How Do You Construct the Dual Basis in Dirac Notation?

[Gumbercules]

Homework Statement

my apologies if this question should be posted in the math forum
3-d space spanned by orthonormal basis: (kets) |1>, |2>, |3>. Ket |a> = i|1> - 2|2> - i|3>. Ket |b> = i|1> + 2|3>.

The question is to construct <a| and <b| in terms of the dual basis (kets 1,2,3)

Homework Equations

given above

The Attempt at a Solution

This is my first time seeing this kind of notation, and I am honestly not quite sure what the question is asking. I read that <a| is a linear function of vectors and that when it acts on a ket it produces a dot product. This also means that the bra can be seen as an instruction to integrate. In order to produce <a| would I integrate a*a?
thanks

 

[kuruman]

Think of the orthonormal basis kets as unit vectors in your 3-d space. A "ket" is a column vector and a "bra" is a row vector. In your example

|a> = i|1> - 2|2> - i|3>

says to me that the ordered "components" of "vector" |a> are (i, -2, -i). [Imagine this as a column 3x1 matrix - I don't know how o make matrices in Latex]. If you wanted to write the same thing as a bra you would say [and this is truly a row 1x3 vector]

<a| = (-i, -2, +i)

Note that the bra is the "complex-conjugate transpose" of the ket.

We write the inner product as a "bra-ket" just like the good-old dot product (matrix multiplication of the 1x3 times the 3x1 which gives a 1x1 or scalar)

<a|a>= (-i)*(i)+(-2)*(-2)+(+i)*(-i) = 1+4+1=6 (stars in this line mean "times" not "complex conjugate")

You can put in functions for |a> = f_a(x), in which case

[tex]<a|a> =\int f^{*}_{a}(x)f_{a}(x)dx [/tex]

where the integration over the appropriate limits.



This should get you started.

 

[Gumbercules]

This makes sense. Thank you!