This is a reference request. I'm looking for a definition of Morita equivalence of *-algebras, as described below. If anyone thinks that this is not the right way to define Morita equivalence of *-algebras, I'd be interested to hear that also.

By a *-algebra, I mean an algebra equipped with an anti-involution ${}^*$, $(ab)^* = b^*a^*$. (I'm mainly interested in finite-dimensional algebras.)

Recall that a Morita equivalence of algebras $A$ and $B$ consists of (1) a pair of bimodules $_AM_B$ and $_BN_A$, and (2) bimodule isomorphisms $f:{}_A(M\otimes N)_A \to {}_AA_A$ and $g:{}_B(N\otimes M)_B \to {}_BB_B$ such that (3) $f$ and $g$ satisfy the zig-zag identities.

If $A$ and $B$ are *-algebras, it seems to me that the definition of Morita equivalence should be modified (enhanced) as follows.

First, note that the *-structure of $A$ allows us to convert between left and right $A$-modules, and similarly for $B$. In particular, to any $A$-$B$ bimodule $_AX_B$ there is associated a $B$-$A$ bimodule $_BX^*_A$. The first modification to the definition of Morita equivalence is to replace $_BN_A$ above with $_BM^*_A$.

Next, note that both $_AA_A$ and $_A(M\otimes M^*)_A$ have involutions coming from the *-structure. The second (and final) modification to the definition is to require that the isomorphism $f:{}_A(M\otimes M^*)_A \to {}_AA_A$ intertwine with these two involutions, and similarly for $g:{}_B(M^*\otimes M)_B \to {}_BB_B$.

Recall that an (ordinary) Morita equivalence gives an isomorphism between 0-th Hochschild homologies $HH_0(A) \cong HH_0(B)$. A *-structure on $A$ gives rise to an involution of $HH_0(A)$. The isomorphism $HH_0(A) \cong HH_0(B)$ coming from an ordinary Morita equivalence need not commute with the involutions on either side, but the above enhanced Morita equivalences do guarantee compatibility with the involutions on $HH_0$.

So, repeating my initial question, where can i find the above definition in the literature?

complex conjugatevector space $\overline{M}$ as its underlying space. Otherwise the right action of $A$ is conjugate-linear rather than linear. +1 Interesting question! – MTS Dec 25 '12 at 0:53