Does the double-slit experiment create path-entangled states

  • #1

Main Question or Discussion Point

If I try to send a vertically polarized photon through one slit and a horizontally polarized photon through the other slit, they actually go through both slits.

But when I measure and find out through which slit the horizontally polarized photon went, I automatically know that the vertically polarized photon went through the other slit.

Mathematically,

(|L>H + |R> H) / √2 is the superposition state for the horizontal photon
(|R>V + |L> V)/√2 is the superposition state for the vertical photon

Because both superpositions interfere behind the double slit, I get an entangled state:

(|L>H|R>V - |R>H|L>V) / √2

Am I right about this or did I do my calculations wrong?
 

Answers and Replies

  • #2
Strilanc
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If you have a pair of entangled photons, it is inconsistent for them to also have individual states. You'll need to be more specific about the situation you have in mind.

If qubit #1 is in the state ##\frac{1}{\sqrt{2}} \left| 0 \right\rangle + \frac{1}{\sqrt{2}} \left| 1 \right\rangle## and qubit #2 is in the state ##\frac{1}{\sqrt{2}} \left| 1 \right\rangle + \frac{1}{\sqrt{2}} \left| 0 \right\rangle##, then the 2-qubit system as a whole is in the state ##\frac{1}{2} \left| 00 \right\rangle + \frac{1}{2} \left| 01 \right\rangle +
\frac{1}{2} \left| 10 \right\rangle + \frac{1}{2} \left| 11 \right\rangle##.

If qubit #1 is entangled with qubit #2 into a singlet state (where they disagree), then the system as a whole is in the state ##\frac{1}{\sqrt{2}} \left| 01 \right\rangle - \frac{1}{\sqrt{2}} \left| 10 \right\rangle##. In this case there's not really a well defined state that each individual qubit is in; the best you can do is marginalize over the other qubit to get a mixed state.
 
  • #3
If you have a pair of entangled photons, it is inconsistent for them to also have individual states. You'll need to be more specific about the situation you have in mind.

If qubit #1 is in the state ##\frac{1}{\sqrt{2}} \left| 0 \right\rangle + \frac{1}{\sqrt{2}} \left| 1 \right\rangle## and qubit #2 is in the state ##\frac{1}{\sqrt{2}} \left| 1 \right\rangle + \frac{1}{\sqrt{2}} \left| 0 \right\rangle##, then the 2-qubit system as a whole is in the state ##\frac{1}{2} \left| 00 \right\rangle + \frac{1}{2} \left| 01 \right\rangle +
\frac{1}{2} \left| 10 \right\rangle + \frac{1}{2} \left| 11 \right\rangle##.

If qubit #1 is entangled with qubit #2 into a singlet state (where they disagree), then the system as a whole is in the state ##\frac{1}{\sqrt{2}} \left| 01 \right\rangle - \frac{1}{\sqrt{2}} \left| 10 \right\rangle##. In this case there's not really a well defined state that each individual qubit is in; the best you can do is marginalize over the other qubit to get a mixed state.
I am afraid I asked something else. Does the double-slit experiment create quantum entanglement between photons with different polarizations by erasing which-path information and making trajectories indistinguishable, OR does it leave them into a separable state after their interference?
 
  • #4
DrChinese
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I am afraid I asked something else. Does the double-slit experiment create quantum entanglement between photons with different polarizations by erasing which-path information and making trajectories indistinguishable, OR does it leave them into a separable state after their interference?
Are you asking about photons that go into the apparatus as separate, distinguishable particles? Those of course do not come out entangled. A V photon does not interfere with an H photon. And of course if you know one is V and the other is H, they are distinguishable and will never become entangled unless you first erase their "markings".
 
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  • #5
Are you asking about photons that go into the apparatus as separate, distinguishable particles? Those of course do not come out entangled. A V photon does not interfere with an H photon. And of course if you know one is V and the other is H, they are distinguishable and will never become entangled unless you first erase their "markings".
But in SPDC, it is the horizontal and vertical photons that become entangled where the emission cones intersect. They do not have indistinguishable polarization from their creation, rather they are emitted into horizontally polarized and vertically polarized cones. Where these cones intersect, the polarization is indistinguishable.
So how it is you say photons of different polarizations cannot interfere when the below photo shows they do?
350px-SPDC_figure.png

" Importantly, the trajectories of the photon pairs may exist simultaneously in the two lines where the cones intersect. This results in entanglement of the photon pairs whose polarization are perpendicular"
 
  • #6
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If I try to send a vertically polarized photon through one slit and a horizontally polarized photon through the other slit, they actually go through both slits.
This is not possible.

If a single photon goes through it goes into a superposition of position/momentum states. It is not possible to create ploarization ( spin-orientation) entanglement like this.
 
Last edited:
  • #7
Strilanc
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Are you imagining that the SPDC happens beforehand and somehow biases the photons so they will end up in a state where they passed through opposite slits? I'm pretty sure that's wrong.

My understanding is that the SPDC is done after the slits, as a mechanism for creating an entangled copy of the which-way information (e.g. for a delayed-choice experiment).
 
  • #8
DrChinese
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But in SPDC, it is the horizontal and vertical photons that become entangled where the emission cones intersect. They do not have indistinguishable polarization from their creation, rather they are emitted into horizontally polarized and vertically polarized cones. Where these cones intersect, the polarization is indistinguishable.
So how it is you say photons of different polarizations cannot interfere when the below photo shows they do?
350px-SPDC_figure.png

" Importantly, the trajectories of the photon pairs may exist simultaneously in the two lines where the cones intersect. This results in entanglement of the photon pairs whose polarization are perpendicular"
The H and V photons are not interfering with each other. They become entangled, and this is because they are indistinguishable. Once entangled, they are in a superposition which gives them properties (as a single system) different than a system composed of an H and a V.

Please keep in mind that your initial question was about sending photons through a double slit apparatus. If you take the two photons per above and send them to a double slit setup, there will be no interference. Entangled photons do not self-interfere. To accomplish double slit interference, they can no longer be entangled. They must be coherent. Speaking in loose terms, that means they must have a point-like source - entangled photons lack this attribute.
 
  • #11
In the present paper we report a double-slit experiment using two photons created by spontaneous parametric down-conversion where we observe interference in the signal photon despite the fact that we have located it in one of the slits due to its entanglement with the idler photon.

http://www.pnas.org/content/109/24/9314
 
  • #12
zonde
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After PBS photons certainly are not polarization entangled. As they do not specify the state after PBS it's hard for me to say in what sense they consider photons to be entangled. Coincidence of two downconverted photons in the same spatial mode is considered classical phenomena as far as I know.

What seems very interesting to me is that upper and lower beam after PBS are coherent as I don't see obvious reason for them to be coherent. Maybe someone else will comment on that.
 

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