Heisenberg's Uncertainty Principle common misconceptions

[Jimmy87]

Hi pf,

Please help me, the more I read about HUP, the more discrepancies I find between different sources. I found a good article that Zapper made on HUP (https://www.physicsforums.com/insights/misconception-of-the-heisenberg-uncertainty-principle/). My confusion is about the way it is explained and came about after watching this short video:



This clip shows two common misconceptions. The second one is about explaining HUP by trying to pinpoint an electron with a high energy photon which kicks the electron giving an uncertainty in its momentum. Both this video and Zapper's post say this is not HUP. However, this is the way it has been taught to me. Also, physics lecturers use this explanation. For example, check 1hr 4mins into this video:

https://www.youtube.com/watch?v=JzhlfbWBuQ8

Leonard Susskind clearly explains HUP through photons kicking electrons which gives an uncertainty in momentum - however Zapper and first video say that this is not the uncertainty in HUP. This video by Nottingham University physics department (look at 5mins 20s into video) also explains it the same as Susskind:



Please can someone help! Are susskind and sixty symbols correct or is this explanation a common misconception as explained by Zapper and first video.

Thanks for any help.

 

[phinds]

This clip shows two common misconceptions. The second one is about explaining HUP by trying to pinpoint an electron with a high energy photon which kicks the electron giving an uncertainty in its momentum. Both this video and Zapper's post say this is not HUP. However, this is the way it has been taught to me.

Then you have been taught incorrectly since that point of view says that the HUP is a measurement problem, which is is not. It is a fundamental characteristic of quantum objects and has nothing to do with the effects of measurement.

Leonard Susskind clearly explains HUP through photons kicking electrons which gives an uncertainty in momentum

This seems very unlikely since Susskind certainly knows better. Are you sure you understood him correctly? Can you give a citation with an exact quote?

 

[PeroK]

Please can someone help! Are susskind and sixty symbols correct or is this explanation a common misconception as explained by Zapper and first video.

Thanks for any help.

The first video deals with the HUP and, from what I watched, appears to be correct in nailing down the common misconceptions.

Susskind is dealing with something else (does he actually say "uncertainty principle"?). He's dealing with the fact that if you have a well-defined momentum and you do a measurement of position, then you lose the well-defined momentum.

Strictly speaking, this is not the HUP. It's effectively a manifestation of the HUP, although you would have to be very careful how you phrased things if you want to say how it is related to the HUP.

 

[PeroK]

In the third video, the guy from Nottingham is (in my opinion) trying to tread a fine line between the maths and a simplistic statement of the HUP. If you listen carefully, once or twice he corrects himself, realising he's been a bit too imprecise. His headline statement is something like:

You cannot know (or measure) the position and momentum of a particle at the same time.

And that's what Susskind was explaining and it is related to the HUP. Although, it's not really a statement of the HUP itself.

 

[Jimmy87]

The first video deals with the HUP and, from what I watched, appears to be correct in nailing down the common misconceptions.

Susskind is dealing with something else (does he actually say "uncertainty principle"?). He's dealing with the fact that if you have a well-defined momentum and you do a measurement of position, then you lose the well-defined momentum.

Strictly speaking, this is not the HUP. It's effectively a manifestation of the HUP, although you would have to be very careful how you phrased things if you want to say how it is related to the HUP.

Thanks. Could you explain why the photon - electron scenario susskind talks about is not the HUP. What is the Nottingham professor trying to avoid saying?

 

[PeroK]

Thanks. Could you explain why the photon - electron scenario susskind talks about is not the HUP. What is the Nottingham professor trying to avoid saying?

It's easier to say what the HUP is. First, we can look at it in a theoretical context. If you have a particle in any given state, then you can look at action of the position and momentum operators on that state. This will theoretically give a range of values for position and momentum according to a probability distribution, based on the specific state of the particle. You can calculate the standard deviation for position and momentum (denoted by ##\sigma_x## and ##\sigma_p##). The HUP says that, for any state:

##\sigma_x \sigma_p \ge \frac{\hbar}{2}##

You can interpret that as a statistical statement about the measurements of position and momentum you will get for a particle in that state. And, loosely, it does indeed say that the smaller the spread of position measurements, then the larger the spread of momentum measurements. In particular, if the particle is in a state where the position is relatively well-defined, then the momentum must be less well-defined and vice-versa.

Now, to verify the HUP you would need to arrange an experiment where a large number of particles was prepared in the same state and you made a measurement of position for half of them and a measurement of momentum for half of them. You would calculate ##\sigma_x## and ##\sigma_p## for the specific data you had gathered and check whether or not the HUP is confirmed in this case. But, like all statistical laws, the HUP requires a large number of experiments and says nothing definite if you simply do one or two measurements.

If we stop there, then that is essentially the HUP in all its glory.

Note, however, that another aspect of QM is that a measurement of positions "destroys" any previous measurement of momentum, and vice versa. Again, you need a large number of particles in the same initial state and you do the following:

1) Measure the momentum. This measurement of momentum sends the particle into a state of relatively well-defined momentum. If you were to repeatedly measure momentum, you would get a similar measurement each time.

2) Measure the position. This will send the particle into a state where the position is relatively well-defined and (by the HUP) the momentum must be relatively not well-defined. You should get a large range of values for your position measurement here.

3) Measure the momentum again. Now, you get a large range of momentum measurements. The well-defined momentum established in step 1) has been destroyed by step 2).

4) Measure position again and now the well-defined position established in step 2) has gone.

This is not the HUP as stated above, but hopefully you can see the relationship between the two. The real issue here is why (physically and practically) must a measurement of position or momentum destroy the previous well-defined state? That's what Susskind was dealing with.

You could ask that IF you could practically measure position without disturbing the momentum and vice versa, then where would be the HUP be? You would effectively have a particle in a state with a well-defined position and momentum. The way I would describe this is that the HUP must be backed up by some physical and practical issues in the simultaneous measurement of position and momentum. But, these issues are not the HUP itself.

 

[vanhees71]

You cannot know (or measure) the position and momentum of a particle at the same time.

And that's what Susskind was explaining and it is related to the HUP. Although, it's not really a statement of the HUP itself.

This is, in my opinion, also misleading. In my opinion (I'm convinced of the minimal statistical interpretation, i.e., that the precise meaning of the quantum state is the probabilistic one given by Born's Rule, which is a basic postulate of QT, and nothing else) the meaning of the uncertainty relations is that a pair of incompatible observables cannot take precisely determined values and that the standard deviations necessarily obey the uncertainty relation,
$$\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}] \rangle|,$$
where the expectation values to evaluate the standard deviations and the expectation value of the commutator can be any pure or mixed state, i.e.,
$$\langle \cdots \rangle = \mathrm{Tr} (\hat{\rho} \cdots).$$
It does NOT tell you in which sense the measurement of one observable affects the precision of another observable. This cannot be said in such a general way but depends on how the measurement is done in detail.

 

[Jimmy87]

It's easier to say what the HUP is. First, we can look at it in a theoretical context. If you have a particle in any given state, then you can look at action of the position and momentum operators on that state. This will theoretically give a range of values for position and momentum according to a probability distribution, based on the specific state of the particle. You can calculate the standard deviation for position and momentum (denoted by ##\sigma_x## and ##\sigma_p##). The HUP says that, for any state:

##\sigma_x \sigma_p \ge \frac{\hbar}{2}##

You can interpret that as a statistical statement about the measurements of position and momentum you will get for a particle in that state. And, loosely, it does indeed say that the smaller the spread of position measurements, then the larger the spread of momentum measurements. In particular, if the particle is in a state where the position is relatively well-defined, then the momentum must be less well-defined and vice-versa.

Now, to verify the HUP you would need to arrange an experiment where a large number of particles was prepared in the same state and you made a measurement of position for half of them and a measurement of momentum for half of them. You would calculate ##\sigma_x## and ##\sigma_p## for the specific data you had gathered and check whether or not the HUP is confirmed in this case. But, like all statistical laws, the HUP requires a large number of experiments and says nothing definite if you simply do one or two measurements.

If we stop there, then that is essentially the HUP in all its glory.

Note, however, that another aspect of QM is that a measurement of positions "destroys" any previous measurement of momentum, and vice versa. Again, you need a large number of particles in the same initial state and you do the following:

1) Measure the momentum. This measurement of momentum sends the particle into a state of relatively well-defined momentum. If you were to repeatedly measure momentum, you would get a similar measurement each time.

2) Measure the position. This will send the particle into a state where the position is relatively well-defined and (by the HUP) the momentum must be relatively not well-defined. You should get a large range of values for your position measurement here.

3) Measure the momentum again. Now, you get a large range of momentum measurements. The well-defined momentum established in step 1) has been destroyed by step 2).

4) Measure position again and now the well-defined position established in step 2) has gone.

This is not the HUP as stated above, but hopefully you can see the relationship between the two. The real issue here is why (physically and practically) must a measurement of position or momentum destroy the previous well-defined state? That's what Susskind was dealing with.

You could ask that IF you could practically measure position without disturbing the momentum and vice versa, then where would be the HUP be? You would effectively have a particle in a state with a well-defined position and momentum. The way I would describe this is that the HUP must be backed up by some physical and practical issues in the simultaneous measurement of position and momentum. But, these issues are not the HUP itself.

Thank you for taking your time to write such an informative answer. So the second part of your last reply (about what Susskind was saying) is talking about collapsing the wave-function by taking a measurement which makes all subsequent measurements completely different to the original state. Whilst this is caused by HUP, it is not HUP because we are talking about taking measurements. HUP is more a statistical statement about position and momentum without taking any measurements. Is that right?

it is argued that the single slit experiment shows HUP because as you reduce the size of the slit the dot spreads out more. I get this but what I don't get is can't you just say the spreading out is caused by classical diffraction as well?

 

[PeroK]

Thank you for taking your time to write such an informative answer. So the second part of your last reply (about what Susskind was saying) is talking about collapsing the wave-function by taking a measurement which makes all subsequent measurements completely different to the original state. Whilst this is caused by HUP, it is not HUP because we are talking about taking measurements. HUP is more a statistical statement about position and momentum without taking any measurements. Is that right?

Essentially, yes. Note that you cannot say you will get different measurements, just that you no longer have the "certainty" about the momentum, say. This is because position and momentum are incompatible. The HUP tells you theoretically why position and momentum are incompatible: their operators do not commute. The Susskind experiment, in a way, tells you practically why they are incompatible. The interesting question, I think, is the extent to which the two are related.

it is argued that the single slit experiment shows HUP because as you reduce the size of the slit the dot spreads out more. I get this but what I don't get is can't you just say the spreading out is caused by classical diffraction as well?

You can say that it's diffraction if you adopt the wave-particle duality. Modern QM, however, tends to avoid the wave-particle duality and prefers a purely probabilistic QM explanation for why the particle behaves as it does.

There was a thread on here recently about why the wave-particle duality is now a historical footnote to modern QM.

 

[PeterDonis]

You could ask that IF you could practically measure position without disturbing the momentum and vice versa, then where would be the HUP be?

The short answer to this is that this hypothesis is contradictory to QM. The position and momentum operators in QM do not commute. That means you can't measure position without disturbing momentum or vice versa.

 

[PeroK]

The short answer to this is that this hypothesis is contradictory to QM. The position and momentum operators in QM do not commute. That means you can't measure position without disturbing momentum or vice versa.

Which makes it a fascinating and wonderful coordination of abstract mathematics and the "nuts and bolts" of experiment physics.

 

[mike1000]

Now, to verify the HUP you would need to arrange an experiment where a large number of particles was prepared in the same state and you made a measurement of position for half of them and a measurement of momentum for half of them. You would calculate ##\sigma_x## and ##\sigma_p## for the specific data you had gathered and check whether or not the HUP is confirmed in this case. But, like all statistical laws, the HUP requires a large number of experiments and says nothing definite if you simply do one or two measurements.

What do you mean when you say "prepared in the same state"? I have run into this wording in many of the articles I am reading. Why do you have to prepare the state?

1) Measure the momentum. This measurement of momentum sends the particle into a state of relatively well-defined momentum. If you were to repeatedly measure momentum, you would get a similar measurement each time.

Why does measurement of the momentum send the particle into a state of well defined momentum? How do you prepare the state where both the position and the momentum are uncertain?

When you "prepare the state" are you putting the system into a state where the superposition of states is what you want? Are you trying to increase the uncertainty intentionally, prior to conducting the experiment? Is that why you can say that before the wave function collapses, particles are not in anyone state but in a superposition of many states?

Basically, what I am asking, how and for what reason do you "prepare the state"?

 

[mikeyork]

First, the third (preferred) description of HUP in your first video is just as mythical as the first two. It is the same nonsense as when people say Schrodinger's cat is "both dead and alive at the same time".

Whether a particle's state is in an eigenstate of momentum or position or neither (but cannot be both) is indeed a matter of how it is prepared and as far as a future observer is concerned, remaining uncertainty is about what information is not available concerning that prepared state in that observer's context.

You ask what does it mean to prepare a state? Well, the short answer is that if you don't prepare a state, then you have no state at all. Every physical system has to be created somehow. That is what "preparation" means. As a result, there exists some limited information concerning that state. When we talk about probability of detecting a state we are talking about the conditional relative frequency of finding further information about that state given whatever information we have about how the system was initially prepared.

The supposed weirdness (from a classical viewpoint) in QM and the HUP then, is not about uncertainty per se, but about how a system in an eigenstate of momentum can be later detected at a definite spatial location, but that location is probabilistic and not simply determined from the momentum and classical mechanics.

There are lots of attempted explanations of this. The only thing everyone agrees on is the math. My explanation is that "observation" has two parts. The first is the state preparation and the second is the state detection. The apparent weirdness comes because the chosen variables ("basis" in Hilbert space) at each stage can be different. In other words, the transition from prepared state to detected state is merely a change in descriptive context implicit in the apparatus used. But, between preparation and detection, there is no chosen variable/context and nothing else can be said other than what was known at the preparation stage. (NB. This is controversial.)

 

[Mentz114]

@mikeyork : I don't think there's much controversial in your post, which seems to be what Ballentine says. The problem is that a state, physically, is not an amplitude. The physical state must be expressed in dynamical variables in space-time. As you say, what we got is a computational tool that has limitations.

[oops, I think this may be off-topic]

 

[mikeyork]

@mikeyork : I don't think there's much controversial in your post, which seems to be what Ballentine says. The problem is that a state, physically, is not an amplitude. The physical state must be expressed in dynamical variables in space-time. As you say, what we got is a computational tool that has limitations.

I don't agree that the physical state must be expressed in space-time. For example that would be impossible if the state is defined by its energy-momentum. Granted momentum has a direction and therefore a spatial orientation, but in the rest frame there is no such direction and no space-time properties at all.

IMO this is a big deal and the reason why people are now talking about space-time as emergent and, in my case, as an observational choice (implicit in the apparatus). This is why I say it is controversial (to some people at least, though not to me).

 

[vanhees71]

The short answer to this is that this hypothesis is contradictory to QM. The position and momentum operators in QM do not commute. That means you can't measure position without disturbing momentum or vice versa.

The very important point of this discussion is that this is not what the HUP is saying. Rather it is saying that you cannot prepare a particle such that both position and momentum are precisely determined. The better the position is determined by the preparation the less precise is the momentum and vice versa. The Heisenberg uncertainty relation does not tell you anything about what can be measured and what not.

According to the minimal statistical interpretation (for me the only one that is really compatible with the mathematical content of QT) the state refers to the statistics of measurements given the preparation of an ensemble of systems prepared in this state. You can as precisely determine the position as you like as well as momentum. Usually, it's not possible to measure both on the same individual system, but you can use different measurement devices to measure position or momentum on ensembles of particles prepared in the state under investigation, and thus you can measure both variables on the ensemble with arbitrary precision. In fact, to check the uncertainty relation you have to measure these quantities with a larger precision than given by the ##\Delta x## and ##\Delta p## due to the state under investigation, because otherwise you measure the uncertainty due to the measurement rather than the one due to the preparation in the state, and the Heisenberg uncertainty relation is about the uncertainty due to the state preparation and not about uncertainties in the measurement. The latter is not a property of the system but the measurement device used to measure the observables.

 

[Mentz114]

I don't agree that the physical state must be expressed in space-time. For example that would be impossible if the state is defined by its energy-momentum. Granted momentum has a direction and therefore a spatial orientation, but in the rest frame there is no such direction and no space-time properties at all.

IMO this is a big deal and the reason why people are now talking about space-time as emergent and, in my case, as an observational choice (implicit in the apparatus). This is why I say it is controversial (to some people at least, though not to me).

We alway use the lab frame in experiments. So if we prepare a beam of particles there is no ambiguity about momentum.

 

[mikeyork]

We alway use the lab frame in experiments. So if we prepare a beam of particles there is no ambiguity about momentum.

In my way of looking at things, there are two lab frames: that for preparation and that for detection. The notion of a single frame seems to derive from the notion of a single "observer". But IMO the preparation apparatus and the detection apparatus are two distinct "observers" and each sees the other in their frame.

Now if you produce a beam of particles in a momentum eigenstate, then your preparation frame is a momentum frame not a spatial frame. And to transform from the momentum frame to a spatial lab frame, you must apply the unitary transformation that takes the momentum basis to a co-ordinate basis.

 

[Mentz114]

In my way of looking at things, there are two lab frames: that for preparation and that for detection. The notion of a single frame seems to derive from the notion of a single "observer". But IMO the preparation apparatus and the detection apparatus are two distinct "observers" and each sees the other in their frame.

Now if you produce a beam of particles in a momentum eigenstate, then your preparation frame is a momentum frame not a spatial frame. And to transform from the momentum frame to a spatial lab frame, you must apply the unitary transformation that takes the momentum basis to a co-ordinate basis.

Well, I can't agree with most of that - but it is off topic.

Regarding the HUP, position and momentum do not commute because they are conjugates in the Hamiltonian sense. In classical mechanic conjugate variables also do not commute. This is a consequence of conservation laws ( like entanglement). If one is disturbed the other must change proportionally.

The question is whether the degree of disturbance is the the only factor in determining the value of the commutator.

 

[mikeyork]

Well, I can't agree with most of that

Now perhaps you see why I said it was controversial.

- but it is off topic.

And that's where I disagree. I think it is very pertinent to the HUP.

 

[mike1000]

I am beginning to think that one of the important problems that the Uncertainty Principle presents is the that there are just some experiments that we cannot ever perform. I think someone said we cannot conduct an experiment where both the momentum and position are well defined in the preparation phase or both are completely uncertain in the preparation phase?

And, if we want to conduct an experiment on a particles position we must first define its momentum and vice versa.

 

[Mentz114]

Now perhaps you see why I said it was controversial.

And that's where I disagree. I think it is very pertinent to the HUP.

Oh dear, should I be seen discussing with you ?

I guess others must decide what's on or off topic. I hope the OP reads my remarks about conjugacy.

 

[PeterDonis]

Granted momentum has a direction and therefore a spatial orientation, but in the rest frame there is no such direction and no space-time properties at all.

This is not correct. "Direction" means direction in spacetime, not space. Every 4-momentum vector defines a direction in spacetime. The rest frame is just the frame in which that direction is also the direction of the timelike basis vector.

 

[PeterDonis]

this is not what the HUP is saying. Rather it is saying that you cannot prepare a particle such that both position and momentum are precisely determined

Let me rephrase this to make the connection between what you are saying and what I was saying clearer. You are saying that there is no state which is both an eigenstate of position and an eigenstate of momentum. That is true. But it is also equivalent to saying that the position and momentum operators do not commute. So even if you insist on applying the label "Heisenberg Uncertainty Principle" only to the first statement (about states), not the second (about operators), the physics still shows a very close connection between them.

 

[mikeyork]

This is not correct. "Direction" means direction in spacetime, not space. Every 4-momentum vector defines a direction in spacetime. The rest frame is just the frame in which that direction is also the direction of the timelike basis vector.

I have no idea what you mean by a "time-like basis vector". The direction of a rest frame 4-momentum is the energy axis. The 3-vector has no direction. This is in the context of Cartesian frames not the basis vectors of Hilbert space.

And you have gone decidedly off-topic.

The context was the HUP and energy-momentum space versus co-ordinate space-time. Of course one can still project onto a space-time basis but the information contained in an energy-momentum eigenstate is free of any information pertaining to the co-ordinate space-time basis.

 

[vanhees71]

Let me rephrase this to make the connection between what you are saying and what I was saying clearer. You are saying that there is no state which is both an eigenstate of position and an eigenstate of momentum. That is true. But it is also equivalent to saying that the position and momentum operators do not commute. So even if you insist on applying the label "Heisenberg Uncertainty Principle" only to the first statement (about states), not the second (about operators), the physics still shows a very close connection between them.

It should be clear that there are neither momentum nor position eigenstates, because according to the HUP both ##\Delta x## and ##\Delta p## cannot vanish. Indeed, look at the position representation ("wave mechanics"). You get a ##\delta## distribution for the generalized position eigenstate and ##\exp(\mathrm{i} \vec{x} \cdot \vec{p})## which both are distributions but not square-integrable functions.

So you cannot prepare a particle to have a determined position or momentum. You can prepare it to have a pretty well determined position at the cost of having a pretty uncertain momentum and vice versa. Of course, the point is that the representing self-adjoing operators of these observable do not commute and that thus there is (usually) no common eigenstate of these operators. In the case where the operators have ony a continuous spectrum there are no eigenvectors at but only generalized ones (in the sense of distributions).

 

[PeterDonis]

I have no idea what you mean by a "time-like basis vector".

Every coordinate chart on 4-d spacetime has four coordinate basis vectors. In the most common type of chart, the kind that is implied by the term "rest frame", one of these is timelike and the other three are spacelike. This is very basic relativity.

The direction of a rest frame 4-momentum is the energy axis.

What you appear to mean by "energy axis" is what I am calling, using standard terminology, the timelike basis vector (or "time axis" if you want to use that term). The timelike/spacelike distinction applies just as well in energy-momentum space as it does in ordinary spacetime.

The 3-vector has no direction.

Spacetime is a 4-d vector space, not a 3-d vector space. So is energy-momentum space. There is no such thing as a "3-vector" in either space.

This is in the context of Cartesian frames not the basis vectors of Hilbert space.

Yes, I'm well aware of that. There is no such thing as timelike vs. spacelike (or null) vectors in Hilbert space.

the information contained in an energy-momentum eigenstate is free of any information pertaining to the co-ordinate space-time basis.

No, it isn't. An energy-momentum eigenstate has a coordinate spacetime description.

 

[mikeyork]

An energy-momentum eigenstate has a coordinate spacetime description.

One which contains no actual space-time information because if ##\delta p_\mu = 0## then ##\delta x_\mu## blows up.

 

[PeterDonis]

One which contains no actual space-time information

Then how can you obtain probabilities for observing the particle at various positions (or, if you want to be mathematically precise, within various small regions of space)?

 

[vanhees71]

There is no momentum eigenstate since the momentum eigenfunction (in position representation) is ##\propto \exp(\mathrm{i} \vec{p} \cdot \vec{x})##, which is not a square-integrable function. Generally the "eigenstates" of self-adjoint operators to "eigenvalues" within the continuous spectrum of these operators are to be seen as distributions. They belong to the dual space of the domain of the operator, which is usually a smaller dense subspace than the Hilbert space. Realized in terms of wave mechanics the Hilbert space is the space of square-integrable functions, while the domain of position and momentum operators is the Schwartz space of quickly falling functions.

 

[mikeyork]

Then how can you obtain probabilities for observing the particle at various positions(or, if you want to be mathematically precise, within various small regions of space)?

First of all, probabilities require an infinite sample. You can only obtain relative frequencies. These will be random because the state has no information that would prefer any spatial location over another. That is what lack of information means in QM. In classical probability theory, lack of information is typically associated with a uniform distribution.

 

[PeterDonis]

First of all, probabilities require an infinite sample.

Fine, substitute "expectation value" for "probability". Same question.

These will be random because the state has no information that would prefer any spatial location over another. That is what lack of information means in QM. In classical probability theory, lack of information is typically associated with a uniform distribution.

A uniform distribution is not the same as no distribution at all, which is what "no spacetime information" implies.

 

[mikeyork]

A uniform distribution is not the same as no distribution at all, which is what "no spacetime information" implies.

In the QM context in which I used the term it clearly means a uniform distribution because it means no information that would distinguish one preferred state over another. Please don't argue over language or we'll get nowhere.

 

[mikeyork]

Fine, substitute "expectation value" for "probability". Same question.

With the same answer. For a finite sample you measure a mean. For randomly distributed events, the mean over a finite range will tend towards the center of that range.

Can you not see how this is getting way off-topic?

I stand by my posts #13, #15 and #18 which were on-topic. If you want to discuss those, fine. But please let's not go down any more languaging rabbit-holes.

 

[PeterDonis]

I stand by my posts #13, #15 and #18 which were on-topic.

And one of the claims you made (in post #15) was:

I don't agree that the physical state must be expressed in space-time.

If that claim is on topic, then disputing it, which is what I've been doing, is also on topic. There are two ways to dispute this claim:

(1) If we're talking about the position vs. the momentum basis, then any state can of course be expressed in either basis (or in any of an infinite number of other possible bases). If your claim just means "we can use the momentum basis instead of the position basis", then of course that's true. But there is no state that can only be expressed in the momentum basis, not the position basis. And in defending your claim, you have been appearing to defend the latter claim; for example, your very next sentence in post #15 was:

For example that would be impossible if the state is defined by its energy-momentum.

This appears to be saying that an energy/momentum eigenstate does not have a position representation, which is false. That's why I objected.

(2) If we're talking about Hilbert space, then Hilbert space is not spacetime, whether we are using the position basis, the momentum basis, or some other basis. But that would mean it is impossible to express any physical state in spacetime, not just an energy-momentum eigenstate--on this view, physical states are expressed in Hilbert space. But you aren't making that objection.

So if you're going to stand by your post #15, you need to clarify exactly what it is you're standing by, so I can tell whether my objections are addressed or not. Is that not on topic?