Is there a ket to correspond to every bra?

[microsansfil]

Hi all

The existence of a scalar product in an Hilbert Space E enable us to show that we can associate, with every ket | V >, an element of E * (Dual space), that is, a bra, which will be denoted by < V |

Is it possible to find bras which have no corresponding kets ?

Best Regards
Patrick

 

[PeroK]

Hi all

The existence of a scalar product in an Hilbert Space E enable us to show that we can associate, with every ket | V >, an element of E * (Dual space), that is, a bra, which will be denoted by < V |

Is it possible to find bras which have no corresponding kets ?

Best Regards
Patrick

See:

https://en.wikipedia.org/wiki/Riesz_representation_theorem

Note the last comment about QM.

 

[microsansfil]

See:

https://en.wikipedia.org/wiki/Riesz_representation_theorem

Note the last comment about QM.

Thank.

Why in the textbook "Quantum Mechanics Vol 1 Cohen Tannoudji" it is show that it's possible to find a bra that can exist without ket corresponding to it ? This example would only be relevant in a specific mathematical context where the Riesz representation theorem doesn't apply ?

Best regards
Patrick

 

[PeroK]

I'm not familiar with that text, so it's a lot of work to go through that. The gist of what is going on is this.

He has a linear functional defined by $\phi(\psi) = \psi(0)$. This is, however, not a continuous linear functional, so the Riesz theorem does not apply.

More specifically, he has a sequence of bras that in the limit pick out the value $\psi(0)$ and concludes that he has a well-defined bra. But, the sequence of kets is a sequence that leads to the Dirac Delta function, which is not a well-defined function.

This leads into the concept of "rigged" Hilbert spaces, which is needed to resolve this - although, to be honest, I have never studied rigged Hilbert spaces myself.

In summary, although he appears to be generating a bra that has no ket, the bra doesn't meet all the conditions of a bra if done rigorously - it's not continuous. So, both the bra and the ket lie outside the normal theory of Hilbert spaces and continuous linear functionals.

I'm tempted to say that this is an example of a physicist trying to be rigorous and creating a bit of a mess!

 

[microsansfil]

Thank you for these clarifications

Best regards
Patrick

 

[George Jones]

Is it possible to find bras which have no corresponding kets ?k

It depends what one means by "ket" and "bra".

If by "ket" we mean any elements of a particular separable Hilbert space $H$, then, as @PeroK noted, the Riesz representation theorem gives a natural bijection between the space of continuous linear functionals, $H'$, on $H$. In other words, if by "bra" we mean any element of $H'$, then there is a natural bijection between the space of bras and the space of kets.

If, however, we take the space of kets to be $S$, a proper subset of $H$, then the set $S'$ of continuous linear functionals on ket space $S$ is "larger" than $H'$. This is because a mapping that is continuous on all of $H$ is automatically continuous on $S$, since $S \subset H$, but a mapping that is continuous on $S$ does not have to continuous on elements in $H$ that are outside of $S$, i.e., we have a rigged Hilbert space (also mentioned by @PeroK, and also called a Gelfand triple)
$$S \subset H = H' \subset S'.$$
In other words, if by "bra" we mean any element of $S'$, then there are "more" bras than kets.

IF this is done in a particular way, then included in the bra space $S'$ are delta functions and plane waves, so this allows physicists to work with distributions as "bras", and these can be used as weak eigenvectors of, e.g., the position and momentum operators.

This is what Cohen-Tannoudji et al. are trying to get at, but I don't have time to work through their technical details.

 

[microsansfil]

This is what Cohen-Tannoudji et al. are trying to get at, but I don't have time to work through their technical details.

Thank you. It's enough clear for my understanding.

Best regards
Patrick