# 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 $$\langle w | \hat{A}v \rangle~=~\langle \hat{A}^{\dagger}w | v \rangle.$$

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$$^{1}$$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.