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I am taking my first steps in learning quantum mechanics and am learning about Dirac's bra-ket notation. I am trying to understand what the inner product is.

My understanding so far: the inner product is an operation between to vectors which returns a scalar. This allows us to define orthogonality: two vectors are orthogonal when their inner product is 0. The inner product is a generalisation of the dot product I have been using up till now, which is essentially the inner product restricted to vectors in $\mathbb{R}^{n}$, always returning real scalers. An inner product space is a vector space for which the inner product is defined.

This is where I get confused: so far, I have been applying the dot product to vectors from the same vector space. Furthermore, from Wikipedia: the inner product "associates each pair of vectors in the [inner product] space with a scalar quantity known as the inner product of the vectors."

However, going through Shankar's Principles of Quantum Mechanics, I have learnt that the vector space of kets has an associated vector space of bras, its dual space. The textbook states the inner product is only defined between bras and kets and hence only between a vector space and its dual space. I haven't found anything about an inner product kets or bras, and my gut feeling is it wouldn't make sense. Are the associated vector spaces to bras and kets not inner product spaces? Or would the inner product just be meaningless?

In summary, is the inner product within a vector space the same as the inner product between bras and kets, or am I confusing two different ideas? What are, in general, the operands on which the inner product acts?

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    $\begingroup$ If you think of the inner product as a matrix multiplication, it should become clearer: bra is a row vector, whereas ket is a column. $\endgroup$ – Vadim Mar 10 at 17:08
  • $\begingroup$ But isn't the dot product, an inner product, defined between any vectors of equal length, regardless of whether they're row vectors or column vectors ? $\endgroup$ – Mr Lolo Mar 10 at 17:24
  • $\begingroup$ You cannot multiply two columns or two rows using matrix multiplication rules. However thinking of it as matrix multiplication saves you from bothering with dual spaces, etc. $\endgroup$ – Vadim Mar 10 at 17:34
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    $\begingroup$ Actually the space of row matrices is simply the dual of the space of column matrices. This is in fact a good way to think of the dual, at least for finite dimensional spaces. The dot product is simply thought of as meaning that you flip a column matrix to a row matrix, and use matrix multiplication, as suggested. This is fine, till you want to do something like deal with vectors in curvilinear coordinates (essential for gtr), when it gets a bit more complicated. $\endgroup$ – Charles Francis Mar 10 at 19:35
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    $\begingroup$ An Hilbert space and its continuous dual (the space of all continuous linear functionals) are the same. This result is known as Riesz's representation theorem. In fact, it is the inner product the way of mapping elements of the Hilbert space to elements of the dual: given $\phi\in\mathscr{H}$, then the corresponding element of the dual is $\langle \phi, \cdot\rangle$. Conversely, every element of the continuous dual can be written as the latter, for some $\phi$. $\endgroup$ – yuggib Mar 11 at 8:18
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The inner product is a map

$$\langle.,.\rangle : \mathcal{H} \times\mathcal{H} \rightarrow \mathbb{C} \\ (\psi,\phi) \mapsto \langle \psi,\phi\rangle$$

that sends two vectors of a vector space $\mathcal{H}$ (in QM this is actually not only a vector space but even a Hilbert space) to the complex numbers. In physics notation, vectors in $\mathcal{H}$ are often written in bra ket notation as a ket $|\psi \rangle$. One reason for that is to distinguish them from "usual" finite dimensional vectors which are written with an arrow above. The point is that vectors in Hilbert space are abstract objects, they are differen from (but often equivalent with) a set of numbers $(\psi_1,\psi_2,...)$ which are the representation of that vector in some basis.

A "bra vector" is a map

$$\alpha^\star : \mathcal{H} \rightarrow \mathbb{C}\\ \psi \mapsto \langle\alpha,\psi\rangle, $$

that sends vectors in Hilbert space to the complex numbers, using the previously defined inner product. Here, $\alpha$ is some vector $\in \mathcal{H}.$ In fact the maps $\alpha^\star$ (read that as a single object, not as $\alpha$ complex conjugated) actually also from a vector space since you can add them and multilply them by a number, which is why they are refered to as bra vectors. Every bra vector also uniquely corresponds to a vector in $\mathcal{H}$ (there is an isomorphism between the two vector spaces) which is why we can denote both the map and the corresponding ket vector in the inner product with $\alpha$.

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  • In summary, is the inner product within a vector space the same as the inner product between bras and kets, or am I confusing two different ideas?

Yes, it is essentially the same thing. Given two kets $|f>$ , $|g>$, we define the inner product $<f|g>$, and this enables us to define the dual space of bras as the space of functionals on ket space $$<f| : |g> \mapsto <f|g> $$

A word of caution, however. This only really works for finite dimensional inner product space. For finite-dimensional Hilbert space it is clear that there is an (anti-)isomorphism between bras and kets (anti- refers to complex conjugation). The Riesz representation theorem extends this result to infinite-dimensional Hilbert space (completeness is required). More generally, for an infinite-dimensional vector space there may be functionals, or bras, in the dual space which do not correspond to kets, and there may be kets such that no corresponding functional is defined according to an inner product.

If you have an infinite dimensional space (as we usually do in qm), then you can get away with pretending that it works most of the time, even when it doesn't. For example, we pretend that we have a Hilbert space spanned by position states $|x>$, but the inner product $<x|y>$ is a delta function. We pretend that this is all covered in distribution theory, but actually it isn't. There are severe restrictions in the theory of distributions. As far as ordinary quantum mechanics is concerned I am not aware of any serious issues which arise, but in relativistic quantum mechanics there are severe problems leading to divergences in qed. Ultimately it means that there is no mathematical definition of quantum fields as generally used in quantum field theory.

I have addressed these problems in A Construction of Full QED Using Finite Dimensional Hilbert Space . Also in Light After Dark III: The Mathematics of Gravity and Quanta, in which I give a mathematically rigorous treatment.

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