# Help with Bra-Kets notation

1. Oct 12, 2012

### sliorbra

hello,

after some struggle,i finally understand these notations, but there is a problem.
i'm studying with "principles of quantum mechanics" [R. Shankar]. in the beginning of the chapter where he discusses 'bra-kets' he mentioned the axioms of inner product.
one of the axioms he mentioned [and used] is the following:

-<V|aW>=a<V|W> [where 'a' is a complex scalar, V and W are vectors]

My problem is this axiom assumes linearity with respect to the second term in the inner product, when until this book i encountered only with axiom who assumes linearity with respect to the first factor.

after reading why it is so comfortable to define the 'bra' space, i can understand the logic of the author, but it seems strange to me that he changed the axiom a little bit for the purposes of quantum mechanics.

Can anyone explain me what i have missed?

2. Oct 12, 2012

### dextercioby

Of all the books on the market, is this only one you could choose ? I don't have this book, but if what you say is right, then wow, it's not to be trusted. A good book at a middle level would be Bransden & Joachain <Quantum Mechanics>.

3. Oct 12, 2012

### George Jones

Staff Emeritus
It's no big deal, it's just a math/physics thing. Almost all math books define inner products to be linear in the first argument; almost all physics books define inner products to be linear in the second argument. It's just notation.

dextercioby, I don't have Bransden and Joachain. For which argument are the inner products in it linear?

4. Oct 12, 2012

### sliorbra

thank to both of you.

George, if I use the axiom like it is written in my book[linearity in the second argument], the whole definition of bra-kets is clear to me and the result i get when I use the definiton of inner product for two vectors [in orthonormal basis] and when I use bra-kets is the same. But if I use the Mathematicians definition,the second result is the complex-conjagate of the first one.

So it seems to me that the definition of bra-kets is useful just if i use the concept of inner product with the axiom like in my book.

hope my point is clear.

5. Oct 12, 2012

### George Jones

Staff Emeritus
sliorbra, welcome to Physics Forums! Have you taken a functional analysis course?

For mathematicians, this is called Riesz representation theorem,

http://en.wikipedia.org/wiki/Riesz_representation_theorem,

which is usually covered in a functional analysis course.

The Riesz representation theorem is independent of the notation used (mathematicians' or physicists'), but physicists' notation exploits the theorem explicitly.

6. Oct 12, 2012

### Fredrik

Staff Emeritus
sliorbra, I will use the notation (x,y) for the inner product of x and y in this post, to avoid confusion with bra-ket notation. Let H be a Hilbert space. The set of continuous linear complex-valued functions on H is denoted by H* and is called the dual space of H. For each x in H, define $\phi_x\to\mathbb C$ by $\phi_x(y)=(x,y)$ for all y in H. This $\phi_x$ is a member of H*. I like to use the alternative notation $(x,\cdot)$ for $\phi_x$, because this notation makes it easy to remember how the function is defined. It's the map that takes y to (x,y).

Bra-ket notation is the convention to write $|x\rangle$ instead of $x$, and $\langle x|$ instead of $(x,\cdot)$. The former is then called a ket, and the latter a bra. The expression $\langle x|y\rangle$ is not defined as an inner product, it's an abbreviated notation for $\langle x|(|y\rangle)$ (the bra is a function that takes the ket as input). In the inner product notation, this can be written as $(x,\cdot)(y)$, which by definition is equal to $(x,y).$ So $\langle x|y\rangle$ is equal to $(x,y)$, but these two expressions are not defined the same way.

If we had started with an inner product that's linear in the first variable, we would have had to define the bra corresponding to x as $(\cdot,x)$ instead of as $(x,\cdot)$. And if we still define $\langle x|y\rangle$ as an abbreviated notation for $\langle x|(|y\rangle)$, then what we end up with is that $\langle x|y\rangle$ is equal to $(y,x)$ not $(x,y)$.

Note that $x\mapsto(x,\cdot)$ is a map from H into H*. It's easily seen to be antilinear. The Riesz representation theorem for Hilbert spaces tells us that this map is also a bijection. This means that every member of H* can be uniquely expressed in the form $(x,\cdot)$.

Edit: This post explains the same things I said here, and a little bit more. If you can tolerate a few typos, you can find a proof of the Riesz representation theorem and a few simple results in post #13 here.

Last edited: Oct 12, 2012
7. Oct 13, 2012

### sliorbra

Thank you Fredrik!
What you have explained in the third paragraph was the thing that bothered me the most, and now it is clear to me.

8. Oct 13, 2012

### vanhees71

I think, in the usual use of bras and kets in physics, it's a bit more complicated than Fredrik told, because at least bras are also used in a generalized way not only on the (topological) dual of H (which in the sense Fredrik explained in detail is isomorphic to H) but to a larger space of linear forms. The reason is that one also likes to describe unbounded operators like the position and momentum operators, and one not only works with the Hilbert space of states but with the rigged Hilbert space. Within this formalism, also the generalized eigenvectors for spectral values of these operators in the continuous part of there spectrum can be formally manipulated with the bra-ket product, but of course one has to be aware of the subtleties of such generalized eigenvectors and distributions. You find a nice introduction to the rigged-Hilbert space formulation in