Let $x_1,x_2\in H$ be two unit vectors in a Hilbert space $H$ and let $t_1,t_2 : B(H)\rightarrow\mathbb{C}$ be the linear functionals given by $t_j(a) =\langle ax_j,x_j\rangle$. Define $p:B(H)\rightarrow\mathbb{C}$ by $p:=t_1-t_2$.
Prove: $|\left|p|\right|=4(1-\left|\langle x_1,x_2\rangle^2\right|)$
To prove this identity we may use the fact that the norm of the linear functional $w:B(H)\rightarrow\mathbb{C}$ given by $w(a)=Tr(ra)$ for all $a\in B(H)$ and $r\in B(H)$ a self-adjoint trace class operator is given by $\|w\|=\sum_{\lambda\in\sigma_p(r)}m_{\lambda}\left|\lambda\right|$ where $\lambda$ are the eigenvalues of $r$ and $m_{\lambda}$ the corresponding multiplicities (i think thats a corolloary of the spectral theorems for compact and self-adjoint operators).
But now back to the identity. Can someone help me with this task? I can't think about a manner to combine what we have to prove and the fact we may use. Thanks a lot.
Disclaimer. this is an answer in tye case of a real Hilbert space. Hopefully, someone will write a proof in the complex case. It would be even better to have a geometrical proof...
I think the correct formula should be $$\Vert p\Vert^{\bf 2}=4(1-\vert\langle x_1,x_2\rangle\vert^2)\, . $$
Note first that for any unit vector $x\in H$, we have $$\langle ax,x\rangle=Tr(ap_x)\, , $$ where $p_x$ is the orthogonal projection onto $[x]$ (the linear span of $x$). One can see this for example by choosing an orthonormal basis $(e_i)_{i\in\mathbb N}$ of $H$ with $e_1=x$ and computing $Tr(ap_x)=Tr(ap_{e_1})$ on this basis.
It follows that for any $a\in B(H)$ : $$p(a)=Tr(a(p_1-p_2))\, , $$ where $p_j$ is the orthogonal projection onto $[x_j]$. So we have to find the eigenvalues of the operator $p_1-p_2$ to arrive at the above formula.
As indicated ine the "disclaimer", I'll do it in the case of a real Hilbert space $H$.
The problem is clearly $2$-dimensional. So we may assume that $H=\mathbb R^2$ and, moreover, that $x_1$ is the first vector of the canonical basis of $\mathbb R^2$, i.e. $x_1=(1,0)$. Write $x_2=(\cos\theta, \sin\theta)$, and for notational simplicity, set $c:=\cos\theta$ and $s:=\sin\theta$. Note that $$\langle x_1,x_2\rangle=c\, . $$ For any $z\in\mathbb R^2$, we have $$(p_1-p_2)(z)=\langle z,x_1\rangle \, x_1-\langle z,x_2\rangle\, x_2\, . $$ Using this, one finds out that after a short computation that the matrix of the linear transformation $p_1-p_2$ is $$M=\left(\begin{matrix}-s^2&sc\\sc&s^2 \end{matrix}\right) $$ This gives $\det(M)=-s^2 $ and $Tr(M)=0$; so the eigenvalues of $p$ are $s$ and $-s$. Hence, from the formula that you recalled in your question, we obtain $$\Vert p\Vert=2\vert s\vert\, .$$ This gives $\Vert p\Vert^2=4s^2=4(1-c^2)=4(1-\langle x_1,x_2\rangle^2)$, as required.