# If I want to measure an electron how would I do it?

If I look at the Electron to see it then a Photon must of hit the Electron. So I have no idea of what the Electron was doing before I looked at it.

To find its position again I need to perform a measurement of some sort. This time I decide to make sure my Electron is in a black box so no Photons can get to it. But to measure something is to scale, to grade, to quantify maybe is a better word. So how do you do quantify an Electrons position with out disturbing it in the first place.

• Do you know about the uncertainty principle? Measuring the position of an electron won't be possible without interfering. Without the measurement process the electrons is describes only as a probability amplitude $\psi$. Dec 17 '19 at 20:45
• There are other ways of 'seeing' an electron. Dec 17 '19 at 20:47
• @Semoi The uncertainty principle doesn't deal with interfering with the electron. It deals with the spread of measurements of non-commuting observables when measuring similarly prepared quantum systems. i.e. it just gives a lower bound on $\Delta x\cdot\Delta p$ for the state before measurement (for the famous Heisenberg Uncertainty Principle at least). Dec 17 '19 at 20:47
• thanks for your comments. Are discussions OK on here? If not then sorry... "Seeing" is a figure of speech. Quantify is a better word. It's a genuine question. @Aaron Stevens and others! I've come across uncertainty principle and I've 100% come across Heisenberg (who can't). Lower bound for delta x and delta p. prior to measurement. Seems like a percentage or proportional change calculation at where something was not and not will be. But you can never know where it will be because you never knew really where it was in the first place. Is there really no way to measure this. Dec 17 '19 at 21:27
• Aaron: Of course you are right. I meant influencing not interfering. @hoboBob: There exists no lower bound for the individual uncertainties of $\Delta x$ or $\Delta p$, but only for their product. So, assuming that this principle is true, we are unable to measure something without influencing it -- it doesn't matter whether you use photons of a magnetic field, the measurement affects the electron. In quantum mechanics the observer (the measurement equipment) becomes part of the system. Dec 17 '19 at 22:07

## 1 Answer

>Bubble chamber photograph of an electron knocked out of a hydrogen atom by $$K^-$$ particle The curly line was produced by an electron that was struck by one of twelve passing beam particles in a liquid hydrogen bubble chamber. It curves in an applied magnetic field and loses energy rapidly, spiralling inwards.

Electrons were first "seen" in cloud chambers, but bubble chambers really "see" elementary particles.

The process is the following, involving zillions of photon exchanges and photons to create the event and to be able to see it with our eyes.

1) The primary generation of an electron: A $$K^-$$ particle ( the beam coming in) hits an electron in the hydrogen of the bubble chamber and a sufficiently energetic electron comes out. All the dots making the tracks visible are very small energy similar scatters, that end up in ionizing the hydrogen in the way, and a bubble forms ( due to the supercooled hydrogen, another story).

2)the electron starts bending in the magnetic field, so its momentum can be measured. ( the $$K^-$$ is also bending but because it has very high energy it cannot be seen by eye, only careful measurements over distance)

3)the electron loses energy with continuous small scatters spiraling inwards

All these reactions are electromagnetic with exchanges of off mass shell photons.

4) a picture is taken and developed by using electronic coincidense for the passing of the beam

5) photons from the environment hit our eye and show the electron path.

Thus we both "see" at second hand the electron, and all the other particles in the link. Photons are always involved.