# Trace norm identity (in bra-ket notation)

I came across the following identity in a paper:

$$\|\hspace{0.3em}|v\rangle\langle v| - |w\rangle\langle w|\hspace{0.3em}\|_{tr}=2\sqrt{1-|\langle v|w\rangle |^2}$$

where the norm on the left is the trace norm, and $|v\rangle,|w\rangle\in\mathbb{C}^d$, for some integer $d$, but I have no idea why it's correct, and can't seem to understand where it comes from. It looks like the kind of thing that should be simple to understand, so any help would be very much appreciated!

Let's define for brevity $\def\tr{\operatorname{tr}}\pi_x = \lvert x\rangle\langle x\rvert$, and $\Delta = \pi_v - \pi_w$, which is the operator the trace norm is taken of.

Note that since $\Delta$ is hermitian, the trace norm of $\Delta$ equals the sum of the absolute values of its eigenvalues. Now $\Delta$ clearly has rank $2$ (except if $\pi_v=\pi_w$; in that case the formula is obviously true), therefore it has two non-zero eigenvalues. Moreover, the trace is $\tr \pi_v - \tr \pi_w = 0$, thus we have a pair of eigenvalues $\lambda$ and $-\lambda$. The trace norm then is $2\lambda$. We therefore have to determine $\lambda$.

Now $$2\lambda^2 = \tr \Delta^2 = \tr\left((\pi_u - \pi_v)^2\right) = \tr(\pi_u^2) - 2 \tr(\pi_u\pi_v) + \tr(\pi_v^2) \tag{*}$$ where in the last step I've used the linearity and the cyclic invariance of the trace.

Now we assume that both states are normalized, therefore $\tr(\pi_v^2) = \tr(\pi_w^2) = 1$. Moreover, $\tr(\pi_v\pi_w) = \tr\left(\lvert v\rangle\langle v\rvert w \rangle \langle w\rvert\right) = \left|\langle v\rvert w\rangle\right|^2$. Inserting in (*), we therefore get $$2\lambda^2 = 2 - 2\left|\langle v\rvert w\rangle\right|^2$$ and therefore $$\|\Delta\|_{\tr} = 2\lambda = 2\sqrt{1-\left|\langle v\rvert w\rangle\right|^2}$$