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The question assumes the standard formalism with projector-valued measures rather than POVMs. Suppose a measurement has two possible outcomes, and the corresponding probabilities are greater than 0 and less than 1. Neither outcome is therefore certain. Then why is it certain that either outcome is obtained (as it seems, if the probabilities add up to 1)?

Added after four answers: All the answers provided so far elaborate on the comment by @Vladimir: "It is not a 'quantum mechanical' feature but a consequence of probability definition." @Lubos and @Mark cast the question into a quantum-mechanical form, e.g., why do the absolute squares of the amplitudes associated with the possible outcomes of a measurement add up to 1? They also explain why the sum remains equal to 1. (However, for a decaying particle the probability of finding it decreases, while the probability of finding its decay products increases. So the "conservation of probability" has something to do with the proper conservation laws.) @David makes it clearest why these answers are insufficient.

Keep in mind that 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 add up to 1, one discards (does not consider) all those experiments in which no outcome is obtained.

So let me follow up with another question.

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It is not a "quantum mechanical" feature but a consequence of probability definition. –  Vladimir Kalitvianski Apr 2 '11 at 9:12
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@Koantum Since you have a new question - or have clarified the old one to the extent it appears new - you might want to start a new thread so that the answers stay relevant to the question at hand. –  Mark Eichenlaub Apr 3 '11 at 3:15
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What could you possibly mean by "no outcome"? The experimenter is transported backwards in time and the experiment never happened? I think this is what Vladimir was alluding to above. At any rate try to define "no outcome" –  sigoldberg1 Apr 3 '11 at 4:00
    
@sigoldberg1, a successful measurement has an outcome. An attempted measurement may or may not be successful. An unsuccessful measurement has no outcome. –  Koantum Apr 3 '11 at 4:58
    
@Mark, as you suggested, I have posted the follow-up question separately. –  Koantum Apr 3 '11 at 5:08
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6 Answers

Suppose a measurement has two possible outcomes

That's why. One of the possible outcomes has to occur - it's the definition of "possible outcomes." Probability theory takes this as an axiom, the axiom of unit measure according to Wikipedia.

Note that we often have to normalize the probabilities so that they add up to 1.

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The probabilities of individual outcomes are given by the squared absolute values of the complex probability amplitudes $a_i$ associated with the individual outcomes. Their sum $$ {\rm Total\,\,probability} = \sum_{i} |a_i|^2 $$ is therefore nothing else than a formula for the squared length of the state vector $|\psi\rangle$: note that it is a complexified version of the Pythagorean theorem.

In quantum mechanics, if the state vector (wave function) has the length equal to one at the beginning, it will have the length equal to one at all times - because of the so-called "unitarity". Unitarity means that the evolution according to Schrödinger's equation is essentially just a rotation around some "axes" in the Hilbert space - an element of $U(N)$ or $U(\infty)$, a complex and/or infinite-dimensional generalization of $O(N)$.

Unitarity means that the evolution of $|\psi\rangle$ - by Schrödinger's equation - preserves the length of the vector, and the squared length of the vector is nothing else than the sum of probabilities of all mutually exclusive outcomes of experiments, regardless of the orthogonal basis of the Hilbert space that we choose.

The unitarity condition is $U U^\dagger={\bf 1}$ for the evolution operators $U$ and may be reduced to $H=H^\dagger$, the hermiticity of the Hamiltonian, whenever the evolution is encoded in a Hamiltonian via $U=\exp(Ht/i\hbar)$, as Mark's answer shows explicitly.

It is important that in quantum mechanics, we don't have to normalize the probabilities in a way that would depend on the evolution: the preservation of the total probability is guaranteed by the equations of motion, namely by the hermiticity of the Hamiltonian that enters these equations of motion.

If an ad hoc normalization were required, quantum mechanics would be spoiled by a new source of non-locality and the "wave function collapse" would become observable. Both of these problems would lead to inconsistencies with the observations - and, from a theorist's viewpoint, internal logical inconsistencies in the theory.

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Suppose we have a quantum state that is normalized at some time. Then it will remain normalized if the Hamiltonian is Hermitian.

$$\frac{\partial}{\partial t}\langle \Psi \mid \Psi \rangle = \left(\frac{\partial}{\partial t} \langle \Psi \right) \mid \Psi \rangle + \langle \Psi \mid \left(\frac{\partial}{\partial t} \Psi \rangle \right)$$

Schrodinger's equations says

$$\frac{\partial}{\partial t}\mid \Psi \rangle = \frac{-i}{h} H \mid \Psi \rangle$$

and

$$\frac{\partial}{\partial t} \langle \Psi \mid = \frac{i}{h} \langle H \Psi \mid = \frac{i}{h} \langle \Psi \mid H^*$$

Substituting in gives

$$\frac{\partial}{\partial t}\langle \Psi \mid \Psi \rangle = \frac{i}{h}\left( -\langle \Psi \mid H \Psi \rangle + \langle \Psi \mid H^* \Psi \rangle\right)$$

As long as $H = H^*$, this is zero. The initial normalization is a postulate, as David said.

This answer is pretty much an explicit way of saying what Lubos wrote - I was half way done when he posted, so check there for more detail.

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Thanks, Mark, +1 for the relevant math supplement. ;-) –  Luboš Motl Apr 2 '11 at 7:47
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Lets look at simple probability theory.

Suppose you toss a coin $N$ times. And you get heads $H$ times and tails $T$ times. When you don't get heads you get tails.

This means $T = N - H$.

Now, probability of getting heads $P_h = \frac{H}{N}$ (by definition)

and probability of getting tails $P_t = \frac{T}{N} = \frac{N-H}{N}$

$P_h + P_t = \frac{H}{N} + \frac{N-H}{N} = \frac{H + N - H}{N} = \frac{N}{N} = 1$

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An outcome is anything that can happen. You might not want to fail to detect a particle, but if you do, it's an outcome. It still causes quantum decoherence (and waveform collapse if you believe the Copenhagen interpretation), in that there will not be interference between waves that are there if the particle is detected and waves that are there if it isn't.

Also, it's entirely possible to predict the probability of not detecting the particle when it's there, and accidentally detecting the particle when it isn't. In case you're wondering, the decoherence is essentially from the detector. There will be interference between part of a waveform that missed the detector and part that just didn't manage to set it off.

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@user2898, 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. You 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 5:17
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What operationally is "no outcome"? As mentioned above this has nothing to to with quantum mechanics, since every quantum experiment must have a classical reading (Bohr's dictum). Perhaps this is the part of which you are unaware.

I have thought about it myself since yesterday. One definition of the absence of outcome is "no stability" or reproducibility. Something like, you think you see blue. You blink or wish or something and the outcome changes, you see green. Blink again and it changes again. Another possibility might be that the experimental apparatus is destroyed.

Also, merely apparent measurements of unphysical quantities may have no outcome, like trying to measure our absolute x,y,z coordinates in the universe, but even this is not operationally so well defined. Some purely statistical experiments like telepathy appear to have no outcome, as they change on reinterpretation.

Do you have other operational possibilities in mind for "no outcome"? Please focus on the meaning of "operational". This was the giant advance in 20th century physics.

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Perhaps it is best to focus on a position measurement using an array of detectors. To simplify further, just two detectors and a state assigning to them a total probability of 1. Since no detector is perfect, no every one of these measurements will have an outcome. All Bohr would require in this case that the unsuccessful measurements be discarded or ignored, if we want to compare the predicted probabilities with the measured ones. –  Koantum Apr 4 '11 at 1:42
    
Do this first with ants, in a Y shaped maze with cameras as detectors. There is the possibility that the ant a) "gets reflected", i.e. turns around, or b) "gets absorbed" in the apparatus, i.e. stops or gets eaten by a spider. To you both of these are "no outcome" results. Similarly, for atoms in a Stern-Gerlach apparatus, there can be poor beam focusing, or poor vacuum, etc. However, to me, in both these circumstances, there was indeed an outcome, namely that neither detector registers an ant or an atom (in the delta t assumed). Another outcome is that both detectors could register. –  sigoldberg1 Apr 4 '11 at 2:30
    
Sorry, ran out of space. My basic criticism is that there is way too much theory in your definition of the "outcome" of an experiment. As an experimentalist, to me, the outcome is whatever comes out, which can more or less be anything. Your question has to do more with the rescalings of probabilities which occur all the time in real experiments, as you indicated. So to me you want to know whether there could be theories which always predict missing probability of detection, like missing energy or spin before neutrinos were discovered? But wouldn't we just call that particle decay? –  sigoldberg1 Apr 4 '11 at 2:43
    
the outcomes isn't whatever "comes out." It's what is represented by a projector. My question is not: does QM "always predict missing probability of detection" but: can QM predict with certainty that a measurement will have one of its possible outcomes (which are represented by orthogonal projectors on an HS)? –  Koantum Apr 5 '11 at 1:56
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