# Does quantum mechanics allow us to formulate causally sufficient conditions for the occurrence of an outcome?

This is a follow-up question to "In QM, why do the probabilities ... add up to 1?".

No actual measurement is perfect. While theorists may ignore this, experimenters know well enough that in many runs of a given experiment no outcome is obtained. (The efficiency of many real-world detectors is rather low.) This means that in order to make the probabilities of the possible outcomes of a measurement add up to 1, one discards (does not consider) all those experiments in which no outcome is obtained.

Quantum mechanics thus allows us to predict the probabilities of measurement outcomes on condition that there is an outcome. But is there anything in quantum mechanics — after all, the theoretical framework of contemporary physics — that allows us to predict that a measurement, which is about to be made, will have an outcome? Does quantum mechanics allow us to formulate causally sufficient conditions for the occurrence of an outcome (no matter which) or the success of a measurement?

-
Maybe you should explain what you mean with "on condition that there is an outcome" . Even zero is an outcome, or inaction. –  anna v Apr 3 '11 at 5:25
@Anna, one needs to distinguish two cases. (i) If a 100% efficient detector doesn't detect a particle, there is no particle. (ii) If a real-world detector (which never is 100% efficient) fails to detect a particle, there may or may not be a particle. One could mock up a 100% efficient detector by ignoring all instances in which it failed to work, but this would beg the question. –  Koantum Apr 3 '11 at 5:27
@Koantum Zero particle is an outcome, and it would have been within the predictions/solutions of the quantum mechanical problem, assuming it is in a region of phase space where quantum mechanics holds. –  anna v Apr 3 '11 at 5:31
@Anna, Zero particle is an outcome only if one is using a perfect (100% efficient) detector. But such a detector does not exist. See my previous comment. –  Koantum Apr 3 '11 at 5:43
Well, if a detector can't measure a quantity reliably, we must know that what the detector shows is not quite the same thing as the value of the quantity that should be measured. This is a problem for the experimenter who tries to measure something, not for quantum mechanics. Despite the fact that it's not a problem for quantum mechanics, quantum mechanics may still help because quantum mechanics predicts what any apparatus - including an "imperfect apparatus" - will do. But of course, one needs to know something about the "imperfection of the gadget" and not just the idealized QM laws. –  Luboš Motl Apr 3 '11 at 6:28
show 1 more comment

I suggest that you need to modify your "Quantum mechanics thus allows us to predict the probabilities of measurement outcomes on condition that there is an outcome." I suggest instead that quantum mechanics allows us to use probability measures to model observed statistics of actual raw data. As an aside, whether this is prediction of future statistics or fitting of past statistics seems to me not relevant to the question you're asking here.

Experimental raw data can take the form of a list of times and whether a thermodynamic transition occurred in an experimental device within some given time tolerance of each time in the list. That could be a list of some analogue signal value (averaged over some time period and converted to n bits), recorded, say, each picosecond. The same list is more-or-less equally well represented, however, by a list of times at which thermodynamic transitions occurred. I say more-or-less because the list of times at which events occurred essentially filters features of the experimental apparatus such as noise and dead-times out of the raw list of signal values, which may well be done in hardware. The rawest possible data is often not recorded because it is too voluminous. Some choices have to be made about what is and is not interesting to record; every now and then someone records something that wasn't expected to be interesting but turns out to be unexpected.

The list of signal conditioning processes that are applied by experimentalists is quite long. To me, qua theorist, the methods used by experimentalists to describe and/or model the signal conditioning processes they use in a particular experiment often seem quite ad-hoc, or at least not clearly unified, which is, I think, a way of more generally formulating the concern you seem to me to be voicing in your Question.

EDIT (after noticing your postscript on your previous Question): Not only is no measurement perfect, nor is any preparation. The process of characterizing both what the preparation apparatus prepares and what the measurement apparatus measures is nonlinear, insofar as we know neither the density operators $\hat\rho_i$ nor the measurement operators $\hat O_j$ in the linear equations that determine the expected values $\mathsf{Tr}\!\left[\hat\rho_i\hat O_j\right]=V_{ij}$, where $V_{ij}$ is the mean of a raw dataset (a statistic) that we obtain when we use the $i$-th state preparation apparatus with the $j$-th measurement apparatus. Both all the density operators $\hat\rho_i$ and all the measurement operators $\hat O_j$ have to be determined from the finite set of experimentally determined statistics $V_{ij}$. Of course things get considerably more nonlinear when we start to model second and higher moments of raw datasets, $W_{ij}$, say, so we also have equations such as $\mathsf{Tr}\!\left[\hat\rho_i\hat O_j^2\right]=W_{ij}$, and still more nonlinear when we introduce unitary or more general state transformations. Note that the experimental statistics also have to determine the dimensionality of the Hilbert space we use. The nonlinearity of the process of characterization is superficially avoided because we know from experience what dimensionality of Hilbert space usually constructs effective models for a given state preparation and we may even have datasheets from the manufacturer that tell us what a given preparation or measurement device prepares or measures to a good approximation.

-

Let us take a simple example of a theory predicting something, Ohms Law from Electricity and Magnetism

I=V/R

I am given the value of R and asked to measure I. I take a voltmeter, measure V and have a value for I.

In the same circuit, same conditions, I take and ampmeter and measure I. The value differs by Delta(I), inevevitably. This is because my measuring instruments have by construction a measurement error, that should always be taken into account when reporting a measurement. Ohm's law is validated within errors.

The above comes from a simple theory.

A quantum mechanical prediction comes from more complicated solutions of quantum mechanical equations describing a system. The measuring apparatus is much more complicated than in my simple example. The measurements are always given with the measurement error estimated. This pdf from the particle data group shows the meticulous estimate and recordings of measurement errors. These have nothing to do with the quantum mechanical prediction, except to allow limits wherein a specific quantum mechanical model can play and be consistent with experiment.

-
Right, +1, Anna. I also feel that the question was just a general question about errors of measurements. Of course, in reality, a gadget tries to imitate a "perfect one", and the deviations of the real gadget from the idealized one is something one must take into account - and the error margin has to be properly estimated. Quantum mechanics allows predictions of outcomes with any systems of particles, whether some of them are organized into gadgets - good or bad ones. A gadget may fail to absorb a particle, in which case the interference may continue etc. All those things are computable. –  Luboš Motl Apr 3 '11 at 8:55
True, @Luboš those things are computable. Using cross sections and coupling constants, one should in principle be able to calculate the efficiency of any detector. Since this is always less than one, the answer to the title question is negative. –  Koantum Apr 4 '11 at 2:05

Yes. A detector isn't ideal, so a model that assumes it is will be flawed. It does, however, obey the laws of physics. Just model it as part of the system, and you can predict the probabilities for any result the detector is capable of giving.

For example, rather than checking if a photon makes it to where the detector is, actually model the detector and see if it ends up going off.

Any error about the probabilities is due to lack of knowledge about the detector, not the detector not being good enough.

-

Looking at both this question and your previous one, I think there's a bit of talking at cross purposes. Theoretically, something has to happen with regard to the outcome of a measurement. Therefore, the total probability of all outcomes must sum to one.

Moving to a real-world scenario, instead of the idealized equations most likely worked up by theorists, one of the possible outcomes of the measurement attempt is to find a value of zero. As you state, because of inefficiencies in a detector, one of the quantum mechanical probabilities in a properly formulated equation that takes into account both the detector and the thing to be measured should be that there is no interaction, i.e. the detector detects nothing. Although the measurement value is zero, the probability for that (non)measurement is non-zero.

I believe the disconnect you're experiencing is from thinking about measuring the particle, for example, rather than the interaction between the particle and the detector. The theoretical equations used to teach QM are simplified by assuming a 100% efficient detector so that the interaction does not need to be computed. In the real world, however, this is necessary. A good example might be neutrino detection, where almost all the probability is for no detection.

-
Agreed. I take this to imply a negative answer to the title question. –  Koantum Apr 4 '11 at 1:45
@Koantum I'm not sure I'd answer the title question with a 'no,' and I don't think the above answer implies a 'no.' I'd say that QM does fully predict outcomes (statistically, of course), and, among those outcomes, is a 'lack of an occurrence, i.e. no interaction. Just because nothing happens doesn't mean we can't predict it. That's what, for example, computation of cross sections is often for: determining the likelihood of a measurable interaction. –  Mitchell Apr 6 '11 at 23:09

I am not sure whether I fully understand this question as it contains some ambiguities I shall discuss below. However in the most simplistic reading of the Title the answer is: YES.

Here I am taking the phrase "the occurrence of an outcome" to mean a non-zero eigenvalue for a given measurement operator. To make this work one needs to mention an axiom of QM, namely that of the existence of a complete set of observables: A, B, C,... As a result the basis of the Hilbert space can be described in terms of the eigenfunctions $\Psi_i$ (usually written in Dirac notation) of that complete set, giving a decomposition $\Psi=\Sigma{_i}a_i\Psi_i$. Now a given operator $P_j$ may have zero and non-zero eigenvalues; if the only eigenvalues are {0,1} then it is a Projection operator. The Spectral decomposition theorem may also be used for Measurement operators to define the outcomes, their probabilities and the resultant eigenspaces too. In particular a zero outcome (for a given measurement $P_j$) can have a non-zero probability.

Clearly it is necessary in an experimental context to ensure that the complete set has been defined and that the algebra is worked through correctly.

The question however introduces more topics than just this simple interpretation, as there is discussion of non-ideal measurement instruments, and any implication from that fact. There will be implications in terms of measurement in practice, and measurement procedures in general as well. However to discuss this further one would need to introduce more clarification into the Question as to exactly what one is concerned about experimentally. To focus on one point. Even although one can calculate that eigenvalues {a1,a2,a3,a4,a5,...} might be observed with different probabilities with an ideal measuring instrument; it might well be that the resolution of the measuring apparatus cannot distinguish between {a1,a2},{a3,a4}, etc. Clearly calculation of quantum probabilities need to take into account this non-ideal resolution. So perhaps your question has to do with whether a complete theory of how to do this exists?

I should also mention, following on from the last point and remarks in the question, that one may need to consider extending the framework of quantum mechanics as described in these questions to deal with classical uncertainty arising from uncertainty in the experimental apparatus or setup. This additional set of probabilities combined with the quantum probabilities results in something called a Density Matrix.

One can ask a range of questions about the formalism of Density Matrices (e.g. what sums to 1 and why) as well as ask about their interpretation and use. However Density Matrices have not been mentioned in these questions so I will leave that topic for further questions.

-
I don't think "no outcome" is the same as a "null outcome." In the first case, all possible outcomes are represented by projectors, and none of them corresponds to the failure of the apparatus to indicate an outcome. In the second case, the failure of the apparatus to respond in any way is one of the possible outcomes and is therefore represented by a projector. In this case every measurement does indeed have an outcome. But this brings up yet another question, which I may ask in another thread... –  Koantum Apr 4 '11 at 2:19
@Koantum, Yes the "no outcome" versus "null outcome" was another ambiguity I (and possibly others looking at the comments) had noticed in the formulation of the question. If one pins down definitions (and maybe an experimental setup definition too) one might find an interesting question here. –  Roy Simpson Apr 4 '11 at 9:46
Note also, of course, that the resolution problem in a non-ideal apparatus (as mentioned in the answer) can make a non-null outcome become a "no outcome" from the experiment perspective. –  Roy Simpson Apr 4 '11 at 9:51