Which Hilbert space is isometrically isomorphism with $B(E)$ for some Banach space $E$. Consider $H$ as a Hilbert space. How can I find a Banach space $E$, for that, $H=B(E)$ where $B(E)$ is the set of bounded linear operator on $E$?
(At least under some conditions on $H$)
Also if the Hilbert space $H$ is Banach algebra, when we can find a Banach space $E$, such that $H$ is a Banach-algebra-isomorphism isometric with some Banach subalgebra of $B(E)$? 
 A: As noted in my comment, $E$ has to be a Hilbert space.
To see this, fix a linear functional $\varphi \in E'$ with $\Vert \varphi \Vert = 1$. For $z \in E$, define
$$
A_z : E \to E, x\mapsto \varphi(x) \cdot z.
$$
It is not hard to see that $E \to B(E), z \mapsto A_z$ is linear and isometric. Thus, if $B(E) = H$ is a Hilbert space, we see that $E$ is isometrically isomorphic to a subspace of a Hilbert space and thus a Hilbert space itself.
Now we show that $B(E)$ can not be a Hilbert space, so that $H = B(E)$ can not hold (as an isometric isomorphism).
EDIT: The idea for this part of the proof is due to @David C. Ullrich, see the comments to the question.
For this, let us assume $\dim(H) > 1$ (otherwise, the claim is trivially true, since $H \cong \Bbb{C} \cong B(\Bbb{C})$). Thus, there are $x,y \in H$ with $x \perp y$ and $\Vert x \Vert = 1 = \Vert y \Vert$. Let $H_1 := {\rm span}(x,y)$ and $H_2 := H_1^\perp$. Let $A,B$ be the bounded operators given in matrix notation (with respect to $H= \langle x \rangle \oplus \langle y \rangle \oplus H_2$) by
$$
A=\left(\begin{matrix}1\\
 & 0\\
 &  & 0
\end{matrix}\right),B=\left(\begin{matrix}0\\
 & 1\\
 &  & 0
\end{matrix}\right).
$$
Put differently, $A = \pi_1$ is the orthogonal projection onto ${\rm span}(x)$ and likewise $B = \pi_2$ is the orthogonal projection onto ${\rm span}(y)$. Since $A+B$ is the orthogonal projection onto $H_1$, we get
$$
\Vert A\Vert = \Vert B \Vert = \Vert A+B\Vert =1.
$$
It is also not hard to see $\Vert A - B\Vert = 1$.
But this  is incompatible with the parallelogram identity
$$
\Vert x + y\Vert^2 + \Vert x-y \Vert^2 = 2 (\Vert x \Vert^2 + \Vert y \Vert^2),
$$
since the left-hand side would yield $2$, while the right-hand side would yield $4$ with $x=A,y=B$.
A: I know a very special case, and I guess it is extensible. 
Suppose that $H$ is Hilbert space such that $H\cong (H_1\hat{\otimes}(H_2)^*)^*$, where $\hat{\otimes}$ is projective tensor product, and $H_1,H_2$ is some Hilbert spaces.
For any Hilbert space $\mathcal{H}$, always $\mathcal{H}^{**}=\mathcal{H}$. Also for two Banach space $E,F$ always $B(E,F^*)\cong (E\hat{\otimes} F)^*$. Hence 
$$H=(H_1\hat{\otimes} (H_2)^*)^*=B(H_1,(H_2)^{**})=B(H_1,H_2)$$
So if $\mathcal{H}$ is a Hilbert space such that $H=(\mathcal{H}\hat\otimes(\mathcal{H})^*)^*$, then $H=B(\mathcal{H})$.
A: For your second question, the answer is always yes.  More generally, given any Banach algebra $A$, let $\tilde{A}=A\oplus \mathbb{C}$ be its unitization (with norm $\|(a,z)\|=\|a\|+|z|$).  Given $a\in A$, let $L_a\in B(\tilde{A})$ be left multiplication by $a$.  Then $a\mapsto L_a$ is an isometric isomorphism from $A$ to a subalgebra of $B(\tilde{A})$.  (Adjoining a unit is necessary because otherwise $\|L_a\|$ might be smaller than $\|a\|$.)
