# What is intuition for the trace distance between quantum states?

Given two mixed states $$\rho$$ and $$\sigma$$, the trace distance between the states is defined by $$\sum_{i=1}^n |\lambda_i|$$, where $$\lambda_i$$'s are eigenvalues of $$\rho - \sigma$$.

I know the definition of eigenvalues, but I don't have intuition on what the sum of eigenvalues represent, and why that signifies as a distance metric. What does distance between two states even mean? Where is trace distance used? Please explain this.

One way to understand the trace distance is to notice that it equals the (classical) trace distance (also referred to as Kolmogorov distance, see this post for some information about it) maximised over all possible POVMs on the states.

To see this, start from the following expression for the trace distance: $$D(\rho,\sigma)\equiv\frac{1}{2}\mathrm{Tr}|\rho-\sigma|=\frac{1}{2}\max_U\mathrm{Tr}[ U(\rho-\sigma)],$$ where the maximum is taken over all unitaries $$U$$.

Being $$\rho-\sigma$$ Hermitian, the maximum is achieved choosing the unitary $$U=\sum_{k:\lambda_k\ge0}\mathbb P(\lambda_k)-\sum_{k:\lambda_k<0}\mathbb P(\lambda_k) =2\sum_{k:\lambda_k\ge0}\mathbb P(\lambda_k)-I,$$ where $$\mathbb P(\lambda_k)\equiv|\lambda_k\rangle\!\langle\lambda_k|$$, $$\lambda_k$$ are the eigenvalues of $$\rho-\sigma$$, and we exploited the fact that the eigenstates of an Hermitian operator form an orthonormal basis for the space. Defining the positive operator $$P$$ as $$P\equiv \sum_{k:\lambda_k\ge0}\mathbb P(\lambda_k),$$ we thus see that $$D(\rho,\sigma)=\mathrm{Tr}[P(\rho-\sigma)]$$ (and one can also show that such $$P$$ is the positive operator that maximises this quantity while satisfying $$P), that is, the trace distance equals the sum of the positive eigenvalues of $$\rho-\sigma$$.

This expression is nice because, being $$P$$ a positive operator, it can be interpreted as a measurement. It encodes one possible answer to some question that can be asked to the states (read, an element of a POVM), and $$\mathrm{Tr}(P\rho)$$ is the probability of getting this answer if the state is $$\rho$$. We, therefore, conclude that $$D(\rho,\sigma)$$ is the maximum difference between the probabilities of getting a given answer when asking a given question to both states. In other words, this tells us how distinguishable the states are: it's the answer to the question among all possible measurements that can be performed on the states, what is the one such that $$\rho$$ and $$\sigma$$ give maximally different answers?

## Maximisation over Hermitians

To see how the maximisation over Hermitian/positive operators is performed, consider an arbitrary Hermitian operator $$Q$$. Then, $$Q=\sum_k q_k\mathbb P(q_k)$$ with $$q_k\in\mathbb R$$, and $$\operatorname{Tr}[Q(\rho-\sigma)]=\sum_{j,k} \lambda_j q_k |\langle \lambda_j|q_k\rangle|^2.$$ For a fixed set of eigenvalues $$q_k$$, it's clear that choosing $$|q_k\rangle=|\lambda_k\rangle$$ gives the maximum value of the trace: $$\sum_j \lambda_j q_j\equiv\langle \boldsymbol\lambda,\boldsymbol q\rangle$$. Without putting further restrictions on the Hermitian $$Q$$, we can get arbitrarily large values of this quantity. One way to work around this is to restrict to the Hermitians $$Q$$ such that $$\|Q\|_2^2\equiv\operatorname{Tr}(Q^2)\le\|\rho-\sigma\|_2^2$$. We can then get a maximum write $$\max_{\text{Hermitians } Q:\|Q\|_2\le\|\rho-\sigma\|_2}\operatorname{Tr}[Q(\rho-\sigma)]=\sum_k\lambda_k^2=\|\rho-\sigma\|_2^2.$$ This way of restricting $$Q$$, however, does not give us an expression similar to the original trace distance. A better way is therefore here to simply impose $$-1\le Q\le1$$, that is, to impose the eigenvalues of $$Q$$ to satisfy $$-1\le q_k\le1$$. With this restriction, $$\sum_k\lambda_k q_k$$ is maximised with the choice $$q_k=\operatorname{sign}(\lambda_k)\equiv\lambda_k/|\lambda_k|$$, and we conclude that $$\max_{-1\le Q\le1}\operatorname{Tr}[Q(\rho-\sigma)]=\sum_k|\lambda_k|=2D(\rho,\sigma).$$

This is telling us that the optimal measurement to distinguish the two states is one that collapses the states in the eigenbasis of $$\rho-\sigma$$, and assigns value $$+1$$ to the outcomes corresponding to $$\lambda_k>0$$ and value $$-1$$ to the outcomes corresponding to $$\lambda_k<0$$.

## Maximisation over positive operators

If we instead impose the restriction $$0\le Q\le 1$$, the expression $$\sum_k\lambda_k q_k$$ is maximised by the choice $$q_k=(\lambda_k+|\lambda_k|)/2$$ ($$q_k=\lambda_k$$ when $$\lambda_k\ge0$$, and $$q_k=0$$ otherwise). This gives $$\max_{0\le Q\le1}\operatorname{Tr}[Q(\rho-\sigma)]=\sum_{k: \lambda_k\ge0}\lambda_k=\frac{1}{2}\sum_k|\lambda_k|=D(\rho,\sigma).$$

• It so happens that the maximum is achieved for U a self-inverse unitary. Isn't the usual characterisation of the trace distance actually for Hermitian operators E, rather than unitary operators U? That would give a natural interpretation in terms of an observable which one measures to distinguish the two possible states. – Niel de Beaudrap Jul 11 '19 at 10:19
• @NieldeBeaudrap isn't that essentially the same as what I mentioned in the last paragraph? Only using positive operators (thus elements of a POVM) using Hermitians/observables – glS Jul 11 '19 at 14:35
• If you get to POVMs in your final paragraph, why do you start from a unitary in your first paragraph? That's what I'm asking about: it happens to be correct but only inasmuch as the extreme will happen to be for a unitary which is also a Hermitian operator. It isn't wrong, but it isn't clear why you choose to maximise over unitaries $U$ rather than some other set of operators. – Niel de Beaudrap Jul 11 '19 at 15:14
• Are POVMs same as unitary operations? – satya Jul 11 '19 at 15:19
• No. POVMs are sets of Positive Operators; the only unitary which is positive is the identity operator. – Niel de Beaudrap Jul 11 '19 at 15:25

Short answer. The trace distance between two states more or less determines how distinguishable they are by any operational means. A trace distance of 0 means that they are indistinguishable (because they're equal); a trace distance of 2 indicates that they can be perfectly distinguished in principle.

Long answer. We will show how, from the objective of distinguishing states, we single out the trace distance as a quantity of interest.$$\def\Tr{\mathrm{Tr}}$$

Suppose that you have a (possibly mixed) quantum state $$\rho$$. You know that it was prepared in one of two different ways: one of which yields the state $$\rho_1$$, and one of which yields the state $$\rho_2$$. So, you either have $$\rho = \rho_1$$ or $$\rho = \rho_2$$. You'd like to determine which is the case (and you don't mind destroying $$\rho$$ to do it). What can you do?

Anything you do will amount to performing some transformation $$U$$, and then performing some measurement. In fact, if you only care about guessing between two outcomes (and you don't mind the fact that you might accidentally guess the wrong outcome), you can reduce your distinguishing procedure to a standard basis measurement, if you use a sophisticated enough unitary. We can suppose that we guess $$\rho_1$$ for the outcome $$|1\rangle$$, and $$\rho_2$$ for the outcome $$|0\rangle$$. We can then ask what the probability of success is for the best possible procedure.

If we don't care about the exact circuit to perform this procedure, and only what the probability of success is, we can think of the unitary $$U$$ as just part of a sophisticated measurement procedure, and consider the two-valued projective measurement $$\{ \Pi_1, \Pi_2 \}$$ given by $$\Pi_1 = U \,\lvert 1 \rangle \! \langle 1 \rvert_a \,U^\dagger = \tfrac{1}{2}U \bigl(\mathbf 1 - Z_a\bigr) U^\dagger, \qquad \Pi_2 = U\, \lvert 0 \rangle \! \langle 0 \rvert_a \,U^\dagger = \tfrac{1}{2}U \bigl(\mathbf 1 + Z_a\bigr) U^\dagger,$$ and describe the distinguishing procedure as a $$\{ \Pi_1, \Pi_2 \}$$ measurement. Supposing that you were given $$\rho_1$$ or $$\rho_2$$ uniformly at random, what you want to maximise is then \begin{aligned} P &= \mathrm{Pr}\bigl[\rho_1\bigr]\,\mathrm{Pr}\bigl[\text{guess 1} \big| \rho_1 \bigr] + \mathrm{Pr}\bigl[\rho_2\bigr]\,\mathrm{Pr}\bigl[\text{guess 2} \big| \rho_2 \bigr] \\[1ex]&= \tfrac{1}{2}\Tr\bigl( \rho_1 \Pi_1 \bigr) + \tfrac{1}{2}\Tr\bigl( \rho_2 \Pi_2 \bigr) \\[1ex]&= \tfrac{1}{2}\Tr\bigl( \rho_1 \bigl[\tfrac{1}{2}U \bigl(\mathbf 1 - Z_a\bigr) U^\dagger\bigr]\bigr) + \tfrac{1}{2}\Tr\bigl( \rho_2 \bigl[\tfrac{1}{2}U \bigl(\mathbf 1 + Z_a\bigr) U^\dagger\bigr] \bigr) \\[1ex]&= \tfrac{1}{4}\Bigl[\Tr\bigl( \rho_1 \bigr) - \Tr\bigl( \rho_1 \bigl[U Z_a U^\dagger\bigr]\bigr) + \Tr\bigl( \rho_2 \bigr) + \Tr\bigl( \rho_2 \bigl[U Z_a U^\dagger\bigr]\bigr)\Bigr]. \end{aligned} Let $$E_U = U Z_a U^\dagger$$: then using the fact that $$\rho_1$$ and $$\rho_2$$ are both density operators, we have \begin{aligned} P &= \tfrac{1}{4}\Bigl[2 - \Tr\bigl( \rho_1 E_U \bigr) + \Tr\bigl( \rho_2 E_U \bigr)\Bigr] \\[1ex]&= \tfrac{1}{2} + \tfrac{1}{4} \Tr\bigl( \bigl[\rho_2 - \rho_1\bigr] E_U \bigr). \end{aligned} We can think of $$E_U$$ as an abstract observable which we measure on either $$\rho_1$$ or $$\rho_2$$: it has eigenvalues $$\pm 1$$ because it is related to $$Z_a$$ by a unitary transformation, and we associate the outcome $$-1$$ with a guess of "1" and the outcome $$+1$$ with a guess of "2". Subject to this restriction, we want to consider the observable $$E_U$$ which maximises the value of $$\Tr( [\rho_2 - \rho_1] E_U )$$; and by convexity, without loss of generality we can consider all operators $$E_U$$ which have eigenvalues bounded within $$[-1,+1]$$ and not just precisely $$\pm 1$$. However, this essentially characterises the trace norm, as $$\max_{\lVert E \rVert \leqslant 1} \Tr\bigl( \bigl[\rho_2 - \rho_1\bigr] E \bigr) = \lVert \rho_2 - \rho_1 \rVert_1$$ where $$\lVert M \rVert_1$$ is the trace norm (see e.g. Theory of Quantum Information, Eqn. (1.173)). You can grasp this on an intuitive level as follows:

• The maximisation problem on the left-hand side is essentially evaluating a large number of inner products of the eigenvectors of $$[\rho_2 - \rho_1]$$ with the eigenvectors of $$E$$.
• In fact, if we consider the trace in the eigenbasis $$|\eta_1\rangle, |\eta_2\rangle, \ldots$$ of $$E$$, we're just considering a sum of inner products $$\pm \langle \eta_k | (\rho_2 - \rho_1) | \eta_k \rangle$$, with a sign determined by the eigenvalue of $$\eta_k\rangle$$.
• The operator $$(\rho_2 - \rho_1)$$ has both positive and negative eigenvalues (assuming that it isn't the zero operator). The way in which we can attempt to maximise the value of the sum above is to choose $$E$$ so that the positive eigenstates $$|\eta_k\rangle$$ of $$E$$ are also positive eigenstates of $$(rho_2 - \rho_1)$$, and the negative eigenstates of $$E$$ to the negative eigenstates of $$(\rho_2 - \rho_1)$$. The effect of this is that $$(\rho_2 - \rho_1) E = \lvert \rho_2 - \rho_1 \rvert$$, where the latter operator is what we get from exchanging the eigenvalues of $$(\rho_2 - \rho_1)$$ with their absolute values.
• The trace of $$\lvert \rho_2 - \rho_1 \rvert$$ then gives us the trace norm, $$\lVert \rho_2 - \rho_1 \rVert_1$$, as you describe it.

Thus we have \begin{aligned} P &= \tfrac{1}{2} + \tfrac{1}{4} \lVert \rho_2 - \rho_1 \rVert_1, \end{aligned} so that the trace norm characterises how distinguishable $$\rho_1$$ is from $$\rho_2$$ by any operational procedure.