Understanding <x' | x> in Quantum Mechanics: Exploring Its Physical Significance

[vanhees71]

This latter idea is kind of a first step of the path-integral formulation. Do not study photons before you haven't fully understood non-relativistic quantum theory. Photons cannot described adequately in the wave-function formalism. It has no position observable to begin with. It's a quite complicated object, even already in the free case and can only described correctly in terms of quantum field theory.

In non-relativistic quantum theory, the meaning (and the only logically consistent meaning!) of the wave function is that
$$P(t,x)=|\psi(t,x)|^2$$
is the probability density function for the position of a single particle, provided that
$$\int_{\mathbb{R}} \mathrm{d} x P(t,x)=1.$$
So the wave function must be square integrable to have a well-defined definition of a physical state in sense of quantum theory.

Of course, not every solution of the Schrödinger equation is a wave function. One example is the above derived propagator for a free non-relativistic particle. It's a solution of the Schrödinger equation, but in the sense of a generalized function. It lives not in the Hilbert space of square integrable wave functions but in the dual of a smaller dense subspace, where the position and momentum operators are defined. Since the subspace is strictly smaller than the Hilbert space, it's dual is larger than the Hilbert space. The dual of the Hilbert space itself is isomorphic to the Hilbert space.

 

[friend]

This latter idea is kind of a first step of the path-integral formulation. Do not study photons before you haven't fully understood non-relativistic quantum theory. Photons cannot described adequately in the wave-function formalism. It has no position observable to begin with. It's a quite complicated object, even already in the free case and can only described correctly in terms of quantum field theory.

Yea, I kind a picked up on that. I think Prof. Strassler must have meant real and virtual particles (not photons) since they have more to do with the Schrodinger equation and the Green's function of that.

 

[vanhees71]

Forget virtual particles. That's pop-sci speach. It describes mathematical terms in the perturbative expansion, which can be efficiently written in terms of Feynman diagrams, where the "virtual particles" are symbolized for the internal lines that stand for propagators, i.e., mathematical expressions, nothing else.

 

[friend]

Forget virtual particles. That's pop-sci speach. It describes mathematical terms in the perturbative expansion, which can be efficiently written in terms of Feynman diagrams, where the "virtual particles" are symbolized for the internal lines that stand for propagators, i.e., mathematical expressions, nothing else.

Yes, they are just mathematical devises, just as are manifolds, and differential lengths. We invent terms like force and mass to give some definition of what we are integrating in classical mechanics. So I'm trying to understand what those little pieces are that we are integrating in the Feynman path integral since that's what we do in classical integrals. And it seems unavoidable since we assume that the whole is constructed from the little pieces. So your advice seems to be to be saying that I should give up on understanding the basic structures of physical things. And the little pieces in the path integral seem to be called "virtual". The only problem left is to discover their properties and to understand the world in terms of them. So you are basically asking me to stop caring at all. Sorry, I can't do that.

 

[friend]

... For the wave function this implies
$$\psi(t,x)=\langle x|\psi,t \rangle=\int \mathrm{d} x' \langle x|\hat{U}(t,t_0)|x' \rangle \langle x'|\psi,t_0 \rangle=\int \mathrm{d} x' U(x,t;x',t_0) \psi_0(x').$$
obviously the propagator $U(x,t;x',t_0)$ satisfies the initial-value problem of the Schrödinger equation,
$$\mathrm{i} \partial_t U(x,t;x',t_0)=\hat{H} U(x,t;x',t_0), \quad U(x,t_0+0^+;x',t_0)=\delta(x-x'),$$
where now $\hat{H}$ is the Hamiltonian in position representation.

It seems what this says is that for the wave function to propagate through time at all means that one must take into account how every point in space interacts with the initial wave function. Or in other words, the wave function cannot ever exist in spacetime at all unless it moves through time, and this in turn means the wave function is always defined in terms of the propagator which takes into account every point in space.

 

[stevendaryl]

Yes, they are just mathematical devises, just as are manifolds, and differential lengths. We invent terms like force and mass to give some definition of what we are integrating in classical mechanics. So I'm trying to understand what those little pieces are that we are integrating in the Feynman path integral since that's what we do in classical integrals. And it seems unavoidable since we assume that the whole is constructed from the little pieces. So your advice seems to be to be saying that I should give up on understanding the basic structures of physical things. And the little pieces in the path integral seem to be called "virtual". The only problem left is to discover their properties and to understand the world in terms of them. So you are basically asking me to stop caring at all. Sorry, I can't do that.

I think you're mixing up two different topics. There is a path-integral formulation for nonrelativistic quantum mechanics, and that in no way involves "virtual particles". Virtual particles appear when one does perturbation theory in quantum field theory. The two subjects are related, in that both of them make use of something called the "propagator", but the expression that is being discussed in this thread, $\langle x| \hat{U}(t, t_0)|x_0\rangle$ is from nonrelativistic quantum mechanics, and doesn't involve virtual particles at all.

As vanhees71 says, a propagator is not a transition amplitude, because it has different normalization rules. However, conceptually, it seems to me that it can be understood as a generalization of transition amplitudes, so I don't agree with him that it's so bad to call them that (as long as the normalization business is made clear).

If you have a quantum system with a discrete number of states $|psi_j\rangle$, then you can compute a transition amplitude:

$A_{ij}(t, t_0)$

which is defined to be the probability amplitude that a system initially in state $|\psi_j\rangle$ at time $t_0$ will be found in state $|\psi_i\rangle$ at time $t$. The mathematical expression for this is: $A_{ij}(t, t_0) = \langle \psi_i | \hat{U}(t, t_0) | \psi_j \rangle = \langle \psi_i | e^{-i \hat{H} (t-t_0)/\hbar} | \psi_j \rangle$. The probability that a system initially prepared in state $j$ will later be observed to be in state $i$ is just $P_{ij} = |A_{ij}|^2$.

The laws of quantum mechanics allow us to relate $A_{ij}$ at different times as follows:

$A_{ij}(t, t_0) = \sum_k A_{ik}(t, t_1) A_{kj}(t_1, t_0)$

where $t_1$ is any time between $t_0$ and $t_1$, and the index $k$ runs over all possible intermediate states.

Obviously, you can continue to expand the amplitude to get something like:

$A_{ij}(t, t_0) = \sum_{k_1, k_2, ..., k_N} A_{ik_N}(t, t_N) A_{k_N, k_{N-1}}(t_N, t_{N-1}) ... A_{k_1, j}(t_1, t_0)$

where $t_n = t + n \epsilon$, where $\epsilon = \frac{n (t-t_0)}{N+1}$

If we define a "path" (not through physical space, but through state space) $p$ to be a sequence of states $\psi_{k_0}, \psi_{k_1}, ..., \psi_{k_{N+1}}$ of length $N+2$, then we can associate an amplitude with such a path:

$\phi(p, t, t_0) = A_{k_{N+1}, k_N}(t, t_N) A_{k_N, k_{N-1}}(t_N, t_{N-1}) ... A_{k_1, k_0}(t_1, t_0)$

Then our formula above can be summarized as:

$A_{ij}(t, t_0) = \sum_p \phi(p, t, t_0)$

where the sum is over all possible paths $p$ that start at $\psi_j$ and end at $\psi_i$. So you can interpret the equation for transition amplitudes as saying: "Take all possible paths from $\psi_j$ to $\psi_i$, and add their amplitudes." I suppose you could say that these paths are "virtual", in that there is no sense in which the system actually takes any of those paths--it's just a calculational device.

The propagator is in some sense the continuum limit of such a transition amplitude, where the states $\psi_j$ are states that are localized in physical space.

 

[friend]

I think you're mixing up two different topics. There is a path-integral formulation for nonrelativistic quantum mechanics, and that in no way involves "virtual particles". Virtual particles appear when one does perturbation theory in quantum field theory. The two subjects are related, in that both of them make use of something called the "propagator", but the expression that is being discussed in this thread, $\langle x| \hat{U}(t, t_0)|x_0\rangle$ is from nonrelativistic quantum mechanics, and doesn't involve virtual particles at all.

It does seem that there are different senses in which the words virtual particles are used. There is the sense used in perturbation theory, as you say. But I think there is also the sense in which virtual particles automatically exist everywhere; they are also called quantum fluctuations. I think it is these that are used when talking about the Casimir effect and Hawking and Unruh radiation. My feeling is that the word virtual is invoked when talking about the little pieces that are being integrated in the Feynman Path Integral. I've even heard of virtual geometries when using the path integrals in quantum gravity when possible geometries of spacetime exist in superposition. You might call each of the infinite number of paths in the path integral virtual sense none of them can be said to really exist but they are nevertheless integrated in superposition. But then again each of those virtual paths consist of even smaller displacements (of what particles?). Since the wave function is a distributed thing, I'm trying to understand the wave function in terms of these tiny, virtual, things. What I suspect is that the wave function can be described in terms of these virtual particles of quantum mechanics. And the virtual particles of quantum mechanics can be explained in terms of the virtual particles of quantum field theory. I imagine it will be easy to confuse the two.

 

[stevendaryl]

It does seem that there are different senses in which the words virtual particles are used. There is the sense used in perturbation theory, as you say. But I think there is also the sense in which virtual particles automatically exist everywhere; they are also called quantum fluctuations. I think it is these that are used when talking about the Casimir effect and Hawking and Unruh radiation.

Those are all cases of the same thing: particles that appear in perturbation theory.

My feeling is that the word virtual is invoked when talking about the little pieces that are being integrated in the Feynman Path Integral.

There might be a sense in which they are related, but I don't think anyone uses the phrase "virtual particle" in talking about Feynman's path integral formulation of nonrelativistic quantum mechanics.

 

[friend]

... There might be a sense in which they are related, but I don't think anyone uses the phrase "virtual particle" in talking about Feynman's path integral formulation of nonrelativistic quantum mechanics.

Well for example, for any wave function to propagate we must have,

<x'|e^-iHt/ħ|x> = ∫∫∫⋅⋅⋅∫<x'|e^-iHε/ħ|x₁><x₁|e^-iHε/ħ|x₂><x₂|e^-iHε/ħ|x₃>⋅⋅⋅<x_n|e^-iHε/ħ|x> dx₁dx²dx₃⋅⋅⋅dx_n

The question is for each <x_i|e^-iHε/ħ|x_j> is there a complex conjugate <x_i|e^-iHε/ħ|x_j>^* ? This would be a kind of virtual antiparticle, perhaps.

 

[friend]

No, it's obviously not. Again, generalized eigenstates of self-adjoint operators in the continuous part of their spectrum are never proper Hilbert-space vectors and thus do not represent (pure) quantum states. They are distributions and belong to the dual space of the domain of the self-adjoint operator. You must not multiply them. In your case of a position eigenvector you have
$$\langle x'|x \rangle=\delta(x'-x).$$
This clearly shows you that you must not take its square!

Shankar, Principles of Quantum Mechanics, page 145, says,
"The other set of improper kets we will use in the same spirit are the position eigenkets |x>, which also form a convenient basis. Again, when we speak of a particle being in a state |x₀> we shall mean that its wave function is so sharply peaked at x=x₀ that it may be treated as a delta function to a good accuracy."
So I think that this means that <x'|x> = δ(x-x') is a very close approximation for the purposes of any calculation to a physical entity, a transition amplitude.

 

[vanhees71]

Well, there are many sloppy quantum mechanics textbooks out there. That's a pity, but you can't help it. Just one last time: It doesn't make sense to claim that non-normalizable distributions represent states in any way. In this example it's obvious. You cannot even square the $\delta$ distribution, let alone integrate over this undefined quantity.

 

[stevendaryl]

So I think that this means that <x'|x> = δ(x-x') is a very close approximation for the purposes of any calculation to a physical entity, a transition amplitude.

Well, you can sort-of approximate a delta function by a very localized, square-integrable function. But the normalization is not right for a wave function. $\delta(x-x')$ is normalized so that $\int \delta(x-x') dx = 1$, while a wave function is normalized so that $\int |\psi(x)|^2 dx = 1$.

 

[friend]

Well, you can sort-of approximate a delta function by a very localized, square-integrable function. But the normalization is not right for a wave function. $\delta(x-x')$ is normalized so that $\int \delta(x-x') dx = 1$, while a wave function is normalized so that $\int |\psi(x)|^2 dx = 1$.

I find it strange to insist that there is no physical content to |x>. For we certainly consider it legitimate to write <x|ψ> = ψ(x). If <x| is meaningful in <x|ψ>, then shouldn't we expect that its dual, |x> is also physically meaningful.

 

[vanhees71]

This is the very point of the entire discussion! A distribution has physical meaning only when folded with a test function. Note that $|\psi \rangle$ must be in the domain of the position operator which is a dense subspace of the Hilbert space, so that $\langle x|\psi \rangle$ is well-defined!

 

[friend]

This is the very point of the entire discussion! A distribution has physical meaning only when folded with a test function. Note that $|\psi \rangle$ must be in the domain of the position operator which is a dense subspace of the Hilbert space, so that $\langle x|\psi \rangle$ is well-defined!

Yea, that all sounds familiar. Thanks for the reminder.

So my understanding from all this is that |x₀> is not a wave-function in the sense that you cannot extract a probability from it because it is not square integrable. That would only mean that we cannot ever predict the probabilities that |x₀> will occur, right? I think that is to be expected. It represents the extreme of the Uncertainty Principle ΔxΔp≥ħ/2. Since |x₀> represents a wave-function located precisely at x₀, then the Δx of it would be zero, which is impossible. Same can be said of |p> for a wave function of exact momentum, p.

 

[vanhees71]

Probabilities predicted in QT are probabilities for finding a certain value when measuring an observable and not that some mathematical auxilliary quantity like $|x \rangle$ occurs. The uncertainty principle clearly tells you that there are no proper eigenvectors of position or momentum operators.

 

[atyy]

I find it strange to insist that there is no physical content to |x>. For we certainly consider it legitimate to write <x|ψ> = ψ(x). If <x| is meaningful in <x|ψ>, then shouldn't we expect that its dual, |x> is also physically meaningful.

This is the very point of the entire discussion! A distribution has physical meaning only when folded with a test function. Note that $|\psi \rangle$ must be in the domain of the position operator which is a dense subspace of the Hilbert space, so that $\langle x|\psi \rangle$ is well-defined!

Well, it could be that position can be exactly measured, so <x|ψ> does make physical sense. The only thing is that after a position measurement, the state cannot collapse to a delta function, since that is not a physical state. So we can have exact position measurement, and we can have collapse after that, but to some other state (consistent with a generalized collapse rule, rather than the textbook projection postulate which cannot deal with continuous variables).

 

[vanhees71]

Position CAN bei exactly measured. It cannot be exactly PREPARED! Since there is no collapse, the no-collapse argument is always true ;-)).

 

[atyy]

Position CAN bei exactly measured. It cannot be exactly PREPARED!

Well said!

Since there is no collapse, the no-collapse argument is always true ;-)).

Here's one way in which one can have collapse after the sharp position measurement:
http://arxiv.org/abs/0706.3526 (edited link)
2.3.2 Ozawa's model of a sharp position measurement"

 

[A. Neumaier]

Position CAN be exactly measured.

How do you measure position exactly (i.e., to an infinite number of decimal places)? I haven't seen any experimental equipment that could do it; not even in a thought experiment. Already the readout of the measurement result would take an infinite time...

 

[friend]

How do you measure position exactly (i.e., to an infinite number of decimal places)? I haven't seen any experimental equipment that could do it; not even in a thought experiment. Already the readout of the measurement result would take an infinite time...

By that logic, we could never confirm theory since we could never measure accurately enough.

Just because QT does not predict |x> doesn't mean that it doesn't happen. Something else is giving rise to particles with which quantum theory works. QFT is giving rise to the particles whose propagation is predicted by QM. And even QFT does not predict where those particles will pop into existence. But it does seem that particles are starting out at some exact location, an atom or nucleus. And the collapse tells us that it is ending exactly somewhere. As I understand it, the collapse is to a point, not to a region.

 

[stevendaryl]

By that logic, we could never confirm theory since we could never measure accurately enough.

Well, and that's true. You never confirm a theory. What you can do is disconfirm (or falsify) a theory by showing that the measured result differs from the predicted result by an amount that is greater than can be plausibly accounted for by measurement imprecision or chance.

 

[friend]

I believe that Quantum Theory predicts that the wave function collapses to a point, not a region. That at least confirms that the <x'| portion of <x'|x> is a physical prediction. So if <x'| is confirmed, then so is its dual |x>

 

[stevendaryl]

I believe that Quantum Theory predicts that the wave function collapses to a point, not a region. That at least confirms that the <x'| portion of <x'|x> is a physical prediction. So if <x'| is confirmed, then so is its dual |x>

No, quantum theory doesn't predict that the wave function collapses to a point. The collapse hypothesis is probably better thought of as a rule of thumb, rather than a necessary postulate of quantum (because there is substantial debate over whether collapse happens at all).

That at least confirms that the <x'| portion of <x'|x> is a physical prediction. So if <x'| is confirmed, then so is its dual |x>

The use of $|\phi\rangle$ and $\langle \psi |$ is just notation. You can't make a deduction about what is sensible based on the fact that you can express it using the notation.

In the mathematics of (single-particle, nonrelativistic, spin-zero, in one spatial dimension for simplicity) quantum mechanics, the possible states of a particle are assumed to be the set of complex-valued functions $\psi(x)$ such that $\int dx |\psi(x)|^2 < \infty$. In the Dirac notation, $|\psi\rangle$ represents such a function.

Those are the "kets" (sort of a joke based on the word "bracket"). A "bra" is any linear function on "kets" that returns a complex number. One example of a "bra" is the function $\langle x|$ that takes $|\psi\rangle$ and returns $\psi(x)$. That is, it takes the function $\psi$ and returns the value of that function at the location $x$. Another example is the Fourier transform, the function $\langle k|$ that takes a function $\psi$ and returns $\int dx \psi(x) e^{-i k x}$. In the Dirac notation, the notation $\langle F|B\rangle$ means the result of applying the bra $F$ to the ket $|B\rangle$.

Now, there is a special type of "bra" which can be obtained from a ket: If $|\phi\rangle$ is any ket, then you can define the corresponding bra $\langle \phi |$ to be that function that takes a ket $|\psi\rangle$ and returns $\int dx \phi^*(x) \psi(x)$.

Note that not every bra comes from a ket. So the existence of a bra $\langle F|$ does not imply that there is a corresponding ket $|F\rangle$. The two most common ones are $\langle x|$ and $\langle k|$. They do not correspond to any state in the Hilbert space.

Now, often people do use the Dirac notation to represent what seems like the duals to $\langle x|$ and $\langle k|$. They define $|k\rangle$ to be the ket corresponding to the function $e^{i k x}$, and they define $|x'\rangle$ to be the ket corresponding to the "function" $\delta(x-x')$. But that's playing loose with the notation. Those are not elements of the hilbert space, and are not possible states of a particle.

 

[friend]

No, quantum theory doesn't predict that the wave function collapses to a point. The collapse hypothesis is probably better thought of as a rule of thumb, rather than a necessary postulate of quantum (because there is substantial debate over whether collapse happens at all).

Whatever you call it, collapse, many worlds, etc., QT predicts a probability distribution, but the result of measurement is only one of those possibilities, even when the spectrum is continuous. Also, the way we prepare a particle in a state is to do a measurement. So measuring the position prepares the particle in that state. We may not be able to prepare a state in an exact position because the probability density to measure in that state is infinitesimal. But that doesn't preclude the fact that a measurement of position will put it in an exact position.

 

[stevendaryl]

Whatever you call it, collapse, many worlds, etc., QT predicts a probability distribution, but the result of measurement is only one of those possibilities, even when the spectrum is continuous.

I disagree with that. Here's a way of thinking of wave function collapse that sort of explains how a measurement works:

Suppose you have a system described by a wave function $|\psi\rangle$. Then you make an observation of the system. Then what you can do is to rewrite the wave function as the sum of two parts: $|\psi\rangle = \alpha |\psi_{yes}\rangle + \beta |\psi_{no}\rangle$, where $|\psi_{yes}\rangle$ and $|\psi_{no}\rangle$ are orthogonal states such that $|\psi_{yes}\rangle$ is consistent with your observation, and $|\psi_{no}\rangle$ is not. Then wave function collapse amounts to the replacement of the full $|\psi\rangle$ by just $|\psi_{yes}\rangle$. You're not going to get a state with a precise value for an observable $O$ unless your observation is only consistent with that one, precise value.

If the observable has a discrete spectrum (such as spin, or the energy of a bound system), then it's possible to have an observation that uniquely determines the observable's value. But if the observable is continuous, then a single observation can't possibly determine the precise value of the observable, and so your observation is consistent with a range of values, and so the "collapse" will result in a state with a range of values.

 

[friend]

But if the observable is continuous, then a single observation can't possibly determine the precise value of the observable, and so your observation is consistent with a range of values, and so the "collapse" will result in a state with a range of values.

Now we're talking about meter accuracy as opposed to what theory predicts. Does theory predict a collapse to only one particular value in the continuous spectrum case? Perhaps this question gets to whether distributions have some inherent existence in and of themselves? Or are distributions always made up of individual samples that are the thing that inherently exist?

 

[A. Neumaier]

Does theory predict a collapse to only one particular value in the continuous spectrum case?

No. Definitely not.

 

[stevendaryl]

Now we're talking about meter accuracy as opposed to what theory predicts. Does theory predict a collapse to only one particular value in the continuous spectrum case?

As I said, collapse is a contentious part of the quantum formalism. Von Neumann's collapse hypothesis was that if you measure $\hat{O}$ and get the value $\lambda$, then the system will collapse into an eigenstate of $\hat{O}$ with eigenvalue $\lambda$. But my claim is that there is no way to measure position. The best you can do is to measure a range of positions. So instead of an operator $\hat{x}$, you might use some operator $\hat{x}_{\delta x}$ which returns a discrete set of possibilities:

• 0, meaning that the particle is somewhere between $x=-\delta x/2$ and $x=+\delta x/2$
• 1, meaning that the particle is somewhere between $x=\delta x/2$ and $x = 3/2 \delta x$
• etc.

Measuring the system to have $\hat{x}_{\delta x} = n$ would then cause the wave function to collapse from $\psi(x)$ to $\psi_n(x)$, where

1. $\psi_n(x) = C \psi(x)$, when $(n-1/2) \delta x < x < (n + 1/2) \delta x$
2. $\psi_n(x) = 0$, otherwise.

where C would be chosen so that $\int_{(n-1/2)\delta x}^{(n+1/2)\delta x} dx C^2 |\psi(x)|^2 = 1$

 

[friend]

Now we're talking about meter accuracy as opposed to what theory predicts. Does theory predict a collapse to only one particular value in the continuous spectrum case? Perhaps this question gets to whether distributions have some inherent existence in and of themselves? Or are distributions always made up of individual samples that are the thing that inherently exist?

I think this is the particle/wave debate. I'm understanding that particles are point particles with 0 radius. So if there are such things as particles, then they exist at a particular point.

 

[stevendaryl]

I think this is the particle/wave debate. I'm understanding that particles are point particles with 0 radius. So if there are such things as particles, then they exist at a particular point.

Saying that something is a point-particle is to say that it has no internal structure (unlike a proton, which has quarks inside it). Whether that means that it exists at a point or not is dependent on the interpretation of QM you are using. But it's never the case that a particle is in a definite state of position. There's no way to put it into such a state, and there is no way to detect it as being in such a state.

 

[A. Neumaier]

Measuring the system to have $\hat{x}_{\delta x} = n$ would then cause

But then the state after the measurement would depend on which accuracy you ascribe to your position measurement...

In reality, the analysis of concrete measurements, the best result to be assigned, and the determination of final state are complicated calibration issues, and the talk about measurement in the foundations is a heavy idealization of the real situation. People in quantum optics who have ot estimate the efficiency and accuracy of what they do always work with density matrices rather than pure states, and employ Lindblad equations or rather than crude collapse arguments to model the dynamics of the state.

 

[friend]

Saying that something is a point-particle is to say that it has no internal structure (unlike a proton, which has quarks inside it). Whether that means that it exists at a point or not is dependent on the interpretation of QM you are using. But it's never the case that a particle is in a definite state of position. There's no way to put it into such a state, and there is no way to detect it as being in such a state.

Which brings up the other question about QM. Does QM tell us what actually is or only what we can measure. Perhaps the "actually is" are the particles. And what we can measure is the distribution of the waves.

What I'm beginning to think is that there are such things as point particles, but we can never measure exactly where they are because their position is continually being passed around between the virtual particles that surround it so that the uncertainty of its position accumulates over time. And in this view particles are true points, and things like charge, mass, and spin are caused by how it interacts with the surrounding virtual particles.

 

[vanhees71]

There's nothing that prevents measuring the position of a quantum as accurately as you like, at least not in principle. It may be difficult to measure a position very accurately, but it's not impossible. To verify the standard deviation $\Delta x$ for a given state you have to measure position on an ensemble of equally prepared particles much more accurate than $\Delta x$. The same holds true for $\Delta p$, and you can never measure momentum and position accurately on the same particle, but you can prepare an ensemble to measure position very accurately and then another ensemble to measure momentum very accurately to check the uncertainty relation $\Delta x \Delta p \geq 1/2$ (this is for non-relativistic QT, for the relativistic case, see the first chapter of Landau Lifshitz vol. 4).

 

[A. Neumaier]

There's nothing that prevents measuring the position of a quantum as accurately as you like

But that's quite different from your earlier claim (in post #53) that one can measure it exactly.