Temporal Impulse response of one electron

[El-Shimy]

Hi,

I am a little bit confused about the impulse response of one electron.

Assume that we have LTI system characterized by impulse response h(t) with unit gain, int{h(t)} = 1.

Let the input is current i(t) [Amp]. So the output current will be i(t)*h(t). We can view it as i(t) is dispersed in time due to passing through system h(t).

- If the current is only one electron with charge q and enters at time t = 0. Therefore i(t) = q delta(t).
- Therefore the output current is i(t)*h(t) = q.h(t)

The question is how one electron is dispersed in time.

 

[berkeman]

Hi,

I am a little bit confused about the impulse response of one electron.

Assume that we have LTI system characterized by impulse response h(t) with unit gain, int{h(t)} = 1.

Let the input is current i(t) [Amp]. So the output current will be i(t)*h(t). We can view it as i(t) is dispersed in time due to passing through system h(t).

- If the current is only one electron with charge q and enters at time t = 0. Therefore i(t) = q delta(t).
- Therefore the output current is i(t)*h(t) = q.h(t)

The question is how one electron is dispersed in time.

The electron wavelength depends on its velocity:

http://hyperphysics.phy-astr.gsu.edu/Hbase/debrog.html

What is the context of your question? What kind of apparatus are you working with?

 

[Okefenokee]

If we only have one charge carrier then do we have a true LTI system? If a single electron moves from one place to another then the place where it was is no longer able to conduct electricity because it no longer has any charge carriers. It's not time invariant.

I don't think any of that LTI stuff will apply at all for a single electron because it assumes that a current is a continuous quantity that can be divided. For example, a system with a gain of 1/2 is a perfectly legit LTI but how do you divide an electron in half?

If you are interested in the wave-like nature of electrons and electricity then you can look up the http://en.wikipedia.org/wiki/Josephson_effect#The_effect".

 

[El-Shimy]

Thank you for the reply.

My question is about Photodetector with impulse response h(t). Where the output current is represented by
i(t) = sum{ h(t-tm) }
where h(t) is the response of one electron such that int{h(t)}=q and tm is the time of released electron.

 

[berkeman]

Thank you for the reply.

My question is about Photodetector with impulse response h(t). Where the output current is represented by
i(t) = sum{ h(t-tm) }
where h(t) is the response of one electron such that int{h(t)}=q and tm is the time of released electron.

A photodetector detects photons, not electrons. Are you thinking of a vacuum photomultiplier tube (PMT)?

 

[El-Shimy]

A photodetector detects photons, not electrons. Are you thinking of a vacuum photomultiplier tube (PMT)?

Thank you for your concern.

Actually, I mean the photodetector. We can assume it as 100% efficiency, where each photon will emit electron and for each electron there is a response function h(t) contains charge q.

 

[Bob S]

Shot noise (the noise from individual electrons) in vacuum photodiodes and similar current sources has been well studied. See equation 6.8 in Section 6.1 in

http://www.qis.ex.nii.ac.jp/qis/documents_YY/y3_02chp6_txt.pdf [Broken]

The electron arrival at the anode is considered to be an impulsive event, so the anode voltage signal in the external circuit is given by Eq. 6.7..

Bob S

 

[El-Shimy]

Shot noise (the noise from individual electrons) in vacuum photodiodes and similar current sources has been well studied. See equation 6.8 in Section 6.1 in

http://www.qis.ex.nii.ac.jp/qis/documents_YY/y3_02chp6_txt.pdf [Broken]

The electron arrival at the anode is considered to be an impulsive event, so the anode voltage signal in the external circuit is given by Eq. 6.7..

Bob S

Thanks a lot for you and for the others reply. I will check the link you provided.