Bra Ket is equivalent to inner product always?

[Kashmir]

We denote a scalar product of two vectors $a, b$ in Hilbert space $H$ as $(a,b)$.

In Bra Ket notation, we denote a vector a in Hilbert space as $|a\rangle$. Also we say that bras belong to the dual space $H$∗ .

So Bras are linear transformations that map kets to a number.

Then it isn't always true that $\langle a \mid b\rangle=(a,b)$

In quantum mechanics do we define bra in such a way so as that the above equality holds? Are there cases when the above equation is not true?

 

[anuttarasammyak]

I agree with what you said. I would just add when Dirac introduced ket, it is for not only discrete but continuous variables, e.g. coordinate, momentum. That makes ket get out of genuine Hilbert space.

 

[PeroK]

We denote a scalar product of two vectors $a, b$ in Hilbert space $H$ as $(a,b)$.

In Bra Ket notation, we denote a vector a in Hilbert space as $|a\rangle$. Also we say that bras belong to the dual space $H$∗ .

So Bras are linear transformations that map kets to a number.

Then it isn't always true that $\langle a \mid b\rangle=(a,b)$

In quantum mechanics do we define bra in such a way so as that the above equality holds? Are there cases when the above equation is not true?

A Hilbert space, by definition, has a complete inner product defined on it; the kets are then defined in terms of that inner product.

For some mathematical justification of how we know this all makes sense, you may be interested in:

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

 

[Kashmir]

A Hilbert space, by definition, has a complete inner product defined on it; the kets are then defined in terms of that inner product.

For some mathematical justification of how we know this all makes sense, you may be interested in:

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

You mean the Bra transformation is defined such that <a|b>=(a, b) ?

 

[PeroK]

You mean the Bra transformation is defined such that <a|b>=(a, b) ?

Yes.

 

[vanhees71]

We denote a scalar product of two vectors $a, b$ in Hilbert space $H$ as $(a,b)$.

In Bra Ket notation, we denote a vector a in Hilbert space as $|a\rangle$. Also we say that bras belong to the dual space $H$∗ .

So Bras are linear transformations that map kets to a number.

Then it isn't always true that $\langle a \mid b\rangle=(a,b)$

In quantum mechanics do we define bra in such a way so as that the above equality holds? Are there cases when the above equation is not true?

It's a bit more subtle as soon as it comes to real-world systems, where you have a true infinitely dimensional (but separable) Hilbert space. The minimal true QT model for me still is not the toy models like a finite set of spins leading to finite-dimensional Hilbert spaces but the realization of the Heisenberg algebra, generated in the usual way by position and momentum operators for at least one particle in 1D space. These operators then are "densely defined" on a Hilbert space, conveniently realized on the Hilbert space of square Lebesgue-integrable functions, i.e., their domain and codomain as self-adjoint operators is a proper but dense subspace of $H=\mathrm{L}^2(\mathbb{R})$ (e.g., the Schwartz space of quickly falling functions), the "nuclear space" $D \subset H$. Then the bras denoting "generalized eigenvectors" of the or position or momentum operator belong to the dual of this sub-space $D^* \supset H^*=H$. Formalizing this "hand-waving-physicists formulation" leads to the G'elfand (aka rigged Hilbert space) formalism. In this formalism an expression like $\langle x|x' \rangle=\delta(x-x')$ or
$$\int_{\mathbb{R}} |x \rangle \langle s|=\hat{1}$$
becomes formally correct. In fact formalizing these intuitive ideas by Dirac lead to the development of Functional Analysis as a new subject in pure math.

 

[Kashmir]

Yes.

Thank you.

 

[Kashmir]

It's a bit more subtle as soon as it comes to real-world systems, where you have a true infinitely dimensional (but separable) Hilbert space. The minimal true QT model for me still is not the toy models like a finite set of spins leading to finite-dimensional Hilbert spaces but the realization of the Heisenberg algebra, generated in the usual way by position and momentum operators for at least one particle in 1D space. These operators then are "densely defined" on a Hilbert space, conveniently realized on the Hilbert space of square Lebesgue-integrable functions, i.e., their domain and codomain as self-adjoint operators is a proper but dense subspace of $H=\mathrm{L}^2(\mathbb{R})$ (e.g., the Schwartz space of quickly falling functions), the "nuclear space" $D \subset H$. Then the bras denoting "generalized eigenvectors" of the or position or momentum operator belong to the dual of this sub-space $D^* \supset H^*=H$. Formalizing this "hand-waving-physicists formulation" leads to the G'elfand (aka rigged Hilbert space) formalism. In this formalism an expression like $\langle x|x' \rangle=\delta(x-x')$ or
$$\int_{\mathbb{R}} |x \rangle \langle s|=\hat{1}$$
becomes formally correct. In fact formalizing these intuitive ideas by Dirac lead to the development of Functional Analysis as a new subject in pure math.

Thankyou but i don't have the maturity enough right now to understand it. Hopefully I'll come back here in future. Thank you Mr Vanhees :)