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You can make an observation of a system whose state is undetermined. By interrogating the system for its state, a state becomes determined. Suppose, for example

You can make an observation of a system whose state is undetermined. By interrogating the system for its state, a state becomes determined. Suppose, for example, that you have a flipped coin and sent it rotating in space, never hitting the floor. Is it heads or is it tails? Until it is looked at, the question doesn't really make sense. And for this case, we'll define "looking at it" to mean sticking out your hand and catching it. There is a probability that it's landed heads up in your hand, and a complimentary probability that it's landed tails up. Once it's in your hand though, you can confidently say which state it's in. Now, you could definitely take issue with this example, because you could argue that the rotation of the coin is well described, so with initial conditions, you can predict its position at any given moment. But imagine a microscopic quantum system, and, for the sake of this simple explanation, believe that its "rotating through the air" state really does not have any precise heads or tails definition. Until something gets in the way of that system, creating an interaction that exchanges information about its observable state, it's not meaningful to say that it's in one of the observable states at all. A superpositition of states, as such, is essentially the representation of a state in terms of a basis set of observables. In the case of the coin, heads and tails are the two observable states, they are orthogonal, and they fully represent the state space of the coin. You could flip the coin, and put its state vector into the form of sqrt(2)/2 * Heads + sqrt(2)/2 * Tails. This state isn't observable, but it can be described in terms of observable components, where the coefficients represent the probabilities that a given observable state will be measured upon observation.

pdonis 11y ago
> You can make an observation of a system whose state is undetermined. By interrogating the system for its state, a state becomes determined. For QM, this is not correct, although it's a common misstatement. The correct statement is this: you can make an observation of a system which is not in an eigenstate of the measurement operator you are using. After the measurement, the system is now in an eigenstate of the measurement operator--i.e., the act of measurement changes the state. Note that this is only true on a collapse interpretation, like Copenhagen. On a no-collapse interpretation, like MWI, the "observation" is just an interaction that entangles the state of the measuring device with the state of the system being measured--it's all just unitary evolution. > You could flip the coin, and put its state vector into the form of sqrt(2)/2 Heads + sqrt(2)/2 * Tails. This state isn't observable* Yes, it is; but it isn't observable by a simple method like looking to see if the coin is heads or tails. But according to QM, every state is an eigenstate of some operator, so there will be some observation that will distinguish sqrt(2)/2 * Heads + sqrt(2)/2 * Tails from the state that is exactly orthogonal to it, which is sqrt(2)/2 * Heads - sqrt(2)/2 * Tails.
- photonic29 11y ago
  >The correct statement is this: you can make an observation of a system which is not in an eigenstate of the measurement operator you are using. I should have distinguished better, but what you're more rigorously calling an eigenstate of a measurement operator, I'm calling an observable state. There is something lost in translation to an audience unfamiliar with terms like eigenstate, but that was my attempt. Would you suggest a better one? >Note that this is only true on a collapse interpretation, like Copenhagen. On a no-collapse interpretation, like MWI, the "observation" is just an interaction that entangles the state of the measuring device with the state of the system being measured The greater point being addressed is that MWI is no more deterministic than Copenhagen. >Yes, it is; but it isn't observable by a simple method like looking to see if the coin is heads or tails. But according to QM, every state is an eigenstate of some operator Some hermitian operator? But more to the point, if looking at the coin is the only operator at our disposal in the simple example, then its eigenstates are the ones we care about.
  - pdonis 11y ago
    > what you're more rigorously calling an eigenstate of a measurement operator, I'm calling an observable state. Yes, but "observable" here is relative to the measurement you are making. If you make a different measurement (i.e., realize a different operator), then the set of "observable states" by your definition is different, because the set of eigenstates of the operator is different. > The greater point being addressed is that MWI is no more deterministic than Copenhagen. But this isn't true. The MWI is completely deterministic, because wave function collapse never occurs, and wave function collapse is the source of all the indeterminism in the Copenhagen interpretation. > Some hermitian operator? Yes. > if looking at the coin is the only operator at our disposal in the simple example, then its eigenstates are the ones we care about. If all you're interested in is that particular experiment, yes. But here we're discussing claims that must apply to all possible experiments and all possible measurements, not just the particular one in the example you chose. So we have to consider all possible operators and all possible sets of eigenstates, not just the ones in your example.
    - 11y ago
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pdonis 11y ago

> You can make an observation of a system whose state is undetermined. By interrogating the system for its state, a state becomes determined. For QM, this is not correct, although it's a common misstatement. The correct statement is this: you can make an observation of a system which is not in an eigenstate of the measurement operator you are using. After the measurement, the system is now in an eigenstate of the measurement operator--i.e., the act of measurement changes the state. Note that this is only true on a collapse interpretation, like Copenhagen. On a no-collapse interpretation, like MWI, the "observation" is just an interaction that entangles the state of the measuring device with the state of the system being measured--it's all just unitary evolution. > You could flip the coin, and put its state vector into the form of sqrt(2)/2 Heads + sqrt(2)/2 * Tails. This state isn't observable* Yes, it is; but it isn't observable by a simple method like looking to see if the coin is heads or tails. But according to QM, every state is an eigenstate of some operator, so there will be some observation that will distinguish sqrt(2)/2 * Heads + sqrt(2)/2 * Tails from the state that is exactly orthogonal to it, which is sqrt(2)/2 * Heads - sqrt(2)/2 * Tails.

photonic29 11y ago
>The correct statement is this: you can make an observation of a system which is not in an eigenstate of the measurement operator you are using. I should have distinguished better, but what you're more rigorously calling an eigenstate of a measurement operator, I'm calling an observable state. There is something lost in translation to an audience unfamiliar with terms like eigenstate, but that was my attempt. Would you suggest a better one? >Note that this is only true on a collapse interpretation, like Copenhagen. On a no-collapse interpretation, like MWI, the "observation" is just an interaction that entangles the state of the measuring device with the state of the system being measured The greater point being addressed is that MWI is no more deterministic than Copenhagen. >Yes, it is; but it isn't observable by a simple method like looking to see if the coin is heads or tails. But according to QM, every state is an eigenstate of some operator Some hermitian operator? But more to the point, if looking at the coin is the only operator at our disposal in the simple example, then its eigenstates are the ones we care about.
- pdonis 11y ago
  > what you're more rigorously calling an eigenstate of a measurement operator, I'm calling an observable state. Yes, but "observable" here is relative to the measurement you are making. If you make a different measurement (i.e., realize a different operator), then the set of "observable states" by your definition is different, because the set of eigenstates of the operator is different. > The greater point being addressed is that MWI is no more deterministic than Copenhagen. But this isn't true. The MWI is completely deterministic, because wave function collapse never occurs, and wave function collapse is the source of all the indeterminism in the Copenhagen interpretation. > Some hermitian operator? Yes. > if looking at the coin is the only operator at our disposal in the simple example, then its eigenstates are the ones we care about. If all you're interested in is that particular experiment, yes. But here we're discussing claims that must apply to all possible experiments and all possible measurements, not just the particular one in the example you chose. So we have to consider all possible operators and all possible sets of eigenstates, not just the ones in your example.
  - 11y ago

deleted 11y ago

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