The system being measured is a spin system with total spin of 12.5 (so any component of the spin have eigenvalues that go from -12.5 to 12.5 by integer leaps.
The hamiltonian of coupling between the measuring dial and the spin is
H= pS
where S is the component of the spin half way between z and x
S=(1/sqrt(2))(Sz+Sx)
and the interaction is switched on for 1 unit of time. The unitary
transformation caused by the measurement is the displacement operator for the
q coordinate, so that the state is phi(q-S) where phi(q) is the initial state
of the measuring apparatus.
The initial state of the spin system is the Sz= 12.5 eigenstate, and the final
state is Sx=12.5.
The state of the measuring apparatus will thus be
< Sz=12.5|phi(q-S)|Sz=12.5>
The following movies will use a variety of functions phi, with "widths" of D,
where D is the standard deviation of q if it exists, or a measure of the width
if the standard deviation does not exist (eg for the Lorentzian). The movies
display the above final function for the various phi as a function fo D.
In the plots, the value of D is the top number on the right side of the
picture, and the total probability for getting the final state to be Sx=12.5
is the lower number listed. In all cases the weak value of the S which is
Weak Value=
is 12.5 sqrt(2)=17.8
All of the functions phi(q) ae normalised so that the integral of the square equals 1. in all of the movies D goes from .03 to 6. Thus at the beginning all are essentialy exact measurements of S.
phi(q)= exp[ -(q^2/4D^2)]/sqrt(sqrt(2 Pi) D)
< >
phi(q)= sqrt{a/(q^2+D^2)/Pi}
Note that the
probability is Lorentzian but the standard deviation is infinite.
Thus a is a measure of the width of the distribution, but is not the standard
deviation.
phi(q)= (sqrt(3) D)^{7/2}/sqrt(Pi*3*sqrt(2)/8) 1/(q^4+(sqrt(3) D)^4)
phi(q)= exp(-|q|/2D}/sqrt(2)D)
phi(q)= (sqrt(2)/sqrt(3*D)) exp(-abs(|q|)/D)*(cos(q/D)+sin(|q|/D))
Copyright William Unruh 2021