# Are the authors saying that the observer effect plays no role in Bohr's thought experiment of the Heisenberg uncertainty principle?

Here is an excerpt from Eisberg & Resnick's Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles. Here is introducing Bohr's though experiment to establish a physical origin for the Hesisenberg Uncertainty Principle.

"Let us say that we wish to measure as accurately as possible the position of a particle, like an electron. For greatest precision we use a microscope to view the electron, as in Figure 3-6. To see the electron we must illuminate it, for it is actually the light photon scattered by the electron that the observer sees. At this stage, even before any calculations are made, we can see the uncertainty principle emerge. The very act of observing the electron disturbs it. The moment we illuminate the electron, it recoils because of the Compton effect, in a way that we shall soon find cannot be completely determined. If we don't illuminate the electron, however, we don't see (detect) it. Hence the uncertainty principle refers to the measuring process itself, and it expresses the fact that there is always an undetermined interaction between observer and observed; there is nothing we can do to avoid the interaction or to allow for it ahead of time. In the case at hand we can try to reduce the disturbance to the electron as much as possible by using a very weak source of light. The very weakest we can get is to assume that we can see the electron if only one scattered photon enters the objective lens of the microscope. The magnitude of the momentum of the photon is $p = \dfrac{h}{\lambda}$. But the photon may have been scattered anywhere within the angular range $2\theta'$ subtended by the objective lens at the electron. This is why the interaction cannot be allowed for. Hence we find the x component of the momentum can vary from $p\sin\theta'$ to $-p\sin\theta'$ and is uncertain after the scattering by an amount $\Delta p=2p\sin\theta'$."

Figure 3-6 is what's shown below. In the last few sentences, what are the authors referring to when they state "interaction cannot be allowed for"? Are they saying that the observer effect plays no role in Bohr's thought experiment? Also, how do they find that $\Delta p=2p\sin\theta$?

• note that what is described here is what Heisenberg had in mind, but not what his famous formula is actually about as I discuss e.g. here: physics.stackexchange.com/q/169730 – Martin Mar 15 '15 at 19:55
• I think this question would be better on History of Science and Mathematics Stack Exchange. It's an excellent question - don't get me wrong - but I think a real historian needs to answer it. What is being spoken about here is clearly purely the observer effect, and my own heresay, flaky knowledge here agrees with @Martin 's comment: as I understand it, this is how Heisenberg did indeed, mistakenly, think about his uncertainty principle at first. Only later did the accurate picture take shape. But I cannot cite primary, historical sources. – WetSavannaAnimal Oct 6 '15 at 23:22

As for the second question, it's fairly simple. For a vertically emitted electron, the horizontal (x-axis) momentum is zero. But, as stated, the horizontal momentum of an electron detected at the edge of the cone is $$p_x = \frac{H}{\lambda}\sin{\theta} = p\sin{\theta}$$ but keep in mind that is simply the maximum possible, since the electron can be anywhere within the cone. Since the total angle of the cone is $2\theta$ the total uncertainty of the momentum $\Delta p$ is the momentum at one edge minus the momentum at the other, or $$\Delta p = p\sin{\theta} - (p\sin{(-\theta})) = 2p\sin{\theta}$$ since $$\sin{(-\theta )} = -\sin{\theta}$$