Is this something like a Wick rotation?

[Heidi]

TL;DR Summary

Young double slits experiment with orthogonal polarizers.

Please look at this YDSE with two orthogonal polarizers.
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Tutorials_(Rioux)/Quantum_Fundamentals/28:_The_Double‐Slit_Experiment_with_Polarized_Light
I wonder if it is not like a Wick rotation.
I recall that there is a symmetry between statistical mechanics and quantum mechanics.
in the former we add probabilities and in qm we add amplitudes.
when we have a system obeying the heat equation
https://en.wikipedia.org/wiki/Diffusion_equation
replacing t by it gives a Schrodinger like equation obeyed by another possible system.
when here we replace the usual double slit device by adding orthogonal polarizers the figure in the link shows thar interfences disappear and that the probabilities are added on the screen. I wonder if behind the slits the particle obeys the diffusion (or heat) equation.

 

[vanhees71]

I'm a bit puzzled how you come to the idea that this has to do with the Wick rotation used in quantum field theory to formulate a QFT in Euclidean rather than Minkowski space, which simplifies some issues in calculating Feynman diagrams. It's just a mathematical trick, and finally you have to analytically continue the result to Minkowski space again, which can be quite complicated.

Another "imaginary-time method" is known in statistical many-body quantum theory. There you use the similarity between the time-evolution operator for a time-independent Hamiltonian
$$\hat{U}(t)=\exp(-\mathrm{i} \hat{H} t)$$
and the canonical statistical operator
$$\hat{\rho}=\exp(-\hat{H} \beta), \quad \beta=1/T.$$
Formally
$$\hat{\rho}=\hat{U}(-\mathrm{i} \beta).$$
In the QFT formulation the field operators have to fulfill the Kubo-Martin-Schwinger condition $\hat{\psi}(0,\vec{x})=\pm \hat{\psi}(-\mathrm{i} \beta,\vec{x})$ with the upper (lower) sign for bosons (fermions). The upshot then is that you can do perturbation theory as in vacuum QFT with the qualification that the $p^0$ (temporal components of the four-momentum) run over the discrete Matsubara frequencies $2 \pi n T$ ($(2n+1) \pi T$) with $n \in \mathbb{Z}$.

I don't see any analogy in the described double-slit experiment with polarizers in the openings.

 

[Heidi]

I'm a bit puzzled how you come to the idea that this has to do with the Wick rotation used in quantum field theory to formulate a QFT in Euclidean rather than Minkowski space, which simplifies some issues in calculating Feynman diagrams. It's just a mathematical trick

I am also puzzled with that thing...
the main question in this thread is whether or not the particle obeys a diffusion equation when
there are orthogonal polarizers.

 

[vanhees71]

Why should they obey a diffusion equation?

 

[Heidi]

Why should they obey a diffusion equation?

your answer is no?

 

[PeterDonis]

I wonder if behind the slits the particle obeys the diffusion (or heat) equation.

The Schrodinger equation is not the same as the diffusion equation. The particle obeys the Schrodinger equation.

Also, the Schrodinger equation and diffusion equation have nothing whatever to do with Wick rotation. Wick rotation is not just "substitute $it$ for $t$". Handwaving analogies like this are not a good idea.

 

[Heidi]

The particle obeys the Schrodinger equation.

No. The state is not pure.

 

[PeterDonis]

The state is not pure.

Sure it is; it's given by a perfectly well-defined wave function, not a density matrix.

 

[Heidi]

Sure it is; it's given by a perfectly well-defined wave function, not a density matrix.

With a pure state you have the equality V +|P1 - P2| = 1
where V is the fringe visibility (here V=0). P1 and P2 are the probablity associated with the
fringe. The only possibility is that one is null and the other equal to one. ie that there is a maximum which path information and a shanon entropy equal to one bit(the entropy of a pure stare is null)

 

[PeterDonis]

With a pure state you have the equality V +|P1 - P2| = 1

Where are you getting this from? What do you think "pure state" means?

 

[PeterDonis]

where V is the fringe visibility (here V=0)

Why do you think the fringe visibility is zero? The experiment you mention detects interference fringes at the detector screen if they are present, and for certain settings of the polarizers they are.

 

[Heidi]

Why do you think the fringe visibility is zero? The experiment you mention detects interference fringes at the detector screen if they are present, and for certain settings of the polarizers they are.

The title of this thread is YDSE with orthogonal polarizers.

 

[PeterDonis]

The title of this thread is YDSE with orthogonal polarizers.

My questions were not about the title of the thread or what it means. If you're not going to answer the questions I ask, this thread will be closed.

 

[Heidi]

In the case of orthogonal polarizers we have a maximally entanglement between the position of the particle and an internal degree of freedom. It is like in the middle of a Stern Gerlach. We have a tensor product $$H \otimes P$$
P (like polarization) contains an orthormal basis {H,V} and H is the hilbert space of the positions.
In the link i gave it is written that with orthogonal polarizers, the vector state in the tensor product is a pure state. it is $$a\otimes H + b\otimes V$$
the squared module of this vector is aa* + bb* (no interference term)
it is possible to erase this entanglement by adding another polarizer at 45 degrees in front of the screen just like we can merge the path after a stern gerlach.
My problem would be to know the formula giving the hamiltonian in this tensor product and what it becomes after tracing out the H,V degrees of freedom. If i call H this "reduced" hamiltonian, what is the equation of motion of the mixed state in function of H?

 

[PeterDonis]

In the case of orthogonal polarizers we have a maximally entanglement between the position of the particle and an internal degree of freedom.

For that particular case, yes. But the experiment you referenced is not limited to that particular case, so your analysis should not be either. The analysis given with the experiment covers the entire possible range of relative orientations of the polarizers.

In the link i gave it is written that with orthogonal polarizers, the vector state in the tensor product is a pure state.

The analysis in the link you gave gives a pure state (wave function) for all possible relative orientations of the polarizers, not just orthogonal.

what is the equation of motion of the mixed state

There is no mixed state (density matrix) at all in the analysis. See above.

 

[vanhees71]

The Schrodinger equation is not the same as the diffusion equation. The particle obeys the Schrodinger equation.

Also, the Schrodinger equation and diffusion equation have nothing whatever to do with Wick rotation. Wick rotation is not just "substitute $it$ for $t$". Handwaving analogies like this are not a good idea.

Well with a constant diffusion coefficient the diffusion equation reads
$$\partial_{\tau} \phi=D \Delta \phi.$$
Setting $\tau=\mathrm{i} t$ you get the Schrödinger equation of a free particle (reinterpreting the constants involved):
$$\mathrm{i} \partial_t \phi=-D \Delta \phi.$$
Setting $D=\hbar/(2m)$ makes it literally the Schrödinger equation for a free particle.

That's a generally valid scheme to describe equilibrium many-body QT as imaginary-time evolution of usual QT. Using QFT, it leads to the Matsubara formalism. For the general case (i.e., off-equilibrium as well as equilibrium), you need extensions of this formalism, known as the time-contour formalism which comes in some varieties. The most common are the Schwinger-Keldysh formalism with a time contour running back and forth along the real time axis (in equilibrium one may add a vertical part at the end running from $0$ straight down to $-\mathrm{i} \beta$) or thermo-field dynamics which chooses the contour to run first forward along the real axis then vertically down to $t_{f}-\mathrm{i} \beta/2$ then back parallel to the real axis to $t_{i}-\mathrm{i} \beta/2$ and then finally vertically down the rest of the extended contour to $t_i-\mathrm{i} \beta$.

For details, see

https://itp.uni-frankfurt.de/~hees/publ/off-eq-qft.pdf

For the non-relativistic case in the Schwinger-Keldysh formalism:

Landau, Lifshitz, vol X.

P. Danielewicz, Quantum Theory of Nonequilibrium Processes
I, Ann. Phys. 152, 239 (1984),
https://doi.org/10.1016/0003-4916(84)90092-7.

P. Danielewicz, Quantum Theory of Nonequilibrium Processes
II. Application to Nuclear Collisions, Ann. Phys. 152, 305
(1984), https://doi.org/10.1016/0003-4916(84)90093-9.

 

[Heidi]

That's a generally valid scheme to describe equilibrium many-body QT as imaginary-time evolution of usual QT. Using QFT, it leads to the Matsubara formalism. For the general case (i.e., off-equilibrium as well as equilibrium), you need extensions of this formalism, known as the time-contour formalism which comes in some varieties. The most common are the Schwinger-Keldysh formalism with a time contour running back and forth along the real time axis (in equilibrium one may add a vertical part at the end running from $0$ straight down to $-\mathrm{i} \beta$) or thermo-field dynamics which chooses the contour to run first forward along the real axis then vertically down to $t_{f}-\mathrm{i} \beta/2$ then back parallel to the real axis to $t_{i}-\mathrm{i} \beta/2$ and then finally vertically down the rest of the extended contour to $t_i-\mathrm{i} \beta$.

For details, see

https://itp.uni-frankfurt.de/~hees/publ/off-eq-qft.pdf

Thanks a lot , I'll study this contour in your pdf.

 

[Heidi]

Thanks a lot , I'll study this contour in your pdf.

I googled "Schwinger" and i found "KMS state" . Has the field behind the orthogonal polarizers something to do with that kind of states?

 

[vanhees71]

No. With thermal states you'll have a hard time to see interference at all since thermal radiation is utmost incoherent.

 

[Heidi]

No. With thermal states you'll have a hard time to see interference at all since thermal radiation is utmost incoherent.

Of course. but i repeat that my question is only about orthogonal polarizers (no interferences). a thermal bath in this case would not lead to "hard time"

 

[Heidi]

Look at the figure in the lind i gave in my firs post. You see 3 curves. they come from
parallel polarizers, orthogonal polarizers, polarizers at pi/4.
the same patterns can be obtained without polarizers but with photons belonging to entangled pairs.
1 you have no interference if each photon belong to a Bell pair. the Bell pair is in a pure state and the state of each subphoton is obtained by tracing out the other in the pair.
2 you have a perfect visibility of the fringes if the pair may be separed in independent photons.
3 the third pattern is obtained by tracing out the other when the degree of intrication in the pair
is not maximall but half of it.
Is there a thermal difference between the 3 ways to produce these entanglement?
Does one of them emit more information in the environment? Is it thes loss on information
that is "seen" in the pattern?

 

[vanhees71]

But all this has nothing to do with Wick rotation. I don't understand, how you come to relate the Wick rotation to this interference phenomena.

Heuristically it's clear that with the quarter-wave plates in $90^{\circ}$ relative rotation you have full which-way information, because the polarization state of the photon is entangled with the path information. So you know with 100% certainty for each photon through which slit it came when measuring its polarization. You don't really need to measure the polarization. It's enough that you can in principle find out throug which slit each photon came to completely destroy the two-slit interference pattern. Having another angle between the quarter-wave plates the which-way information is not certain but if the relative angle isn't 0, it's a bit more probable that a particle with a given polarization came through one slit than to come through the other. That's why you see some two-slit interference fringes but not at a contrast as high as without quarter-wave plates. If you make the relative angle $0$ then it's again impossible to decide through which slit each photon came, because then everything is symmetric again, i.e., all photons have the same polarization no matter through which slit they came and thus in this case you get a gain the two-slit interference pattern at full contrast.

 

[Heidi]

But all this has nothing to do with Wick rotation. I don't understand, how you come to relate the Wick rotation to this interference phenomena.

Heuristically it's clear that with the quarter-wave plates in $90^{\circ}$ relative rotation you have full which-way information, because the polarization state of the photon is entangled with the path information. So you know with 100% certainty for each photon through which slit it came when measuring its polarization. You don't really need to measure the polarization. It's enough that you can in principle find out throug which slit each photon came to completely destroy the two-slit interference pattern. Having another angle between the quarter-wave plates the which-way information is not certain but if the relative angle isn't 0, it's a bit more probable that a particle with a given polarization came through one slit than to come through the other. That's why you see some two-slit interference fringes but not at a contrast as high as without quarter-wave plates. If you make the relative angle $0$ then it's again impossible to decide through which slit each photon came, because then everything is symmetric again, i.e., all photons have the same polarization no matter through which slit they came and thus in this case you get a gain the two-slit interference pattern at full contrast.

and what about the purity of the state behind the polarizers?

 

[Heidi]

and what about the purity of the state behind the polarizers?

I had a false idea about this question of pure or impure state. let a vertical polarizer be on slit 1 and an horizontal one on slit 2. The light is linearly polarized the imput is the pure state aV + bH
let us take a = 1 and b = 0 then behind the polarizers the state is given by the pure density matrix
1 0
0 0
there is no interference because the off diagonal terms are null.
same thing when a = 0 and b = 1 with
0 0
0 1
the thing is different with a = b = cos Pi/2
then the density matrix is
1/2 0
0 1/2
it is a mixed state and the lack of interference come from the null off diagonal terms.î