# Is measuring energy with arbitrary precision inherently impossible?

To see where this question comes from, consider a time independent Hamiltonian $$H$$ and an initial wave function $$\psi(t=0,x)$$. We can express time dependant wave function $$\psi(t,x) = \sum_j e^{-iE_jt/\hbar} \phi_j(x) c_j$$ where $$\{\phi_j(x)\}$$ is the set of eigen functions of H and $$c_j=\int \psi(t=0,y) \phi_j^*(y) dy$$.

Let's say we strongly measure energy at some arbitrary time. This causes wave function to collapse to $$\psi(t,x)=e^{-iE_kt/\hbar}\phi_k(x)$$, and we get $$E_k$$ as the energy of the system, where k is a random integer. Since $$[H(t), H(t')] = 0$$, measuring energy over and over again at any time always result in $$E_k$$ and wave function stays untouched. This is of course only true by assuming that the very act of measurement is not associated with a time dependant Hamiltonian $$H_m(t)$$ such that $$[H_m(t), H] \neq 0$$. If it does, then it would imply that measuring eigenvalues of $$H$$ with arbitrary precision in principle is impossible. There is always some noise coming from measurement itself, and the energy we are measuring corresponds to $$H_m(t) + H$$, not $$H$$ itself.

Let's assume otherwise and $$H_m(t)=0$$. At first, it may seem odd that, on one hand we can measure energy over and over and expect to get $$E_k$$ and indeed get $$E_k$$, so $$\Delta E = 0$$ but then we would have $$\Delta E \Delta t= 0$$. However, I see no issue, since t is not well defined, and in this context, the time between measurement is not necessarily $$\Delta t$$ anyway. Perhaps $$\Delta t$$ refers to inverse rate of changes of some operators that goes to infinity.

So either I am to believe that "measuring energy with arbitrary precision inherently impossible", which seems weird to say the least. Consider spin operator, as long as I measure spin on a fixed axis (say, z) over and over again, I always get the same value no problem, this is the case for all time independent operators, leaving $$H$$ out is a bit odd. For example, imagine a two state energy system as follows

$$|\psi\rangle = \frac{1}{\sqrt 2} (|e_0 \uparrow \rangle + |e_1 \downarrow \rangle)$$

Measuring spin gives full information about energy. So any limitation set on $$H$$ measurement should be also set on $$S$$ measurement. Or accept that $$\Delta t = \infty$$ somehow and I should not ask why. Or perhaps there is better explanation that I didn't think about, please let me know.

Edit:

Consider an ideal single-photon light emitter that uses a battery as it source. We can monitor the battery and measure its energy with arbitrary precision. Let say we measure battery's energy with 100% precision at arbitrarily small time intervals $$\Delta t$$. Now as soon as this emitter emits a single photon, we see a jump in battery's energy, indicating that how much energy was used for producing that photon. Since we know photon's energy, we know its exact momentum (the emitter emits photons in 1d). On the other hand, we roughly know when the photon was fired too by simply looking at the time of jump in the energy of the battery. Since a time interval $$\Delta t$$ exists between each battery's energy measurement, the uncertainty in the location of photon is $$\Delta x = c\Delta t$$ but given $$\Delta p = 0$$, we have violated uncertainty principle. Again, one of these should be true

1. All measurement devices are noisy, we can never truly measure energy without noise.

2. Or $$\Delta t$$ DOES refer to the time interval between energy measurements, and it goes to infinity for measuring energy value exactly.

The problem is, neither of this conclusions comes from standard texts in QM. Measurement devices can collapse wave function and measure energy exactly. $$\Delta t$$ is not introduced as time between measurements either, it can be anything, really.

Some related topics:

Interpretation of the energy-time uncertainty

What is $\Delta t$ in the time-energy uncertainty principle?

Energy-time uncertainty and pair creation

• Interesting question. As you correctly mention yourself it has to do with the time-energy uncertainty relationship which has always been more wishy-washy than the standard HEP. The reason is essentially due to the fact that time is not an operator in standard QM. The time-energy uncertainty has been investigated first by Mandelstam-Tamm and later by Margolus-Levitin and later by many others. You can look here en.m.wikipedia.org/wiki/Quantum_speed_limit_theorems
– lcv
Commented Mar 26 at 10:09
• @Icv out of current answers, yours is closest to an actual answer. The only missing point is that, as long as we keep measuring the wave-function, we stop its evolution, the so called Quantum Zeno effect. So I don't think that theorem would hold here. en.wikipedia.org/wiki/Quantum_Zeno_effect Commented Mar 26 at 22:49
• Thanks, which theorems are you talking about? The quantum speed limit theorems?
– lcv
Commented Mar 26 at 23:19
• @Icv Indeed. The unitary time evolution of wave-function is used to prove Margolus–Levitin theorem. When we have very rapid consecutive measurements, the time evolution is no longer a unitary. It freezes up to the eigenbasis that we have previously measured. I have somewhat explained this myself in my post, second paragraph. Commented Mar 26 at 23:34

Saying that a quantity can be measured with arbitrary precision means that you can measure it with any non-zero level of precision. So if you pick some particular non-zero number $$\Delta E$$ there is some measurement that could measure the energy of whatever process you're interested in with that precision. The fact that you have to measure energy with time dependent processes implies that $$\Delta E$$ must be non-zero, but that doesn't imply that measuring the energy with arbitrary precision is impossible.

Real measurements are interactions that produce records. Those interactions suppress interference: a process called decoherence:

https://arxiv.org/abs/quant-ph/0306072

Decoherence produces states that are narrowly peaked in position, momentum and other quantities such as energy on the scales of everyday life

https://arxiv.org/abs/0903.1802

https://arxiv.org/abs/1111.2189

Depending on what kind of interaction you choose you can make the widths of the states in terms of energy or momentum or whatever narrower or wider, but you can't break the uncertainty principle. And the limits imposed by the uncertainty principle are very small on the scale of everyday life. For example, if you have a particle with $$\Delta x = 10^{-7}m$$, the lower bound on the uncertainty in momentum is $$\Delta p > 10^{-27}kgms^{-1}$$.

Nor can you make the uncertainty in position zero since real measurement devices are governed by differential equations of motion whose solutions are differentiable and so aren't infinitely narrowly peaked in position.

• Can you explain more since I don't understand. By arbitrary precision here I meant measuring the value of energy with arbitrary amount uncertainty. For exact measurements, $\Delta E$ becomes zero. Commented Mar 26 at 22:42
• @Paradoxy Arbitrary accuracy does not mean zero. For example, in this paper core.ac.uk/download/pdf/216161148.pdf at the top of p. 345 a measurement is described as arbitrarily accurate when a quantity tends to zero which isn't the same as saying there is a measurement that realises the zero limit. By saying the zero limit is required you are deviating from standard usage and eliminating a necessary distinction. If the authors of the paper wanted to say "with zero inaccuracy" they could do that but they didn't. Commented Mar 27 at 9:31
• I think we were not discussing terminologies here, do we? What difference does it exactly make in my question even? Ok, $\Delta p$ goes to zero, whereas $\Delta x$ does not go to infinity. Still we have the violation of uncertainty principle. Commented Mar 28 at 16:27
• I have changed my answer to explain more fully. Commented Mar 29 at 0:09
• You are actually correct, and this is one of the first things that people learn in QFT. Nevertheless, since projective measurments are in principle possible in QM, I was wondering about self-consistency of QM theoretically. If no better answer comes up I will accept yours. Commented Mar 30 at 18:24

I assume that by a "strong measurement" you mean the standard "projective" measurement.

In your second paragraph, the state that the measurement process needs to be time independent is incorrect. We can always represent a measurement unitarily on an enlarged ("dilated") Hilbert space due to Stinespring's dilation theorem and Kraus's second representation theorem. The additional degrees of freedom in this dilated Hilbert space encode the state of the apparatus. Following von Neumann, we can think of this unitary as capturing time evolution under some "pointer" Hamiltonian $$H_\text{M} (t)$$ acting on the system and apparatus. It's fine if $$H_\text{M} (t)$$ is time dependent, provided that it commutes with $$H$$. Generally speaking, $$H_\text{M} (t) \propto H \otimes P,$$ where $$H$$ is the system's Hamiltonian that we want to measure and $$P$$ acts only on the apparatus (detector). For a general measurement of observable $$O$$, we replace $$H$$ with $$O$$. The coefficient of proportionality above may be time dependent.

This requires the ability to engineer evolution under $$H_\text{M}(t)$$, and you don't end up measuring $$H_\text{M} (t) + H$$, but simply $$H$$. Essentially, if $$P$$ is the momentum operator of a pointer particle, which we initialize in some Gaussian wavepacket $$\Phi_0 (X)$$ centered around $$X=X_0$$, then the system evolves under $$H_\text{tot} = H \otimes \mathbb{1} + \lambda (t) H \otimes P,$$ where the two terms correspond to evolution of the system alone under $$H$$ and the interaction that realizes the measurement, respectively. Because $$H$$ and $$H_\text{tot}$$ commute, and assuming for convenience that $$\lambda$$ is time independent, evolving the system + detector for time $$t$$ leads to $$\psi(x) \phi_0 (X_0) \mapsto \sum_j c_j \, e^{-i E_j t} \, \phi_j(x) \times \Phi_0 (X - \lambda E_j t ),$$ so that the shift in the position of the "pointer" wavepacket is proportional to the energy eigenvalue. We control $$\lambda$$ and $$t$$ is determined as in the Stern-Gerlach experiment (i.e., the pointer particle travels at constant speed $$v$$ in the $$Y$$ direction so that we know how long it will take to reach a detector screen, whereupon its position is registered). We can recover the energy eigenvalue $$E_j$$ from the observed location of the pointer. Importantly, all of the above works for a time dependent $$H_\text{tot}$$, and the measurement is of $$H$$, not $$H_\text{tot}$$.

As for an uncertainty principle for time, there is debate on how or whether this would work. This is because, in contrast to position and momentum, while energy is associated with an operator, time is simply a parameter. As far as I know, the only agreed-upon application of an energy-time uncertainty principle is to radioactive decay and similar processes. In the situation you describe, I would argue that there is no sense in which you are (quantumly) measuring time. I agree with your assessment that there is no issue.

At this point, I see no obstacle to measuring energy with arbitrary precision in principle. Although I would also note that nothing is arbitrarily precise in real experiment.

EDIT stuff [note that I have now edited this part of my answer].

In the second question you linked, the second answer gives an argument that $$\Delta t$$ does not reflect the time required to measure energy to precision $$\Delta E$$, which agrees with what I know about the experimental measurement of photon number / energy / intensity / etc. Because there is no "time operator," there is no canonical commutation relation to violate. The Heisenberg uncertainty principle is derived from canonically conjugate operator, but there is no time operator.

I'm not sure how many of the precise details of the scenario you propose are necessary. Based on your comment to my original answer, it seems like you're maybe ok with the $$\Delta E \Delta t \geq \hbar /2$$ relation not applying in this case due to the lack of a time operator. And that rather, your main point is that we can detect a photon's energy $$E$$ by absorbing it, from which we infer its momentum, despite also knowing its position at the time of measurement.

So the concern is that we may have a violation of the standard $$\Delta x \Delta p \geq \hbar / 2$$ uncertainty relation. However, the likely resolution is that, much like there isn't a time operator canonically conjugate to the Hamiltonian, there are not canonically conjugate position and momentum operators for photons --- at least, not in the usual sense. Dirac shows that no position operator exists for the photon in Principles of Quantum Mechanics (4th revised edition, p. 267, end of Sec. 70), and I also suggest this question and this one for more details.

This specifically refers to massless particles (e.g., photons) with integer spin (likely spin one). Importantly, these quantum systems described by an annihilation operator $$a_{\vec{k}}$$ at wavevector $${\vec{k}}$$ conjugate to position $$\vec{x}$$. Importantly, while $$a + a^\dagger \propto X$$ is a position operator for the harmonic oscillator, sums of the form $$a^{\vphantom{\dagger}}_{\vec{k}} + a^{\dagger}_{\vec{k}}$$ are instead related to the components of the electric field $$\vec{E}(\vec{x})$$, whose canonical conjugate is the vector potential $$\vec{A}(\vec{x})$$. In both QM and QFT, the wavevector $$k$$ is simply a label on those states.

While the momentum operator can be defined as a sum over wavevectors $$\vec{k}$$ of $$\hbar {\vec{k}} a^\dagger_{\vec{k}} a^{\vphantom{\dagger}}_{\vec{k}}$$, so Dirac's argument seems reasonable. There is no way to write down the position operator of a photon. We only ever know where they are when we create one or destroy one, because we know where we created or destroyed it. So I think that, in the case you describe (and for photons generally), you don't really measure the photon's momentum, but instead measure the number of photons at a particular momentum. This is how photodetection works in experiment (up to unimportant but complicated details). And there is no position operator from which an uncertainty relation for position and momentum could derive, in contrast to massive particles, where you use scattering and don't know where the scattering event took place, just the momenta of the outgoing scattered particles.

So for photons, it is possible to know the momentum with arbitrary precision while also having finite uncertainty in where the photon was. For example, by knowing the energy of an emitted photon and knowing where it was absorbed. This does not involve the measurement of noncommuting operators, and so there is no uncertainty principle. Per Dirac's arguments.

So I think the resolution to your question (both parts) is that there are not canonically conjugate operators from which an uncertainty relation could derive, so there is no violation of an uncertainty principle in either case. Let me know if I'm still missing / misunderstanding anything from your question!

• Thanks. I didn't understand your answer for Edit part fully. It seems that you are suggesting $\Delta t$ refers to the time that it takes for the photon to reach the detector. But this is not what I was asking. We measure battery's energy projectively over and over again. due to the conservation of energy, any energy shift observed in the battery corresponds to photon's energy, from which we find photon's momentum. So If nothing stops me from making $\Delta p = 0$(knowing photon's momentum exactly), we have a violation of uncertainty because $\Delta x = length of system$ which is not infinite. Commented Mar 30 at 18:15
• Oh sorry, I misunderstood the "edit" part. But I don't see how measuring the energy of a battery is a quantum measurement? But also there are subtleties to defining position operators for photons due to gauge redundancy, etc. There might not be canonically conjugate position and momentum operators for photons (an operator of the form $a + a^\dagger$ corresponds to the $E$ field, not position, e.g.). Commented Mar 30 at 18:55
There are many reasons for that, one is your statement 1. but of course the concept of noise should be generalized to include quantum uncertainty. Regarding statement 2., you have defined $$\Delta t$$ as the interval between measurements. However you have disregarded the time, say $$\tau$$, that it takes to carry out one measurement. Trying to achieve arbitrary precision implies $$\tau \rightarrow \infty$$ and since $$\Delta t > \tau$$ statement 2. becomes true as well.
• "It is impossible to measure anything with arbitrary precision" Why? What does in principle stop us from say, measuring exact position of a particle? Every standard QM text-book has at least a few problems that assumes such measurement is possible. "you have disregarded the time, say τ, that it takes to carry out one measurement" It is included in $\Delta t$ Commented Mar 26 at 22:37
• The uncertainty principle $\Delta x \Delta p \ge \hbar$ should be included in every standard QM text-book. Commented Mar 27 at 7:32