Timeline for Why is addition of observables in quantum mechanics commutative?
Current License: CC BY-SA 2.5
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| Sep 29, 2015 at 19:18 | review | Late answers | |||
| Sep 30, 2015 at 20:05 | |||||
| Jan 3, 2012 at 17:54 | comment | added | Toby Bartels | While Theo's comments are excellent, yours (Jonathan) get to the heart of the difference between addition and multiplication: states are expectation values, expectations values add, they don't multiply. (Obviously there is more to be said.) | |
| Jul 29, 2010 at 8:47 | comment | added | Jonathan L Long | As for a purely physical definition of addition of observables, I don't have one, and neither does Strocchi; he even notes that the introduction of the sum leads to an extension of the physically defined observables. The situation for multiplication seems even worse, so I am not sure how much more physical motivation can be given at the purely algebraic level. | |
| Jul 29, 2010 at 8:41 | comment | added | Jonathan L Long | If you say $\omega(A + B)$ is the expected value of preparing the universe in state $\omega$, then measuring $A$, then measuring $B$ in the state the universe was left in after the first measurement, then summing the results, then indeed $\omega(B + A)$ may be different. Instead, $A + B$ represents a single measurement we can do that results (in expectation) in the sum of $A$ and $B$. That is a more sensible requirement than the corresponding one for multiplication, since even if $A$ and $B$ are somehow linked (as they are in QM), expectation is still linear. | |
| Jul 28, 2010 at 10:23 | comment | added | Andrea Ferretti | Basically, a state is the current situation of the universe, an observable is some physical quantity we have some instrument to measure. They have an obvious pairing: in a given state, apply your instrument and read the result. If you now define sum and multiplication of observables by doing the two measures one after the other, the result may depend on the order. And indeed it does for the multiplication. Why not for the addition? | |
| Jul 28, 2010 at 10:20 | comment | added | Andrea Ferretti | I followed myself the course from Strocchi, and it is exactly his argument I'm trying to fill. Defining a state as a linear functional seems as arbitrary as defining it as a multiplicative linear functional. The question is: how con we derive these properties from the description of the process of measure alone? I have tried to explain the full reasoning in my question. | |
| Jul 28, 2010 at 0:33 | history | answered | Jonathan L Long | CC BY-SA 2.5 |