# Bra-ket notation confused

1. Jan 8, 2014

### iScience

so i'm fine with the kets, e.g, |a>.. but i don't understand what the bras are. the kets are basically just a column vector right? ie the components (with the direction) of the vector being described.
but what is the bra?

this was given to us in class:
<a|=a1<e1|+a2<e2|= (a1* a2*)

(where e1 and e2 are the unit basis vectors)

but why the complex conjugates? where'd they come from?

and the professor gave us this in class:

<a|b>=a1*b1<e1|e1>+a1*b2<e1|e2>+a2*b1<e2|e1>+a2*b2<e2|e2>

i understand the pattern but i am still at a loss as to where this all came from

2. Jan 8, 2014

### WannabeNewton

Given a vector space $V$ over the desired field there exists a dual space $V^{*}$ consisting of all linear functionals mapping elements of $V$ into said field. Furthermore given an inner product $< , >$ on $V$ there exists a bijection between $V$ and $V^{*}$ such that for any $v\in V$, there exists a unique $f \in V^{*}$ such that $f(w) = <v, w>$ for all $w\in V$, which is normally called the Riesz representation theorem. As a result, given any $v\in V$ we can denote it by $|v\rangle$ (called a "ket vector") and since the Riesz representation theorem says there exists a unique $f\in V^{*}$ associated with $v$ such that $f(w) = <v,w>$, we can simply denote $f$ by $\langle v|$ (called a "bra vector") and denote $f(w)$ by $\langle v|w\rangle$ so that $\langle v|w\rangle= <v,w>$. So the Riesz representation theorem is the reason why bra-ket notation can be used.

3. Jan 8, 2014

### ModusPwnd

Since this poster is probably just starting with quantum mechanics, do you actually think that any of that is going to be useful to him/her? Looks to me like you are showing off instead of helping the poster. I would like to see some other responses to the original post, something more digestible for a beginner. Maybe I'm wrong and your post cleared it up perfectly to the original poster (It sure didn't for me)... Often people asking for help around here get replies that are correct, but not useful at all.

4. Jan 8, 2014

### iScience

Yah.. ModusPwnd is correct.. the answer you gave WannabeNewton, is something i could have easily found on wikipedia or on some other web site, but, although embaressing, i still do not understand mathematical statements like the one you just gave. can you put it in terms of easer.. lingo/format?

Last edited: Jan 9, 2014
5. Jan 8, 2014

### strangerep

iScience,

Consider a simple system whose states can be spanned by a finite number of basis states. (I.e., any state can be expressed as a linear combination of those basis states.) This is modelled in basic QM by an inner product space (i.e., vector space with an inner product defined thereon). IF it were a real vector space, then the bras would simply be the transposes of the column vectors. But in QM, we use complex vector spaces in general, so the bras are conjugate-transposes of the column vectors. This corresponds to the 1st equation you wrote in your original post.

This choice is made so that the inner product on the space, can be used to define a positive-definite norm. E.g., $\def\<{\langle} \def\>{\rangle}$
$$\|a\| ~:=~ \<a|a\> ~=~ \pmatrix{a_1^* & a_2^*} \pmatrix{a_1 \\ a_2} ~=~ |a_1|^2 + |a_2|^2 ~\ge~ 0~.$$I can't say much more than that without knowing more of the context of your lecture. (I'm guessing it's in the context of a system with 2 independent states.)

Btw, WBN's reply is also applicable to the infinite-dimensional Hilbert spaces that are quite common in QM.

Last edited: Jan 8, 2014
6. Jan 8, 2014

### strangerep

It helps to keep in mind that this forum is really intended for graduate-level questions. Homework and textbook-level questions should probably go in the homework forum.

7. Jan 8, 2014

### Staff: Mentor

No, it's not. Undergraduate-level conceptual questions are fine here. Even high-school level, although you don't get too many of those in QM!

The homework forums are for getting help with working through specific exercises, e.g. from textbooks.

8. Jan 9, 2014