- #1
Student149
- 58
- 0
We know that Bell States follow the Rotational Invariance property i.e. the probability of results on measurement of two entangled particles do not change if the initial measurement basis (say ##u##) is rotated by an angle θ to a new basis (to say ##v##).
Lets take the Bell State ##\psi = \frac 1 {\sqrt 2}(\uparrow \uparrow + \downarrow \downarrow)##. Here both the terms have the probability ##50##% in basis ##u## or rotated basis ##v##.
How does the Rotational Invariance property get affected if the probability of two terms is unequal i.e. if the entangled state is of the form ##\psi = (α\uparrow \uparrow + β\downarrow \downarrow)##, ##α^2+β^2=1##, ##α≠β##?
My guess is they remain unchanged as it is a generic case of Bell States.
Lets take the Bell State ##\psi = \frac 1 {\sqrt 2}(\uparrow \uparrow + \downarrow \downarrow)##. Here both the terms have the probability ##50##% in basis ##u## or rotated basis ##v##.
How does the Rotational Invariance property get affected if the probability of two terms is unequal i.e. if the entangled state is of the form ##\psi = (α\uparrow \uparrow + β\downarrow \downarrow)##, ##α^2+β^2=1##, ##α≠β##?
My guess is they remain unchanged as it is a generic case of Bell States.
Last edited: