Why the bra vector is said to belong in the dual space?

[patric44]

Homework Statement

why the bra vector lives in the dual space ?

Relevant Equations

bra vectors belongs to the dual space ?

hi
i was recently introduced to the Dirac notation and i guess i am following it really well , but can't get my head around the idea that the bra vector
said to live in the dual space of the ket vectors , i know about linear transformation and the structure of the vector spaces , and i realize that the bra should live in another space basically because its a raw vector as it comes from transposing the ket vector , but why it lives in the dual space ?
and comparing with the formal definition of the dual space : the space of all linear transformation of v to R ... , where the bra vector fit in that definition ?

 

[PeroK]

The bra associated with a ket $|a>$ is the linear transformation that maps any ket $|b>$ to the complex number $<a|b>$.

 

[patric44]

The bra associated with a ket $|a>$ is the linear transformation that maps any ket $|b>$ to the complex number $<a|b>$.

oh, so i can thought of the bra vector as a functunal i may denote by $L(v_{i}) = (v_{i}^{T})^{*}.v_{i}$ , which just take the inner product of the vector with its transpose conjugate and is just "fancy" denoted by $<v_{i}|$ , am i getting this right?

 

[PeroK]

oh, so i can thought of the bra vector as a functunal i may denote by $L(v_{i}) = (v_{i}^{T})^{*}.v_{i}$ , which just take the inner product of the vector with its transpose conjugate and is just "fancy" denoted by $<v_{i}|$ , am i getting this right?

I'm not sure i understand that. In linear algebra if you take a vector and generate the inner product of that vector with all other vectors then that is a libear functional from the vector space to the complex numbers.

 

[PeroK]

The dual space is the space of all linear functionals. And it can be shown that every linear funtional is the inner product corresponding to some vector as above. I.e. the dual space is the set of these functionals defined by the inner product.

 

[patric44]

I'm not sure i understand that. In linear algebra if you take a vector and generate the inner product of that vector with all other vectors then that is a libear functional from the vector space to the complex numbers.

but that seems just like what the bra vector do ? doe's the inner product requires the creation of a bra vector to be completed ?

 

[PeroK]

but that seems just like what the bra vector do ? doe's the inner product requires the creation of a bra vector to be completed ?

It's assumed there is an inner product defined on the vector space. A Hilbert space is an inner product space. It must have an inner product defined on it.

 

[patric44]

The dual space is the space of all linear functionals. And it can be shown that every linear funtional is the inner product corresponding to some vector as above. I.e. the dual space is the set of these functionals defined by the inner product.

ok i get it now thank you so much

 

[patric44]

It's assumed there is an inner product defined on the vector space. A Hilbert space is an inner product space. It must have an inner product defined on it.

i have a little bit of a stupid question : suppose that i have a vector space , now i can perform linear operations on that space to map it to a specific field and correspondingly i have a dual space associated with that space , it seems now that i can define an inner product for every vector space i could have ? , which would mean that every vector space is a Hilbert space ?!

 

[PeroK]

i have a little bit of a stupid question : suppose that i have a vector space , now i can perform linear operations on that space to map it to a specific field and correspondingly i have a dual space associated with that space , it seems now that i can define an inner product for every vector space i could have ? , which would mean that every vector space is a Hilbert space ?!

Yes, finite dimensional vector spaces are all the same. But, you do have a choice of how to define the inner product. So, you have a choice of which bra vector a given functional corresponds to. In QM this is usually not something you need to worry about.

 

[patric44]

Yes, finite dimensional vector spaces are all the same. But, you do have a choice of how to define the inner product. So, you have a choice of which bra vector a given functional corresponds to. In QM this is usually not something you need to worry about.

oh , every thing is clear now thank you so much

 

[PeroK]

oh , every thing is clear now thank you so much

Take $\mathbb R^2$ as an example. For any basis we can define the inner product of the two basis vectors as $0$. That makes them orthogonal with respect to that inner product even if they are not orthogonal with respect to the usual inner product.

Moreover, as Hilbert spaces they are isomorphic. So, there is no unique way to associate a specific vector with a linear functional.

 

[patric44]

Take $\mathbb R^2$ as an example. For any basis we can define the inner product of the two basis vectors as $0$. That makes them orthogonal with respect to that inner product even if they are not orthogonal with respect to the usual inner product.

Moreover, as Hilbert spaces they are isomorphic. So, there is no unique way to associate a specific vector with a linear functional.

so in that some how fake orthogonal space ${R}^{2}$ i can now expand any vector in terms of all the other vectors as they are all orthogonal using my new definition of the inner product i associated the space with . wow
so i have to be careful now with defining the basis expansion of my vectors as the basis may not be orthogonal with respect to the inner product i choose in the first place .

 

[vanhees71]

The dual space is the space of all linear functionals. And it can be shown that every linear funtional is the inner product corresponding to some vector as above. I.e. the dual space is the set of these functionals defined by the inner product.

Here one must be a bit careful. The Hilbert space is a pretty special vector space. By definition it is a metric vector space with the metric induced by the scalar product, and you have to distinguish between different kinds of dual spaces used in quantum mechanics.

Within the Hilbert space itself there is the the dual in the restricted sense of continuous linear forms, and these build a vector space isomorphic to the Hilbert space itself, i.e., for each such linear form there's a (true normalizable) vector $|\psi \rangle$ such that the linear form is given by the adjoint, i.e., $\langle \psi|$.

Nevertheless QT doesn't work with this kind of linear forms only, but as soon as you have to describe observables by an unbound essentially self-adjoint operator, and this is already the case for the most simple case of a single classical particle quantized by "canonical quantization", i.e., defining position and momentum operators obeying the canonical commutator relations, defining the Heisenberg algebra. They are only defined on a dense subspace of the Hilbert space. As self-adjoint operators this domain must also be their co-domain. This "nuclear space" is a proper subspace, and now you also have to consider the linear forms defined on this restricted subspace, defining its dual, and this dual is larger than the dual of the Hilbert space. This allows you to handle also "generalized eigenstates" of a self-adjoint operator in a formal manner, and this enables constructs like the eigenvectors of the position operator $|\vec{x}' \rangle$ and their (formal) dual. In position representation you explicitly see that this implies the use of generalized functions (distributions), in this case the Dirac $\delta$ distribution, $u_{\vec{x}'}(\vec{x}) =\langle \vec{x}|\vec{x}' \rangle=\delta^{(3)}(\vec{x}-\vec{x}')$. This is of course not defined on the entire Hilbert space of square-integrable functions (which is the concrete realization of the Hilbert space in terms of the position representation, which is nothing else than "wave mechanics") but only on the corresponding set of test functions (like the Schwartz space of "rapidly-enough vanishing functions").

All this hand-waving can beformalized mathematically in terms of the socalled "rigged-Hilbert space formalism", which is much closer to the physicists "robust mathematics" invented by Dirac than the equivalent more conventional formalization of it by von Neumann.

For a gentle introduction see

L. E. Ballentine, Quantum Mechanics, World Scientific,
Singapore, New Jersey, London, Hong Kong (1998).

For a more thorough treatment see

A. Galindo and P. Pascual, Quantum Mechanics, Springer
Verlag, Heidelberg (1990), 2 Vols.

of de la Madrid's PhD thesis:

http://galaxy.cs.lamar.edu/~rafaelm/webdis.pdf

 

[Kashmir]

The bra associated with a ket $|a>$ is the linear transformation that maps any ket $|b>$ to the complex number $<a|b>$.

Thank you. Is <a|b> the inner product here?

 

[PeroK]

Thank you. Is <a|b> the inner product here?

Yes.