Twin slits; state preparation

[cesiumfrog]

Before the results of an experiment have been obtained, how does one determine how to write down the initial state of the quantum system?For the example of simple slit experiments (following Marcella's "Q.I. w/ slits") a particle emerging from slit A is in a position eigenstate, eg. $|A> = \delta (y - y_A)$. For the double slit we always take the superposition $|\psi > = (|A>+|B>)/ \sqrt 2$. Interference fringes are obtained by measuring the momentum of this prepared state (or the position, after letting it evolve).

In contrast, the (more complex) quantum eraser experiments tend to assume a different initial state described by $|\psi > = (|A>+e^{i \triangle \phi}|B>)/ \sqrt 2$. Various measurements then give rise to interference fringes ($|A>+|B>$), anti-fringes ($|A>-|B>$) and non-interference ($|A>$, or alternatively $|B>$).

To me this seems to assume that the photon could not only have emerged from either part of the down-conversion crystal, but that it could have done so at an earlier (or later) time, and those four possible results or superpositions correspond to the different possible real/imaginary parts of $|\psi >$? If so, it would seem to demand an explanation of why the photon couldn't also have emerged earlier (or later) from the second slit in the simple double-slit experiment? Is there a simpler way to prepare a state such as $|\psi > = (|A>-|B>)/ \sqrt 2$?

 

[cesiumfrog]

I'm still stuck trying to understand the contrast between those examples above, but I am guessing that the anti-fringes state could be simply prepared by putting a half wave plate (or a little perspex) behind one slit in Young's experiment.

After further thought.. when we integrate over all paths that a particle could have taken, it still seems that we must also somehow include paths that differ in time. For coherent light the intensity (in some sense) seems periodic such that the most probable paths always differ roughly by a multiple of the wavelength (producing no effect) whereas if we suddently double the wavelength then some fraction of the probable paths could now be 180 degrees out of phase (so both must be accounted for).

Does that make any sense?

 

[Edgardo]

In contrast, the (more complex) quantum eraser experiments tend to assume a different initial state described by $|\psi > = (|A>+e^{i \triangle \phi}|B>)/ \sqrt 2$. Various measurements then give rise to interference fringes ($|A>+|B>$), anti-fringes ($|A>-|B>$) and non-interference ($|A>$, or alternatively $|B>$).

Which quantum eraser experiment (or paper) are you referring to?

To me this seems to assume that the photon could not only have emerged from either part of the down-conversion crystal, but that it could have done so at an earlier (or later) time, and those four possible results or superpositions correspond to the different possible real/imaginary parts of $|\psi >$? If so, it would seem to demand an explanation of why the photon couldn't also have emerged earlier (or later) from the second slit in the simple double-slit experiment? Is there a simpler way to prepare a state such as $|\psi > = (|A>-|B>)/ \sqrt 2$?

Could you elaborate on what you mean?

 

[cesiumfrog]

Which quantum eraser experiment (or paper) are you referring to? [...] Could you elaborate on what you mean?

For example, equation 1 of Kwiat et. al. `Three proposed "quantum erasors"' PRL v.75 p.3034 (1995), or also of PRA v.49 p.61 (1994). My thinking is closely tied to the `Delayed "Choice" Quantum Eraser', PRL v.84 p.1, for which the similarity to Young's two-slit experiment is more explicit.

(Unsure which part to elaborate on, so I'll try to clarify my question:)

I'd like to know why phase information ($e^{i \triangle \phi}$) is written in the initial state for a quantum erasor but not in the initial state for a traditional two-slit experiment.