I can prove that there are no non- zero multiplicative linear functionals on $M_n(\mathbb{C})$.
It is said that if dim $H > 1$. Since $B(H)$ contains an embedded copy of $M_2(\mathbb{C})$, such a functional would have to restrict to a multiplicative linear functional on $M_2(\mathbb{C})$, which doesn’t exist.Hence $B(H)$ has non non- zero multiplicative linear functionals.
Can anyone tell how $B(H)$ contains an embedded copy of $M_2(\mathbb{C})$,?
The answer linked by Bruno is not too clear, in my opinion.
Here are two easy ways of doing it. Let $\{e_n\}$ be an orthonormal basis (I'll assume countable for notational simplicity, but there is no real need).
A key notion to talk about this stuff is that of matrix units. The matrix units coming from the orthonormal basis $\{e_n\}$ are the operators $E_{kj}$ given by $$ E_{kj}\,z=\langle z,e_j\rangle\,e_k. $$ They satisfy $$ E_{kj}E_{ab}=\delta_{j,a}\,E_{kb}. $$
If you don't care about the embedding being unital, you can consider the "upper left corner": $$M=\big\{T\in B(H): \ker T=\ker T^*=\{e_1,e_2\}^\perp\big\}. $$ You also have $M=PB(H)P$, where $P$ is the orthogonal projection onto $\operatorname{span}\{e_1,e_2\}$.
If you want the embedding to be unital, you can take $p=\sum_kE_{2k-1,2k-1}$ and $q=\sum_kE_{2k,2k}$. Then $p,q$ are orthogonal projetions with $p+q=1$. If we take $v=\sum_k E_{2k-1,2k}$, then $v$ is a partial isometry with $v^*v=p$, $vv^*=q$. You can now identify $N=\operatorname{span}\{p,q,v,v^*\}$ with $M_2(\mathbb C)$ (as a $*$-isomorphism) by mapping $$ p\longmapsto e_{11},\qquad v\longmapsto e_{12},\qquad v^*\longmapsto e_{21},\qquad q\longmapsto e_{22}. $$