# Does a measurement always require the exchange of energy?

Can you measure something about something without adding or removing energy from it?

This comment has got me wondering if there is a way to measure a property or a quantity of something without some kind of energy exchange or energy conversion. It asks me for an example, and I can't think of any right now.

So the answer here needs some, any, well though out example if the answer is yes.

• @ACuriousMind - fixed the link and the note, thanks! – uhoh Jun 16 '16 at 12:51
• I'm confused on your example. You have a neutron beam, do something with it, and have some (possibly changed) neutron beam leaving your apparatus. But, you have not performed a measurement yet - you have a neutron beam with some property you want to measure to see if it is different. So, how are you going to measure neutron polarization (for example) without doing something that interacts with (exchanges energy with) at least some of the neutrons? – Jon Custer Jun 16 '16 at 13:18
• @JonCuster Well, I said "That's just a possible example that I've started to think about.." In that beginning of a possible example, the measurement would be the magnetic field difference between the two paths the neutron can take. There is energy in the field, but changing the phase of some expectation value associated with a neutron not polarized in the field direction might not involve an exchange of energy with the field. – uhoh Jun 16 '16 at 13:23
• <An unpolarized / polarized beam of neutron passing through a magnetic field and simultaneously passing through a different field, and then their interference is related to the difference in the fields, but so that the expectation value of the energy removed from the field (or the magnet) is still zero>. - in the above example when you pass the particle beam from the field -the field does some work involving energy transfer therefore to say that no changes in the field is involved does not stand,,,any interaction involves transfer of energy in one way or other. – drvrm Jun 16 '16 at 14:12
• Well one thing that might be helpful to point out: if you start with a quantum state with well-defined energy, and you then measure an operator that doesn't commute with the Hamiltonian, then you necessarily change the energy of the system. – Ruben Verresen Jun 18 '16 at 16:11

Yes. The Aharonov-Bohm effect is an example where something can be measured without exchanging energy.

A long solenoid inside the two paths of an electron two-slit experiment induces a phase shift $\Delta \phi=q \Phi_B / \hbar$, even if there is no external electromagnetic field. The only thing that matters is that the integrated magnetic potential around the path is nonzero [1]. So here we can measure magnetic flux without adding or removing energy to it.

I think interaction-free measurements in quantum mechanics also fits [2]. The Elitzur–Vaidman bomb-testing problem is likely the most famous example [3]. Here measurements are using counterfactual interactions, where information is gained even when there is no energy exchange (and clever setups can make it arbitrarily unlikely to happen).

[2] Elitzur, A. C., & Vaidman, L. (1993). Quantum mechanical interaction-free measurements. Foundations of Physics, 23(7), 987-997. https://arxiv.org/abs/hep-th/9305002

• This might be a good answer, but considering it starts with a question, I couldn't accept it. Can you make a more definitive statement? Like "yes" or "no" for example? One way would be for you to post this separately as a question. "Can the AB effect be used to measure the strength of a magnet without any energy transfer from/to the magnet?" Notice that I've said magnet and not magnetic field. Fields have to come from somewhere, and I suspect there are some subtitles in the assumption that the field just happens to be there for no apparent reason. – uhoh Aug 6 '17 at 7:00
• If you skew the distribution of electrons to one side or the other, does that not look like they will have some net transverse momentum, and would that not also mean they have gained some energy, and if so, where did it come from? So if you ask this as a question you may get some help addressing the entire system, and not just the AB effect. As you've said elsewhere, "Momentum conservation is a headache." :) – uhoh Aug 6 '17 at 7:02
• Turned the answer into more of an answer. In the AB effect, the solenoid generates the magnetic field: it could be powered by a battery. Also, the electrons are not pushed: there is no change in wavefunction amplitude, only phase. – Anders Sandberg Aug 6 '17 at 10:05
• OK, a very cheerful "yes" now, thanks! The use of the Mach-Zehnder interferometer in [2] gets rid of the double slit and things going "left" or "right" so I don't have to worry about that anymore. I'll keep reading on the topic, thank you for this! – uhoh Aug 6 '17 at 14:12

In theory there is no lower limit on the amount of energy that must be exchanged to make a measurement, at least directly. But you are constrained by Landauer's principle when you initialize the memory you need to record the measurement. Effectively the measurement is the replication of information about the measured system's state in the measurement system's state. This means that some information about the measurement system's state - that which the measurement "displaces" - must wind up encoded in the environment and ultimately one needs to do work to expel this excess entropy.

So you will need to expend work $k\,T\,\log 2$ to initialize the memory you need to record each bit of information measured.

For some time after Szilard thought up his famous Szilard engine thought experiment (1929) he believed the mechanism whereby the engine complied with the second law was exactly that it takes an amount of work equal to $k\,T\,\log 2$ to measure one bit of information about a system's state.

However this was shown not to be so by several thought experiments, most notably the Fredkin-Toffoli Billiard Ball Computer. In this device, internal register states can be polled without expenditure of energy. An excellent review of these and other ideas is to be found in

Charles Bennett, "The Thermodynamics of Computation: A Review", Int. J. Theo. Phys., 21, No. 12, 1982

As discussed in the Bennett paper, the work input to the Szilard engine that the second law would require is needed to "forget" bits of information, as I summarize in my answer here. When you make a measurement, you need to "make room" for it by "forgetting" the former state of the system you encode that measurement in.

Some further explanation connecting the frictionless computer with measurement. The Toffoli computer, if I understand the history correctly, was the first accepted demonstration of the error in Szilard's assumption that it was the measurement that required the amount $k\,T\,\log 2$ of energy to decide whether a molecule were travelling fro- or to-wards the door in his own version of the Maxwell Daemon. The polling of one bit of a computer's memory is exactly the same thing as the molecule measurement: it is the inference of one bit of information about an observed physical system's state and, at least in this case, the Toffoli experiment shows that this inference can be done without energy expenditure.

Now, if you are worried about friction in the Toffoli experiment, then include it in the thought experiment, and imagine decreasing it through some engineering measure: better machining, magnetic levitation, whatever. As the friction is decreased through these measures, there is no fundamental physical principle encountered which halts the process. It may be impracticable to further the process, but there is no fundamental physical reason why the friction cannot be lowered. You can in principle make it arbitrarily small. This is quite different from the realization that to infer one bit of information about a system, you have to write that information in the physical state of some system, and the state that you write over has to be recorded elsewhere. This follows from an assumption that the microscopic physical laws are reversible, and, if true, the limit Landauer limit is a fundamental one.

• The four links seem to focus on computation. I'm not asking if measurement requires energy, I'm asking if it requires exchanging energy with the system being measured. It's a generously relaxed form of "do you have to convert something to energy to measure it?" Maybe you've explained why this energy must come from the thing (particle, field, gravity wave) being measured and I just don't see it yet. If so can you help me out? – uhoh Jun 16 '16 at 14:24
• c.f. the first sentence: "Can you measure something about something without adding or removing energy from it?" – uhoh Jun 16 '16 at 14:25
• @unoh the point about the links is that states of computers can be in principle polled without energy expenditure and without removing energy from the computer. The thought experiments show how this is done explicitly. So I think this answers your question can you measure something without adding or removing energy from it: the answer is in principle yes. You're simply replicating some information about its state in your measurement device, and this latter device is where the expenditure must take place in accordance with Landauer's principle. – WetSavannaAnimal Jun 16 '16 at 14:28
• So if the answer is in fact "Yes", then could you add that to your answer explicitly? At least something of a boolean or binary nature? By the time I get to the second sentence: "But you are constrained..." it sounds like you are building up to a foil. If in this example, the memory is the thing to be measured, then the one-sentence paragraph: "So you will need to expend work $k\,T\,\log 2$ to initialize the memory you need to record each bit of information measured." doesn't help reassure me at least that you don't have to add or remove energy from the thing to be measured. – uhoh Jun 16 '16 at 14:43
• And if it's not, then it's just distracting. (well, distracting me at least). – uhoh Jun 16 '16 at 14:45

If we measure length by a scale there is no change of energy. So it depends on subjects to be measured.

• You can't use a scale in the dark, and light is a form of energy. – Dmitry Grigoryev Aug 4 '17 at 10:37
• But it doesn't require change of energy – Bivas Das Aug 4 '17 at 19:43
• If you shine a light on something, you will add kinetic energy both thermally and kinematically (recoil). The act of seeing something changes it. – uhoh Aug 6 '17 at 6:55

Everything is always receiving and emitting energy. If we observe a distant star to estimate its temperature then we are receiving energy from it. However we are not causing any additional energy exchanges.

IMO all particles are standing waves that continuously receive and emit spherical waves at the Compton frequency of the particle. This understanding explains the properties of particles including de Broglies equations. Look up WSM for more.

Yes. Time. We can think of an experiment where a black box sits in a reference frame together with a clock. Everthing that happens inside the box, happens in the same frame of reference as the clock, however to measure the ticks of time inside the box, doesn't require to take energy from the box. Synchronisation of clocks would be another problem though....