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B Measuring other observable

  1. Oct 24, 2016 #1
    Can we say a piece of marble is in an eigenstate of the observable position? If you try to measure other observable like momentum, the other eigenstate of position is supposed to be erased. So how come we can't cause a marble to vanish by measuring momentum. Would you know other objects (macroscopic) where you can make its position vanish by measuring momentum?
     
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  3. Oct 24, 2016 #2

    mfb

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    Marbles are way too large to observe quantum effects: the planck constant is small.

    But even if your replace the marbles by atoms: you don't make the atoms disappear. A precise momentum just means the position is less precise. If you follow-up with another position measurement it will be somewhere, but you don't know in advance where.
     
  4. Oct 24, 2016 #3
    Would it make sense to measure the energy Hamiltonian of the marble? What would be the effect of this?
     
  5. Oct 25, 2016 #4

    PeterDonis

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    No. First of all, position eigenstates even for single quantum particles aren't physically realizable (because of the uncertainty principle). Second, even if they were, an object like a marble having a definite position (at least to the accuracy of our measurements) is not the same as a quantum object being in a position eigenstate. When we measure the marble's position, we certainly don't measure the exact position of every single atom. We only measure a sort of "average" position of all of them (heuristically, we measure the center of mass position and the marble's radius from its center of mass). This does not pin down a single quantum state for the entire marble; it only restricts it to some subspace of all possible states, the subspace that is consistent with the measurement result I just described.

    Measuring the marble's momentum has the same issues as above. Momentum eigenstates aren't physically realizable even for single quantum particles (again because of the uncertainty principle), and measuring the momentum of an object like a marble doesn't measure the exact momentum of every single atom. All it measures is the momentum of the center of mass (and, heuristically, it restricts the momenta of individual atoms to some reasonable range around the center of mass momentum). Again, this doesn't pin down a single quantum state for the entire marble, just a subspace consistent with the measurement result. But this subspace has plenty of overlap with the subspace that is consistent with the marble as a whole (its center of mass + size) being in a position consistent with a series of position measurement results that lie along the path in space that is consistent with our momentum measurement results. So the marble doesn't vanish when we measure its momentum, and it doesn't become blurry when we measure its position; there is plenty of room for the marble to be in states which are consistent with both sets of measurement results being definite.
     
  6. Oct 25, 2016 #5

    PeterDonis

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    Measuring its momentum is equivalent to measuring its energy (assuming we have already measured its rest mass). But even if we use some different method to measure the marble's energy, all the things I said in my previous post apply.
     
  7. Oct 25, 2016 #6
    Is the marble energy related to the temperature? Can you change the Hamiltonian (or energy) of the marble without changing the temperature? But momentum and energy are separate observables.. how can energy (is this potential energy) and momentum be one?
     
  8. Oct 25, 2016 #7

    PeterDonis

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    Yes. More precisely, the expectation value of the energy operator (the Hamiltonian--see below) is related to the temperature.

    You are conflating two very different things here. The Hamiltonian is not the "energy" we measure; it is the operator that represents, in the math, the process of measuring energy. (All measurement processes are represented by operators.) The "energy of the marble" is the result of the measurement; it's just a number, not an operator.

    Yes, but that doesn't mean they are unrelated. The momentum and energy operators are distinct, but they have the same eigenstates--i.e., a state with a definite momentum is also a state with a definite energy, and vice versa.

    However, a large object like a marble is not in an eigenstate of energy any more than it is in an eigenstate of momentum, nor can you "measure the energy" of the marble by measuring the state of every single atom, for the reasons I gave in post #4. Even trying to define a Hamiltonian operator at all for an object like a marble is problematic; nobody has ever written one down. The Hamiltonians you see in textbooks are for much, much, much simpler systems.

    One (heuristic) way of seeing that no object you will ever observe can be in an eigenstate of energy or momentum is to consider that an object which is in such an eigenstate can never change: nothing can ever happen to it. The reason is that the Hamiltonian operator is also the operator that describes "time evolution", i.e., the way things change with time. Being in an eigenstate of that operator means not changing at all with time. But no real objects are like that. Even a marble which is just sitting there on a table is changing with time; air molecules are bouncing off of it, dust particles are adhering to it, etc., etc.

    The "energy" whose measurement process is represented by the Hamiltonian includes potential energy, but it also includes kinetic energy (and, if we are being relativistic, rest energy).
     
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