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I am a bit confused as to what the superposition principle actually means. Does it merely mean that you can express any given state as the linear combination of two vectors? If so, isn't that kind of obvious. I mean, mathematically, you could express any given vector in nature as a combination of two other vectors. Or does it additionally mean that a quantum state before measurement, for all intents and purposes, exist in every possible state? For example, does it mean that before detection, a photon has all possible polarizations?

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  • $\begingroup$ That thing about having all possible states is certainly not correct and not what anyone means when that mention the superposition principle. $\endgroup$ – DanielSank Jul 19 '16 at 22:55
  • $\begingroup$ Why would it be obvious that a natural system obeys such a rule? $\endgroup$ – CuriousOne Jul 19 '16 at 23:57
  • $\begingroup$ It's obvious that vectors add. It's not so obvious that quantum states behave like vectors. But a theory based on a description of states as vectors works extremely well, so we conclude that quantum states do behave like vectors, and the same superposition principle applies. $\endgroup$ – garyp Jul 20 '16 at 3:06
  • $\begingroup$ Quantum states are described by rays in a Hilbert space; rays do not form a vector space. See physics.stackexchange.com/a/157028/109928. $\endgroup$ – Stéphane Rollandin Jul 20 '16 at 10:26
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It depends a bit on the context. The superposition principle in its general form states that for a linear system the sum of two possible solutions is also a solution -- or, to quote Wikipedia:

"the net response at a given point in space and time due to two or more stimuli is the sum of the responses that would have been caused by each stimulus individually

In the case of quantum mechanics, the solution of the Schroedinger equation allows for various states to exist. Superposition says that the actual state of a system, at a given moment in time, can always be expressed as a linear sum (superposition) of those states. But from QM we also know that when the system is observed, it will in fact be in just one of these states.

This doesn't mean that a photon, before detection, has "all possible polarization states" - if you know how the photon was generated, you may know its polarization. If that is the case, then you can calculate the probability of the photon passing a polarizer - with zero probability if the polarizer is at right angles to the initial polarization. Which implies that, just before detection, not all polarization states were available to that particular photon.

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  • $\begingroup$ is there any zero probability with a polarizer ? experimentaly not. See constructors curves $\endgroup$ – user46925 Jul 20 '16 at 0:22
  • $\begingroup$ @igael - A "real" polarizer is not ideal; but for an ideal polarizer at right angles to the known angle of polarization, the transmission ought to be zero. Of course you can argue that you never knew the initial state of polarization perfectly to begin with... $\endgroup$ – Floris Jul 20 '16 at 2:27
  • $\begingroup$ @floris - So are you saying that even before measurement, a particle is in a definite state (i.e. it definitely has a certain polarization), but that that state can be described as a superposition of different orthogonal states? $\endgroup$ – Clement Decker Jul 20 '16 at 16:45
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It is of course a trivial observation that the sum of vectors is a vector and that every vector can be expressed (in a great many different ways) as a sum of other vectors.

But in QM, you're often interested not just in the state of a system, but the way that state evolves over time. So you're looking at a function from the real numbers (representing time) to your vector space (or more precisely to the projectivization of that vector space). Only certain such functions are allowed, namely those that satisfy the Schrodinger equation. It is somewhat less obvious that the sum of two allowable functions is an allowable function. In many contexts, that somewhat less obvious statment is what's meant by the "superposition principle".

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In a simple way, it means a point (x,y) in x-y plane has both x component and y component.

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  • $\begingroup$ No. You should rephrase this answer or delete it. Components or coordinates do not explain superposition. See thee other answers. $\endgroup$ – Aziraphale Jul 22 '16 at 10:31

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