QM: Changing Basis | Why Use T_{σa,σb}?

  • Thread starter Thread starter Niles
  • Start date Start date
  • Tags Tags
    Basis Qm
Niles
Messages
1,834
Reaction score
0

Homework Statement


Hi

Say I have the kinetic energy operator denoted by T(ri) for the particle i. I wish to represent it in some \left| \sigma \right\rangle-representation. My book says it is given by

<br /> T = \sum\limits_{\sigma _a ,\sigma _b } {T_{\sigma _a ,\sigma _b } \left| {\psi _{\sigma _a } \left( {r_i } \right)} \right\rangle \left\langle {\psi _{\sigma _b } \left( {r_i } \right)} \right|},<br />

where the first part of the sum denotes the matrix elements of T. My question is why the author is using

<br /> \hat 1 = \sum\limits_{\sigma _a ,\sigma _b } {\left| {\psi _{\sigma _a } \left( {r_i } \right)} \right\rangle \left\langle {\psi _{\sigma _b } \left( {r_i } \right)} \right|} ,<br />

when in fact it is given by

<br /> \hat 1 = \sum\limits_\sigma {\left| {\psi _\sigma \left( {r_i } \right)} \right\rangle \left\langle {\psi _\sigma \left( {r_i } \right)} \right|} <br />

I hope you will shed some light on this.


Niles.
 
Physics news on Phys.org
He is, but part of the expression has been contracted to form the matrix elements of T. Specifically, start with

T = \hat{1} T \hat{1} = \left( \sum\limits_{\sigma_a} {\left| {\psi _{\sigma_a} \left( {r_i } \right)} \right\rangle \left\langle {\psi_{\sigma_a} \left( {r_i } \right)} \right|}\right) T \left( \sum\limits_{\sigma_b} {\left| {\psi _{\sigma_b} \left( {r_i } \right)} \right\rangle \left\langle {\psi_{\sigma_b} \left( {r_i } \right)} \right|} \right)

= \sum\limits_{\sigma_a,\sigma_b } {\left| {\psi _{\sigma_a} \left( {r_i } \right)} \right\rangle \left\langle {\psi_{\sigma_a} \left( {r_i } \right)} \right|} T {\left| {\psi _{\sigma_b} \left( {r_i } \right)} \right\rangle \left\langle {\psi_{\sigma_b} \left( {r_i } \right)} \right|} = \sum\limits_{\sigma_a,\sigma_b } {\left| {\psi _{\sigma_a} \left( {r_i } \right)} \right\rangle T_{\sigma_a\sigma_b \right\rangle \left\langle {\psi_{\sigma_b} \left( {r_i } \right)} \right|}
 
You are right, thanks!
 
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