Particle interchange and z component of spin


by LostConjugate
Tags: component, interchange, particle, spin
LostConjugate
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#1
Jun6-11, 10:34 AM
P: 842
The one thing that is NOT identical for different particles in a state is the z component of spin, having a probability of being positive or negative.

So when interchanging particles how do you handle the z component of the spin, it simply can't be interchanged along with the position and total spin as you could end up with a different state.
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clem
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#2
Jun6-11, 10:49 AM
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If it is [tex]\uparrow\downarrow[/tex], you do get a different state.
Symmetric and antisymmetric combinations are eigenstates of total spin.
LostConjugate
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#3
Jun6-11, 11:44 AM
P: 842
Quote Quote by clem View Post
If it is [tex]\uparrow\downarrow[/tex], you do get a different state.
Symmetric and antisymmetric combinations are eigenstates of total spin.

So how can the original hypothesis hold. What happens to the z component of spin, it must be transfered to the swapped particle.

clem
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#4
Jun6-11, 07:19 PM
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Particle interchange and z component of spin


Quote Quote by LostConjugate View Post
So how can the original hypothesis hold. What happens to the z component of spin, it must be transfered to the swapped particle.
What is "the original hypothesis"?
When interchanging particles, all properties are interchanged.
SpectraCat
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#5
Jun6-11, 07:30 PM
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Quote Quote by clem View Post
What is "the original hypothesis"?
When interchanging particles, all properties are interchanged.
His point was that if the spin directions are opposite, then the particles are not identical. However of course that is not an issue, because any fermionic wavefunction must be properly (anti)symmetrized. So you never have a state like, [tex]\uparrow\downarrow[/tex] rather you have states like [tex]\frac{1}{\sqrt{2}}(\uparrow\downarrow - \downarrow\uparrow)[/tex] which are properly symmetrized.

[EDIT] that is a very simple example of how the Pauli exclusion principle is applied, and also why it is required.
matonski
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#6
Jun6-11, 07:39 PM
P: 166
Quote Quote by LostConjugate View Post
So when interchanging particles how do you handle the z component of the spin, it simply can't be interchanged along with the position and total spin as you could end up with a different state.
Everything is interchanged. Indeed, naively, you could end up with a different state. However, those states are just not allowed. The (anti) symmetrization requirements of identical particles severely restricts the possible multiple particle states in the Hilbert Space.
LostConjugate
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#7
Jun7-11, 11:07 AM
P: 842
Oh I do see here that not all states satisfy the interchange hypothesis. Only combinations of states. I still do not see how this just makes it OK to swap particles when there is some probability that they have different spin up/down states, physically it is not logical.
SpectraCat
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#8
Jun7-11, 11:43 AM
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Quote Quote by LostConjugate View Post
Oh I do see here that not all states satisfy the interchange hypothesis. Only combinations of states. I still do not see how this just makes it OK to swap particles when there is some probability that they have different spin up/down states, physically it is not logical.
You are not thinking about it correctly then ... the whole process of symmetrizing the states is to create wavefunctions where you CANNOT tell when the particles have been exchanged. If you have an equal mix of |up>|down> and |down>|up>, and you flip the spins, you will have and equal mix of |down>|up> and |up>|down> ... nothing has changed.
LostConjugate
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#9
Jun7-11, 11:58 AM
P: 842
Quote Quote by SpectraCat View Post
You are not thinking about it correctly then ... the whole process of symmetrizing the states is to create wavefunctions where you CANNOT tell when the particles have been exchanged. If you have an equal mix of |up>|down> and |down>|up>, and you flip the spins, you will have and equal mix of |down>|up> and |up>|down> ... nothing has changed.
Ok, so the idea is to carefully select only wave functions where particles can be exchanged.

An equal mix of up/down states for a large number of particles may work well.

What happens when you only have two particles that are involved in the wave function?
SpectraCat
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#10
Jun7-11, 12:04 PM
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Quote Quote by LostConjugate View Post

What happens when you only have two particles that are involved in the wave function?
The case I have been describing IS a two particle wavefunction.
LostConjugate
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#11
Jun7-11, 12:06 PM
P: 842
Quote Quote by SpectraCat View Post
His point was that if the spin directions are opposite, then the particles are not identical. However of course that is not an issue, because any fermionic wavefunction must be properly (anti)symmetrized. So you never have a state like, [tex]\uparrow\downarrow[/tex] rather you have states like [tex]\frac{1}{\sqrt{2}}(\uparrow\downarrow - \downarrow\uparrow)[/tex] which are properly symmetrized.

[EDIT] that is a very simple example of how the Pauli exclusion principle is applied, and also why it is required.
This puts each particle in a perfect superposition with equal probabilities which then makes "swapping" them non-trivial. Isn't that just a mathematical loop hole? Physically each particle could have opposite spins if measured.
SpectraCat
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#12
Jun7-11, 12:19 PM
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Quote Quote by LostConjugate View Post
This puts each particle in a perfect superposition with equal probabilities which then makes "swapping" them non-trivial. Isn't that just a mathematical loop hole? Physically each particle could have opposite spins if measured.
The "mathematical loophole" I described demonstrates how a wavefunction for a two-particle state with paired spins can conform to the Pauli exclusion principle, which we know from experimental evidence must be true. Identical fermions obey Fermi-Dirac statistics (another experimental fact) ... therefore writing a wavefunction for a fermion state that doesn't conform to those experimental facts doesn't make any sense. I cannot prove to you that the properly symmetrized wavefunction has any experimental reality .. I can only say that it is consistent with available experimental results, and has never been shown to be wrong.

For that example, they *will* have opposite spins when measured ... what does that have to do with anything? Once you measure the state, you "collapse" the wavefunction, so only one of the two possible components of the superposition can be observed. That is why we describe fermions in states like the one I gave as "spin-entangled".
LostConjugate
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#13
Jun7-11, 01:05 PM
P: 842
Quote Quote by SpectraCat View Post
The "mathematical loophole" I described demonstrates how a wavefunction for a two-particle state with paired spins can conform to the Pauli exclusion principle, which we know from experimental evidence must be true.

Identical fermions obey Fermi-Dirac statistics (another experimental fact) ... therefore writing a wavefunction for a fermion state that doesn't conform to those experimental facts doesn't make any sense.
That helps clear it up.


Quote Quote by SpectraCat View Post

For that example, they *will* have opposite spins when measured ... what does that have to do with anything? Once you measure the state, you "collapse" the wavefunction, so only one of the two possible components of the superposition can be observed. That is why we describe fermions in states like the one I gave as "spin-entangled".
I suppose because it is uncomfortable knowing that even without measurement you could be swapping spins that are not identical.


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