Given an orthogonal projection $\Pi$, is $\|\Pi(\sigma-\rho)\Pi\|_1\le\|\sigma-\rho\|_1$ true?

Question

Suppose I have an arbitrary orthogonal projector $\Pi$ and two density operators $\rho, \sigma$. Is it true that:

$$ ||\Pi (\sigma - \rho) \Pi||_1 \le || \sigma - \rho ||_1 $$

where $||\cdot||_1$ denotes the trace norm?

Answer 1

Yes, you can use Cauchy interlacing theorem to prove that.

Let $M = \sigma - \rho$ and dimension of the space is $n$.

In an appropriate basis $\Pi = I_k \oplus 0_{n-k}$. Assume $k=n-1$ (in general situation we can use an induction).

Then $\Pi M \Pi$ is the principle submatrix of $M$ of size $n-1 \times n-1$ (plus a row and a column of $0$s, but we discard that). So that, its eigenvalues $\mu_i$ are interlaced by eigenvalues $\lambda_i$ of $M$, that is $$ \lambda_1 \le \mu_1 \le \lambda_2 \le \mu_2 \le \lambda_3 \le \dots \le \mu_{n-1} \le \lambda_n. $$

Now, if $\mu_i \ge 0$ then $|\mu_i| \le |\lambda_{i+1}|$. Otherwise, $|\mu_i| \le |\lambda_{i}|$. Hence $$ \sum_{i=1}^{n-1}|\mu_i| \le \sum_{i=1}^n|\lambda_i| $$

Answer 2

Here is a slightly more general alternative to Danylo's answer. The trace norm satisfies the inequality $$ \| X Y Z \|_1 \leq \|X\| \|Y\|_1 \|Z\| $$ for all choices of operators $X$, $Y$, and $Z$ that can be composed as $XYZ$ (so nothing needs to be Hermitian). Note that the norms $\|X\|$ and $\|Z\|$ on the right-hand side refer to the spectral (or operator) norm.

Hence, in the case at hand, we have $$ \|\Pi(\sigma - \rho) \Pi\|_1 \leq \|\Pi\| \|\sigma - \rho\|_1 \|\Pi\| \leq \|\sigma - \rho\|_1 $$ because the spectral norm of any orthogonal projector is equal to 1 (unless it is zero).

The inequality $$ \| X Y Z \|_1 \leq \|X\| \|Y\|_1 \|Z\| $$ is actually true for any unitarily invariant norm in place of the trace norm. One way to prove this is to first use the fact that every contraction (i.e., operator having spectral norm at most 1) can be expressed as a convex combination of unitary operators, and then use the convexity and unitary invariance of the norm.