# Quantum mechanics: probability of being at a location, or probability that we observe it there?

It is said that an unobserved electron is at no particular place but has a probability of being at various places.

Is this probability a function of the electron or of our observation?

That is, is there a certain probability of the electron being at x, or is there a certain probability of our making a measurement at x?

Pardon me if this is a naive question.

• what's the difference? – AccidentalFourierTransform Nov 8 at 22:57
• @AccidentalFourierTransform - I suppose this is what I'm asking. – PeterJ Nov 9 at 13:32
• @PeterJ The question seems to list three options: (1) the probability that a measurement of the electron's position will result in the outcome $x$, (2) the probability that the electron is at $x$ even if we don't measure its position, (3) the probability that we decide to measure its position. QM gives (1). Various theorems demonstrate that (1) does not imply (2), and learning why it doesn't is an important part of learning QM. Regarding (3), trying to use QM to anticipate a person's decisions would be impractical, to say the least. That's not how we use QM. We use QM to get (1). – Chiral Anomaly Nov 9 at 13:58
• @ChiralAnomaly - Thanks for a very clear comment. But does it answer the question? To me it seems not. – PeterJ Nov 9 at 14:06
• @PeterJ I don't know if it answers the question, because I don't know what the question is trying to ask. That's why I wrote a comment instead of an answer. It was an implicit request for clarification. – Chiral Anomaly Nov 9 at 15:19

It is not easy to directly answer, because you are implicitly assuming that the electron is a classical object. So my answer will be necessarily a bit articulate.

The crucial difference between classical and quantum objects is that for a classical object all its properties always have definite values. This requirement is usually called realism. When instead dealing with quantum objects, their properties may have undefined values. These values are however fixed when (after) one measures them (through an interaction between the quantum object and the measurement apparatus).

Quantum theory provides the probability to fix each value of each property after a measurement of it (if I decide to measure it).

A quantum state is nothing but that assignment of all those probabilities for every property and every value of it (this is the physical content of the famous Gleason theorem). When a property has a definite value, its probability is $$1$$.

There are pairs of quantum properties which are mutually incompatible: they cannot have definite values simultaneously and this explains why there are properties which are not defined in some state. They cannot have definite values because other, incompatible, properties have definite values since I have just measured them. Notice that therefore measurements change the state of the system.

In summary, the position of the electron can be undefined: the electron has literally no position (!). However Quantum Theory gives us the probability that if I measure the position it results to belong to a given interval. If I perform such a measurement, the position localizes, but the state (the probabilities of the outcomes of other measurments) changes and other properties, which were defined before the measurement, become undefined.

One could be tempted to assume some hidden form of realism: the electron always has a position and every property is actually always defined, but, for some reason, we do not know it and we know it only up to a probabilistic estimate (exactly as, e.g., in classical statistical mechanics).

As a consequence of several theoretical investigations (with also several experimental confirmations) generally known as the Kochen-Specker theorem (actually based on previous results by Gleason and Bell), this viewpoint is untenable, unless assuming some weird (strongly non classic) behavior of the "classical" properties called contextuality.

• QM absolutely does not imply that "measurement changes the state of the system". This is a misleading explanation that does years if not decades of damage to someone's understanding of QM and promotes all sorts of ridiculous "quantum magical thinking", like the sort of questions we see all the time about using entanglement to transmit information. – R.. Nov 8 at 23:58
• @R.. If you don't like the wording "measurements change the state of the system," can you suggest a better wording that is still consistent with the Kochen-Specker theorem (and with the fact that the CHSH bound is violated, and so on)? Those are important results that are difficult to convey in colloquial terms. Regardless of how it is worded, if this answer inspires a person to go learn about results like the Kochen-Specker theorem, then it has served a good purpose. A person who reads this answer and still doesn't bother to go learn about those results has missed the point of the answer. – Chiral Anomaly Nov 9 at 3:44
• Thanks for this answer. I'm not sure it quite addresses the question, but this may be my denseness. . . – PeterJ Nov 9 at 13:31
• @ChiralAnomaly: The non-misleading way to state it that's accessible to non-experts is that there is a high degree of complex structure on correlations between observations. This is true regardless of interpretation and does not suggest that the theory predicts causal state-change relationships that don't actually exist. – R.. Nov 9 at 16:32

That is, is there a certain probability of the electron being at x, or is there a certain probability of our making a measurement at x?

Both? It certainly depends on the measurement, but your latter statement makes it seem like you are saying QM predicts where I put my detector.

I would say QM predicts the probability of detecting a particle at a certain location (more precisely, within a certain region of space). If we detect the particle to be at some location, we then we can say we "found it there" I suppose. If we don't detect it at some location, we have still "made a measurement", but it's just that we don't observe the particle to be there. But you are correct in saying that before this there is no definite position of the particle.

• This answer may be short, but I don't think there is anything wrong with it? – Aaron Stevens Nov 8 at 19:05
• Nor me. The answer 'both' seems to nail it as I can't see how to decide which it is, But on reflection I see my question may be rather trivial for just this reason. – PeterJ Nov 9 at 13:30