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Every vector space $|\vec{v}\rangle$ over the field $\mathbb{R}$ or $\mathbb{C}$ contains a dual space, and so if we make an identification between elements in the dual space and the original vector space, it seems that every vector space naturally comes equipped with an inner product, called the dual space inner product.

For example in quantum field theory we have a representation of the Poincare group where our vector space can be denoted as $|p^{\mu},\sigma\rangle$ where $\sigma$ denotes little group indices. Without any physical assumption we can say that there exists a space dual to this space, and that this gives rise to an inner product on our original vector space $\langle p,\sigma|p',\sigma'\rangle=\delta(p-p')\delta_{\sigma\sigma'}$. Now this is AN inner product but not necessarily THE inner product against which the representation of the Poincare group is unitary.

Question:Am I correct in saying that for multiparticle states the dual space inner product is \begin{equation} \langle \lbrace p,\sigma\rbrace|\lbrace p',\sigma'\rbrace\rangle=\sum_{\text{all possible pairings of primed states with unprimed states}}\,\,\,\,\prod_{\text{pairs}}\delta(p_i-p_{i^{'}}')\delta_{\sigma_i\sigma_{i'}'} \end{equation} whereas there is another, distinct, inner product given by the amplitude: \begin{equation} \langle \lbrace p,\sigma\rbrace|\lbrace p',\sigma'\rbrace\rangle=\delta(\sum p -\sum p')\mathcal{M}(\lbrace p,\sigma\rbrace,\lbrace p',\sigma'\rbrace) \end{equation} We want the representation of the Poincare group to be unitary against both inner products. TLDR: Are the dual space inner product and the inner product against which we want our representation to be unitary distinct?

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  • $\begingroup$ The correspondence between a Hilbert space and its dual is given by the Riesz representation theorem, which isn't known nearly as widely as it should be. $\endgroup$ – leftaroundabout Aug 20 at 17:50
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Making an identification between the dual space and original space is entirely equivalent to choosing an inner product. There are infinitely many ways of identifying $V$ with $V^*$, so there are infinitely many possible inner products.

You may think that, given a basis ${\bf e}_a$ for $V$ and a dual basis ${\bf e}^{*a}$for $V^*$ such that ${\bf e}^{*a}({\bf e}_b)=\delta^a_b$, you can naturally identify ${\bf e}_a$ with ${\bf e}^{*a}$. You can do this of course, but there are infinitely many choices of basis, and each one gives a different identification and a different inner product. In quantum mechanics we make a choice of inner product by a choice of the antilinear dagger map $\dagger :V\to V^*$ in which $\dagger: |n\rangle \mapsto (|n\rangle)^\dagger =\langle n|$. By choosing to identify the "$|p\rangle$" (momentum) basis with the its dual, your recipe makes a particular choice of inner product.

I think that rather than speaking of the dual space inner product, you should speak of the dual basis inner product.

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  • $\begingroup$ Thanks! So in non-relativistic quantum mechanics, where we examine a representations of the translation and rotation groups, it is a physical assumption to take the identification between the vector space and its dual to be the antilinear dagger map? Perhaps given the representation, this is the only pairing against which the representation is unitary, in which case the only physical assumption is which representation you take. $\endgroup$ – Luke Aug 20 at 17:45
  • $\begingroup$ I think so. If we use the $x$ basis, then requiring translation invariance pretty well determines $\langle \psi|\phi\rangle =\int \psi^*(x)\phi(x) dx$ and hence $\dagger:\phi\mapsto \phi^*$. $\endgroup$ – mike stone Aug 20 at 17:48
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You can check that \begin{align} \hat {\cal L}_z= -i\frac{d}{d\varphi}\, ,\quad \hat {\cal L}_+= -e^{i\varphi}\left(\lambda\mathbb{I}+i\frac{d}{d\varphi}\right)\, ,\quad \hat {\cal L}_-= -e^{-i\varphi}\left(\lambda\mathbb{I}-i\frac{d}{d\varphi}\right)\, . \tag{1} \end{align} satisfy the same commutation relations as $\hat L_z, \hat L_\pm$. Assume the operators in (1) act on functions of the form $f_m(\varphi)=e^{i m\varphi}/\sqrt{2\pi}$.

The "natural'' inner product is $\langle{m}\vert{m'}\rangle= \int d\varphi \, e^{-i m\varphi}e^{i m'\varphi}/(2\pi)$ but if you use this you will find that the matrix representation of $\hat {\cal L}_x$ and $\hat{\cal L}_y$ acting on the states $f_{m}(\varphi)$ are not hermitian matrices, so does not exponentiate to a unitary representation.

In other words, there is no reason to believe that the natural "inner product" for states will produce a unitary representation.

It's not hard to be uncomfortable with (1) since the "usual" representation by differential operators acts not on the 1-torus but on $S^2/U(1)$ (the spherical harmonics); it's intuitively weird to have a kind of coordinate-based representation of $SU(2)$ depending on only one angle.

In the case of compact group (such as $SU(2)$ above), what you can say is that the matrix representation of (1) is equivalent (by a similarity transformation) to a unitary one. There are systematic ways of finding the similarity transformations. In the case of non-compact groups, it is more delicate to establish such an equivalence.

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I think the following example shows that I am correct that inner product against which a representation of a group should be unitary does not necessarily coincide with the dual space inner product, but I would still appreciate feedback.

Consider the $(\frac{1}{2},0)$ representation of $SL(2,\mathbb{C})$. This acts on the space of two dimensional complex vectors $\begin{pmatrix}a\\ b\end{pmatrix}$. Now this vector space is naturally paired with a dual space $(a^{\star},b^{\star})$ and hence we have the inner product on our original vector space as \begin{equation} \bigg( \begin{pmatrix}a\\ b\end{pmatrix}, \begin{pmatrix}c\\ d\end{pmatrix}\bigg)_{\text{Dual space inner product}}=ac^{\star}+bd^{\star} \end{equation} Now the representation is not unitary with respect to this inner product, because for example in some conventions the Lie Algebra elements associated to boosts are non-Hermitian. However, there does exist an inner product on this vector space which is unitary, namely the determinant of the two vectors \begin{equation} \bigg( \begin{pmatrix}a\\ b\end{pmatrix}, \begin{pmatrix}c\\ d\end{pmatrix}\bigg)_{\text{Determinant inner product}}=ad^{\star}-bc^{\star} \end{equation} So, the inner product against which a representation is unitary need not coincide with the dual space inner product.

Edit 1: In view of Mike Stones response, one could correct what I said as follows. One can view the determinant inner product as the dual space inner product if one chooses the association between the vector space and dual space to be: \begin{equation} \begin{pmatrix}a\\ b\end{pmatrix}\rightarrow \bigg(b^{\star},-a^{\star}\bigg) \end{equation}

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  • $\begingroup$ Sorry I accidently edited your answer instead ow my own! $\endgroup$ – mike stone Aug 20 at 17:51

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