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

Question

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 $p_{\sigma}$?

Answer 1

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.