Visualising the Conjugate Transposition of a Vector

[H Smith 94]

Hi there!

As you might have already guessed, I'm referring primarily to the 'geometrical' difference (is there such geometry in Hilbert space?) between ##n##-dimensional state vectors
[tex] | \psi \rangle = \left( \begin{matrix} \psi_1 \\ \psi_2 \\ \vdots \\ \psi_n \end{matrix} \right) [/tex]
and their corresponding basis vectors
[tex] \langle \psi | = \left( \begin{matrix} \psi_1^* & \psi_2^* & \cdots & \psi_n^* \end{matrix} \right). [/tex]
What would these (particularly the latter) look like on a graph? (What kind of graph could one even represent these on?)

Additionally, if the inner product of two vector spaces represents the projection of one vector onto another, the inner product of these two vector spaces would equal 1, meaning they are parallel (which is true). But what does that mean for two non-equal states? For example,
[tex] \langle \phi | \psi \rangle [/tex]
where ##|\phi\rangle \ne |\psi\rangle##. How does this inner product represent the probability amplitude of the wavefunction from two separate states? I've been imagining this as an ##n##-dimensional generalisation of the dot (scalar) product of two vectors thus far!

I'm looking mainly for conceptualisations/visualisations (personal interpretations invited) of how this process works but also corrections on where I'm misunderstanding the theory.

Thanks in advance,

Harry

 

[bhobba]

Additionally, if the inner product of two vector spaces represents the projection of one vector onto another, the inner product of these two vector spaces would equal 1,

I don't know what you mean by the inner product of two different spaces. If you have two spaces you need to form their product space |a>|b> and the inner product will not in general be one in that product space:
http://en.wikipedia.org/wiki/Tensor_product#Tensor_product_of_vector_spaces

In fact with the usual way of doing it the inner product will be zero ie you get a vector (a1, a2, ...an,b1, b2,...bm). A vector with all the bj zero will always have an inner product of zero with a vector that has the ai zero.

Thanks
Bill

 

[H Smith 94]

I don't know what you mean by the inner product of two different spaces. If you have two spaces you need to form their product space |a>|b> and the inner product will not in general be one in that product space:
http://en.wikipedia.org/wiki/Tensor_product#Tensor_product_of_vector_spaces

Hi Bill, thanks for your response!

I apologise for the ambiguity - I didn't mean the inner product of two vector spaces; I meant the inner product of two vectors, [itex]|a\rangle[/itex] and [itex]|b\rangle[/itex] within the same vector space. I realize now that is an important distinction to make!

 

[ddd123]

Liboff - Introductory Quantum Mechanics offers a graphical representation similar to what you're thinking when it discusses the Hilbert space.

 

[H Smith 94]

Liboff - Introductory Quantum Mechanics offers a graphical representation similar to what you're thinking when it discusses the Hilbert space.

Thank you! Yes that is very helpful!

 

[vanhees71]

A "ket" represents a vector in Hilbert space, a "bra" a covector. If you take the topological dual space, you have the Hilbert space again (or more precisely the topological dual of the Hilbert space is equivalent to the Hilbert space again).

In quantum theory, however, you extend this notion of the dual space to an extended object, i.e., to distributions on the Hilbert space. The reason is that then you can deal with "generalized eigenvectors" for undbound self-adjoint operators that are defined only on a dense subspace of the Hilbert space, and you consider the dual space of this subspace. This is called the "rigged Hilbert-space formalism). You find a good physicists' introduction in

L. Ballentine, Quantum Mechanics, Addison Wesley.