If a quantum particle can take an unlimited number of paths to get from point A to point B wouldn't a quantum particle never get from point A to point B?

A quantum particle takes every path at the same time to get from A to B? How is that even possible? Can anyone really explain what is going on? And maybe shed some light on the math?

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    $\begingroup$ The wave function explores/traverses all possible paths, but it is a parallel process, not sequential. Thus, you get the various patterns from single and double slits (and 'infinite' slits like a diffraction grating) from single particles. $\endgroup$
    – Jon Custer
    Oct 15, 2015 at 14:36

3 Answers 3


A quantum object is not something that can "take a path". It only has a definite position if its position is being measured.

The reason it is said that the quantum object "takes every possible path at once" is that the probability for finding a particle that was at position $q_i$ at time $t_i$ at position $q_f$ at time $t_f$ is given by the Feynman path integral $$ \langle q_f,t_f \vert q_i,t_i\rangle = \langle q_f \vert \mathrm{e}^{\mathrm{i}H(t_i - t_f)}\vert q_i \rangle = \int_{q(t_i) = q_i}^{q(t_f) = q_f} \mathrm{e}^{\mathrm{i}S[q]}\mathcal{D}q$$ where $S[q] = \int_{t_i}^{t_f} \left(\dot{q}(t)^2 - V(q(t))\right)\mathrm{d}t$ is the action evaluated on the continuous path $q(t)$ and the integral is over all continuous paths with startpoint $q_i$ and endpoint $q_f$, sometimes also called the conditional Wiener space and $\mathcal{D}q$ together with the canonical kinetic term in $S[q]$ is the (conditional) Wiener measure.

Since the integral is over all paths from $q_i$ to $q_f$ and every $\mathrm{e}^{\mathrm{i}S[q]}$ for every path contributes to the value of the integral (that is, we must not neglect any such possible path1), it is said that the Feynman path integral represents the quantum object "talking all possible paths" from $q_i$ to $q_f$.

The actual quantum time evolution by $\mathrm{e}^{\mathrm{i}H(t_i - t_f)}$ does not look like this, it just traces out one path in the quantum space of states which has nothing to do with the classical paths.

1As GlenTheUdderboat points out in a comment, this is not strictly true. Just as a usual one-dimensional integral can neglect individual discrete points, here, too, we may neglect a zero measure set. Nevertheless, the path integral is defined as the integral over all (continuous) paths, and there is no distinguished zero measure set of paths which we might want to take out.

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    $\begingroup$ You say that "every path contributes to the value of the integral (that is, we must not neglect any such possible path)", but is that (mathematically, literally) true? I think there's some infinity of paths that may be ignored. Isn't this (at least usually) a key property of integrals? $\endgroup$
    – Řídící
    Oct 15, 2015 at 19:18
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    $\begingroup$ @GlenTheUdderboat: You're correct, it is not literally true. We may neglect a set of paths of zero Wiener measure. Funnily enough, e.g. the set of differentiable paths is such a zero measure set, so one might say that the quantumly relevant paths are exactly those which don't occur as the solution of a classical (differential) equation of motion. ;) $\endgroup$
    – ACuriousMind
    Oct 15, 2015 at 19:22
  • $\begingroup$ Thanks. I didn't realise that the whole set of differentiable paths is an example of being that "small". But it makes a lot of sense, now that you have stated it. Funny indeed. $\endgroup$
    – Řídící
    Oct 15, 2015 at 19:28
  • $\begingroup$ @ACuriousMind Now I am a bit confused, as I fail to link your answer in comments to Glen with your footnote: in the footnote you say that "there is no distinguished zero measure set of paths..", but in comments you named one, namely "set of differentiable paths ...". Maybe I don't understand what type of measure Wiener represents and this then is causing my confusion. Thanks for any added clarifications. $\endgroup$
    – user929304
    Oct 16, 2015 at 9:46
  • $\begingroup$ @user929304: Well, but it isn't distinguished in the sense that it would be somehow natural to declare that the integral is not to be taken over those paths. Consider the usual one-dimensional Lebesgue integral: The rationals have Lebesgue measure zero, but no one (to my knowledge) thinks it would be natural to think of the integral as only coming from the values of the function at the irrational points. $\endgroup$
    – ACuriousMind
    Oct 16, 2015 at 14:03

Let me try a low-tech explanation, expanding on @JonCuster comment.

Think about waves. If you drop a pebble in water, a wave spreads out in a circle. If there is a short wall in the water, the wave will curve around the wall and go the other way.

If you simultaneously drop two pebbles a small distance apart, each starts its own wave in a circle. These waves interfere, so in some places they reinforce, and in other places they cancel, but you could say in all places both waves are there, and they are just combining.

Suppose you drop 100 pebbles in a line 1 inch apart. They all start their own waves, but what do you see? Out to the sides, the waves pretty much always cancel, and perpendicular to the line of pebbles the waves reinforce along a line that moves away. At intermediate angles, they partially reinforce. If you drop 200 pebbles 1/2 inch apart, you'll see that the effect is stronger - the waves at intermediate angles are weaker.

Light is the same way, when you consider it as a wave. When you shine a laser, you can think of it as a very large number of pebbles all in a line being dropped together, so the wave is very coherent in the forward direction. It still spreads out a bit at the edges.

Well, the power in that wave at a given place is nothing more the probability of seeing a number of photons in that place, over a given time.

So to try to answer your question, waves take all possible paths, but waves interfere so you only see photons in places where the waves interfere constructively.

You might be tempted to say the photon took all possible paths, but that's not right. The waves take all possible paths, just as they do with pebbles in water.

  • $\begingroup$ But we are talking about one quantum particle taking all paths at once from A to B. Really don't see your point for the waves as you said were generated by laser and a number of photons. $\endgroup$
    – StarDrop9
    Oct 15, 2015 at 22:10
  • $\begingroup$ @StarDrop9: Take 1 photon. Its quantum "reality" is that it is a probability wave, exactly equivalent to its electromagnetic wave. (Other particles also "are" waves - just not electromagnetic.)The energy in that wave at a place and time is the probability that the photon can be observed at that place and time. That wave spreads everywhere it can go, but it must interfere, and where its energy is smallest, the photon can basically not be seen. $\endgroup$ Oct 15, 2015 at 22:26
  • $\begingroup$ Are you saying the photons wave goes out in all spatial dimensions at once but directionally? And have you ever seen a simulation of this wave action or an approximation? Still very fuzzy. $\endgroup$
    – StarDrop9
    Oct 16, 2015 at 13:39
  • $\begingroup$ @StarDrop9: Go to YouTube and search for "water wave motion physics". Here's one result that shows it very well. $\endgroup$ Oct 16, 2015 at 13:56
  • $\begingroup$ commons.wikimedia.org/wiki/File%3APath_integral_example.webm $\endgroup$
    – StarDrop9
    Oct 16, 2015 at 14:28

Let's take a simple thought experiment which involves no quantum effects and is quite easy to visualise.

Consider a table on which I mark two points (on its surface); and I ask you to draw the straight line between these two points.

This is easy to do, you merely take a piece of chalk and by eye draw the straight line between them; a moments work.

But there is an alternative method: you draw every possible line connecting the two points, measure it's length, and then chose the shortest one. This is a lot more work, but is exactly equivalent.

This is a cautionary tale in that one must be judicious in interpreting mathematics in a physical situation.

For example, in your situation there is an alternative interpretation: the de Broglie-Bohmian perspective where a pilot wave is posited, and it is it's motion that 'goes everywhere', so to speak, that is described by the description you've given and it acts causally on the particle to guide it - which then follows a classical path.

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    $\begingroup$ This is wrong. In the Feynman integral the mathematical measure of the stationary (e.g., minimal; classical) path is exactly $0$. It is the close neighbourhood of such path (i.e. the 'sum' of paths close to it) that has significant positive (if not close to $1$) measure. (Because $\hbar \neq 0$.) $\endgroup$
    – Řídící
    Oct 15, 2015 at 18:23
  • $\begingroup$ @glen the udderboat: true - and what has that got to do with what I've written? $\endgroup$ Oct 15, 2015 at 18:28
  • $\begingroup$ In your answer you refer to the classical path, whereas you could have referred to the immediate 'neighbourhood' of such path. The classical path itself, in the integral, has zero measure. This makes the "exactly equivalent" statement suspicious (educationally). Methinks. I mean: classical physics is false. $\endgroup$
    – Řídící
    Oct 15, 2015 at 18:33
  • $\begingroup$ @glen the udderboat: Its the pilot wave that probes the 'local neighbourhood'. $\endgroup$ Oct 15, 2015 at 18:39
  • $\begingroup$ I don't know enough about it, so I cannot comment, except that pilot waves are a non-mainstream interpretation. $\endgroup$
    – Řídící
    Oct 15, 2015 at 18:42

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