Is $\Phi_A$ meaningful for a non-commutative Banach Algebra $A$? For a commutative Banach Algebra $A$ we can define $\Phi_A$, the character space of $A$, to be the set of all characters of $A$. (A character is a non-zero ring homomorphism $\phi:A\to \Bbb C$.) 

What happens if we allow $A$ to be non-commutative?

An issue that I can think of is that the Gelfand Representation Theorem might not hold (although I am not sure if its proof relies on commutativity in any non-trivial way). Nevertheless, is there a merit in talking about, says, the character space of $\mathcal B(H)$, the algebra of bounded linear operators on $H$? 
 A: I'm not sure about how much of this goes through for general Banach algebras, but I can talk about the $C^*$ case.  
For C$^*$-algebras (including $B(H)$), we can try to generalize the character space by using representation theory.  If $A$ is an abelian C$^*$-algebra, then any character gives an irreducible representation of $A$.  In fact, the character space is precisely the collection of all irreducible representations of $A$.  So to generalize the character space (also sometimes called the spectrum of $A$), we gather up all of the irreducible representations of our algebra ($B(H)$ for example), and study that (after suitably topologizing it).  
We can also study a related space, called the primitive spectrum of $A$, where we collect all of the kernels of irreducible representations instead of the representations themselves. Obviously there's a surjection from the spectrum of $A$ to the primitive spectrum, since we can just associate to each irreducible representation its kernel.
For commutative C$^*$-algebras, the character space, spectrum, and primitive spectrum all agree.  So we really are generalizing the character space, at least in the C$^*$ setting.
One nice thing that we learn from Gelfand duality is that we can fully recover a commutative C$^*$-algebra $A$ from its spectrum, so the spectrum plays the role of a sort of dual object to $A$.  Unfortunately, we can't recover C$^*$-algebras in general from their spectra: The spectrum of $K(H)$ is just a single point.  In particular, we can't distinguish $M_n(\mathbb{C})$ from $M_m(\mathbb{C})$ by using their spectra.
For more, check out the wikipedia page here.
