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# Sequential measurements of conjugate observables

1. Apr 7, 2014

### atyy

There is an argument that accurate sequential measurement of conjugate observables A and B on the same state is possible if the state is an eigenstate of one of the observables. When the state is an eigenstate of A, an accurate measurement of A will not disturb the state, so B can then be accurately measured on the same state.

However, that seems to contradict Eq 15 of http://arxiv.org/abs/quant-ph/0207121:

ε(A)η(B) + ε(A)σ(B) + σ(A)η(B) ≥ |<ψ|[A,B]|ψ>|/2

where ε(A) is the error in the measurement of A, η(B) is the disturbance caused by A to the subsequent measurement of B, and σ(A) and σ(B) are the respective intrinsic uncertainties of the state ψ when A or B are accurately measured separately. For noncommuting observables A and B, ε(A) and η(B) cannot both be zero. So presumably there is a mistake in the argument made at the top. What is the mistake?

Last edited: Apr 7, 2014
2. Apr 7, 2014

### universal_101

I think there is NO meaning to the words, accurate measurement of an observable by a single measurement, therefore one must use several measurements to specify the accuracy of a measurement, in other words QM uses 'variance' of several measurements to specify the accuracy or uncertainty.

Now it is important to see that even if the state is an eigenstate it would still have some variance practically(experimentally), that is we won't have zero variance measured even on an eigenstate.

Now if we do measurements on B, a single measurement won't settle it, we would again need variance of the measurement on B(which would certainly be huge compared to A). Now it is this variance which should follow HUP, that is, not only the ΔA and ΔB are inversely proportional(which is also correct classically for conjugated variables) but that their product cannot be lower than a certain quantum mechanical constant.

Last edited: Apr 7, 2014
3. Apr 7, 2014

### Simon Bridge

I don't think hes talking about experimental uncertainty - but about how a measurement of one observable may make it's conjugate uncertain ... HUP and all that.

There are two forms of the HUP - or, rather, two descriptions which are frequently given the same name.
iirc. One applies to a canonical ensemble of identically prepared systems and the other to an individual system.

iirc. (also) The usual text book treatment says that even if the system is prepared in an eigenstate of A, the subsequent action of measurement (of the conjugate observable) is to destroy the eigenstate... i.e. on measuring B the system is no longer in an eigenstate of A.

The arxiv reference does, indeed, seem to be talking about uncertainties though...

So to know what is going on, we need a reference for that "argument" that seems to contradict the arxiv paper.

4. Apr 7, 2014

### Ken G

It seems to me that inequality is not particularly useful for eigenstates of A, because isn't it true that we still expect sigma(A)*sigma(B) to obey a similar inequality? Thus, the second term on the LHS of the inequality is already superceded by the relation I'm referring to if you are treating an eigenstate of A, because then surely epsilon(A) will be much larger than sigma(A) (as the latter is essentially zero). So we already know that inequality holds if the system is in an eigenstate of A, as it is not close to the "=" version of the inequality.

As to your initial statement, I don't think anything is wrong with it as written. You can indeed always imagine a measurement with epsilon(A)=0, and then a subsequent one with epsilon(B)=0. That latter quantity does not even appear in either version of the HUP, it's not relevant, it depends on the instrument not the state. In other words, I don't think eta(B) plays any role in epsilon(A) or epsilon(B), nor do the latter combine in any uncertainty relation. But you already know you cannot have sigma(A)*sigma(B) violate the HUP, if A and B are complementary, so disturbances don't enter into that issue. In other words, if you do an accurate preparation in an eigenstate of A, it doesn't matter how many times you measure it, you won't disturb it, and you can do an arbitrarily accurate (noiseless) measurement of B, but you still can't predict what you'll get to any decent precision, because of the HUP.

It seems to me the importance of this new inequality is rather esoteric, in fact. It does seem to give a valid violation of the Heisenberg noise-disturbance relation, by first justifying that it is possible (with the inequality you give), and then by constructing a case that does indeed violate it. But the case it constructs is not an eigenstate of either A or B, so it's not clear what importance that has. If you start with an eigenstate of B (not A), the inequality gets very interesting, because then sigma(B)=0, and the inequality says [epsilon(A)+sigma(A)]*eta(B) obeys the HUP, which means that sigma(A) gets added to epsilon(A) in Heisenberg's own noise-disturbance relation. In that situation, it's not a noise-disturbance relation, it's a noiseplusuncertainty-disturbance relation. That means that when you do very noiseless measurements of A on an eigenstate of B, you do not necessarily need to mess up the B eigenstate much, because the large sigma(A) will insure eta(B) need not be any larger than sigma(B). In principle, you could imagine preparing the state to be extremely close to an eigenstate of B, and you could in principle get essentially zero eta(B), even if you do a noiseless measurement of A, without violating the inequality, which would be of seismic importance if you could really get a low eta(B). However, his analysis does not tell you that you will actually get a result close to the "=" value in that case, and if you get a result much larger than that, it really doesn't do anything useful! In other words, just because you don't necessarily violate the inequality if you use an eigenstate of B, it doesn't necessarily imply you will achieve a low eta(B) with a large sigma(A) and a zero epsilon(A). He doesn't explore in general when you get the all-important =, so necessary if you want to do things like detect gravitational waves, he only quantifies what you get in a case that shows you can violate the Heisenberg version, not a case that actually makes epsilon(A), eta(B), and sigma(B) all low. Only the latter ends up violating the spirit of the HUP, and help us find gravitational waves.

Last edited: Apr 7, 2014
5. Apr 7, 2014

### universal_101

Seems like the sigma is the experimental uncertainty(S.D) or noise or inherent uncertainty in measurement, and they are all the same. Whereas eta is the disturbance caused by the act of measurement, or experimental uncertainty or noise or inherent uncertainty, they are all the same as far as current QM formulation is considered. And can be seen in the section 1 of referenced arxiv preprint, where there are three formulation of HUP.

And just because the measurement distribution is same as the theoretical probability distribution, does not make the measurements accurate and therefore the uncertainty intrinsic, instead as shown by three ways of expressing the HUP, i.e. it does not matter if we consider the distribution of measurement noise, disturbance, or intrinsic.

But the author seems to treat them differently when he writes,
$\epsilon(A)\eta(B)+\epsilon(A)\sigma(B)+\sigma(A)\eta(B) ≥ \frac{\hbar}{2}$

6. Apr 7, 2014

### kith

An eigenstate of A has σ(A)=0, so the HUP tells us that σ(B) cannot be finite for non-zero <[A,B]>. So ε(A)σ(B) is undefined if ε(A) is identical to zero.

7. Apr 7, 2014

### atyy

Yes, that seems to be right. It's obviously right for position and momentum, where the eigenstates are not physical. My confusion was I somehow thought it was different for a single spin - let me do some homework with Pauli matrices ...

8. Apr 7, 2014

### Ken G

Eigenstates are never physical, they are idealizations. I don't see that taking sigma to be exactly zero really matters, it suffices to take it as small as you need. What matters is sigma << epsilon, that is as good of an eigenstate as matters here. Is not the real problem here that it is easy to confuse sigma for epsilon? The HUP does not constrain two epsilons, it constrains pairs chosen from sigma, epsilon, and eta, where if you repeat sigma you have one version of the HUP, but you cannot repeat any of the others in a pair in a HUP relation.

Last edited: Apr 7, 2014
9. Apr 7, 2014

### kith

Just a side note: you were talking specifically about conjugate pairs of observables while the spin projection observables don't have conjugate partners. This doesn't matter much because your argument doesn't need conjugate pairs in the first place. A non-zero <[A,B]> for the specific eigenstate is enough.

10. Apr 7, 2014

### kith

I agree. The question of this thread is rather academic because you cannot perform measurements with infinite precision anyway.

11. Apr 7, 2014

### Ken G

They are still idealizations, since epsilon is never zero, you can never test a preparation such that you could know it's really in a state of definite spin. But that's OK, we only need sigma << epsilon, where epsilon is already small, and we can as well say that we have a quasi-definite spin state. I'm not saying we can't use eigenstates as conceptual entities, we desperately need them for that, I'm just saying we needn't worry about apparent indeterminacies of when sigma is essentially zero, because it would always have some finite value we could use instead.
It would be more precise to specify some particular yet very small epsilon, and show how the noise-disturbance relation can be violated for that finite epsilon. The author is essentially taking it for granted that this will be true, and I think he's right about that, but showing it would remove all doubt.

12. Apr 7, 2014

### Matterwave

If you do this for spin, you will notice that the right hand side of the inequality is 0 because the expectation value of Sz in an eigenstate of Sx or Sy is 0 etc. So you have a non-issue saying that the errors cannot be negative is all.

13. Apr 7, 2014

### atyy

Yes, thank you. So the basic argument that if the system is in an eigenstate, then one can do sequential accurate measurements is correct and not in conflict with the inequality.

In the case of position and momentum, where the eigenstates are not physical, regarding the issue of whether arbitrary accuracy is possible, the strict answer is no, since ε and η cannot both be zero for the physical states. However, the answer can be reasonably yes or no depending on how one asks the question:

(1) If we set ε=0 and η>0, no matter how small η is, we can find a state such that the inequality is satisfied - so in this sense arbitrary accuracy on sequential measurements of position and momentum is possible.

(2) But for any given state, if we set ε=0, we can find δ<η<0, such that when δ is small enough the inequality is not satisfied - so in this sense arbitrary accuracy on sequential measurements of position and momentum is not possible.

Does all that seem correct?

14. Apr 7, 2014

### Ken G

What I don't understand is why you are saying it matters to put the state in an eigenstate. You can put it in any state, and do sequential measurements of arbitrary accuracy, and it doesn't invoke the HUP because the accuracy of the two measurements, epsilon(A) and epsilon(B), never appear in the same HUP. The value of eta(B) will matter, but you are talking about epsilon(A) and epsilon(B) here, not eta(B).
No, arbitrary accuracy on sequential measurements of position and momentum is always possible, that is epsilon(x) and epsilon(p). Those two epsilons never appear in a HUP. If you start with an eigenstate of A, and set epsilon(x) to be zero (or really, smaller than any value someone cares to name), it is the term epsilon(x)sigma(p) that satisfies the inequality. This is true for any epsilon(x), even in the limit as epsilon(x) goes to zero, it just makes sigma(p) get larger. The value of epsilon(x) is not constrained in any way, and the value of epsilon(p) never matters.

15. Apr 7, 2014

### atyy

The question is about arbitrary accuracy on the same state, ie. in order to be considered accurate the distribution of results produced by each measurement in the sequential measurement must be the same as when that accurate measurement is performed on the initial state. So this is not about the HUP, which has nothing to do with sequential measurements. The idea here is the sequential version of whether simultaneously accurate momentum and position measurements can be performed.

16. Apr 7, 2014

### Ken G

But these are sequential measurements, which must be distinguished from joint measurements as is done in that paper. That paper alludes to the key issue that simultaneous measurements are a rather different animal, because of the absence of any guarantee that the two measurements don't interfere with each other (a point I raised above-- two simultaneous measurements have limits on their epsilons by virtue of being simultaneous, whereas sequential observations do not, and the inequality you cite from that paper is about sequential observations only). Is not the accuracy of the two measurements epsilon(A) and epsilon(B)? So we see that epsilon(B) never appears, whether we call this the HUP or not (I have been referring to any equality with h on the RHS as a type of HUP, but we can restrict to your meaning, that's fine it doesn't affect my point). I think what you are saying is that if you have an eigenstate of A, you can make epsilon(A) very small and still have a small eta(B), so you can make epsilon(A)*eta(B) small, but if you have an eigenstate of A, it is the term epsilon(A)sigma(B) that you must worry about, it will be doing all the heavy lifting there.

17. Apr 7, 2014

### atyy

The paper is about sequential measurements. The last paragraph of p1 describes the setup "Suppose that the A measurement using the apparatus A is followed immediately by a measurement of the observable B using a noiseless measuring apparatus B."

Let's say the state is an eigenstate of A. Then measuring A accurately will mean ε(A)=0. Since the state is an eigenstate of A, then η(B) will be zero, since the measurement will not disturb the state, leaving it available for an accurate measurement of B. The inequality will be satisfied because in the eigenstate of A, σ(B) is infinite, so σ(B)ε(A) will "do the heavy lifting", as you and kith pointed out (and it's in the paper too - I was confused by the spin case, which matterwave has pointed out is fine).

So heuristically, if you get to choose your state to be an eigenstate of A, then accurate sequential measurements are possible. Physically, eigenstates of position and momentum are not possible, so the statement becomes that if one sets ε=0 and η>0, no matter how small η is one can always choose a state so that the inequality can be satisfied. So in this sense arbitrarily accurate sequential measurements are possible.

On the other hand, for any given state, since σ(A) and σ(B) are fixed, if ε=0, then one can find η>0 sufficiently small that the inequality is violated. In this sense arbitrarily accurate sequential measurements are not possible.

18. Apr 7, 2014

### Ken G

Yes, I agree there.
This is what I do not see. Why would the inequality be violated, even for eta=0? I thought we agreed the inequality is maintained by the epsilon(x)sigma(p) term, for a sequential x-then-p measurement on an arbitrarily precise x eigenstate. If it is maintained by epsilon(x)sigma(p), it makes no difference what epsilon(p) is, and can be arbitrarily small. It is epsilon(p) that says how accurately we are measuring p, and is a function of our measuring apparatus, whereas sigma(p) is a function of the initial state.

19. Apr 7, 2014

### atyy

In the second case, I was talking about a given initial state. The idea is that since an eigenstate is not strictly possible, if we specify any given initial state, no matter how close it is to an eigenstate, since the state is specified, σ(x) and σ(p) are both fixed and finite. If we set ε(x)=0, then the inequality will only contain η(p)σ(x)>CAB. Since σ(x) is fixed and finite, η(p) cannot be arbitrarily small.

Last edited: Apr 7, 2014
20. Apr 7, 2014

### atyy

http://arxiv.org/abs/0911.1147 is an interesting paper by Ozawa. The first (Theorem 13) is basically the argument given in the OP that conjugate observables can be simultaneously measured in an eigenstate of one of them. So presumably he doesn't think that as contradicting his inequality, as we have agreed. The second claim (Theorem 14) is even more interesting. I think here he is talking about joint measurements, whereas the OP talked about sequential measurements, but I think the ideas are similar. Technically, the claimed theorems are for finite dimensional vector spaces.

Theorem 13. In any Hilbert space, every pair of observables are simultaneously measurable in any eigenstate of either observable.

Theorem 14. In any Hilbert space with dimension more than 3, there are nowhere commuting observables that are simultaneously measurable in a state that is not an eigenstate of either observable.

I guess Theorem 14 has to do with an EPR-like setup, which Ozawa describes in the discussion of the paper "In the EPR state of two particles, I and II, the momentum of particle I can be measured by directly and locally measuring the momentum of particle II taking into account the EPR correlation; this follows from the EPR original argument stating that the locality of measurement ensures that the predicted correlation determines the value of momentum of particle I. The locality of the momentum measurement of particle II also concludes that it does not disturb the particle I, and hence we can simultaneously measure the position of particle I by a direct measurement on particle I. Thus, the momentum and position of particle I are simultaneously measurable, so that both the measured values corresponds to elements of reality."

Last edited: Apr 7, 2014


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# Help!

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Suppose that $f$ is a function and $f^{-1}$ is the inverse of $f$. If $f(3)=4$, $f(5)=1$, and $f(2)=5$, evaluate $f^{-1}\left(f^{-1}(5)+f^{-1}(4)\right)$.

Jul 15, 2018

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Lightning,

You are presenting high level questions so you are advance enough to learn how to present them properly.

I cannot be bothered interpreting this and I have read other people make similar comments.

The stuff between the dollars signs is in Latex.

Open the latex box in the ribbon, past the latex into there and press ok. Do not include the dollar signs.

Practice a little and you will find it easy.

If you need better instructions you can ask.

Jul 15, 2018

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