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For instance, assume the particle rests on the ground state $\psi_0 (x)$ of a one-dimensional simply harmonic oscillator around the origin of axes $ x=0 $, and once we measure the position of the particle and happen to obtain a definite value $ x=0 $. Now, we immediately measure the energy on this eigenstate of the position $\delta (x)$. The energy is divergent. To see it, let us expanding this delta function in terms of the stationary states $\psi_n (x)$ of the harmonic oscillator, we obtain the expansion coefficients that are the stationary states at the point $ x=0 $, $\psi_n (0)$. The energy is thus $\sum_n |\psi_n (0)|^2 (n+1/2)\hbar \omega $ which is divergent.

There is an interpretation of the absurd result: Once one measures the position of the particle with great accuracy, he drastically changes its momentum. Thus kinetic energy is transferred to the particle.

In order to remove the infinitely large energy, I prefer an alternative interpretation: In quantum mechanics, the measurement of the energy immediately after the measurement of the position is illegal. Am I right?

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  • $\begingroup$ Do you think measuring the position of a particle with infinitely high precision is reasonable? How much energy would you need to do so? $\endgroup$ – Mark Mitchison Oct 3 '16 at 15:00
  • $\begingroup$ Notice also that the thing you really take issue with is the fact that the expected energy after the position measurement is infinite. This has nothing to do with whether or not you actually measure the energy after measuring the position. In fact, an energy measurement will be sure to reduce the expected energy back down to a finite value (since the state after an energy measurement will be an energy eigenstate, and those all have finite energy). $\endgroup$ – Brian Moths Oct 3 '16 at 15:13
  • $\begingroup$ We need a huge amount of the energy, in a realistic experimental setup. However, we can always do so in thought experiment otherwise the probability density can not be defined. In fact, we can pose this question in a measurement of the energy immediately after the measurement of the momentum, without necessary to transfer energy to the particle. @Mark Mitchison $\endgroup$ – Quanhui Liu Oct 3 '16 at 15:16
  • $\begingroup$ It's perfectly reasonable to define the probability density as describing the limit of a sequence of increasingly fine position measurements, without ever discussing the absurdity of infinite-resolution measurements. I'm just pointing out that when you talk about impossible procedures it's no surprise to find impossible outcomes. $\endgroup$ – Mark Mitchison Oct 3 '16 at 15:24
  • $\begingroup$ I attempted the calculation of the position measurement as a sequence of increasing resolution, and found that in the limit of infinite accuracy, the results converge to the known one. Where had I made mistake(s)? @MarkMitchison $\endgroup$ – Quanhui Liu Oct 3 '16 at 15:45
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For this answer I will (at the OP request) make two assumptions:

  1. Standard Quantum Mechanics is a complete theory of nature
  2. When I do quantum measurements there are no classical experimental uncertainties i.e. I am as good an experimental physicists as quantum mechanics itself allows

Now the OP ask me to perform a position measurement on a particle in a potential well. To do this I will need to arrange for some form of interaction with the particle, say I send a photon into the well which interacts with the particle. I then set up photographic plates around the well (or whatever - I'm really a philosopher - don't bore me with all this) and by "measuring" the photon I obtain knowledge about the position of the original particle. In effect I have "measured" the original particle.

Now we've got the messy experimental stuff out the way, we can get back to quantum theory! Now I have some knowledge of the position of the particle I can use the knowledge in my Schrodinger equation to predict its future evolution. Where before the measurement I had a wavefunction spread out all over the well (I had no knowledge where it was), now I know where it is: I can perform "wavefunction collapse" and continue the evolution of the particle from that new wavefunction with localised position. Key point: this wavefunction collapse is not itself a physical process - this is just me obtaining more knowledge of the situation I am now dealing with.

Now you can ask what is wrong with your argument: how localised can you really make the wavefunction after the measurement? Even as a perfect experimental physicist you had to perform an interaction of the particle with a photon. From a different viewpoint, we can think of this interaction not as a measurement but simple as a quantum mechanical interaction. If you could do all this (I can't) you would find the couplings that are produced between the states of the photon and the position states of the original particle at not so well differentiated that doing a precise measurement of the photon state will not allow you to infer the exact position of the particle.

So in summary, the mathematical structure of quantum mechanics does allow you to meaningfully talk about a position delta function (with infinite uncertainty in momentum and hence infinite energy) but analysis of the complete interaction theory of quantum mechanics reveals that no interaction process would carry away information to allow you to represent a particle using a position delta function. In short you can't, even in theory, measure a particle with infinite precision. In detail what you find is that the higher frequency (shorter wavelength) of the photon the greater precision, but as you know to get an infinite frequency photon would need infinite energy. So the infinite energy constraint comes in at the start, prohibiting the measurement in the first place.

Clarification

As with everything in quantum mechanics, we should not talk about the results of the single measurements but consider an ensemble of identical measurements and the distribution of results we expect to get. In this situation suppose the particle is initially described by a Gaussian wavepacket with variance $V_x$. Then by Fourier transform or applying Heisenberg uncertainty principle the variance of momentum $V_p = \frac{\hbar^2}{4V_x}$. Now we design and perform a photon probe experiment which is designed to increase our knowledge of the particles position (apparently measuring the photon phase change or the reflection time are good ways to measure the particle position but the argument here is general). So, by design, our physical measurement process results in a new particle wavefunction with smaller $V_x$ and therefore necessarily larger $V_p$. Now the momentum expectation value is unchanged by the measurement, in other words the change in momentum is completely random. However the average kinetic energy is proportional to the square of the momentum, so the measurement increases the kinetic energy on average. i.e. on average the photon necessarily adds energy to the particle (this change is called the back-action of the measurement). This is what actually happen with real finite energy experiments. Now we can use this calculation to argue that it is impossible to design (even in principle) an experiment which results in $V_x = 0$ (a delta function). For to do so would require the photons, again on average, to add an infinite amount of energy to the particle. And you cant do infinite energy (average or no average).

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  • $\begingroup$ I do not understand why infinite uncertainty implies infinite energy. To me it implies completely unknown energy: the momentum being wholly undefined, it could very well be subsequently measured as very small; in that case, what would the infinite energy put in the system become? $\endgroup$ – Stéphane Rollandin Oct 5 '16 at 12:12
  • $\begingroup$ @StephaneRollandin Simple answer: if you have a momentum distribution which has expectation value 0 but infinite standard deviation (+ and -) then energy is momentum squared so it will have infinite expectation value. I stand to be corrected on the details of this. $\endgroup$ – Bruce Greetham Oct 5 '16 at 13:03
  • $\begingroup$ Right, but the expectation value is statistical. For a single measurement, any momentum can arise; my question stands. $\endgroup$ – Stéphane Rollandin Oct 5 '16 at 13:05
  • $\begingroup$ @StephaneRollandin See if my clarification works for you. $\endgroup$ – Bruce Greetham Oct 5 '16 at 16:58
  • $\begingroup$ Well, I do not agree with the standpoint taken in the clarification (the ensemble interpretation of QM). I guess it deserves to be added to your initial assumptions, as item 3. $\endgroup$ – Stéphane Rollandin Oct 6 '16 at 9:43
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Nothing is wrong here. The exact measurment of a particle's position requires an infinite energy "probe", e.g. a photon who's energy is $\lim_{\omega \rightarrow \infty} \hbar \omega $ infinte, which will in turn result in an infinitly high momentum transfere, and thus an infinite particle energy.

However, if you believe that no such infinite energies exist, there is no way to measure an exact position in the first place. What this all amounts to is that there is no way for an experimentalist to test the Heisenberg uncertainty principle $$ \Delta x ~ \Delta p \ge \frac{\hbar}{2} $$ in the limit $\Delta x \rightarrow 0$. So the question is rather philosophical. Which of the properties are disallowed or allowed, e.g. infinite probe energy or the resulting infinite particle energy or the exact position measurment, is not important since in the context of quantum mechanics and experimental limitations (finite energy), they can never be tested. As such, these limits should be avoided when calculating measurable results and only done in the last step as an approximation or simplification.

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Thanks to @Mark Mitchison, @Mikael Kuisma and @NowIGetToLearnWhatAHeadIs for discussions. I now present a brief answer to the question.

From theoretical consideration, by the position measurement on a quantum state $ \psi (x) $ defined in a spatial interval, we mean that, one of the positions, say, $x’$, is obtained, and the quantum state is collapsed into the state $\delta (x-x’)$. It is theoretically true. Next, if one tries to measure the energy of the particle on the collapsed state $\delta (x-x’)$, the energy is infinitely large. So theoretical prediction shows that the measurement of the energy on the state $\delta (x-x’)$ is impossible.

From point of experiment, the position measurement results in an approximate "collapsed" state, say, $1/\epsilon$ defined over small interval $ x\in (x’-\epsilon/2, x’+\epsilon/2)$, thus the immediately measuring the energy of the particle on this state is possible because of the predicted energy is finite with finiteness of the $\epsilon$.

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    $\begingroup$ Well, you assume the impossible, that the state collapses to a delta function, and then conclude that it is impossible to measure it's ENERGY. Wouldn't following argumentation be more reasonable: assume the collapse to a delta function. Deduce that it leads to infinite energy. Conclude that such measurement OF POSITION is impossible. $\endgroup$ – Mikael Kuisma Oct 4 '16 at 13:12
  • $\begingroup$ @QuanhuiLiu Just for fun (but actually a serious argument in quantum gravity): if you try to measure too small a distance, you put so much energy in a small space that the mass-energy would collapse into a black hole. So not only is collapse to delta function impossible but you can see things go horribly wrong with QM even at small finite length scales (the Planck length) $\endgroup$ – Bruce Greetham Oct 4 '16 at 17:46
  • $\begingroup$ @Mikael Kuisma I assume nothing, but the fundamental hypothesis in quantum mechanics does it----"A measurement always causes the system to jump into an eigenstate of the dynamical variable that is being measured, the eigenvalue this eigenstate belongs to being equal to the result of the measurement (P.A.M. Dirac )". Moreover, I do not care what operational definition of the measurement would be, and I do care "It is the theory which decides what can be observed ( Einstein)". $\endgroup$ – Quanhui Liu Oct 5 '16 at 0:00
  • $\begingroup$ @Bruce Greetham I am a theoretical physicist and we must start our discussion from the fundamental principles of quantum mechanics. Please tell me where I made the mistake(s) from the fundamental principles. $\endgroup$ – Quanhui Liu Oct 5 '16 at 0:12
  • $\begingroup$ @QuanhuiLiu I did not want to say you had made any mistake. I was just pointing out that quantum mechanics is not the final theory of nature. If I can I will try to give a more direct answer to your question. $\endgroup$ – Bruce Greetham Oct 5 '16 at 3:11

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