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In studying the math of Quantum Mechanics, I came across the idea of the Dual Space, and how we need it to take scalar products. If a Hilbert space of kets represents state vectors/wavefunctions, what does the dual space of that Hilbert space represent in a physical sense? What exactly does a ket's corresponding bra represent?

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  • $\begingroup$ I'm not sure how detailed an answer you were looking for, but a space of bras also represents state vectors. An inner product $\langle \psi|\phi\rangle$ can be thought of as the action of a linear functional ($\langle \psi$) on the ket $|phi\rangle$. When you write $|\psi\rangle\langle\phi|$, the fact that $\langle \phi|$ belongs to the dual Hilbert space is important. In a matrix representation, the ket is represented by a column vector and a bra by a row vector (so that $|\psi\rangle\langle\phi|$ is a matrix). $\endgroup$ – leastaction Sep 22 '16 at 4:02
  • $\begingroup$ I am not writing an answer because I am not sure of this, but something that could hint a solution to your answer would be to consider the Non-Hermitian case. In the Non-Hermitian case, the dual space is not the ket space of the Hamiltonian itself but the ket space of the the Hermitian conjugate of it. $\endgroup$ – Chetan Waghela Dec 5 '16 at 8:23
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Kets do not represent wave functions, so this may contribute to your idea of there being an imbalance between bras and kets. Instead, we do not get a wave function associated with a state (ket, $\left|\alpha\right>$) until we project it onto a basis, such as the position- or momentum-space bases, revealing that the wave function is actually a scalar:

$\psi(x) = \left<x|\alpha\right>$, $\phi(p) = \left<p|\alpha\right>$.

This can be likened to the scenario of a vector, $\vec{v}$, which has an existence and meaning independent of any basis, as can be seen geometrically with an arrow. It is only when we choose a particular basis, such as $\hat{i}, \hat{j}, \hat{k}$, that we can specify the vector by its (scalar) components. Similarly, the wave functions (in $x$, $p$, or some other basis) merely specify the components of a given ket (for a nice discussion related to this, see the opening chapter of Shankar's Principles of Quantum Mechanics). There is nothing stopping us from projecting a bra onto a basis:

$\psi^*(x) = \left<\alpha|x\right> = \left<x|\alpha\right>^*,\ \phi^*(p) = \left<\alpha|p\right> = \left<p|\alpha\right>^*$.

What this gives us, though, is the complex conjugate of our previous wave functions. The physical meaning of why the complex conjugate shows up with the bra here is related to our conception of probability being a non-negative measure: since quantum mechanics necessitates the use of complex numbers, in order to guarantee that $\left<\alpha|\alpha\right> \ge 0$, the dual to a ket has to involve complex conjugation ($z^* z \ge 0$ in general). As previous comments have indicated, the only remaining difference is that between column vectors and row vectors, which is arguably a convention.

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The bra also represents the state. Each bra is associated with exactly one ket, so the two contain exactly the same information.

I think it's easiest to see this when we consider finite-dimensional systems. Say we have a ket $| \psi \rangle = [a, b]$. The corresponding bra is the object which brings $| \psi \rangle$ to its norm: in this case, $\langle \psi | = [a*, b*]^T$. I.e. for finite-dimensional systems the kets are just the (conjugate) transpose of the bras!

So no new information is added. We need nevertheless to distinguish between the bras and the kets in order to coherently form inner products.

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  • $\begingroup$ So would it be right to say that there is a symmetry in the sense that quantum mechanics could have been formulated by interchanging the bras and kets? $\endgroup$ – Sreekar Voleti Sep 23 '16 at 3:50
  • $\begingroup$ Well I wouldn't exactly call it a "symmetry", but sure you could do that. This is sort of like saying that Newtonian physics could equally well be formulated using either row or column vectors. $\endgroup$ – AGML Sep 23 '16 at 23:51
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The dual space concept describes the Hilbert space, which is by definition, a complete inner product space. The duality is a property of the vector spaces. That particular is essential to define our Hilbert space to be complete. That is it helps us to make any measurements we are allowed to ask in the vector space.

A definite state of a physical system is represented as a state ket (or a state vector) in the complex vector space. It can be visualized as a vector in the ket space. Corresponding to that particular state ket in the ket space, there exists a mirror image of the vector (mathematically and more precisely called a dual) in some other space called the bra space, so that there is a one-to-one (linear) mapping between any vector in the ket space with it's dual in the bra space.

Together with these bras and kets, we could make physical information about the system and these information lies in the Hilbert space. For example, you have a state ket $|\alpha\rangle$, which belongs to the ket space. Then there exists a corresponding dual- a state bra $\langle\alpha|$- in the bra space, so that we could know physically relevant information about the system by combining them somehow. In short, a state ket as such, even though represent the physical system and is postulated to be containing all the physical information about the system that are allowed to ask, we cannot derive the information from it unless we make use of it's dual in the bra space.

In short, the state ket as such is of no importance if it cannot give us any relevant information about the system. It becomes physically meaningful if it could give values of the measurement of observables, which is by taking the inner product of the ket and bra vectors. That means, any measured value should be contained in the space where we do the operation. A dual space guarantees that. Also, if you operate with your state ket some operator corresponding to a physical observable, it is postulated that the result will be a state ket. Hence our dual space must satisfy the completeness or closure property.

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