kleinwolf
Nov17-05, 11:58 PM
From the Copenhagen Interpretation we learned that the measurement disturbs the syste m in a "fuzzy" way : System=X..Measurement op.=A...End state=Y Y=Eig(A), prob(Y)=|Proj(X,Y)|^2.
Hence after the measurement, the system is Y...
But in my mind, the opposite way is even clearer : The system disturbs the measurement operator...It is even more indicating what a measurement process is : e.g. you put a thermometer, and the thermometer state (operator configuration some kind of, still unknown in QM)..changes...
To recover usual spin QM, let's see this phenomenon in "classical" QM :
the correlation of two observation in A and B is given by the defintion :
C(A,B)=<A\otimes B>-<A\otimes\mathbb{I}><\mathbb{I}\otimes B>
Then it is clear that the non-averaged operator (we remember : <A>=<\Psi|A|\Psi> is then by omitting the \textrm{\emph{exterior}} average :
K(A,B)=A\otimes B-(A\otimes\mathbb{I})\underbrace{|\Psi\rangle\langl e\Psi|}_ {nlin-link}(\mathbb{I}\otimes B)
It's clear that this operator K contains a linear term and a non-linear one. THis means that the measurement operator depends on the system it measures...
Does this make any sense ? (There is always a sense, but most of the time the one we wouldn't like, or the one we don't understand).
Hence after the measurement, the system is Y...
But in my mind, the opposite way is even clearer : The system disturbs the measurement operator...It is even more indicating what a measurement process is : e.g. you put a thermometer, and the thermometer state (operator configuration some kind of, still unknown in QM)..changes...
To recover usual spin QM, let's see this phenomenon in "classical" QM :
the correlation of two observation in A and B is given by the defintion :
C(A,B)=<A\otimes B>-<A\otimes\mathbb{I}><\mathbb{I}\otimes B>
Then it is clear that the non-averaged operator (we remember : <A>=<\Psi|A|\Psi> is then by omitting the \textrm{\emph{exterior}} average :
K(A,B)=A\otimes B-(A\otimes\mathbb{I})\underbrace{|\Psi\rangle\langl e\Psi|}_ {nlin-link}(\mathbb{I}\otimes B)
It's clear that this operator K contains a linear term and a non-linear one. THis means that the measurement operator depends on the system it measures...
Does this make any sense ? (There is always a sense, but most of the time the one we wouldn't like, or the one we don't understand).