rangle. But this new mixture has exactly the same density operator as the previous one:92 \begin{aligned} \sum_j|\widetilde\phi_j\rangle\langle\widetilde\phi_j| &= \sum_{ijl} U_{ij}|\widetilde\psi_i\rangle\langle\widetilde\psi_l|U^\star_{lj} \\&= \sum_{il} \underbrace{\left(\sum_j U_{ij}U^\star_{lj}\right)}_{\delta_{il}}|\widetilde\psi_i\rangle\langle\widetilde\psi_l| \\&= \sum_i|\widetilde\psi_j\rangle\langle\widetilde\psi_j|, \end{aligned} which is exactly \rho^\mathcal{A}. So does it really matter whether Bob performs the measurement or not?

It does not.

After all, Alice and Bob may be miles away from each other, and if any of Bob’s actions were to result in something that is physically detectable at the Alice’s location that would amount to instantaneous communication between the two of them.

From the operational point of view it does not really matter whether the density operator represents our ignorance of the actual state (mixtures) or provides the only description we can have after discarding one part of an entangled state (partial trace).93 In the former case, the system is in some definite pure state but we do not know which. In contrast, when the density operator arises from tracing out irrelevant, or unavailable, degrees of freedom, the individual system cannot be thought to be in some definite state of which we are ignorant. Philosophy aside, the fact that the two interpretations give exactly the same predictions is useful: switching back and forth between the two pictures often offers additional insights and may even simplify lengthy calculations.

6.8 Partial trace, yet again

The partial trace is the only map \rho^{\mathcal{AB}}\mapsto\rho^{\mathcal{A}} such that94 \operatorname{tr}[X\rho^{\mathcal{A}}] = \operatorname{tr}[(X\otimes\mathbf{1})\rho^{\mathcal{AB}}] holds for any observable X acting on \mathcal{A}. This condition concerns the consistency of statistical predictions. Any observable X on \mathcal{A} can be viewed as an observable X\otimes\mathbf{1} on the composite system \mathcal{AB}, where \mathbf{1} is the identity operator acting on \mathcal{B}. When constructing \rho^{\mathcal{A}} we had better make sure that for any observable X the average value of X in the state \rho^\mathcal{A} is the same as the average value of X\otimes\mathbf{1} in the state \rho^{\mathcal{AB}}. This is indeed the case for the partial trace.

For example, let us go back to the state in Equation (9.7.1) and assume that Alice measures some observable X on her part of the system. Technically, such an observable can be expressed as X\otimes \mathbf{1}, where \mathbf{1} is the identity operator acting on the subsystem \mathcal{B}. The expected value of this observable in the state |\psi_{\mathcal{AB}}\rangle is \operatorname{tr}(X\otimes\mathbf{1})|\psi_{\mathcal{AB}}\rangle\langle\psi_{\mathcal{AB}}|, i.e. \begin{aligned} \operatorname{tr}[(X\otimes \mathbf{1}) \rho^{\mathcal{AB}}] &= \operatorname{tr}\left[ (X\otimes\mathbf{1}) \left( \sum_{ij} |\widetilde\psi_i\rangle\langle\widetilde\psi_j| \otimes |b_i\rangle\langle b_j| \right) \right] \\&= \sum_{ij} \left[ \operatorname{tr}\left(X |\widetilde\psi_i\rangle\langle\widetilde\psi_j|\right) \right] \underbrace{\left[\operatorname{tr}\left(|b_i\rangle\langle b_j|\right)\right]}_{\delta_{ij}} \\&= \sum_i \operatorname{tr}\big[X |\widetilde\psi_i\rangle\langle\widetilde\psi_i|\big] \\&= \operatorname{tr}\left[ X \underbrace{\sum_i p_i|\psi_i\rangle\langle\psi_i|}_{\rho^{\mathcal{A}} = \operatorname{tr}_{\mathcal{B}}\rho^{\mathcal{AB}}} \right] \\&= \operatorname{tr}[X\rho^{\mathcal{A}}] \end{aligned} as required.

!!!TODO!!! The uniqueness of the partial trace, for now see Nielsen & Chuang Box 2.6.

6.9 Remarks and exercises

6.9.1

Consider two qubits in the state |\psi\rangle = \frac{1}{\sqrt2}\left( |0\rangle\otimes\left( \sqrt{\frac23}|0\rangle - \sqrt{\frac13}|1\rangle \right) + |1\rangle\otimes\left( \sqrt{\frac23}|0\rangle + \sqrt{\frac13}|1\rangle \right) \right).

  1. What is the density operator \rho of the two qubits corresponding to the state |\psi\rangle? Write it in both Dirac notation and explicitly as a matrix in the computational basis \{|00\rangle,|01\rangle,|10\rangle,|11\rangle\}.

  2. Find the reduced density operators \rho_1 and \rho_2 of the first and second qubit (respectively). Again, write them in both Dirac notation and explicitly as a matrix in the computational basis.

6.9.2 Purification of mixed states

Given a mixed state \rho, a purification of \rho is a pure state |\psi\rangle\langle\psi| of some potentially larger system such that \rho is equal to a partial trace of |\psi\rangle\langle\psi|.

  1. Show that an arbitrary mixed state \rho always has a purification.

  2. Show that purification is unique up to unitary equivalence.

  3. Let |\psi_1\rangle and |\psi_2\rangle in \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}} be two pure states such that \operatorname{tr}_{\mathcal{B}}|\psi_1\rangle\langle\psi_1| = \operatorname{tr}_{\mathcal{B}}|\psi_2\rangle\langle\psi_2|. Show that |\psi_1\rangle = \mathbf{1}\otimes U|\psi_2\rangle for some unitary operator U on \mathcal{H}_{\mathcal{B}}.

6.9.3

Two qubits are in the state described by the density operator \rho = \rho^\mathcal{A}\otimes\rho^\mathcal{B}. What is the partial trace of \rho over each qubit?

6.9.4

Write the density matrix of two qubits corresponding to the mixture of the Bell state \frac{1}{\sqrt 2}\left(|00\rangle + |11\rangle\right) with probability \frac12 and the maximally mixed state of two qubits ((4\times 4) matrix in \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}}) with probability \frac12.

6.9.5 The trace norm

The trace norm of a matrix A is defined as \|A\|_{\operatorname{tr}} = \operatorname{tr}\left(\sqrt{A^\dagger A}\right).

  1. Show that, if A is self-adjoint, then its trace norm is equal to the sum of the absolute values of its eigenvalues.

  2. What is the trace norm of an arbitrary density matrix?

The distance induced by the trace norm is called the trace distance, defined as d_{\operatorname{tr}}(\rho_1,\rho_2) = \frac12\|\rho_2-\rho_1\|_{\operatorname{tr}}.

  1. What is the trace distance between two arbitrary pure states?

6.9.6 Distinguishability and the trace distance

Say we have a physical system which is been prepared in one of two states (say, \rho_1 and \rho_2), each with equal probability. Then a single measurement can distinguish between the two preparations with probability at most \frac12\big(1+d_{\operatorname{tr}}(\rho_1,\rho_2)\big).

  1. Suppose that \rho_1 and \rho_2 commute.95 Using the spectral decompositions of \rho_1 and \rho_2 in their common eigenbasis, describe the optimal measurement that can distinguish between the two states. What is its probability of success?

  2. Suppose that you are given one of the two, randomly selected, qubits of the state |\psi\rangle = \frac{1}{\sqrt2}\left( |0\rangle\otimes\left( \sqrt{\frac23}|0\rangle - \sqrt{\frac13}|1\rangle \right) + |1\rangle\otimes\left( \sqrt{\frac23}|0\rangle + \sqrt{\frac13}|1\rangle \right) \right) from above. What is the maximal probability with which you can determine whether it is the first or second qubit?

6.9.7 Spectral decompositions and common eigenbases

!!!TODO!!!


  1. If we choose a particular basis, operators become matrices. Here I will use both terms (density operators and density matrices) interchangeably.↩︎

  2. A self-adjoint matrix M is said to be non-negative, or positive semi-definite, if \langle v|M|v\rangle\geqslant 0 for any vector |v\rangle, or if all of its eigenvalues are non-negative, or if here exists a matrix A such that M=A^\dagger A. (This is called a Cholesky factorization.)↩︎

  3. A subset of a vector space is said to be convex if, for any two points in the subset, the straight line segment joining them is also entirely contained inside the subset.↩︎

  4. The rank of a matrix is the number of its non-zero eigenvalues.↩︎

  5. If M is one of the orthogonal projectors P_k describing the measurement, then the average \langle P_k\rangle is the probability of the outcome k associated with this projector.↩︎

  6. A pure state can be seen as a special case of a mixed state, where all but one the probabilities p_i equal zero.↩︎

  7. Physicists usually still refer to the Bloch ball as the Bloch sphere, even though it really is a ball now, not a sphere.↩︎

  8. One might hope that there is an equally nice visualisation of the density operators in higher dimensions. Unfortunately there isn’t.↩︎

  9. Take any of the Bell states, write its (4\times 4)-density matrix explicitly, and then trace over each qubit. In each case you should get the maximally mixed state.↩︎

  10. The U_{ij} are the components of a unitary matrix, hence \sum_k U_{ik}U^\star_{jk}=\delta_{ij}.↩︎

  11. The two interpretations of density operators filled volumes of academic papers. The terms proper mixtures and improper mixtures are used, mostly by philosophers, to describe the statistical mixture and the partial trace approach, respectively.↩︎

  12. One can repeat the same argument for \rho^{\mathcal{AB}}\mapsto\rho^{\mathcal{B}}: the partial trace is the unique map \rho^{\mathcal{AB}}\mapsto\rho^{\mathcal{B}} such that \rho^{\mathcal{B}} satisfies \operatorname{tr}[Y\rho^{\mathcal{B}}] = \operatorname{tr}[(1\otimes Y)\rho^{\mathcal{AB}}] for any observable Y on \mathcal{B}.↩︎

  13. The commutativity assumption makes this problem essentially a special case of a purely classical one: distinguishing between two probability distributions.↩︎