Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is meant actually by superposition?

  1. Jun 14, 2015 #1

    ftr

    User Avatar

    I am confused by the my readings on the subject. Does superposition says that the system has all the values at the same time or it has one of the possible values but we don't know which one until "measured". And is there different answers for different observables. Moreover, how do these systems know how to interact without carrying specific values.

    Also in the calculation the whole of the wavefunction is used, doesn't that tell us that all these values are real, ie. the electron does exist in all those positions as in the hydrogen, which means there is no such single electron in that case.
     
    Last edited: Jun 14, 2015
  2. jcsd
  3. Jun 14, 2015 #2

    jtbell

    User Avatar

    Staff: Mentor

    It says that the system potentially can have any of those values when we measure it.

    As for what is "really happening" before we measure it, QM does not address that question. This is the subject of interpretations of QM, of which there are some that give different answers. However, they are constructed so they all make the same predictions for actual experiments, so there's no way to distinguish between them by experiment.
     
  4. Jun 14, 2015 #3

    bhobba

    User Avatar
    Science Advisor
    Gold Member

    Superposition simply expresses the vector space structure of quantum states. That's all there is to it as far as the formalism goes - interpretations however often have a different take

    Thanks
    Bill
     
  5. Jun 14, 2015 #4

    radium

    User Avatar
    Science Advisor
    Education Advisor

    Superposition must refer to a specified basis. For example, a state that is in a spin z eigenstate is in a superposition of spin up and spin down if it is written in the sx or sy basis.
    If operators can be simultaneously diagonalized in the same basis, you can have a state which is an eigenstate of both operators. However, if you cannot do this, (the operators don't commute), an eigenstate of one operator will not be an eigenstate of the other. If a system is in a position eigenstate for example, it is most definitely in a superposition of momentum eigenstates.

    We don't think of the state of the hydrogen atom as a superposition of energy eigenstates but instead as being in a superposition of position eigenstates since the electron has no specified position. This is because we really never see the hydrogen atom in superposition of energy eigenstates. This is because of the way it is coupled to the environment reflected in the hamiltonian. Decoherence is the process in which a quantum system starts to behave classically over time because of interaction with the environment. This selects eigenstates which can effectively describe the system classically since decoherence makes a superposition decay exponentially to a chosen eigenstate at a decoherence time scale.
     
  6. Jun 15, 2015 #5

    Strilanc

    User Avatar
    Science Advisor

    What it actually means is that a valid quantum state can be a weighted sum of classical states, that you get the right predictions if you do the math with those weighted sums, and you get wrong predictions if you try to avoid them.

    For example, if you apply a Hadamard operation to a qubit in the OFF state, then the qubit is now in the (OFF + ON)/sqrt(2) state. If you measure the qubit, you'll get 50% ON results and 50% OFF results. If you apply the Hadamard operation again (without measuring in between), the qubit goes back to the OFF state and you get 100% OFF results when measuring. But if the second operation you applied was a Z gate instead, the qubit would be in the (OFF - ON)/sqrt(2) state, which has a different relative phase but still 50/50 ON/OFF results when measuring, and when you applied the final Hadamard operation it would end up in the ON state instead of the OFF state.

    Try coming up with ways for an operation to randomize the values when applied once, yet undo itself when applied twice. It's hard to do without using a weighted sum. And as you add in all the various other things you can do with qubits, like HZH = X and HXH = Z and XYZ = iI and quantum teleportation and superdense coding and bell tests and the no cloning theorem and the no communication theorem and..., well, you start to realize you need the superposition.
     
  7. Jun 15, 2015 #6

    ftr

    User Avatar

    Thanks for all the responses. However, it seems my real question is not getting across. Specifically, we think of the electron as a particle( even though its "behavior" might be strange), when in hydrogen atom its position is a "cloud". So, when not measured either have a particle with undefined position or an entity compromising all these positions. It seems to be a no brainer as the wavefunction is indicating the latter. Now, if we insist that it is a particle that is in "superposition" then that sound ultimate contradiction,i.e. how can something unique carry multiple values, that sound very illogical and un intuitiveness does not seem to have anything to do with it.

    So my question is why this view has not been considered, in all the interpretations the particle view before measurement seem to be a must , WHY.
     
  8. Jun 15, 2015 #7

    radium

    User Avatar
    Science Advisor
    Education Advisor

    Well for one, the electron in a hydrogen atom is not a free electron, it is in a bound state with the proton. The electron itself is not in a superposition, it's just not localized. This is the uncertainty principle. If you considered energy or angular momentum eigenstates, the electron is most definitely not in a superposition. But you can't have the electron in an eigenstate of both since L and p don't commute.

    In quantum field theory you would think of the electron as a fermionic field not as a point like particle. An election is not defined to be a point like particle, the formal definition is that Each type of particle corresponds to an irreducible representation of the Poincare group classified by it's four momentum and spin.
     
  9. Jun 15, 2015 #8

    Nugatory

    User Avatar

    Staff: Mentor

    Yes, but do remember that we're using the word "particle" in the sense of a quantum particle, not in the colloquial English sense of something like a very small grain of sand or tiny lump of matter that you can say is here or there or somewhere even if we don't know where. A quantum "particle" is something that behaves in certain ways, and one of its more important properties is that it is has no position before it is measured - not "it is somewhere but we don't know exactly where", not "it's all fuzzy and spread out in a cloud" but rather that it makes no more sense to talk about the particle's position than it does to talk about the position of an idea. Thus....
    Yes, and it's the first of these. We have an entity that for historical reasons is called a "particle", its position is undefined, and it will acquire a position when we measure it in the same way that I will acquire a lap when I sit down, or that I will acquire a fist when I curl my fingers back into my hand.

    The wave function is indicating the former. The wave function says nothing about the position of the particle before measurement, it just tells us what the results of a measurement might be if we do measure it. Now, it is natural to think that that has to imply something about the position before we measured, but it doesn't - at the risk of sending you off on a long and confusing digression I'm going to suggest that you google for "counterfactual definiteness" (the idea that something not measured must still have a value) and "Einstein EPR paradox" (a classic paper from 1935 in which Einstein and two colleagues developed the idea that there had to be something there before the measurement) and "Bell's theorem" (the 1965 discovery, followed by several decades of experimental confirmation) that this view is not how the world works.

    It does sound like a contradiction, and it is unquestionably counterintuitive. However, that's because your intuition is trained to deal with classical objects, and it doesn't want to accept the logical but very different rules that apply to quantum particles.
     
  10. Jun 15, 2015 #9

    Nugatory

    User Avatar

    Staff: Mentor

    It's not in a superposition of energy, ##L^2##, or ##L_z##, but it certainly is in a superposition of position eigenstates.
     
  11. Jun 16, 2015 #10

    ftr

    User Avatar

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: What is meant actually by superposition?
  1. What is superposition? (Replies: 1)

Loading...