Every *-isometric isomorphism of $B(\mathcal{H})$ keep compact operators? Let $\mathcal{H}$ be a Hilbert space, $B(\mathcal{H})$ denotes the $\mathcal{C}^*$-algebra consisting of bounded linear transformation on $\mathcal{H}$ ($*$ is the adjoint). Now consider a isometric *-isomorphism of $B(\mathcal{H})$ onto it self namely $\phi$. If $K$ is a compact operator, then $\phi(K)$ is also compact. Is this claim true? 

Here are my thoughts: (I guess it is right)
1.Maybe we can consider finite-rank operators. If we can prove $\phi$ keeps rank-one operators, then it will keep finite-rank operators, and thus keep compact operators by the density and continuity. 
2.The ideal $\mathcal{K}$ formed by all compact operators is a minimal nonzero closed ideal of $B(\mathcal{H})$, and $\phi(\mathcal{K})$ is a closed ideal of $B(\mathcal{H})$. So if we can prove $\phi(\mathcal{K})\cap \mathcal{K}\not= \varnothing$, then $\phi(\mathcal{K})=\mathcal{K}$ and the claim is true. 

I think this question is not too difficult, but I just sticked. It also helpful if you can give me some hints or references. Thanks.

Note that a $*$-isomorphisms is automatically isometric.


*

*If $p$ is a rank-one projection, then it is minimal (and viceversa). Then $\phi(p)$ is also a minimal projection, so it is also a rank-one projection. A selfadjoint compact operator $x$ is a norm-limit of linear combinations of rank-one projections, and this will be preserved by $\phi$. So $\phi$ maps compact selfadjoints to compact selfadjoints. Now any compact $x$ is, by the polar decomposition, $x=vr$, with $r$ positive. Since $r=(x^*x)^{1/2}$, it is compact. So $\phi(x)=\phi(v)\phi(r)$ is compact. 

*Much easier than the above, when $H$ is separable: since $\phi$ is a $*$-isomorphism, it preserves ideals. And $K(H)$ is the only nontrivial (closed, bilateral) ideal. 
