# FeaturedI On the Heisenberg uncertainty relation

Tags:
1. Jan 22, 2017

### A. Neumaier

Are there fundamental limits on the accuracy for measuring both position $q$ at time $t$ and momentum $p$ at time $t+\Delta t$, with tiny $\Delta t$?

If yes, why?

If no, why can't one then measure (in principle) both $q$ and $p$ arbitrarily well at the same time $p$ (which is not allowed by Heisenberg's uncertainty relation), by taking $\Delta t$ sufficiently small and noting that any measurement takes time?

2. Jan 22, 2017

### Mordred

Inherent fuzziness as the state of a particle isn't distinct as defined by the no go theorem.

It is impossible to prepare states in which position and momentum are simultaneously arbitrarily well localized. (B) It is impossible to measure simultaneously position and momentum. (C) It is impossible to measure position without disturbing momentum, and vice versa.

Quote from the following arxiv.

Its one of the more rigorous treatments but I'll let you judge it on its own merits. Particularly since your far more skilled at QFT than I am. Though I'm also not sure if you've blocked me or not.

Another good article I enjoyed is

Violation of Heisenberg’s Measurement-Disturbance Relationship by Weak
Measurements.

I haven't heard of any further means to reduce the HUP beyond the last link. I do recall the last paper was discussed previously on this forum back when I was active a few years back

Last edited: Jan 22, 2017
3. Jan 22, 2017

### Strilanc

The qubit analogy of what you're asking is "What if I prepare a qubit to point upward, then very quickly turn it to point rightward? Does that give the qubit a well-defined X and Z axis values at the same time, even though those two measurements don't commute and there's no unit vector with $x=z=1$?".

The answer is no. Even if you could do the rotation literally infinitely fast after $t=0$, the qubit would just be pointing upward for times satisfying $t \leq 0$ and pointing rightward for times satisfying $t > 0$. You moved the uncertainty around really quickly, but it's still there. There is no time $t$ where you know with certainty what both the X and Z axis measurements would be.

Analogously, even if you could transform the position state $q=0$ into the momentum state $p=0$ instantaneously after $t=0$, the uncertainty relationship is still satisfied at every time $t$.

Last edited: Jan 23, 2017
4. Jan 22, 2017

### Mentz114

I think perhaps you are having a little joke, because you refer to the publication

P. Busch, P.J. Lahti and P. Mittelstaedt, The quantum theory of measurement,
2nd. ed., Springer, Berlin 1996.

in your 2003 paper with the axiomatic construction ( which I am still sudying and enjoying). I imagine some of the conclusions in the paper cited above by @Mordred will be in this book?

The 2007 measurement paper confirms my prejudices, so naturally, I think it's great.

5. Jan 22, 2017

### rubi

One way to discuss this is by looking at Heisenberg observables $\hat O(t)=e^{i t \hat H} \hat O e^{-i t \hat H}$. Then ($\hbar = 1$, $\hat H = \frac {p^2} {2m} + V$, sloppy application of Hadamard rule):
$$\left[\hat x(t),\hat p(t+\Delta t)\right] = \left[\hat x(t), \hat p(t) + \left[i\Delta t \hat H, \hat p(t)\right]\right] + O(\Delta t^2) = i + O(\Delta t^2)$$
So for small enough $\Delta t$, the uncertainty bounds don't change much. Corrections first appear in quadratic order. But even if you include many orders, the dynamics will usually not evolve an $\hat x$ eigenstate into a $\hat p$ eigenstate, not even after a long time.

6. Jan 23, 2017

### A. Neumaier

I explicitly asked about the non-simultaneous version, where there is a slight time difference.

7. Jan 23, 2017

### A. Neumaier

You are answering an unrelated question.

The correct qubit analogue of my question is: I have a qubit in an arbitrary state. I measure its up-ness, and after a very short time, I measure its right-ness. In between there is presumably a collapse of the state (or whatever you wish to assume in your measurement model).

According to the eigenvalue-measurement link, I should in both measurements obtain a 100% exact answer (up or down in the first, right or left in the second). Thus shouldn't the answer be yes?

Fundamental limitations can only come from an argument showing that it is impossible to make the two measurements.

8. Jan 23, 2017

### A. Neumaier

But the Heisenberg equation is not valid at all times as the measurement induces an uncontrolled perturbation of the system at the moment of measurement. Or do you argue in a framework where there is no collapse?

9. Jan 23, 2017

### A. Neumaier

This is not needed as a system is never in a position or momentum eigenstate (these are not normalizable, hence invalid states). Position and momentum measurements can nevertheless be made.

Last edited: Jan 23, 2017
10. Jan 23, 2017

### dextercioby

The HUP is nothing but a mathematical statement (whose rigorous deduction within the framework of functional analysis requires care) about the statistical spread of expectation values around the mean. I like an interpretation of QM in terms of virtual statistical ensembles of identically prepared systems. There one can easily relate expectation values to experimentally accurate values of measured observables.

11. Jan 23, 2017

### A. Neumaier

What do you mean by ''virtual statistical ensembles''?

12. Jan 23, 2017

### dextercioby

Last edited: Jan 23, 2017
13. Jan 23, 2017

### A. Neumaier

14. Jan 23, 2017

### rubi

After a measurement of $\hat x(t)$, the system will be in a state $P_{\hat x(t)}(B)\Psi$, where $P_{\hat x(t)}$ is a projector of $\hat x(t)$. If you perform a measurement on this state at time $t'$, the question is whether the projectors $P_{\hat p(t')}(O')$ of $\hat p(t')$ commute with $P_{\hat x(t)}(B)$, because otherwise, the state after the second measurement will not lie in the subspace onto which $P_{\hat p(t)}(B)$ projects. So the question is whether the projectors of $\hat x(t)$ and $\hat p(t')$ commute and (up to technicalities about unbounded operators) this is equivalent to asking whether $\hat x(t)$ and $\hat p(t')$ commute.

(This is already without loss of generality, since a POVM measurement can be realized (in a bigger Hilbert space) as a PVM followed by unitary evolution and we can assume this extra unitary evolution to be included in $e^{-i\Delta t \hat H}$.)

Well, I ignored these technicalities, because it doesn't really change the conceptual problem. Let's make a concrete example. Let's say, we have a Hamiltonian $H=\frac{p^2}{2m}$ and we have just measured the position. The quantum state will be sharply peaked on some definite position (let's say $0$). We know that sharply peaked functions correspond to broadly spread functions in momentum space. If the wave function is a Gaussian $\Psi(x,t=0)=e^{-\frac{x^2}{4\sigma_x}}$, then the momentum wave function will be a Gaussian too with $\sigma_x \sigma_p \approx 1$. Now, our concrete Hamiltonian will just evolve $\Psi$ into a Gaussian with $\sigma_x(\Delta t)=\sigma_x + \Delta t \xi$, hence after a short time $\Delta t$, the Gaussian will still be sharply peaked. But this means that in momentum space, it will still be broadly spread: $\sigma_p(\Delta t)\approx\frac{1}{\sigma_x+\Delta t\xi}\approx\frac 1 {\sigma_x}-\frac{\Delta t\xi}{\sigma_x^2}$. So after a short time, the state won't have evolved into a state of definite momentum.

15. Jan 23, 2017

### A. Neumaier

But why should this be needed? After the second measurement we simply have the state $P'P\psi$, which is as good as any other.

Thus your argument tells nothing about how accurate $q$ and $p$ are measured - which are properties encoded into $P$ and $P'$.

From this perspective, there seems to be no restriction on the precision with which $q$ at time $t$ and $p$ at a slightly later time $t'$ can be measured.

16. Jan 23, 2017

### A. Neumaier

Even after a long time it won't have evolved into one, assuming the free motion. But surely we can make a macroscopic time later a momentum measurement and will get a definite measurement result approximating the momentum.

17. Jan 23, 2017

### rubi

What I'm trying to explain is the following: If you have measured position with an accuracy $\sigma_x(0)$, then the state of the system is $P\Psi$. Now the question is the following: What is the variance $\sigma_p^2(\Delta t)$ of momentum in the state $e^{-i\Delta t H}P\Psi$? In order for the variance to be small, the state must be close to being a momentum eigenstate. But if $\sigma_x(0)$ is small, then $\sigma_p(\Delta t)$ will not be small in general. Hence, we get a trade-off between position uncertainty $\sigma_x(0)$ and momentum uncertainty $\sigma_p(\Delta t)$. There is $c$ with $\sigma_x(0)\sigma_p(\Delta t)\geq c$.

18. Jan 23, 2017

### Strilanc

If you're picturing a situation where the measurements take time, then once $\Delta t$ is lower than that time your results will become junk. The specific way in which they become junk depends on how you're doing the measurement. There are many ways to model this, creating many varieties of junk.

For example, suppose we're on a quantum computer that performs up/down measurement by performing a continuous controlled-NOT of the target qubit onto a fresh spare qubit over $1us$ (then it properly measures the spare qubit using some slower process). Furthermore, the computer performs left/right measurement by applying a Hadamard operation arbitrarily fast (rotating the qubit so its old left/right axis is its new up/down axis), then performing an up/down measurement, then rotating the qubit back with another Hadamard.

If we start the up/down measurement at $t_0$ and the left/right measurement at $t_0 + \text{us}/2$ then we are effectively doing half of a CNOT onto an ancilla, Hadamarding, then the other half of the CNOT as well as a full CNOT onto a second ancilla, then Hadamarding, and taking our time to measure the ancilla properly. Like this circuit:

If you pass an upward or downward qubit into this circuit, the qubit ends up leftward or rightward and you don't really learn that the qubit was upward or downward. The circuit does something, but that something certainly isn't violating the HUP. I won't go into the various other behaviors of this particular circuit (I will mention that what's happening the the entangled partner is kinda neat). The main points I wanted to make with it are:

1. You can make explicit models of the "measurements happening so close that they overlap" situation.
2. Those models fail to distinguish between upward and rightward with certainty.

If your measurements are arbitrarily fast, then having them happen near each other isn't different from doing one then the other. If your measurements aren't arbitrarily fast, then they will overlap and generally break each other's effects. The way they break depends on how the measurements are implemented.

19. Jan 23, 2017

### Mordred

Doesn't lemma c I posted state that you disturb the system when you make the first measurement? So when you make the make the second measurement the system is not the same?

I don't know about anyone else but I would call that a fundamental limit.

(C) It is impossible to measure position without disturbing momentum, and vice versa.

So now you have induced additional inaccuracy before you take the second measurement

20. Jan 23, 2017

### A. Neumaier

But this is applying different standards to the two measurements. You consider the accuracy of the measured position but the accuracy of the predicted momentum. Whereas I was asking about limitations in the accuracy of the measured position and the measured momentum.

The usual HUP is conventionally derived as (i) a constraint on the predicted uncertainty of both position and momentum at the same time, but then interpreted (with somewhat vague arguments) as (ii) a constraint on the measured uncertainty of both position and momentum in a simultaneous measurement. I am interested in the (ii)-part - except with the time delay.

Last edited: Feb 3, 2017