# What is the relationship between trace distance and total variation distance?

Consider two quantum states $$\rho$$ and $$\sigma$$ and the probability distributions induced by measuring both of them in the standard basis. Let’s call the probability distributions $$p_{\rho}$$ and $$p_{\sigma}$$ respectively. What is the relation between the trace distance between $$\rho$$ and $$\sigma$$ and the total variation distance between $$p_{\rho}$$ and $$p_{\sigma}$$?

• By the data processing inequality you will have that the total variation is no larger than the trace distance. I'm not sure if one can say anything more than that in general. Sep 1, 2020 at 10:16

## A bound on the total variation distance

Rammus already provided a short answer, but I'd like to elaborate a bit on why this is the case. This is basically the proof of theorem $$9.1$$ on page $$405$$ of Nielsen & Chuang. Note that they call the total variation distance the (classical) trace distance, to draw the connection to the quantum trace distance.

For any two states $$\rho$$ and $$\sigma$$, let the trace distance be $$D(\rho, \sigma) = \frac{1}{2}\mathrm{tr}|\rho - \sigma|$$ (eq. $$9.11$$, page $$403$$). Furthermore, let $$\{E_{m}\}$$ be any POVM such that $$p_{m} = \mathrm{tr}(\rho E_{m})$$ and $$q_{m} = \mathrm{tr}(\sigma E_{m})$$, where I've slightly changed your notation to ease the analysis. Note that in your case $$\{E_{m}\}$$ is just the collection of projectors on the standard basis.

Note that $$|\rho - \sigma|$$ means $$\sqrt{(\rho - \sigma)^{\dagger}(\rho - \sigma)}$$ and is not a number; if we take the trace of this operator it of course becomes a number.

The classical trace distance between $$\{p_{m}\}$$ and $$\{q_{m}\}$$ is (eq. $$9.1$$, page $$400$$): $$\begin{equation} D(p_{m},q_{m}) = \frac{1}{2} \sum_{m}|p_{m} - q_{m}| = \frac{1}{2} \sum_{m}|\mathrm{tr}(E_{m}(\rho - \sigma))|. \end{equation}$$

The most straightforward answer to your question: Theorem $$9.1$$ can be rephrased to say:

$$\begin{equation} D(p_{m},q_{m}) \leq D(\rho, \sigma) \end{equation}$$

## Why 'exactly'?

The classical trace distance looks a bit like the quantum trace distance, but not completely. The important detail to note is that for any $$E_{m}$$: $$\begin{equation} |\mathrm{tr}(E_{m}(\rho - \sigma))| \leq \mathrm{tr}(E_{m}|\rho - \sigma|). \end{equation}$$ This holds, because (and I again quote Nielsen & Chuang, eq. $$9.25-9.27$$ (page $$405$$)), we can always write $$\rho - \sigma = Q - S$$ for some positive operators $$Q$$ and $$S$$ with orthogonal support, meaning that $$|\rho - \sigma| = Q + S$$.

## Proof of the above statements

We know that both $$\rho$$ and $$\sigma$$ are positive and Hermitian; this means that $$A = \rho - \sigma$$ is also Hermitian, now with both positive and negative (but real!) eigenvalues. That is, $$A$$ has eigenvalue-eigenvector pairs $$(\lambda_{i},|\psi_{i}\rangle)$$, with some $$\lambda_{i} \geq 0$$ and the others less than zero. Also, $$\langle \psi_{i}| \psi_{i'}\rangle = \delta_{ii'}$$. Splitting the spectrum of $$A$$ into the positive part $$\{\lambda_{+}\}$$ and negative part $$\{\lambda_{-}\}$$, we can write for $$A$$:

$$\begin{equation} \begin{split} A = \sum_{i}\lambda_{i} |\psi_{i}\rangle \langle \psi_{i} | &= \sum_{i\in +}\lambda_{i} |\psi_{i}\rangle \langle \psi_{i} | + \sum_{i \in -}\lambda_{i} |\psi_{i}\rangle \langle \psi_{i} | \\ &= \sum_{i\in +}\lambda_{i} |\psi_{i}\rangle \langle \psi_{i} | - \sum_{i \in -} |\lambda_{i}| |\psi_{i}\rangle \langle \psi_{i} | \\ &= Q - S, \end{split} \end{equation}$$ where both $$Q = \sum_{i\in +}\lambda_{i} |\psi_{i}\rangle \langle \psi_{i} |$$ and $$S = \sum_{i \in -}|\lambda_{i}| |\psi_{i}\rangle \langle \psi_{i} |$$ are positive operators, with completely orthogonal support. It is now easy to verify that $$|\rho - \sigma| = Q + S$$. Moreover, by construction we have $$\rho - \sigma = Q - S$$.

We can now write:

$$\begin{equation} \begin{split} |\mathrm{tr}(E_{m}(\rho - \sigma))| &= |\mathrm{tr}(E_{m}(Q - S))|\\ &= |\mathrm{tr}(E_{m}Q) - \mathrm{tr}(E_{m}S)| \\ & \leq \mathrm{tr}(E_{m}Q) + \mathrm{tr}(E_{m}S) = \mathrm{tr}(E_{m}(Q+S)) \\ &= \mathrm{tr}(E_{m}|\rho - \sigma|)), \end{split} \end{equation}$$

which proves our equation above.

## Back to the problem at hand

In the end, we can now combine the two equations: $$\begin{equation} \begin{split} D(p_{m},q_{m}) &= \frac{1}{2} \sum_{m}|p_{m} - q_{m}| = \frac{1}{2} \sum_{m}|\mathrm{tr}(E_{m}(\rho - \sigma))| \\ & \leq \frac{1}{2} \sum_{m}\mathrm{tr}(E_{m}|\rho - \sigma|) = \frac{1}{2} \mathrm{tr}|\rho - \sigma| = D(\rho, \sigma). \end{split} \end{equation}$$ where the last identity holds because $$\{E_{m}\}$$ is a POVM and thus $$\sum_{m} E_{m} = I$$.

Of course there is a particular POVM $$\{E_{m}\}$$ for which the bound is saturated, but it is highly dependent on both $$\rho$$ and $$\sigma$$; it is very unlikely that this is exactly the standard basis.

• I don't understand the part where you say $|\rho - \sigma| = Q+S$. The left hand side is a number while the right hand side is an operator. I don't have a copy of the book but would you mind clarifying this bit so the answer is self-contained? Sep 1, 2020 at 14:49
• @user1936752 I've added some extra discussion; note that $|\rho - \sigma| = \sqrt{(\rho - \sigma)^{\dagger}(\rho - \sigma)}$, which is not a number.
– JSdJ
Sep 1, 2020 at 15:13
• Ah I see, thank you! Sep 1, 2020 at 16:12