How does a linear operator act on a bra?

If $A:V\to V$ is a linear operator on a vector space $V$ with basis $|m\rangle$, Dirac defined the action $\langle n|A$ of $A$ on the bra vector $\langle n|$ (an element of the dual space $V^*$)as a new bra vector that is evaluated on a ket $|m\rangle$ to give the same answer as the evaluation of $\langle n|$ on $A|m\rangle$. In other words he set $$ (\langle n|A)|m\rangle = \langle n|(A|m\rangle). $$ Since it does not matter where one puts the brackets Dirac just writes $$ \langle n|A|m\rangle $$ for both things. Note that the dagger operation $\dagger: |m\rangle\mapsto \langle m|= (|m\rangle)^\dagger$ that provides an antilinear map between $V$ and $V^*$ (and so defines the inner product on $V$) is not needed here. So $A$ acting on bras has nothing to do with the hermitian conjugate $A^\dagger$ which needs the inner product. As $A$ acting to the left on bra vectors is really an operator $V^*\to V^*$ rather than an operator $V\to V$ it usual in linear algebra to regard it as different operator $A^*:V^*\to V^*$ that is called the "conjugate," or the "transpose." The latter name is probably best as no complex conjugation is involved, and in the dual basis $A^*$ is representated by the transpose (but not the complex conjugate transpose) of the matrix representing $A$.

Note added: I just saw Qmechanic's answer. What he says: $A|\phi\rangle = |\chi\rangle \Leftrightarrow \langle \phi|A^\dagger =\langle \chi|$ is quite correct, but you really don't need to use the dagger to define left action of $A$.

  1. Using Shankar's notation from his eq. (1.6.14), an operator acting on a bra$^1$ $$\langle \phi | ~=~ \langle \psi |\hat{A} $$ is equivalent to the Hermitian adjoint operator acting the corresponding ket $$|\phi \rangle ~=~ \hat{A}^{\dagger}|\psi \rangle . $$

  2. The Hermitian adjoint operator fulfills the relation $$ \langle w | \hat{A}v \rangle~=~\langle \hat{A}^{\dagger}w | v \rangle. $$


$^{1}$We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.

There is a standard definition for the action of a linear operator on a bra, although many authors (including Shankar) sloppily skip defining it. $\langle \phi| A$ is defined to be the linear functional such that $\big( \langle \phi| A \big) |\psi\rangle \equiv \langle \phi| \big(A |\psi\rangle\big)$. In other worlds, you first act the linear operator on the ket that's getting "eaten" and then act the bra $\langle \phi |$ on the output of that.

Similarly, we can define the action of the adjoint of a linear operator on a ket. $$A^\dagger | \psi \rangle$$ is defined by how an arbitrary bra $\langle \phi |$ acts on it: $$\langle \phi | \big( A^\dagger | \psi \rangle \big) := \big( \langle \phi | A^\dagger \big) | \psi \rangle = \langle A \phi | \psi \rangle.$$ Note that unlike the former case, this requires an inner product structure so that we can define the adjoint.

The inner product of a ket $A^\dagger | \psi \rangle$ with all possible kets $|\phi\rangle$ completely specifies the ket, because the positive-definiteness of the inner product implies that it's nondegenerate. It's not often appreciated that the adjoint of an operator only has a unique action on kets because of the positive-definiteness requirement on the inner product.