# Square integrable functions - Hilbert space and light on Dirac Notation

1. Jul 16, 2012

### esornep

Square integrable functions -- Hilbert space and light on Dirac Notation

I started off with Zettilis Quantum Mechanics .... after being half way through D.Griffiths ..... Now Zettilis precisely defines what a Hibert space is and it includes the Cauchy sequence and convergence of the same ..... is there any proof for the same for square integrable functions as he skips the same in the book .....
The second question he says Ʃ|θn><θn| = I where I stands for the operator .... now is this not going to infinity and please suggest some text on operators from scratch .... Thanks

2. Jul 16, 2012

### esornep

Re: Square integrable functions -- Hilbert space and light on Dirac Notation

The Dirac Notation says
|si> is ket and its conjugate is the the bra ??? Is this consistent ...... please explain

3. Jul 16, 2012

### vanhees71

Re: Square integrable functions -- Hilbert space and light on Dirac Notation

To your first question: The space of square integrable functions, describing a scalar non-relativistic particle, is defined as the vector space of functions $\psi:\mathbb{R}^3 \rightarrow \mathbb{C}$ with a sesquilinear form
$$\langle{\psi_1}|{\psi_2} \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \; \psi_1^*(\vec{x}) \psi_2(\vec{x}).$$
This scalar product induces a metric on this space
$$\|\psi \|=\sqrt{\langle \psi|\psi \rangle}.$$
To make this a pre-Hilbert space, you have to identify functions, for which
$$\|\psi_1-\psi_2\|=0.$$
I.e., you don't distinguish between such two functions. One can show that this pre Hilbert space is in fact a true Hilbert space, i.e., it is complete in the sense of the topology induced by the norm.

To your second question: Since you have a non-degenerate scalar product you can identify any continuous linear form with a vector and vice versa. If you have a Hilbert-space vector $|\psi \rangle$ the corresponding linear form is written as $\langle \psi|$ and this already suggests how it acts on another Hilbert-space vector: $|\phi \rangle \mapsto \langle \psi|\phi \rangle$.

In quantum mechanics you need a bit more than that, namely also distributions since you have to deal with unbounded (mostly self-adjoint) operators (like the position and momentum operators) that have a restricted domain, but that should be explained in your book.

4. Jul 16, 2012

### Fredrik

Staff Emeritus
Re: Square integrable functions -- Hilbert space and light on Dirac Notation

See this post.

If you're wondering what linear functionals have to do with conjugates, you will need to study the relationship between linear operators and matrices. It's explained in post #3 in this thread. (Ignore the quote and the stuff below it).

"Linear algebra done right", Sheldon Axler. (This one only deals with finite-dimensional vector spaces).
"Introductory functional analysis with applications", Erwin Kreyszig.

Most people only study finite-dimensional vector spaces and hope that there's some way of making sense of what they're doing in the context of infinite-dimensional vector spaces.

Last edited: Jul 16, 2012
5. Jul 16, 2012

### micromass

Re: Square integrable functions -- Hilbert space and light on Dirac Notation

This is actually a very tricky question. The answer relies crucially on what we mean with square integrable. Sure, you can say that it are functions such that

$$\int |\psi|^2$$

is finite. But what is that integral?? If you interpret that integral as just a Riemann integral, then the space of square integrable functions will not be complete!! In order for it to be complete, we need the more advanced notion of Lebesgue integrals. With that integral, it can be shown (but it is a bit tricky to do so) that the square integrable functions are complete.

The sum is not infinity by Bessel's inequality.
Fredrik suggested the wonderful texts by Axler and Kreyszig. I second these texts.
If you want to know more about the Lebesgue integral, then the text "Lebesgue integral on euclidean space" by Jones is excellent and not very hard.