Talk:Inelastic electron tunneling spectroscopy

Physics Low‑importance

	Physics portal This article is within the scope of WikiProject Physics, a collaborative effort to improve the coverage of Physics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.PhysicsWikipedia:WikiProject PhysicsTemplate:WikiProject Physicsphysics articles
Low	This article has been rated as Low-importance on the project's importance scale.

Engineering

	Engineering portal This article is within the scope of WikiProject Engineering, a collaborative effort to improve the coverage of engineering on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.EngineeringWikipedia:WikiProject EngineeringTemplate:WikiProject EngineeringEngineering articles
???	This article has not yet received a rating on the project's importance scale.

[Untitled]

I have just addes this page, but I am not an expert on Wikipedia, so I hope for support from other users. Some questions I have now: Why can't I redirect IETS to this page? What is the name of this page anyway? Why does the search function not find this page when looking for the exact words "Inelastic Electron Tunneling Spectroscopy"? Erwin (talk) 09:38, 4 April 2008 (UTC)[reply]

Diagrams for current, conductivity and its derivative
The diagram d²I/dV² of the image "Second_derivative.gif" is not correct, since it should be zero everywhere except for the peak! Otherwise the conductivity dI/dV would increase linearly and that is not the case. (Njeorp (talk) 16:47, 6 July 2008 (UTC))[reply]

Finite rise-times will affect the output of the measurments

You are not exactly correct in what you are saying. In a real world experiment there will be a finite rise-time when dI/dV changes from one value to another, particularly if it is measured as a voltage going from one value to another. If d2I/dV2 were being correctly represented based on the dI/dV graph shown, it would be a very short square pulse-like wave-form. However, in a real world set-up again there will be a finite time in which the voltage signal representing d2I/dV2 will take to reach its value, and again to return to zero. The duration spent at the value is very short hence the waveform looks like a triangular peak rather than a square pulse.

If the graph for dI/dV were represented ideally as a square step, then d2I/dV2 would be a single point with zero everywhere else. This diagram is more to illustrate an idea, that you will get a peak in your d2I/dV2 signal when you have a change in your measured conductance. These are real world phenomena which will be subject to the effects detailed above.

Please feel free to point out where this argument may be wrong. John —Preceding unsigned comment added by 143.239.65.170 (talk) 19:21, 29 September 2008 (UTC)[reply]

Of course, in the real world, the point at which the slope of the current changes is not a single point, but changes gradually. This is mostly due to the energy of the electrons at a finite temperature, and has nothing to do with time, unless you mean with time by definition that which is on the x-axis. This broadening of the point of change of slope results in a broader step in the first derivative and a broader (Lorentzian- or Gauss-shaped) peak in the second derivative.

The first comment is right that the second derivative should be zero everywhere, except for at the peak, but I have drawn it slightly above it to make it visible at all. It all depends on where you define the zero level on the y-axis of course whether the graph is correct or not.

Erwin (talk) 16:04, 26 November 2008 (UTC)[reply]

That's what I meant to say. The peak in d2I/dV2 itself clearly is only a simplified illustration (of a more or less gaussian curve in reality), I absolutely agree on that; however the signal should be zero for energies far from the peak. Where you put the zero in the diagram is of course arbitrary - still, the way it is now might be confusing. Maybe a different color for the signal could help!?

Njeorp (talk) 23:41, 10 December 2008 (UTC)[reply]