Projector in an arbitrary basis, howto compute?

  • Thread starter Thread starter keen23
  • Start date Start date
  • Tags Tags
    Basis Projector
keen23
Messages
9
Reaction score
0
Hello!
once again I got trouble with Dirac-notation:

Homework Statement


Given an non-orthonormal basis. Measurement via the projection operator should not give an definit answer, in which state the system was due to the overlap. Geometrical that's clear, but I'm unable to compute that... :(

I tried it for the following states:
|\psi_0\rangle=|0\rangle
|\psi_1\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)

Homework Equations



The Attempt at a Solution


My projector looks the following way (I am not sure about the coefficients):

P=|0\rangle\langle0|+\frac{1}{2} (|0\rangle+|1\rangle)(\langle 0|+\langle 1|)
And now I am confused. When I plug state psi1 on both sides I get a sum of several 0s and 1s I don't know how to interprete.

Any hints? Thank you!
 
Physics news on Phys.org
What exactly are you trying to do?
 
keen23 said:
My projector looks the following way (I am not sure about the coefficients):

P=|0\rangle\langle0|+\frac{1}{2} (|0\rangle+|1\rangle)(\langle 0|+\langle 1|)

Does P^2 = P? If it doesn't (I don't think that it does), then P is not a projector.

Avodyne said:
What exactly are you trying to do?

I, too, would like to know the answer to this.
 
At the moment I am reading about general measurements and POVMs.
They say, that you can't definately distinguish two non-orthonormal states by a "normal" measurement, meaning by a projection onto the states (since each has an overlap with the other).
(That's why you need the POVM to do that.)

I just would like to see the math.

Or is that the point, that I can't construct an projection-operator from my non-orthonormal states? But actually you can always project a vector onto its constructing basis vectors, no matter if they are orthogonal or not.

I hope this makes my confusion a bit clearer.
 
##|\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