Time evolution of a detected particle

[hokhani]

TL;DR

What is the time evolution of a quantum particle which is already detected at a point?

Suppose that we detect a quantum particle at the position $x_0$. Then, we would like to know about its position at a next time. Does the particle stay at $x_0$ while the detector is turned on?

 

[anuttarasammyak]

No. Uncertainty of momentum is infinite so the particle has almost light speed. Furthermore particle anti particle pairs with similar speed are created by precise measurement of position which requires high energy. We will not be able to distinguish which particle is our own among these dispersing anti particle particle juice.

 

[renormalize]

No. Uncertainty of momentum is infinite so the particle has almost light speed. Furthermore particle anti particle pairs with similar speed are created by precise measurement of position which requires high energy. We will not be able to distinguish which particle is our own among these dispersing anti particle particle juice.

While that's true if you insist on zero uncertainty in location, you may instead use a medium like photographic film to continuously determine the particle positions to within a finite resolution. Recently, in another thread I posted this bubble-chamber image of elementary particles in relativistic motion:

(https://cds.cern.ch/record/39469)
This represents a record over time of the positions of the particles that is accurate to within about the film grain size. Moreover, it should be possible, at least in principle, to replace the single frame of film by a digital motion-picture camera with a femtosecond frame-rate that would record the particle tracks as they form. If each frame is time-stamped, this would yield a moment-by-moment measurement of the particle positions (accurate to within the camera pixel size) versus time (accurate to within femtoseconds) as they traverse the frame.

 

[hokhani]

No. Uncertainty of momentum is infinite so the particle has almost light speed.

Thanks, how does the infinite momentum uncertainty predict light speed for the particle whereas it seems that the expectation value of momentum is zero?

 

[PeterDonis]

Uncertainty of momentum is infinite so the particle has almost light speed.

No, the particle's speed is extremely uncertain. It could be almost light speed or it could be zero. And it could be moving in any direction.

 

[anuttarasammyak]

Thanks, how does the infinite momentum uncertainty predict light speed for the particle whereas it seems that the expectation value of momentum is zero?

The probability that we get momentum in finite region say [-p,p] is infinitesimally smaller than we get momentum in the outside region. Thus velocity close to c,-c which correspond to infinite momentum would be frequently observed.

 

[QuarkyMeson]

The probability that we get momentum in finite region say [-p,p] is infinitesimally smaller than we get momentum in the outside region. Thus velocity close to c,-c which correspond to infinite momentum would be frequently observed.

An exact position eigenstate is a generalized, non-normalizable state; i.e., not a physically preparable one. Its divergent momentum spread is not a literal prediction for what a detector would observe. A real position measurement has finite resolution and prepares some finite-width state set by that resolution, with a correspondingly finite momentum distribution. As the localization scale approaches the reduced Compton wavelength $\hbar/(mc)$, relativistic effects become unavoidable and a fixed one-particle description ceases to be adequate.

The divergence is mainly telling you then that the idealized state has been pushed outside the domain of the theory, not that a real particle would simply be observed moving at ~$\pm c$.

 

[hokhani]

The probability that we get momentum in finite region say [-p,p] is infinitesimally smaller than we get momentum in the outside region. Thus velocity close to c,-c which correspond to infinite momentum would be frequently observed.

Sorry I don't understand. Could you please tell where I am wrong!
We have the expansion of the wave function as $\delta(x-x_0)=1/2\pi \hbar\int exp(ip(x-x_0)/\hbar)dp$ and so the expansion coefficients for all values of momentum are identical no matter inside or outside the region you said.

 

[hokhani]

As far as I know, from elementary quantum mechanics, while we are measuring the position, the position would be one of the eigenstate of position operator. Knowing time evolution requires knowing the Hamiltonian which disturbs the position. So, while we kept the position measurement (and so don't disturb the position by measuring the Hamiltonian) we don't know the Hamiltonian and so it seems that we shouldn't know time evolution. Please correct me if I am wrong.

 

[QuarkyMeson]

Sorry I don't understand. Could you please tell where I am wrong!
We have the expansion of the wave function as $\delta(x-x_0)=1/2\pi \hbar\int exp(ip(x-x_0)/\hbar)dp$ and so the expansion coefficients for all values of momentum are identical no matter inside or outside the region you said.

The problem again is that this is not a normalized momentum probability distribution, its integral over p diverges. So you can't directly assign probabilities to the finite interval $\left[-P, P\right]$ or to its complement for the exact delta state.

It does become well-defined as a limit of normalized states. For a packet of position width $\sigma$, the momentum probability inside a fixed $\left[-P,P\right]$ tends to 0 as $\sigma \to 0$. That's a statement about the limit, not about the position eigenstate itself.

Likewise $\langle p \rangle$ is undefined rather than 0. A symmetric regularization gives 0 at every finite stage, but $\langle x_0|\hat{p}|x_0\rangle$ isn't an expectation value in the first place, since $|x_0\rangle$ is not a Hilbert-space vector. $\langle p^2 \rangle$ makes the point more clear, for normalized states, $\Delta x \to 0$ forces $\Delta p \to \infty$ by the uncertainty relation, so it diverges under every regularization.

 

[QuarkyMeson]

As far as I know, from elementary quantum mechanics, while we are measuring the position, the position would be one of the eigenstate of position operator. Knowing time evolution requires knowing the Hamiltonian which disturbs the position. So, while we kept the position measurement (and so don't disturb the position by measuring the Hamiltonian) we don't know the Hamiltonian and so it seems that we shouldn't know time evolution. Please correct me if I am wrong.

The Hamiltonian isn't something you measure, it's part of the specification of the system.

Measuring energy is a separate operation that would indeed collapse the state, but time evolution doesn't require it. You need $\hat{H}$ as an operator, not one of its eigenvalues. Measure position at $t=0$, get a localized state, propagate with $U(t) = e^{-i\hat{H}t/\hbar}$.

The real problem is again the exact eigenstate. $\delta(x-x_0)$ isn't in the Hilbert space, so it isn't a physical state to begin with. For a free particle the formal answer gives $|\psi(x,t)|^2 = m/2\pi\hbar t$, uniform over all space and non-normalizable for any $t > 0$. Narrow normalized packets evolve perfectly well.

 

[QuarkyMeson]

TL;DR: What is the time evolution of a quantum particle which is already detected at a point?

Suppose that we detect a quantum particle at the position $x_0$. Then, we would like to know about its position at a next time. Does the particle stay at $x_0$ while the detector is turned on?

Back to this, howwww are we detecting the particle? You suggest that the detector is left on, so I'm assuming you want continuous measurements. In that case it would also be quite silly if the detector destroyed the thing you're measuring, so it has to be non-destructive measurement I'm also assuming.

This is kind of the seminal paper on that topic: https://journals.aps.org/pra/abstract/10.1103/PhysRevA.36.5543

 

[anuttarasammyak]

Sorry I don't understand. Could you please tell where I am wrong!

Another approach.
$$<\frac{v^2}{c^2}>=<\frac{p^2}{m^2c^2+p^2}>=\frac{<p^2>}{m^2c^2+<p^2>}=\frac{(\triangle p)^2+<p>^2}{m^2c^2+(\triangle p)^2+<p>^2} \rightarrow 1-0$$ for $$ \frac{\triangle p}{mc} \rightarrow +\infty $$ or from uncertainty relation $$ \frac{\triangle x}{\frac{\hbar}{mc}} \rightarrow +0 $$
Thus v should be almost -c, c, if single particle picture were right in the limit

 

[pines-demon]

Well (assuming free particle) starting at $|\psi(t=0)\rangle=|x_0\rangle$ you need only to carry
$$|\psi(t)\rangle=e^{-i\hat H t/\hbar}|x_0\rangle=\int_{-\infty}^{\infty} \mathrm dx\,\sqrt{\frac{m}{2\pi i\hbar t}}\exp\!\left(\frac{i m(x-x_0)^2}{2\hbar t}\right)|x\rangle$$
so it acquires a non-zero value at every $x$ instantaneously (and it is not normalizable because $|x_0\rangle$ is non-normalizable).

Note that you can ask the same question for an initial Gaussian wavefunction centered at $x_0$. In that case the wavefunction spreads as time goes by.

 

[Nugatory]

As far as I know, from elementary quantum mechanics, while we are measuring the position, the position would be one of the eigenstate of position operator.

Not quite.
We need to consider the resolution of our position-measuring apparatus. That resolution is not infinite, so the result of a position measurement does not leave the particle in an "eigenstate" of the position operator (scare-quotes because as others have pointed out, those position delta functions aren't really eigenstates). Instead the post-measurement wave function is peaked around the measured value, and the higher the resolution the sharper the peak.

So why do we even mess with these physically unrealizable "eigenstates"? It is because we can write the physically realizable states as superpositions of these "eigenstates", and it is particularly easy to calculate the time evolution of the wave function when it is written in that form.

 

[Roberto Pavani]

If each frame is time-stamped, this would yield a moment-by-moment measurement

As an electronics engineer, I would add that at very high frame rates the timestamp itself becomes a nontrivial measurement. Clock jitter is already a major limitation in high-speed instruments, so a "femtosecond movie" is not simply a matter of making the camera faster. The timing chain introduces its own uncertainty.

 

[QuarkyMeson]

As an electronics engineer, I would add that at very high frame rates the timestamp itself becomes a nontrivial measurement. Clock jitter is already a major limitation in high-speed instruments, so a "femtosecond movie" is not simply a matter of making the camera faster. The timing chain introduces its own uncertainty.

I think the point of @renormalize post was that finite-resolution measurements can provide a time-resolved record of particle motion without requiring exact localization at every instant. I'm sure if you were careful you could build an ultrafast experiment here, you just aren't going to do it with a quartz crystal.

 

[Roberto Pavani]

Probabily your knowledge on PLL phase noise is better than mine. I only have a patent on synchronization telecom equipments (100Gbps), so little knowledge of keeping track of timing on labs for physics experiments CERN, LIGO. etc where 10-12 precision is probably not enough.

 

[QuarkyMeson]

I guess when you patent a hammer everything starts looking like nails.

 

[Roberto Pavani]

I guess when you patent a hammer everything starts looking like nails.

Only when you are talking of nails.

 

[QuarkyMeson]

The eye diagram makes my point rather than yours. It's a good tool for the problem it was built for. It's the wrong tool for the one under discussion, which is what "not with a quartz crystal" meant.

Ultrafast imaging doesn't derive frame times from an electronic clock at all. STAMP encodes them in pulse sequence and dispersion: https://www.nature.com/articles/s41377-018-0044-7. This group, https://opg.optica.org/josab/abstract.cfm?uri=josab-35-11-2822, set them with mirror-array path-length differences, where 1 µm of path is 3.34 fs. T-CUP and CUSP use temporal shearing and spectral–temporal mapping: https://www.nature.com/articles/s41467-020-15745-4. What you won't find anywhere are PLLs.

So, no nails here. The main problem to me, as a lowly incoming physics graduate student, is that the bubble nucleation in the bubble chamber might not happen fast enough. What I'm not worried about is image timing, because it's been done.

Not really the point of this thread though, and otherwise the post was relevant to the discussion.

 

[renormalize]

The main problem to me, as a lowly incoming physics graduate student, is that the bubble nucleation in the bubble chamber might not happen fast enough.

That's a good point, which may well limit the temporal resolution of the particle positions. Moreover, depending on the size of the created bubbles, the spatial resolution could be limited more by the bubbles than the camera pixels.

 

[Roberto Pavani]

I just meant that capturing a single femtosecond snapshot is one thing. Constructing a sequence of frames at $t_0$, $t_0+1\,\mathrm{ps}$, $t_0+2\,\mathrm{ps}$, $t_0+3\,\mathrm{ps}$, ... while demonstrating that each interval is known to within $1\,\mathrm{ps}\pm10\,\mathrm{fs}$ is a significantly harder metrology problem.

I'm not saying it cannot be done. I'm saying that in that regime, calibration, path-length stability, drift, vibration, and uncertainty propagation become part of the experiment itself.