linear algebra - Does bra-ket notation work for all inner product spaces?

My quantum computation instructor keeps referring to the vector space in which he is using Dirac's bra-ket notation as an "inner product space", but doesn't it need additional properties to use that notation? In particular don't we need to

1. specify an implementation of the inner product in terms of some other vector space for the first argument; and




2. require that the inner product be linear in that argument.

The first requirement seems to be necessary to get bras in the first place (I gather there are some theorems that guarantee we can use the inner product to do this) and the second seems necessary to allow us to identify the notation with something like "multiplication" of a bra and a ket.

Does bra-ket notation work for all inner product spaces, or are additional properties required? If so, do these properties have names; does the space that has them?

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Forgive the naive formulation. I may not have the language quite right.

Answer

Given an inner product space $V$, you can imagine that there are two different copies of $V$, say $V_1$ and $V_2$, in which each vector $v\in V$ corresponds to a bra $\langle v|\in V_1$ and a ket $|v\rangle\in V_2$. To multiply a bra and a ket together, $\langle v|$ times $|w\rangle$ will by definition be $\langle v,w\rangle$ via the inner product.

Another way to think about this is as $V$ and its Hilbert space dual $V^*$ being identified together; each vector $v\in V$ is afforded the covector $v^*$ which is the linear mapping $v^*(w):=\langle w,v\rangle$ afforded by the given inner product. In this setting the covectors / dual vectors / linear functionals $v^*$ are denoted as bras $\langle v|$ and the usual vectors as kets $|v\rangle$, & multiplication is evaluation $\langle v||w\rangle=v^*(w)=\langle w,v\rangle$.

The reason for $v^*(w):=\langle w,v\rangle$ having $v$ in the second argument is so that each bra is a complex-linear functional of the argument $w$. This is related to a Hilbert space $V$ and its dual $V^*$ being anti-isomorphic; see Riesz representation theorem.