# Repeated Position Measurements

I'm reading Griffith's Introduction to Quantum Mechanics and came across the extract below:

"What if I made a second measurement, immediately after the first? Would I get C again, or does the act of measurement cough up some completely new number each time? On this question everyone is in agreement: A repeated measurement (on the same particle) must return the same value. Indeed, it would be tough to prove that the particle was really found at C in the first instance, if this could not be confirmed by immediate repetition of the measurement."

So the text implies that a second position measurement on a particle will return the same value as the first. However if position is known, then the momentum (and hence velocity) should be uncertain according to the uncertainty principle, and therefore shouldn't the position of the particle have changed between the first and second measurement (due the particle having an uncertain and likely non-zero velocity)?

## 2 Answers

The key idea here is that, just after measurement, a quantum system is always in an eigenstate of the observable (operator) that represents the measurement. Therefore, if the same measurement is applied again straight away, before significant quantum state evolution governed by the Schrödinger equation has happenned, then the probability of observing the system in that eigenstate is unity, because the projection of the state onto the eigenstate has no effect, whereas there is zero projection onto any other eigenstate, because eigenstates of self adjoint operators (which observables all are) are orthogonal.

You seem worried that the uncertain momentum implies the particle has to move between measurements and indeed you are partially correct. But the assertion is being made with an assumption of negligible time between measurements. No, or negligible, quantum state evolution can happen in between the measurements to allow the assertion claimed.

It can be shown that the evolution projection of the position measurement onto the other position eigenstates after the measurement is quadratic in time, so the situation changes only relatively slowly. Normally projections change linearly with time over short timescales: this unusual situation in the position eigenstate arises because the projection onto the eigenstate corresponding to the measurement has reached a maximum, and therefore linear terms in the evolution equations vanish. This unusual quadratic dependence gives rise to the quantum Zeno effect, whereby one can keep a quantum system frozen in an eigenstate of an observable by repeatedly making the observation, often enough so that significant quantum state evolution has no time to happen before the state is "reset" back (almost certainly) to the nearest eigenstate.

Sorry, I dislike Griffith's approach to QM.

About the question near the bar, an operator $$\hat{A}$$ which does not commute with the Hamiltonian varies in time. Position does not commute with the Hamiltonian of a free quantum system. Therefore, if one allows time to pass between two measurements of position it is expected that the results will be different.

• I expect that's what the word "immediately" means in the quote from Griffiths... – Philip Aug 20 '20 at 10:39
• @Philip, I don't understand you and I don't know Griffith as rigorous about quantum mechanics. The rigorous approach is the one indicated by the quantum formalism. For the behavior in time of an operator Â we have the formula: dÂ/dt = ∂Â/∂t + (i/ħ)[Ĥ,Â]. The position operator, x, does not depend explicitely on time, however, does not commute with the free particle Hamiltonian. Therefore, no matter how "immediately" you measure it a second time, you won't find the particle in the same place. I advice, leave Griffith and look at the quantum formalism. - Best regards! – Sofia Aug 21 '20 at 18:58
• I have my own problems with Griffiths, I just meant that in this case I always assumed immediately to mean "separated by an infinitesimal time interval", and while I see your argument about any time interval -- however small -- changing the state, I assume that the smaller the interval, the more likely you are to get back the original state, so in that sense I find it a useful argument, but point taken! :) – Philip Aug 22 '20 at 12:22
• @Philip, maybe you'd tell me what is in fact your problem. It seems that you want to defend something else by insisting on this issue. So, you'd better tell me what is the other issue. You see, we are not in classical mechanics here to get dx = vdt. There is no v. Thus, the word "immediate" (i.e. dt -> 0) has no meaning. Now, I would recommend to look at the Ehrenfest theorems. Maybe they can help you, but, to your attention, they recover the classical mechanics when the wave-packet is very small - which happens at high velocities. Best regards! – Sofia Aug 23 '20 at 22:19