Is the photon pair generated from the electron-positron annihilation entangled?
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Is the photon pair generated from the electron-positron annihilation entangled?
And would they work as a source of entangled photons suitable for experiments in quantum optics?
quantum-mechanics particle-physics photons quantum-entanglement
edited Nov 19 at 17:56
Peter Mortensen
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asked Nov 19 at 11:23
E.phy
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Is the photon pair generated from the electron-positron annihilation entangled?
And would they work as a source of entangled photons suitable for experiments in quantum optics?
quantum-mechanics particle-physics photons quantum-entanglement
edited Nov 19 at 17:56
Peter Mortensen
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asked Nov 19 at 11:23
E.phy
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add a comment |
up vote
17
down vote
favorite
up vote
17
down vote
favorite
Is the photon pair generated from the electron-positron annihilation entangled?
And would they work as a source of entangled photons suitable for experiments in quantum optics?
quantum-mechanics particle-physics photons quantum-entanglement
edited Nov 19 at 17:56
Peter Mortensen
1,91211323
asked Nov 19 at 11:23
E.phy
1857
Is the photon pair generated from the electron-positron annihilation entangled?
And would they work as a source of entangled photons suitable for experiments in quantum optics?
quantum-mechanics particle-physics photons quantum-entanglement
quantum-mechanics particle-physics photons quantum-entanglement
edited Nov 19 at 17:56
Peter Mortensen
1,91211323
asked Nov 19 at 11:23
E.phy
1857
edited Nov 19 at 17:56
Peter Mortensen
1,91211323
asked Nov 19 at 11:23
E.phy
1857
edited Nov 19 at 17:56
Peter Mortensen
1,91211323
edited Nov 19 at 17:56
Peter Mortensen
1,91211323
edited Nov 19 at 17:56
Peter Mortensen
1,91211323
1,91211323
asked Nov 19 at 11:23
E.phy
1857
asked Nov 19 at 11:23
E.phy
1857
asked Nov 19 at 11:23
E.phy
1857
1857
add a comment |
add a comment |
2 Answers
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Yes, they are definitely entangled. Their combined energy will exactly equal the combined energy of the original electron-positron pair, for example. The same is true for combined momentum and combined spin.
answered Nov 19 at 11:40
S. McGrew
5,2812923
"Combined spin" should also include any orbital angular momentum if the electron and positron briefly orbited each other (positronium) before annihilating.
– Andreas Blass
Nov 19 at 16:08
That is correct--
– S. McGrew
Nov 19 at 16:13
And from what I understand about these things (only a layman as far as physics goes), the angular momentum of the electron-positron pair will almost always be non-zero.
– EvilSnack
Nov 20 at 0:57
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The photon-pair is definitely entangled. It is produced from an $S$-wave state of the electron-positron system and as such has orbital angular momentum $L=0$. But it's not useful for quantum optics, the photons' energy is too high for your usual mirrors to reflect. See fig 33.15 on p. 23 of this http://pdg.lbl.gov/2018/reviews/rpp2018-rev-passage-particles-matter.pdf to get an idea what happens to a 511 keV photon from two-photon annihilation of an electron-positron pair at rest once it hits matter: ionization by Compton scattering dominates the interaction, that's not what you want to happen in a mirror or lense.
It's worth keeping in mind though, that the electron-positron system (Positronium) can decay to any number of photons $>1$, though numbers higher than three are very rare. An even number of photons can be produced for decays of the singlet ground state $^1S_0$ (where the electron and positron are in an anti-symmetric spin state
$frac1sqrt2left( left|uparrow downarrow rightrangle - left|downarrow uparrow rightrangleright)$, "Parapositronium"). This defines the spin entanglement in the case you had in mind, and it has actually been measured and found to agree with full entanglement, see e.g. here http://adsabs.harvard.edu/abs/2009APS..HAW.GB108S
An uneven number can be produced from the triplet $^3S_0$ ("Orthopositronium"). The triplet is much longer-lived than the singlet state (roughly thousand times), but since it consists of three states and because symmetries ($CP$) prevent the positronium from going from the triplet to the singlet state, an appreciable number of electron-positron pairs decays to three photons.
In fact, also the $S$ states of higher energy levels can decay to photons directly, but usually they will decay to the ground state first, emitting photons of energies $O(10 - 100 emathrmV)$ before the actual decay to the high-energy pair or to the three continuous spectrum photons.
answered Nov 20 at 6:54
tobi_s
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protected by David Z♦ Nov 20 at 9:58
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2 Answers
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
22
down vote
Yes, they are definitely entangled. Their combined energy will exactly equal the combined energy of the original electron-positron pair, for example. The same is true for combined momentum and combined spin.
answered Nov 19 at 11:40
S. McGrew
5,2812923
"Combined spin" should also include any orbital angular momentum if the electron and positron briefly orbited each other (positronium) before annihilating.
– Andreas Blass
Nov 19 at 16:08
That is correct--
– S. McGrew
Nov 19 at 16:13
And from what I understand about these things (only a layman as far as physics goes), the angular momentum of the electron-positron pair will almost always be non-zero.
– EvilSnack
Nov 20 at 0:57
add a comment |
up vote
22
down vote
Yes, they are definitely entangled. Their combined energy will exactly equal the combined energy of the original electron-positron pair, for example. The same is true for combined momentum and combined spin.
answered Nov 19 at 11:40
S. McGrew
5,2812923
"Combined spin" should also include any orbital angular momentum if the electron and positron briefly orbited each other (positronium) before annihilating.
– Andreas Blass
Nov 19 at 16:08
That is correct--
– S. McGrew
Nov 19 at 16:13
And from what I understand about these things (only a layman as far as physics goes), the angular momentum of the electron-positron pair will almost always be non-zero.
– EvilSnack
Nov 20 at 0:57
add a comment |
up vote
22
down vote
up vote
22
down vote
Yes, they are definitely entangled. Their combined energy will exactly equal the combined energy of the original electron-positron pair, for example. The same is true for combined momentum and combined spin.
answered Nov 19 at 11:40
S. McGrew
5,2812923
Yes, they are definitely entangled. Their combined energy will exactly equal the combined energy of the original electron-positron pair, for example. The same is true for combined momentum and combined spin.
answered Nov 19 at 11:40
S. McGrew
5,2812923
answered Nov 19 at 11:40
S. McGrew
5,2812923
answered Nov 19 at 11:40
S. McGrew
5,2812923
answered Nov 19 at 11:40
S. McGrew
5,2812923
5,2812923
"Combined spin" should also include any orbital angular momentum if the electron and positron briefly orbited each other (positronium) before annihilating.
– Andreas Blass
Nov 19 at 16:08
That is correct--
– S. McGrew
Nov 19 at 16:13
And from what I understand about these things (only a layman as far as physics goes), the angular momentum of the electron-positron pair will almost always be non-zero.
– EvilSnack
Nov 20 at 0:57
add a comment |
"Combined spin" should also include any orbital angular momentum if the electron and positron briefly orbited each other (positronium) before annihilating.
– Andreas Blass
Nov 19 at 16:08
That is correct--
– S. McGrew
Nov 19 at 16:13
And from what I understand about these things (only a layman as far as physics goes), the angular momentum of the electron-positron pair will almost always be non-zero.
– EvilSnack
Nov 20 at 0:57
"Combined spin" should also include any orbital angular momentum if the electron and positron briefly orbited each other (positronium) before annihilating.
– Andreas Blass
Nov 19 at 16:08
"Combined spin" should also include any orbital angular momentum if the electron and positron briefly orbited each other (positronium) before annihilating.
– Andreas Blass
Nov 19 at 16:08
That is correct--
– S. McGrew
Nov 19 at 16:13
That is correct--
– S. McGrew
Nov 19 at 16:13
And from what I understand about these things (only a layman as far as physics goes), the angular momentum of the electron-positron pair will almost always be non-zero.
– EvilSnack
Nov 20 at 0:57
And from what I understand about these things (only a layman as far as physics goes), the angular momentum of the electron-positron pair will almost always be non-zero.
– EvilSnack
Nov 20 at 0:57
add a comment |
up vote
4
down vote
The photon-pair is definitely entangled. It is produced from an $S$-wave state of the electron-positron system and as such has orbital angular momentum $L=0$. But it's not useful for quantum optics, the photons' energy is too high for your usual mirrors to reflect. See fig 33.15 on p. 23 of this http://pdg.lbl.gov/2018/reviews/rpp2018-rev-passage-particles-matter.pdf to get an idea what happens to a 511 keV photon from two-photon annihilation of an electron-positron pair at rest once it hits matter: ionization by Compton scattering dominates the interaction, that's not what you want to happen in a mirror or lense.
It's worth keeping in mind though, that the electron-positron system (Positronium) can decay to any number of photons $>1$, though numbers higher than three are very rare. An even number of photons can be produced for decays of the singlet ground state $^1S_0$ (where the electron and positron are in an anti-symmetric spin state
$frac1sqrt2left( left|uparrow downarrow rightrangle - left|downarrow uparrow rightrangleright)$, "Parapositronium"). This defines the spin entanglement in the case you had in mind, and it has actually been measured and found to agree with full entanglement, see e.g. here http://adsabs.harvard.edu/abs/2009APS..HAW.GB108S
An uneven number can be produced from the triplet $^3S_0$ ("Orthopositronium"). The triplet is much longer-lived than the singlet state (roughly thousand times), but since it consists of three states and because symmetries ($CP$) prevent the positronium from going from the triplet to the singlet state, an appreciable number of electron-positron pairs decays to three photons.
In fact, also the $S$ states of higher energy levels can decay to photons directly, but usually they will decay to the ground state first, emitting photons of energies $O(10 - 100 emathrmV)$ before the actual decay to the high-energy pair or to the three continuous spectrum photons.
answered Nov 20 at 6:54
tobi_s
1812
add a comment |
up vote
4
down vote
The photon-pair is definitely entangled. It is produced from an $S$-wave state of the electron-positron system and as such has orbital angular momentum $L=0$. But it's not useful for quantum optics, the photons' energy is too high for your usual mirrors to reflect. See fig 33.15 on p. 23 of this http://pdg.lbl.gov/2018/reviews/rpp2018-rev-passage-particles-matter.pdf to get an idea what happens to a 511 keV photon from two-photon annihilation of an electron-positron pair at rest once it hits matter: ionization by Compton scattering dominates the interaction, that's not what you want to happen in a mirror or lense.
It's worth keeping in mind though, that the electron-positron system (Positronium) can decay to any number of photons $>1$, though numbers higher than three are very rare. An even number of photons can be produced for decays of the singlet ground state $^1S_0$ (where the electron and positron are in an anti-symmetric spin state
$frac1sqrt2left( left|uparrow downarrow rightrangle - left|downarrow uparrow rightrangleright)$, "Parapositronium"). This defines the spin entanglement in the case you had in mind, and it has actually been measured and found to agree with full entanglement, see e.g. here http://adsabs.harvard.edu/abs/2009APS..HAW.GB108S
An uneven number can be produced from the triplet $^3S_0$ ("Orthopositronium"). The triplet is much longer-lived than the singlet state (roughly thousand times), but since it consists of three states and because symmetries ($CP$) prevent the positronium from going from the triplet to the singlet state, an appreciable number of electron-positron pairs decays to three photons.
In fact, also the $S$ states of higher energy levels can decay to photons directly, but usually they will decay to the ground state first, emitting photons of energies $O(10 - 100 emathrmV)$ before the actual decay to the high-energy pair or to the three continuous spectrum photons.
answered Nov 20 at 6:54
tobi_s
1812
add a comment |
up vote
4
down vote
up vote
4
down vote
The photon-pair is definitely entangled. It is produced from an $S$-wave state of the electron-positron system and as such has orbital angular momentum $L=0$. But it's not useful for quantum optics, the photons' energy is too high for your usual mirrors to reflect. See fig 33.15 on p. 23 of this http://pdg.lbl.gov/2018/reviews/rpp2018-rev-passage-particles-matter.pdf to get an idea what happens to a 511 keV photon from two-photon annihilation of an electron-positron pair at rest once it hits matter: ionization by Compton scattering dominates the interaction, that's not what you want to happen in a mirror or lense.
It's worth keeping in mind though, that the electron-positron system (Positronium) can decay to any number of photons $>1$, though numbers higher than three are very rare. An even number of photons can be produced for decays of the singlet ground state $^1S_0$ (where the electron and positron are in an anti-symmetric spin state
$frac1sqrt2left( left|uparrow downarrow rightrangle - left|downarrow uparrow rightrangleright)$, "Parapositronium"). This defines the spin entanglement in the case you had in mind, and it has actually been measured and found to agree with full entanglement, see e.g. here http://adsabs.harvard.edu/abs/2009APS..HAW.GB108S
An uneven number can be produced from the triplet $^3S_0$ ("Orthopositronium"). The triplet is much longer-lived than the singlet state (roughly thousand times), but since it consists of three states and because symmetries ($CP$) prevent the positronium from going from the triplet to the singlet state, an appreciable number of electron-positron pairs decays to three photons.
In fact, also the $S$ states of higher energy levels can decay to photons directly, but usually they will decay to the ground state first, emitting photons of energies $O(10 - 100 emathrmV)$ before the actual decay to the high-energy pair or to the three continuous spectrum photons.
answered Nov 20 at 6:54
tobi_s
1812
The photon-pair is definitely entangled. It is produced from an $S$-wave state of the electron-positron system and as such has orbital angular momentum $L=0$. But it's not useful for quantum optics, the photons' energy is too high for your usual mirrors to reflect. See fig 33.15 on p. 23 of this http://pdg.lbl.gov/2018/reviews/rpp2018-rev-passage-particles-matter.pdf to get an idea what happens to a 511 keV photon from two-photon annihilation of an electron-positron pair at rest once it hits matter: ionization by Compton scattering dominates the interaction, that's not what you want to happen in a mirror or lense.
It's worth keeping in mind though, that the electron-positron system (Positronium) can decay to any number of photons $>1$, though numbers higher than three are very rare. An even number of photons can be produced for decays of the singlet ground state $^1S_0$ (where the electron and positron are in an anti-symmetric spin state
$frac1sqrt2left( left|uparrow downarrow rightrangle - left|downarrow uparrow rightrangleright)$, "Parapositronium"). This defines the spin entanglement in the case you had in mind, and it has actually been measured and found to agree with full entanglement, see e.g. here http://adsabs.harvard.edu/abs/2009APS..HAW.GB108S
An uneven number can be produced from the triplet $^3S_0$ ("Orthopositronium"). The triplet is much longer-lived than the singlet state (roughly thousand times), but since it consists of three states and because symmetries ($CP$) prevent the positronium from going from the triplet to the singlet state, an appreciable number of electron-positron pairs decays to three photons.
In fact, also the $S$ states of higher energy levels can decay to photons directly, but usually they will decay to the ground state first, emitting photons of energies $O(10 - 100 emathrmV)$ before the actual decay to the high-energy pair or to the three continuous spectrum photons.
answered Nov 20 at 6:54
tobi_s
1812
edited Nov 20 at 9:31
answered Nov 20 at 6:54
tobi_s
1812
answered Nov 20 at 6:54
tobi_s
1812
answered Nov 20 at 6:54
tobi_s
1812
1812
add a comment |
add a comment |
protected by David Z♦ Nov 20 at 9:58
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Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
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