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?

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


1 Answer 1


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.

  • $\begingroup$ You say "To multiply a bra and a ket together ... will by definition be ⟨v,w⟩ via the inner product." But that's not an integral part of the definition of the inner product, right? That's an additional piece. We're saying: One way to get an inner product is to multiply a bra and a ket together. Is that right? $\endgroup$
    – orome
    Mar 27, 2013 at 15:12
  • $\begingroup$ @raxacoricofallapatorius In your question you were talking about the inner product and bra-ket structures as if we already have the inner product and from it we need to form the bra-ket structure. Please inform me if I am misunderstanding you. This answer explains how bras, kets and their multiplication is understood via an already-given inner product. So, no, you do not need a bra-ket view of a space to have an inner product, and we're saying: the way to multiply a bra and ket together is to use the inner product. $\endgroup$
    – anon
    Mar 27, 2013 at 15:23
  • $\begingroup$ My question might need editing (my terminology is probably not clear): I'm just taking as given a linear vector space $L$, with an inner product — that is, some unspecified $f:L\times L\rightarrow\mathbb{C}$ that has three required properties: conjugate symmetry, linearity in the first argument, and positive-definiteness. The question is, is that enough to let us write the inner product as multiplication of bras and kets? $\endgroup$
    – orome
    Mar 27, 2013 at 15:33
  • $\begingroup$ @raxacoricofallapatorius yes, because the multiplication of bras and kets is defined to be the inner product, that's what my answer says. $\endgroup$
    – anon
    Mar 27, 2013 at 15:34
  • $\begingroup$ Ah, so (sorry I'm slow on the uptake) can the logic be thought of something like this: We start with an inner product, as yet undefined. Clearly I can't use bra-ket notation at that point. We note that the inner product is anti-linear in its first argument, so we can't just implement the inner product as multiplication. But, it turns out (I'd like a better understanding than that) that we can an always find a partner for every vector that lets us implement the inner product as multiplication by that partner. We label the partner as a bra, we define the inner product as bra-ket multiplication. $\endgroup$
    – orome
    Mar 27, 2013 at 15:52

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