# Why/how does energy change when measuring position?

1. Sep 11, 2011

### nonequilibrium

Hello,

As a concrete model to refer to, I'll work with the one-particle system in a potential box with infinite edges. Inside the box is a potential barrier of height V. Let's say we've measured the energy at one moment to be E < V. The probability of measuring the particle inside the barrier is not zero. Say we measure it there, then $E \geq V$ and energy has changed as we would indeed expect from the fact that the Hamiltonian and position operator do not commute.

Now my question is: why/how does the energy change when I perform a position measurement? I realize it is dangerous to ask such a question, because the mathematics doesn't give reasons or mechanisms. I suppose my question is: doesn't it?
Am I right in supposing that the isolated system of, well, the described system together with the measuring apparatus has a total conserved energy? If so, the energy that the (sub)system gains when being seen inside the potential barrier, must've come from the measuring apparatus.

So is it true that if we accept the mathematical formalism of QM, we must also accept that whenever I succesfully measure such a particle in the potential barrier, the measuring apparatus must be built so such that it will impart to it an amount of energy which is at least as great as how much the particle previously lacked to go over it?

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Another experiment: take the diffraction of a wave packet through a slate with a slit in it. Say the initial state is a plane wave approaching from the far left (with a known impulse), and say the final state is a position measurement on the far right. Consistent with the idea of diffraction, the momentum will most likely have changed from beginning till end. Can I regard the reason for this being that the slate in the middle can be see as a delta function-esque potential, that interacted with the particle (better: wave packet?) as to change its momentum? Or is this a fishy way of looking at things?

Also, will the interaction of the particle with the wall/potential also be able to modify the particle's energy from beginning to end, or can it be assumed constant?

2. Sep 12, 2011

### The_Duck

You're of course right that energy is conserved, so if the overall system of particle+measuring apparatus starts with a definite energy, it must end up with the same definite energy. So if the energy of the particle changes during the measurement, the energy of the measuring apparatus must change by the opposite amount, as you suggest.

Examining some realistic cases may suggest to you that this is not so strange. Suppose I want to measure the position of an electron by shining light at it and tracking back the light scattered by the electron to find the electron's position. In this case the price of localizing the electron is that it absorbs uncertain amounts of momentum and energy from the photons that scatter off it. Higher frequencies of light correspond to shorter wavelengths, which tend to let you locate the electron better, but induce greater uncertainties in the electron's energy and momentum (recall that for a photon E = h*(freq) and p = E/c).

Here is another way one might conceivably measure the position of an electron in a square well. Acquire some number of very thin negatively charged plates (so that they repel the electron) and insert them into the well at regularly spaced intervals. Now the electron will be confined to the gap between some particular pair of plates, and we can determine which region it is in by removing the plates one by one and observing when the electron comes flying out. As you observe, if the electron had some definite energy state before this process we expect its energy to have changed as a result of its position being measured. We can envision the physical process responsible for this by considering what happens when the plates are inserted: with some probability the electron will happen to be located near one of the plates as it slides into place, and will receive a kick of kinetic energy as it is violently repelled from the plate. A measurement device that is going to operate by inserting plates in this way must be prepared to impart an extra energy to the plates equal to the amount of energy delivered to the electron when it is kicked away from a nearby plate.

It's fine. After all, a wall is just a region where the potential is so large that the particle cannot reasonably hope to penetrate it to any significant distance. Also, there's no reason to talk about delta functions if you find it more appealing to consider the barrier to have some small but finite width (and thus as a region of large but finite potential).

Energy is conserved, and if the wall is fixed and collisions with the wall are elastic there is no way for the wall to absorb energy from the particle. So you should expect a particle sent through the slit to have the same energy is started with (though only to the extent that it had a definite energy in the first place!)

3. Sep 13, 2011

### nonequilibrium

Hello Duck, thank you for your reply. I have two remarks/questions in reply to your post, and I'd really appreciate it if you could answer them because I'm not sure where I'd have to go to find an answer.

(1) regarding the influence of the apparatus
Okay, you state it's not weird that the apparatus can, for example, change the energy, and I get what you're saying, but it seems that the amount of influence the apparatus has depends on the apparatus: different ways of measuring something might influence different quantities in different amounts. On the other hand, however, the quantum formalism requires no information what so ever about the incidental apparatus used, so how can it take into account how much influence the apparatus has?

(2) regarding the physical reasoning itself
In your whole post you're working somewhat with a physical reasoning (as opposed to with the QM formalism that only says something about the result of a measurement) and don't get me wrong: I'd like to be able to do that too; I'm not a pragmatist so I'm not content with knowing "what number appears on my apparatus", but regardless of that: are you sure that you're "allowed" to do that? Is it consistent?
And regardless of how (or even if) you answer on that more general question, let's take a specific example:

You say: "You're of course right that energy is conserved" but am I? Isn't this a classical reasoning? My point is: what do you mean after all (--hear me out--) with energy being conserved when it comes to QM systems? The thing that comes closest as far as I know is that for the system used in my OP (the first, with the potential barrier) "if I measure energy E at some moment in time and I don't measure anything else in the meanwhile, then if I were to measure energy again, I'd get the same result E". You then seem to take this to mean that the energy had been E at every moment in between (and I agree it seems but a harmless and obvious extrapolation). This reasoning is consistent with the other physical reasoning (that you make) that it's the apparatus itself that makes the system change in energy.
But listen to this story: say we measure energy in the beginning and that E < V where V is the height of the potential barrier. First we let the system develop somewhat and then we measure position and see it is inside the barrier. Obviously simply by being there, E >= V now. Physically this is argued by the measuring apparatus having imparted energy to the system. However, right before the position measurement, the particle was already in the barrier (again, according to the physical reasoning, by continuity of motion), so E >= V already at that moment. But energy is conserved, since the apparatus hadn't interacted yet, so both E < V and E >= V. Hence the physical reasoning seems to be (sadly) quite unreliable?

4. Sep 13, 2011

### sweet springs

Hi, mr.wodka

Say the particle is charged one e.g. electron, you should go into valley of potential V with fog chamber. Measurement of position of electron by fog chamber causes uncertainty in momentum thus energy also undertake uncertainty.

Regards.

5. Sep 13, 2011

### The_Duck

A change in language may help to see what's going on here. Consider the set of all possible states of an electron. In QM, there *is no state* that simultaneously has a definite position and a definite momentum. There are states that have a somewhat definite position and a somewhat definite momentum, and the amount to which the position and momentum can both have definite values in the same state is quantified by the uncertainty principle. This statement does not refer to any measuring apparatus.

But suppose we have an electron that is initially in a state of perfectly definite momentum, and we measure its position very precisely with some apparatus. After the measurement, the electron is in a state of fairly definite position. The only such states have poorly defined momentum. Thus, entirely independent of how the position-measuring apparatus is constructed, its operation affects the momentum of the electron. If you have some device that for some reason cannot affect a particle's momentum in any way, then this device cannot possibly be a position-measurer, because a position-measurer must necessarily turn states of definite momentum into states of uncertain momentum. Again, this is because there *are no* states of definite momentum that have a well-defined position. We can make statements about the behavior of any conceivable measuring apparatus because they are really statements about what kind of states exist for the system under consideration, independent of any measuring apparatus.

[I discussed position and momentum above, but the same discussion applies to any pair of noncommuting observables, such as energy and position, with the difference that there may not be quite as simple an uncertainty relation between them. But in general, the fact that two observables do not commute means that there is no complete set of states with definite values of both.]

I disagree with your statement that QM "only says something about the result of a measurement." To the contrary, measurement is a physical process that consists of the interaction of a measuring device and a measured system. "Physical reasoning" applies to this physical process as to any other. And it can be treated mathematically according to quantum mechanics. That is, we can set up the quantum mechanical problem where we have an overall system composed of two sub-systems, the measured thing and some specific kind of measurement apparatus. The overall Hamiltonian will have terms that allow these two subsystems to interact, and we can watch the time evolution of this overall system to see what is physically happening during the measurement.

There is maybe a philosophical difficulty when you wonder why we perceive a single world when quantum mechanics only ever seems to output superpositions of possibilities. This is the so-called "measurement problem," leading to various interpretations of quantum mechanics which attempt to explain this [without modifying the math of QM--these are interpretations, not modifications]. However, I would advise you not to worry about this right now; there is a reason QM textbooks tend to leave any discussion of interpretation to the end. One is better off considering the philosophical interpretation of the math after one has worked with it for a while.

No, conservation of energy in an isolated system is exactly true in QM and not an iffy "classical" idea.

That definition sounds OK to me. The idea is that performing some measurement in the meantime involves causing the system to interact with some measuring device, so the system is not isolated and may exchange energy with the device. Indeed, part of what we are discussing is that if the apparatus measures a variable that does not commute with the Hamiltonian, it must necessarily be constructed in a way that lets it exchange energy with the system. If this happens you're not guaranteed to measure the same energy afterwards as before.

I don't think I said this, but this particular assertion does indeed seem harmless. The quantum state is at all times between the measurements an eigenstate of energy, so it seems reasonable to say that during the intervening time the system has energy E.

No!

Here you appeal to a principle that does not appear in the math of QM. This is inapplicable classical reasoning. In no sense was the particle in some region at a certain time just because you happened to measure it there at a slightly later time. To the contrary, just before the position measurement, the particle's wave function was quite spread out. Your language suggests that you visualize a definite but unknown path traversed by the particle between measurements. This is false. If you want to know what the particle is doing, you look at the wave function, and the wave function is spread out over a region. As a consequence of interacting with the measuring device, it becomes concentrated in some narrow region, but just before the measurement it was spread out much more.

Above I allowed that you might consider the particle to have a certain energy even at times when its energy wasn't being measured. This was because even during those intermediate times, the wave function is an eigenfunction of energy and so describes a state with a definite energy. By contrast, during times when the particle's position is not being measured the wave function does /not/ describe a state with a definite position and so it is not correct to speak or think as if the particle has a definite position at those times.

Intuition can lead you astray, but "the great thing about physical intuition is that it can be adjusted to fit the facts." Of course it is not a mathematical theory to make up little verbal stories about what is happening in various physical processes. The only way to be perfectly sure of getting the exact right answer is to grind through the math. But there are certain bedrock principles enshrined in the math, like conservation of energy. By appealing to these you can often get a good sense of what is going on without writing down any equations. Conversely, appealing to principles that appear nowhere in the math will cause you problems.

Last edited: Sep 13, 2011
6. Sep 13, 2011

### nonequilibrium

I'm familiar with the mathematical formalism and what it states, but that's not really what my question was about, it was in response to your "Examining some realistic cases may suggest to you that this is not so strange. Suppose I want to measure the position of an electron by [...]" where the reason for the change in momentum when measuring the position was due to the specific 'construction' of the position measurement apparatus; it seems one can easily think of another measurement apparatus that will destort the momentum more or less, but as you say yourself the math does not imply such things, so it's questionable that the math and your physical example are talking about the same principle (why do you think it's obvious they are?).
A side remark: any measurement of position will make it collapse into a position eigenstate (and thus all information about momentum will go away), yet in your physical reasoning of the experiment, there seems to be a gradual increase of the removal of momentum-information, based on how well you measure the position. Does this not raise doubt on the issue whether your physical conception strikes the heart of what the math tells us about a measurement?

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" I disagree with your statement that QM "only says something about the result of a measurement." [...] and we can watch the time evolution of this overall system to see what is physically happening during the measurement."

Well I think this might be semantics: you seem to be calling the wave function physical, as in really there in some sense. This might very well be, but I don't see this as an appendix of the mathematical formalism: all the math seems to be telling me is the probability on some outcome when performing a measurement (even if another measurement apparatus is part of the investigated system, as in your subsequent example). If I choose to see the wave-function (when not measured) to be a physical representation or a statement of ignorance is interpretation.

The rest of your post argues against the piece of physical reasoning I typed out in my last post, and of course I welcome you to do that (not sarcastic in any way), but I wonder if you realized that the intent of my post was to indeed argue that the use of such physical reasoning is easily inconsistent? Well, I can already hear you say "but no, correct physical reasoning is allowed, but the continuity you used is simply not correct physical reasoning" and I again would agree with this, or at least that nobody has any reason to believe that it would be correct, because it is indeed not a part of the math, but my point: you did use it in your previous post: in your previously quoted example where you measure the position of a particle, you're working with the physical idea of a particle moving continuously, having a well-defined position and momentum, aspects you are now telling me I cannot use. So did I misunderstand your earlier use of it or what is the deal?

7. Sep 13, 2011

### The_Duck

I guess I have mistaken your misgivings somewhat; I will try to say something new to address them better instead of repeating myself unhelpfully.

You are correct that it is possible to construct various measuring apparatuses for some observable, some of which will cause more or less disturbance to the incompatible observables. The minimum disturbance is determined by the commutation relations and the corresponding uncertainty relations. But nothing prevents a measurement from causing *more* uncertainty in incompatible variables than the minimum specified by the uncertainty relations.

Consider: I have an electron traveling along a tube of 1 meter in diameter and many meters long. Suppose the electron's initial state is a wave packet with a sharply defined momentum, but spread out over a significant portion of the tube in position space. I now perform the following crude position measurement: I irradiate a 1 meter long section of the tube with electromagnetic plane waves of enormously high frequency. If I detect any scattered photons I have localized the electron to this cylindrical region of 1 meter in diameter and 1 meter in length. This is a very large uncertainty in position and the corresponding uncertainty in momentum demanded by the uncertainty principle is quite small. But in fact the electron has received, from recoils, a potentially huge amount of momentum in an unknowable direction, so after the measurement the electron has a very large uncertainty in momentum.

This raises a point you should think about when considering these questions: the measurements we most like to think about when doing the math are highly idealized (as what mathematical physics is not?). A position eigenstate represents a particle localized to a volume of size literally zero, which is manifestly unrealistic. No physical process can really be said to measure the position observable we use so often in quantum mechanics. Instead consider as a slightly more realistic example the observable

$$\lfloor \hat{x} \rfloor$$

That is, the floor of the position operator (let's confine ourselves to one spatial dimension, and also fix some unit of length so that the floor of the position makes sense). It's effect, for instance, on the position eigenstate localized at x=2.5 is

$$\lfloor \hat{x} \rfloor | x=2.5 \rangle \;\; = \;\; 2 | x = 2.5 \rangle$$

and its effect on some wave function psi(x) is

$$\lfloor \hat{x} \rfloor \psi(x) \;\; = \;\; \lfloor x \rfloor \psi(x)$$

Any wave function in position space entirely contained between x=n and x=n+1 for some integer n is an eigenfunction of this operator (note that there are infinitely many such eigenfunctions for any n; the spectrum of the operator is infinitely degenerate). A measurement of this operator has the following effect on the wave function

$$\psi(x) \to \left\{ \begin{array}{lr} \psi(x) & : x \in [n,n+1)\\ 0 & : otherwise \end{array} \right.$$

on obtaining eigenvalue n.

This is a crude representation of an approximate position measurement, which specifies the approximate position in discrete increments of size 1. This observable is not too different from x, but obviously it has limited precision. It is more or less the observable that is measured when you follow the plate-insertion measurement procedure I mentioned a couple posts above.

Now, if I measure this observable I know that my wave function is contained within some position interval of size one, and accordingly there is some minimum dispersion in momentum space of the wave function after the measurement. This dispersion increases if I make the plates closer together. Of course, if I do this my measurement corresponds to a slightly different operator, perhaps floor(Ax)/A, which represents a measurement when the plates are separated by a distance 1/A. By increasing A I make an increasingly precise position measurement, and correspondingly the minimum momentum uncertainty after a measurement increases. In the limit where A goes to infinity and the plates have zero spacing, floor(Ax)/A goes to x and I am conducting infinitely precise position measurements, which correspond to measuring the operator x, and which induce infinite uncertainty in momentum.

In consideration of the above you might consider it an abuse of language how much I have talked about "a fairly precise measurement of position x" or the like. Really, perhaps, I ought to say, "A measurement of the observable floor(Ax)/A where A is fairly large but finite."

I'm not sure what you mean exactly by this, and maybe this is your main point. Can you expand on it and say whether or not what I wrote above touches on it at all?

All right; I was just trying to make the point, which maybe was unnecessary, that a measurement is not some extra-physical process that proceeds by different rules than normal physical processes.

To an extent, yes. But you might note that interpretations in which the wave function is an expression of ignorance of so-called "local hidden variables" are in contradiction with the QM formalism, and experimentally ruled out. ("Non-local hidden variables" are allowed and realized in the Bohmian interpretation.)

Right; I tried to acknowledge this in the very last part of my post and to provide a strategy for reasoning physically while being less likely to stumble into error. But when you are not confident what the equations imply, the only way to be sure is to write out the equations.

As an aside, I'm not sure if maybe you are using "physical reasoning" to mean something like "classical reasoning." If so, I haven't been interpreting it this way.

Rereading my initial examples, particularly the charge-plate-insertion example, I realize I have a confession to make. To determine the origin of the change in the electron's energy in the charge plate insertion position measurement technique, I first reasoned internally about wave functions and then translated my thoughts into classical language. I think there is a tendency in QM to do this often, maybe because relatively few effects are purely quantum mechanical; many have analogs in classical mechanics, and discussing them in classical language is easier and keeps us from obscuring this connection. But we are talking about the incompatibility of observables, which is purely quantum mechanical, so use of classical language is perhaps inexcusable. Here, then is a quantum mechanical discussion of plate insertion.

Say we have an electron in some energy eigenstate of a potential well. We are going to insert a number of plates that the electron cannot penetrate, dividing the well into a number of small regions. A plate can be modeled as a very narrow region in which the potential spikes to a very high value, making it impenetrable to the electron. Let's imagine inserting the plates somewhat gradually, so that the potential spikes grow continuously from zero to their large (but finite) size. Imagine the shape of a potential spike as some continuous function like a Gaussian.

What happens in the region of a plate as we insert it? Initially the potential is flat at zero. Suppose we are considering a small enough region so that the wave function in this region is approximately constant. A bump in the potential appears and starts growing. Now, a classical particle located on the left slope of the bump would feel a force to the left (F = -dV/dx). Quantum mechanically, the wave function in the region of the left slope develops a momentum component to the left. Inversely, the wave function in the region of the right slope develops a momentum component to the right. Said another way, the probability current, which was zero in the stationary state, starts to point away from the growing bump in the potential. As a result, the magnitude of the wave function begins to decrease at the spike. As the spike grows larger these effects only get stronger, and the wave function falls to zero in the region of the spike. However, we clearly saw that as the spike was growing, it change the momentum components of the wave function.

Another way to reason about this process is: Suppose we insert the plates fairly quickly, so that overall wave function does not change much during the insertion process, except in the region of each plate. At each plate, the wave function must fall to zero, since the electron is being excluded from these regions. But since the overall wave function was largely unaffected, near the border of a plate the wave function still has its original nonzero value. Accordingly the wave function varies quite rapidly in space as you move away from the edge of a plate. But rapid spatial variation of the wave function corresponds to large kinetic energy. So the wave function has acquired significant high-momentum and high-energy components as a result of the plates driving the wave function to zero in certain regions.

Probably the way to insert the plates that creates the minimum disturbance to the particle's momentum and energy is to insert the plates very very slowly ("adiabatically"). If we do this, the "adiabatic theorem" tells us that if we start out in, say, the ground state of the square well, we will end up, after the very slow modification of the potential by plate insertion, in the ground state of the new potential. So the new wave function will have the particle in the ground state of each of the new little infinite square wells formed by the plates. Clearly this state has higher energy and greater momentum dispersion, and we can see where the energy and momentum come from by considering the processes described two paragraphs above, which occurred slowly over the process of insertion.

Last edited: Sep 13, 2011
8. Sep 15, 2011

### nonequilibrium

Hello, thanks for replying once again, I don't have a lot of time at the moment, so I'll give a short reply now with only a few (but important) remarks; I'll come back to the rest later.

There are many things I'd like to comment upon, but I think one of your statements makes clear why you make other statements that I don't really tend to agree with (or understand) so it might be smart to put emphasis on it:

I have never heard this view before. As I understood it, a measurement always drops it into an actual eigenstate, it's just that the humans might not know it to such precision (i.e. I separate the concept of 'determination of the quantity' and 'experimental error'). I also think your view would lead to some inconsistencies, but before I go into a new lengty segment, I think it's just best to ask you: can you give any references to support that view? (to be precise: the view that the 'amount of collapse', for want of a better term, depends on the apparatus; of course I agree QM is a mathematical idealization, in the sense any theory is, but not to the extent you see it)

I think a good characterization of what I mean with "physical" is "any reasoning using the idea of a point particle in between measurements" (it might need some nuanciation)