Author Topic: Inner product (Read 2817 times) Tweet Share

/0 · « **on:** November 25, 2009, 01:02:10 pm »

Ahmad · « **Reply #1 on:** November 25, 2009, 01:34:11 pm »

Not a big fan of your notation

. What book do you use?

/0 · « **Reply #2 on:** November 25, 2009, 01:36:24 pm »

lol I thought you might not be... "Principles of Quantum Mechanics" - Shankar

Ahmad could you recommend a good book where I can read about this?

Ahmad · « **Reply #3 on:** November 25, 2009, 01:42:46 pm »

Physics, that explain it

, but great to see you reading ahead anyhow!

The idea is that we have linearity in the second slot of the inner product, in this case. But what do we have in the first slot?
$\langle aW+bZ|V\rangle = a^*\langle W|V \rangle +b^*\langle Z|V \rangle$ which you can prove by applying (1) then (3) then (1) back again. So you have conjugate linearity in the first slot and linearity in the second slot. Plug in your expression for W and V and expand (using linearity). I'd give a more detailed explanation but not a fan of the notation

Ahmad · « **Reply #4 on:** November 25, 2009, 01:44:15 pm »

Quote from: /0 on November 25, 2009, 01:36:24 pm

Ahmad could you recommend a good book where I can read about this?

My favourite linear algebra book which I believe is very readable is Linear Algebra Done Right by Sheldon Axler. The notation is a little different there (fairly standard in pure maths), but the concepts are the same, of course.

humph · « **Reply #5 on:** November 25, 2009, 01:56:19 pm »

AHHHHH BLOODY QM NOTATION!!!!!

It's shocking, don't get into the bad habit so early

/0 · « **Reply #6 on:** November 25, 2009, 02:05:42 pm »

Quote from: Ahmad on November 25, 2009, 01:42:46 pm

Physics, that explain it , but great to see you reading ahead anyhow!

The idea is that we have linearity in the second slot of the inner product, in this case. But what do we have in the first slot?
$\langle aW+bZ|V\rangle = a^*\langle W|V \rangle +b^*\langle Z|V \rangle$ which you can prove by applying (1) then (3) then (1) back again. So you have conjugate linearity in the first slot and linearity in the second slot. Plug in your expression for W and V and expand (using linearity). I'd give a more detailed explanation but not a fan of the notation

Thanks, I was able to do it! But wow the summation notation + expansion really kicked my head in

Quote from: Ahmad on November 25, 2009, 01:44:15 pm

Quote from: /0 on November 25, 2009, 01:36:24 pm
Ahmad could you recommend a good book where I can read about this?

My favourite linear algebra book which I believe is very readable is Linear Algebra Done Right by Sheldon Axler. The notation is a little different there (fairly standard in pure maths), but the concepts are the same, of course.

A Maths book written by Sheldon!! Awesome

I'll have a read

Quote from: humph on November 25, 2009, 01:56:19 pm

AHHHHH BLOODY QM NOTATION!!!!!

It's shocking, don't get into the bad habit so early

lol yeah already it seems a bit awkward to use

edit: Ok now I hate it

QuantumJG · « **Reply #7 on:** November 26, 2009, 10:56:28 pm »

Quote from: /0 on November 25, 2009, 01:02:10 pm

Can someone help me understand the inner product?

The axioms of the inner product are

1. $\langle V|W\rangle = \langle W|V\rangle^*$

2. $\langle V|V\rangle \geq 0\ \ \ \ \ 0 \ \ iff\ \ |V\rangle = |0\rangle$

3. $\langle V|(a|W\rangle +b|Z\rangle ) \equiv \langle V|aW+bZ\rangle = a\langle V|W \rangle +b\langle V|Z \rangle$

Given that $|V\rangle$ and $|W \rangle$ can be expressed in terms of their basis vectors,

$|V \rangle = \sum_i v_i |i \rangle$

$|W \rangle = \sum_j w_j|j \rangle$

How can the axioms be used to obtain:

$\langle V|W \rangle = \sum_i \sum_j v_i^*w_j\langle i|j \rangle$

thanks

I don't really like your notation!

An inner product is an expression that obey's the following axioms:

say I have three vectors $(\textbf{u,v,w}) \because \textbf{u,v,w} \in V (\textit{where V is a vector space}) and \alpha \in \mathbb{R}$

1) $\langle \textbf{u,v} \rangle = \langle \textbf{v,u} \rangle$

2) $\alpha \langle \textbf{u,v} \rangle = \langle \alpha \textbf{u,v} \rangle$

3) $\langle \textbf{u,v + w} \rangle = \langle \textbf{u,v} \rangle + \langle \textbf{u,w} \rangle$

4)

a) $\langle \textbf{u,u} \rangle \geqslant 0$

b) $\langle \textbf{u,u} \rangle = 0 \textit{iff} \textbf{u} = \textbf{0}$

The most simple inner product that is seen (when you do specialist maths) is the "dot product".

Anyway this is the notation that is used in linear algebra, as for your notations I have yet to see them.

You can use the inner product to do all the same things you do with the dot product,

i.e.

$\parallel \textbf{u} \parallel = \langle \textbf{u,u} \rangle$

angle $\theta$ between u and v is,

$\theta = \dfrac{\langle \textbf{u,v} \rangle}{\parallel \textbf{u} \parallel \parallel \textbf{v} \parallel}$

Please keep in mind that these are angles and norm's with respect to the given inner product and they will not yield the same results as the dot product if the inner product used isn't the dot product.

You could try proving that:

if $\textbf{u,v} \in \mathbb{R} ^{2}$

prove that,

$\langle \textbf{u,v} \rangle = {u_{1} v_{1} + 2 u_{2} v_{2}$

is an inner product.

/0 · « **Reply #8 on:** November 26, 2009, 11:19:59 pm »

lol thanks Quantum, when they started using $\langle i|\Omega | j\rangle$ to denote tranformations I abandoned ship completely

QuantumJG · « **Reply #9 on:** November 26, 2009, 11:27:01 pm »

Quote from: /0 on November 26, 2009, 11:19:59 pm

lol thanks Quantum, when they started using $\langle i|\Omega | j\rangle$ to denote tranformations I abandoned ship completely

Lol. That notation is just weird.

Search the forums now!

We have moved!

Author Topic: Inner product (Read 2817 times) Tweet Share

/0

Inner product

Ahmad

Re: Inner product

/0

Re: Inner product

Ahmad

Re: Inner product

Ahmad

Re: Inner product

humph

Re: Inner product

/0

Re: Inner product

QuantumJG

Re: Inner product

/0

Re: Inner product

QuantumJG

Re: Inner product

Recent Posts

User:
Password: