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I Quantum superposition and its physical interpretation

  1. May 26, 2016 #1
    I understand that if we have a quantum mechanical system, then its state at some given time ##t## is fully described by a state vector ##\lvert\psi(t)\rangle## in a corresponding Hilbert space. This state vector containing all possible information about the distributions (of all possible values) of the observables of the system. When we talk of quantum superposition this arises by projecting the state onto an eigenbasis of one of its observables, and hence it is in a superposition of all possible eigenstates of this given observable. I get that superposition follows mathematically from the linearity of the Schrödinger equation, but what is the physical interpretation of it? Is it simply that before making any measurement the system is not in a well-defined eigenstate of the given observable that we are measuring and it is only by making an observation that the quantum state of the system collapses into a particular eigenstate of the observable measured in the experiment? (Clearly the quantum system itself is in some well-defined state ##\lvert\psi\rangle## before measurement, but is the point that unless the state is prepared to be in a given eigenstate at the start of the experiment then there is no way, even in principle, a priori to predict with certainty that it is in a given eigenstate and in general it will be in a superposition of all possible eigenstates available to it, "pointing" in some definite "direction" in Hilbert space.)
     
    Last edited: May 26, 2016
  2. jcsd
  3. May 26, 2016 #2
    "Clearly the quantum system itself is in some well-defined state before measurement".

    Why would you think so? Superposition is real, the clearest experiment I can think of proving that superposition is real is the Quantum Zeno experiment performed by Itano et al in 1990, also dubbed "the quantum pot which never boils if it is repeatedly observed". I can not think of any explanation to it other than invoking quantum superposition.

    The details are a bit long so I will not post them right now, if you have trouble checking about it just let me know and I will post, but it's going to be a bit long...
     
  4. May 27, 2016 #3
    I meant this in the abstract sense, that the system is represented by some state vector in a Hilbert space.
    Quantum superposition is never directly observable, although one can observe the effects of it in terms of interference patterns (à la double slit experiment).
     
  5. May 27, 2016 #4

    blue_leaf77

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    When you have a superposition state ##|\psi\rangle## before a measurement, it does not mean that the system is in one of the superposing states before any measurement is done. Before any measurement, the system is in that exact state ##|\psi\rangle##. Superposition tells you about the probability distribution of observables, not the true state of a system before we make any contact with it. If the initial state before measurement becomes a question due to it having possibility to be in some possible states, then the system is said to be in a mixed state.
     
  6. May 27, 2016 #5
    This is my point. Is the idea of superposition not that the system isn't in some well-defined state ##\lvert\psi\rangle## before measurement, but that its observables do not have well-defined values before they are measured, i.e. they are given by probability distributions?! As such, superposition only arises when we expand our state ##\lvert\psi\rangle## in the eigenbasis of a particular observable - since the observable doesn't have a well-defined value, but only a statistical distribution of "allowed" eigenvalues of its associated operator, the state ##\lvert\psi\rangle## will be in a superposition of the eigenstates of the operator. It is only once a measurement is made that the state de-coheres and collapses into a particular eigenstate of the operator (associated with the observable).
     
  7. May 27, 2016 #6

    blue_leaf77

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    That's correct, if ##|\psi\rangle## is a superposition state, then before measurement the system is in a well defined state ##|\psi\rangle##. But the values of observables is not well defined.
     
  8. May 27, 2016 #7
    Great, I think I have a clearer picture in my head now about the intuitive meaning of superposition.

    Relating this to the Schrödinger's cat paradox, is the resolution to the paradox that the system is in a superposition state before any measurement is made (the observable (dead or alive) is not well-defined at this point), but that this measurement doesn't have to be made by opening the box; by simply interacting with the environment the state of the "cat-box" system becomes entangled with the environment state and de-coheres into either a "dead-state" or an "alive-state", and hence there is no paradox about the cat being in some kind of permanent superposition of dead-alive?
     
  9. May 27, 2016 #8

    Nugatory

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    Yes. A pure state may be written as a superposition in one basis yet not be a superposition in another basis. For example, the particles in the upper beam of a vertically oriented Stern-Gerlach device have been prepared in the unsuperimposed state ##|V\rangle## - but in a different basis that state would be written as ##\frac{\sqrt{2}}{2}(|L\rangle+|R\rangle)## which is a superposition. Which basis I choose depends on which is more convenient for the problem at hand.
     
  10. May 27, 2016 #9
    If you are looking for an intuitive understanding of superposition you could always consider the vibration of some sting (like in a guitar) which can be represented as a sum of different weighted harmonics of the string. This could then be described as the vibration is a superposition of harmonic oscillations. Of course this does not transfer directly to quantum mechanics as you dont have collapse of the system when it is observed, but the same principle holds.
     
  11. May 27, 2016 #10

    bhobba

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    Its not that - its that states form a vector space. Or rather while what you said is true, these days Schrodinger's equation is developed from symmetries in an already assumed Hilbert space.

    This relates to your other query about why we have Hilbert spaces. That's a very deep question we don't quite have an answer to yet - but we are very very close - almost there.

    Thanks
    Bill
     
    Last edited: May 27, 2016
  12. May 27, 2016 #11
    Ah ok. Would the rest of the stuff I wrote about superposition be correct at all?
     
  13. May 27, 2016 #12

    bhobba

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    It looks all right.

    But really its in the math - not semantics.

    Thanks
    Bill
     
  14. May 27, 2016 #13
    Yes, sure. I like to try and develop an intuitive picture to go along with the mathematics if possible :smile:
     
  15. May 27, 2016 #14
    What about what I put in post #7 about the Schrödinger's cat paradox? Would what I said be correct all?
     
  16. May 27, 2016 #15

    bhobba

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    I would say no.

    There are a number of ways of resolving Schrodingers cat.

    My way is to simply note that from decoherence macro objects like cats etc always have a definite position. This means a cat can never be in a superposition of alive and dead - the two states have constituent parts with different positions ie the alive cat has a beating heart - the dead cat doesn't. The observation occurs at the particle detector - everything is common sense classical from that point.

    Thanks
    Bill
     
  17. May 27, 2016 #16

    bhobba

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    In my experience with QM that's not always possible - sometimes - yes - but mostly no. That's why popularizations resort to half truths and even on occasion downright lies.

    Thanks
    Bill
     
  18. May 28, 2016 #17

    vanhees71

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    One should also stress that a system doesn't need to be in a pure state, if it isn't prepared in such a state. This is due to entanglement.

    E.g., take a quantum system consisting of two particles which may fly appart from each other originating from a decay of some unstable other particle. Suppose these are indistinguishable particles (e.g., fermions). Then they may be in an antisymmetrized product state,
    $$|\Psi \rangle=\frac{1}{\sqrt{2}} (|\psi_1 , \psi_2 \rangle - |\psi_2 ,\psi_1 \rangle).$$
    We assume that ##\langle \psi_i|\psi_j \rangle=\delta_{ij}##.

    Now you forget about one of the particle, looking only at the other. So, what's the state of this one particle now?

    To describe this you need to "forget" about the other particle. This is done by taking a "partial trace" over one of the particles. So let ##|n \rangle## a complete set of single-particle states. The Statistical Operator of the two-particle system is the projector (because it's a pure state)
    $$\hat{\rho}_{12} = |\Psi \rangle \langle \Psi|=\frac{1}{2} (|\psi_1,\psi_2 \rangle \langle \psi_1,\psi_2| -|\psi_1,\psi_2 \rangle \langle \psi_2,\psi_1 | - |\psi_1,\psi_2 \rangle \langle \psi_2,\psi_1 |+ |\psi_2,\psi_1 \rangle \langle \psi_2,\psi_1|).$$
    Now one defines
    $$\mathrm{Tr}_2 (|\psi_1,\psi_2 \rangle \langle \psi_3,\psi_4|)=\sum_n \langle n|\psi_2 \rangle \langle \psi_4 | n \rangle |\psi_1 \rangle \langle \psi_3| =\langle \psi_4|\psi_1 \rangle |\psi_1 \rangle \langle \psi_3|.$$
    Thus we get as the statistical operator for particle 1 alone
    $$\hat{\rho}_1 = \mathrm{Tr}_2 \hat{\rho}_{12} = \frac{1}{2} (|\psi_1 \rangle \langle \psi_1|+|\psi_2 \rangle \langle \psi_2|).$$
    So you know, particle ##1## is either in state ##|\psi_1 \rangle## or in state ##|\psi_2 \rangle##, but it's not in a pure state, and there are not even any correlations, i.e., although the total system (two particles) is in a pure state, looking at one of the particles only, you have prepared a system with maximum entropy given the information that it must be either in state ##|\psi_1 \rangle## or in state ##|\psi_2 \rangle##.
     
  19. May 28, 2016 #18

    naima

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    You think of a peculiar paper?
     
  20. May 28, 2016 #19

    vanhees71

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    Well, there's no answer in physics, why nature behaves as she does. It's just an observation that quantum theory with its formulation on Hilbert space works very well in describing nature. Physics is an empirical science. Any answer to a question, why the one or the other physical theory works well, is no more physics but philosophical speculation, which is of very little if not even no use for physics itself.
     
  21. May 28, 2016 #20
    To follow up on Nugatory's message, I don't understand why there is so much focus on superposition in QM where IMHO it is only a manner of speaking ? It's just a word expressing the idea that a given vector can be decomposed in a predefined basis ( for example ## (|0 \rangle, |1 \rangle ) ## for a two dimensional case).

    Pure quantum states are not vectors they are rays (one dimensional subspaces or equivalently one dimensional projectors or even better functionals on observables). The fact is a set of rays (a projective space) does not have a nice algebraic structure that allow to describe an element algebraically relatively to others whereas a vector space does. So for computational reasons it is sometimes easier to select an arbitrary basis of vectors (considered as states) and to describe other states in term of coordinates (as if it was a linear combination). But there is nothing really physical in that. Nevertheless in a lot of explanation and books we keep reading about the cruxial importance of superposition. Isn't that misplaced and confusing ? (in particular leading to the false conclusion that they are superposed and non-superposed states) Am I missing something ?
     
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