Suppose the state of the system is [itex]|\psi>[/itex]. Measurement of the position means to apply the operator X to the state [itex]|\psi>[/itex] to obtain the eigenvalue of position x, i.e. X[itex]|\psi>[/itex] = x[itex]|\psi>[/itex]. The system then falls into an eigenstate of position which we can write as [itex]|x>[/itex].Palindrom said:What else?
O.K., so there it is: I must be having a really stupid misconseption, so be gentle.
X and P can never have rigorous (that is, without delta functions and infinity barriers) eigenstates, right?
So, when I measure, for example, X, to what state does the system collapse?
Since the new state is very confined in position space, how does the timeGeorge Jones said:After a measurement on a system in state [itex]|\psi>[/itex], the state of the system is [itex]P|\psi>[/itex], where [itex]P[/itex] is an appropriate projection operator. This is true for observables that have non-degenerate spectra, degenerate spectra, continuous spectra, etc.
No apparatus that measures an observable with a continuous is selective enough that a precise value is singled out.
Consider a position measurement. The apparatus used to make the measurement will only be able to ascertain that the position lies within a certain interval, say [itex]x[/itex] is in the interval [itex]\left[a , c \right][/itex]. Then [itex]P = \int_{a}^{c} dx |x><x|[/itex] and after the measurement the state of the system is [itex]|\psi '> = \int_{a}^{c} dx |x><x|\psi>[/itex].
Regards,
George
OK.pmb_phy said:Suppose the state of the system is [itex]|\psi>[/itex]. Measurement of the position means to apply the operator X to the state [itex]|\psi>[/itex] to obtain the eigenvalue of position x, i.e. X[itex]|\psi>[/itex] = x[itex]|\psi>[/itex]. The system then falls into an eigenstate of position which we can write as [itex]|x>[/itex].
OK.On the other hand if you were to measure x-momentum you'd get something similar, i.e. measurement of x-momentum means to apply the operator P_{x} to the state [itex]|\psi>[/itex] to obtain the eigenvalue of position p_{x}, i.e. P_{x}[itex]|\psi>[/itex] = p_{x}[itex]|\psi>[/itex]. The system then falls into an eigenstate of mometum which we can write as [itex]|p_x>[/itex].
That's what I've been thinking for a while, too. However, the preceding statements above have made me have second thoughts. If you simultaneously measure both [itex]x[/itex] and [itex]p_x[/itex], then the system must at that point in time be in a simultaneous eigenstate of both of the corresponding operators. But there can be no such state because the position and momentum operators don't commute, right? Or does that restriction apply only to stationary states?None of this means that you can't simultaneously measure p_{x} and x exactly.
To take a free particle for example, the new state is a superposition of waves of the form [itex]A \exp [i(kx - \omega t)][/itex]. These waves have different phase velocties, so their phase differences change as time goes on, and the combined wave changes shape as a result.Ratzinger said:Since the new state is very confined in position space, how does the time
evolution operator make it spread again?