Prove that $\|p^{\otimes n} - q^{\otimes n}\| \leq n \|p-q\|$ for density operators $p,q$

Question

I've been trying to figure this out for a while and I'm totally lost.

My goal is to show that for two density operators $p$, $q$, that $$||p^{\otimes n} - q^{\otimes n}|| \leq n ||p-q||$$

So far I have constructed a weak inductive hypothesis over n to get

$$||p^{\otimes (n-1)} - q^{\otimes (n-1)}|| \leq (n-1) ||p-q||$$ However, I'm having a great deal of difficulty manipulating my tensor products on the LHS to be able to use my IH. What first steps can be taken?

Note: The norm being used here is the trace norm.

Answer 1

Hint: To make your induction work, write

$$\eqalign{p^{\otimes n} - q^{\otimes n} & = & \left(p^{\otimes(n-1)}\otimes p \right)-\left(q^{\otimes (n-1)} \otimes q\right)\\ & = & \left(p^{\otimes(n-1)}-q^{\otimes (n-1)} \right)\otimes p+\left(q^{\otimes (n-1)} \right) \otimes (p-q)}$$

Then, use triangle inequality and finally the fact that your norm is the trace norm, i.e. you will need $\text{Tr}(A\otimes B) = \text{Tr}(A) \text{Tr}(B)$. Then, note that $p$ and $q^{\otimes (n-1)}$ are density operators with have unit trace by definition.

Answer 2

Marsl is correct, and his "hint" is really more a sketch of a solution than a hint. Rather than viewing the question or its solution as just formal algebra, you can also approach his same solution more conceptually. The conceptual reasoning is really identical to the algebra, just phrased differently.

You can rely on the following two facts:

1) Trace norm is a type of distance between states; in particular it satisfies the triangle inequality. Sometimes it is called trace distance.

2) Since trace norm represents distinguishability of states, it satisfies the relation $\mathrm{dist}(p \otimes r,q \otimes r) = \mathrm{dist}(p,q)$. In other words, Alice's state $p$ and Bob's state $q$ become no more and no less distinguishable if you also give Alice and Bob each a copy of another state $r$. You are allowed to have a copy of the state $r$ in your brain if you are asked to distinguish $p$ from $q$.

Now the argument is that using (1) and (2), you can get from $n$ copies of $p$ to $n$ copies of $q$, i.e., from $p^{\otimes n}$ to $q^{\otimes n}$, by replacing the states one by one. You thus get an upper bound of $n$ times $\mathrm{dist}(p,q)$.