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Preparation of electron spin directions
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ben...@hotmail.com
Feb 26, 2019, 2:30:39 PM2/26/19
to
I have just written the following in a draft commentary on (someone
else's) Bell model.
"However, I do not believe that electrons can be aligned exactly in a
given direction. That would require measuring along three base axes
simultaneously. That cannot happen as only one measurement axis is
permitted at a time. So preparation of an electron in the direction of
vector a is a fuzzy process not an exact one. Particles can change spin
axes on detection to reverse the direction of their spin axes. That
suffices to point them in the general direction of (say) vector a. That
is, clearly pointing more towards say vector a rather than towards
vector a*, but not pointing exactly in line with vector a."
This is broadly-speaking true, isn't it?
[Moderator's note: Due to possibly garbled characters, the vector names
might not be what was intended. It seems that the point is to
distinguish two vectors, above denoted by a and a*. -P.H.]
else's) Bell model.
"However, I do not believe that electrons can be aligned exactly in a
given direction. That would require measuring along three base axes
simultaneously. That cannot happen as only one measurement axis is
permitted at a time. So preparation of an electron in the direction of
vector a is a fuzzy process not an exact one. Particles can change spin
axes on detection to reverse the direction of their spin axes. That
suffices to point them in the general direction of (say) vector a. That
is, clearly pointing more towards say vector a rather than towards
vector a*, but not pointing exactly in line with vector a."
This is broadly-speaking true, isn't it?
[Moderator's note: Due to possibly garbled characters, the vector names
might not be what was intended. It seems that the point is to
distinguish two vectors, above denoted by a and a*. -P.H.]
ben...@hotmail.com
Feb 26, 2019, 5:18:54 PM2/26/19
to
Sorry, if that garbling of text was my fault.
The a* in the last line of the quotation should have been -a.
To put the question another way: if an electron with a hypothetical
exact spin vector p is subsequently prepared by presenting it to
to a detector with vector a (Alice's detector), then does the
electron exact spin axis p become changed to be exactly aligned
with vector a (or vector -a, whichever is appropriate)? IMO the
electron spin vector after preparation is either along exact vector
p or along exact vector -p. The outcome of p or -p alignment depends
on which of p and -p is nearest in alignment to vector a.
The a* in the last line of the quotation should have been -a.
To put the question another way: if an electron with a hypothetical
exact spin vector p is subsequently prepared by presenting it to
to a detector with vector a (Alice's detector), then does the
electron exact spin axis p become changed to be exactly aligned
with vector a (or vector -a, whichever is appropriate)? IMO the
electron spin vector after preparation is either along exact vector
p or along exact vector -p. The outcome of p or -p alignment depends
on which of p and -p is nearest in alignment to vector a.
Tom Roberts
Feb 28, 2019, 2:27:32 AM2/28/19
to
On 2/26/19 1:30 PM, ben...@hotmail.com wrote:
> "However, I do not believe that electrons can be aligned exactly in
> a given direction. [...]
> "However, I do not believe that electrons can be aligned exactly in
In the decay of a pi- to anti_nu_e and e-, the helicity of the electron
must be exactly +1, so its spin is exactly parallel to its 3-momentum.
No other value will satisfy the conservation of angular momentum.
Similarly for the much more frequent decay of pi- to anti_nu_mu and mu-.
Note that helicity is a scalar, so this applies in any inertial frame
(not just the pi- rest frame).
Tom Roberts
Jos Bergervoet
Feb 28, 2019, 9:49:34 PM2/28/19
to
Well, it was more the actual meaning that was somewhat puzzling..
Why do you not believe the electron to be polarized in direction
p, if it has been polarized in direction p (by having its spinor
associated to direction p via the Hopf fibration, I mean!)
Do you not believe that it can have a spinor which is some norm
times [ cos(th/2) exp(i phi), sin(th/2) ], with (th, phi) the
polar coordinate angles of p? Why would than not be possible?
Or do you believe that this spinor can indeed be the state of the
electron spin, but that is not equivalent to saying that it is
polarized in direction p? Why would you reject that definition?
> ..
who could answer it in that case?) If you accept quantum mechanics
then you don't have to ask. We know that it will *of course not*
become an exact spinor in direction a (nor -a). Instead we know
that it will be an entangled spinor, entangled with the state of
the detector that it passed through.
> IMO the
> electron spin vector after preparation is either along exact vector
> p or along exact vector -p.
Your opinion then differs from quantum mechanics, which tells us
that it isn't a spinor any more. Instead we have an entangled
state in a larger space than the spinor.
> The outcome of p or -p alignment depends
> on which of p and -p is nearest in alignment to vector a.
The entangled state which is the outcome would usually depend on
that alignment, but since you wrote (above) that after preparation
your state was "either along exact vector p or along exact vector
-p," you did not do an unambiguous preparation at all! If you'd
start with a random electron you'd have exactly what you claim to
have "prepared".
--
Jos
Why do you not believe the electron to be polarized in direction
p, if it has been polarized in direction p (by having its spinor
associated to direction p via the Hopf fibration, I mean!)
Do you not believe that it can have a spinor which is some norm
times [ cos(th/2) exp(i phi), sin(th/2) ], with (th, phi) the
polar coordinate angles of p? Why would than not be possible?
Or do you believe that this spinor can indeed be the state of the
electron spin, but that is not equivalent to saying that it is
polarized in direction p? Why would you reject that definition?
> ..
> To put the question another way: if an electron with a hypothetical
> exact spin vector p is subsequently prepared by presenting it to
> to a detector with vector a (Alice's detector), then does the
> electron exact spin axis p become changed to be exactly aligned
> with vector a (or vector -a, whichever is appropriate)?
If you reject quantum mechanics you can ask this question (but
> exact spin vector p is subsequently prepared by presenting it to
> to a detector with vector a (Alice's detector), then does the
> electron exact spin axis p become changed to be exactly aligned
> with vector a (or vector -a, whichever is appropriate)?
who could answer it in that case?) If you accept quantum mechanics
then you don't have to ask. We know that it will *of course not*
become an exact spinor in direction a (nor -a). Instead we know
that it will be an entangled spinor, entangled with the state of
the detector that it passed through.
> IMO the
> electron spin vector after preparation is either along exact vector
> p or along exact vector -p.
that it isn't a spinor any more. Instead we have an entangled
state in a larger space than the spinor.
> The outcome of p or -p alignment depends
> on which of p and -p is nearest in alignment to vector a.
that alignment, but since you wrote (above) that after preparation
your state was "either along exact vector p or along exact vector
-p," you did not do an unambiguous preparation at all! If you'd
start with a random electron you'd have exactly what you claim to
have "prepared".
--
Jos
Lawrence Crowell
Feb 28, 2019, 9:49:45 PM2/28/19
to
This can be done by heating a cathode and using a positive charged grid
to move the electrons and then accelerate with RF cavities. To prepare
the spin state one can have this electron beam pass through a highly
asymmetric magnetic field. The spinning electron with its magnetic moment
has a magnetic field. This will then force the electrons to be aligned or
anti-aligned. Then with magnetic means one controls one of these split
beams with spin up or down.
This is the Stern-Gerlach apparatus. This is usually discussed with respect
to measurement of spin. However, the preparation of a quantum state is really
much the same as measurement. The alignment of the electron spins in principle can be made as exact as possible. For the region of magnetic field having
the size d and the electrons with momentum vector k the angle uncertainty of
the beam or cone of the electron beam one prepares is an angle @ ~ (kd)^{-1}.
So for E ~ 1ev then k ~ 10^{7}cm^{-1} and if the apparatus magnetic port is
d ~ 1cm then the angle uncertainty is about 10^{-7}rad. The uncertainty in
the direction of the spins of the electrons will have a comparable value.
LC
ben...@hotmail.com
Mar 1, 2019, 4:09:33 AM3/1/19
to
On Thursday, February 28, 2019 at 7:27:32 AM UTC, Tom Roberts wrote:
Thank you, Tom.
I am not clear that the electron spin axis in your post has been aligned
exactly with a given direction. I agree that it is aligned in the
direction of its linear momentum, but that direction was not (could not
be?) a given or pre-selected direction.
Suppose the electron subsequently traveled on to meet Alice's detector.
In general, Alice's detector vector (a) could be at any angle to the
electron spin axis vector (p).
IMO what happens when the electron is detected by Alice is that either
the electron is already in its lowest energy state wrt the detector and
so will retain its spin vector p or else it will switch to a lower
energy state by emitting a photon and changing its own spin axis to one
along spin vector -p. (S-G detectors can give a reading of +1 or -1
rather than null or +1, but I used the easier version for convenience of
description.)
What IMO the electron does not do is change its exact spin axis to
vector +a or -a. But someone's model for a Bell test implies that I am
wrong.
Referring to your post, I am not sure where you write "the helicity of
the electron must be exactly +1," I thought that the helicity of an
electron cannot exceed 0.5? Unless you mean 100% projection onto the
direction of linear momentum. Or I wondered if you meant e- + anti_ต
had +1 helicity in total?
Commenting on the rest of your post to see if I understand ... The
anti_ต must have spin (both helicity and chirality) = +0.5 as it has
restricted handedness. That forces the electron to have chirality -0.5,
so that the chirality of the pair is zero which matches the original
chirality before decay. So the anti_ต moves off in one direction and
the electron moves in the opposite direction with chirality -0.5 but
with helicity +0.5.
I note that as helicity is defined wrt the linear momentum vector, then
helicity is tied to the direction of motion of the particle. So it is
not observer dependent. However I have always thought of chirality as
the more ph ysically important as it gives spin direction in the
particle's rest frame, despite difficulties in observing chirality
directly.
I am not clear that the electron spin axis in your post has been aligned
exactly with a given direction. I agree that it is aligned in the
direction of its linear momentum, but that direction was not (could not
be?) a given or pre-selected direction.
Suppose the electron subsequently traveled on to meet Alice's detector.
In general, Alice's detector vector (a) could be at any angle to the
electron spin axis vector (p).
IMO what happens when the electron is detected by Alice is that either
the electron is already in its lowest energy state wrt the detector and
so will retain its spin vector p or else it will switch to a lower
energy state by emitting a photon and changing its own spin axis to one
along spin vector -p. (S-G detectors can give a reading of +1 or -1
rather than null or +1, but I used the easier version for convenience of
description.)
What IMO the electron does not do is change its exact spin axis to
vector +a or -a. But someone's model for a Bell test implies that I am
wrong.
Referring to your post, I am not sure where you write "the helicity of
the electron must be exactly +1," I thought that the helicity of an
electron cannot exceed 0.5? Unless you mean 100% projection onto the
direction of linear momentum. Or I wondered if you meant e- + anti_ต
had +1 helicity in total?
Commenting on the rest of your post to see if I understand ... The
anti_ต must have spin (both helicity and chirality) = +0.5 as it has
restricted handedness. That forces the electron to have chirality -0.5,
so that the chirality of the pair is zero which matches the original
chirality before decay. So the anti_ต moves off in one direction and
the electron moves in the opposite direction with chirality -0.5 but
with helicity +0.5.
I note that as helicity is defined wrt the linear momentum vector, then
helicity is tied to the direction of motion of the particle. So it is
not observer dependent. However I have always thought of chirality as
the more ph ysically important as it gives spin direction in the
particle's rest frame, despite difficulties in observing chirality
directly.
ben...@hotmail.com
Mar 2, 2019, 3:45:26 AM3/2/19
to
Hi Lawrence and Jos
Thank you both very much for the information. I am clearly wrong in my
understanding of electron spin preparation and I bow to your expertise.
I am very surprised though that I am wrong.
The main reason this issue arose is that I am comparing a geometric
algebra model wrt Bell's Theorem against what I know from QM. In QM when
measuring along a single axis I always assumed that a measurement only
indicated that the axis pointed somewhere in the correct hemisphere,
rather than pointing exactly parallel to the detector vector. (I
suppose this means that I never quite trusted quantisation enough.) I
did follow 40 hours of Susskind's online "theoretical minimum" courses
on Entanglement and on QM, some years ago, which are quite mathematical.
I also know that a vector in 3D is not sufficient to completely define
a spin which requires QM and complex numbers etc. But in geometric
algebra the idea of a spin bivector pointing in normal 3D (or what only
at face value appears to be 3D space as it has three orthogonal basis
vectors.) implies an exact spin direction in 3D. So when the geometric
algebra model had every one of Alice's particles having its spin axis
exactly aligned parallel or antiparallel with Alice's detector vector, I
smelt a rat ... but I was wrong.
I also had thought that ... given an electron precessing in a varying
magnetic field, then at some stage the particle axis could align exactly
parallel to the detector vector. And maybe be detected at that point.
But as the electron is precessing, its axis's average direction would
not have been changed. Ie it would continue to precess about the
original axis after detection. But presumably that is wrong also. I
had assumed that one could precess an axis about its original vector,
but that it was extremely difficult, if not impossible, to change the
original vector.
Finally, for Jos, as the problem concerned a model disagreeing with
entanglement, I could not simply assume entanglement to be correct, even
if it is correct.
Thanks again.
Thank you both very much for the information. I am clearly wrong in my
understanding of electron spin preparation and I bow to your expertise.
I am very surprised though that I am wrong.
The main reason this issue arose is that I am comparing a geometric
algebra model wrt Bell's Theorem against what I know from QM. In QM when
measuring along a single axis I always assumed that a measurement only
indicated that the axis pointed somewhere in the correct hemisphere,
rather than pointing exactly parallel to the detector vector. (I
suppose this means that I never quite trusted quantisation enough.) I
did follow 40 hours of Susskind's online "theoretical minimum" courses
on Entanglement and on QM, some years ago, which are quite mathematical.
I also know that a vector in 3D is not sufficient to completely define
a spin which requires QM and complex numbers etc. But in geometric
algebra the idea of a spin bivector pointing in normal 3D (or what only
at face value appears to be 3D space as it has three orthogonal basis
vectors.) implies an exact spin direction in 3D. So when the geometric
algebra model had every one of Alice's particles having its spin axis
exactly aligned parallel or antiparallel with Alice's detector vector, I
smelt a rat ... but I was wrong.
I also had thought that ... given an electron precessing in a varying
magnetic field, then at some stage the particle axis could align exactly
parallel to the detector vector. And maybe be detected at that point.
But as the electron is precessing, its axis's average direction would
not have been changed. Ie it would continue to precess about the
original axis after detection. But presumably that is wrong also. I
had assumed that one could precess an axis about its original vector,
but that it was extremely difficult, if not impossible, to change the
original vector.
Finally, for Jos, as the problem concerned a model disagreeing with
entanglement, I could not simply assume entanglement to be correct, even
if it is correct.
Thanks again.
Lawrence Crowell
Mar 2, 2019, 4:01:31 AM3/2/19
to
On Friday, March 1, 2019 at 3:09:33 AM UTC-6, ben...@hotmail.com wrote:
>
> Thank you, Tom.
>
> I am not clear that the electron spin axis in your post has been aligned
> exactly with a given direction. I agree that it is aligned in the
> direction of its linear momentum, but that direction was not (could not
> be?) a given or pre-selected direction.
The electron with a mass has spin with projection along the momentum given
>
> Thank you, Tom.
>
> I am not clear that the electron spin axis in your post has been aligned
> exactly with a given direction. I agree that it is aligned in the
> direction of its linear momentum, but that direction was not (could not
> be?) a given or pre-selected direction.
by m = 1/2, -1/2. I indicated above what the uncertainty in momenta is with
measurement, and since the z-component of spin is projected along the
momentum the above uncertainty in angle will pertain to the spin of the electron.
LC
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