How Do You Construct the Dual Basis in Dirac Notation?

Gumbercules
Messages
11
Reaction score
0

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
 
Physics news on Phys.org
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> = fa(x), in which case

&lt;a|a&gt; =\int f^{*}_{a}(x)f_{a}(x)dx

where the integration over the appropriate limits.



This should get you started.
 
This makes sense. Thank you!
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top