« previous next »
Pages: [1]   Go Down

Author Topic: transmission eigenstate  (Read 1943 times)

0 Members and 1 Guest are viewing this topic.

Offline bubble

  • Heavy QuantumATK user
  • ***
  • Posts: 59
  • Country: cn
  • Reputation: 0
    • View Profile
transmission eigenstate
« on: July 28, 2017, 02:50 »
Hi,
   What is the difference between the left and right contributions for the transmission eigenstate (TE) calculation? 
   For example, If the bias voltage is finite, I think that, for a given energy, the electron can be viewed as incident from the left and right electrodes at the same time, and the final current is the result of, e.g.  'right contribution minus left contribution'. And based on this consideration,  we can obtain the left and right contributions for the TE?
  Am I right?
Thank you !
   
Logged

Offline Jess Wellendorff

  • QuantumATK Staff
  • Supreme QuantumATK Wizard
  • *****
  • Posts: 933
  • Country: dk
  • Reputation: 29
    • View Profile
Re: transmission eigenstate
« Reply #1 on: July 28, 2017, 08:23 »
As writtin in the online documentation for TransmissionEigenstate analysis, http://docs.quantumwise.com/manuals/Types/TransmissionEigenstate/TransmissionEigenstate.html#transmissioneigenstate-c,
the analysis method will by default compute the transmission eigenstates for electrons going from left to right, i.e. Bloch states in the left electrode that are transmitted to Bloch state in the right electrode. Perhaps the sign of the finite bias (forward/reverse bias) determines if you should consider Left or Right moving electrons, but I'm not sure.
Logged
Pages: [1]   Go Up
« previous next »