spinor representation Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)What are spinors mathematically?Fundamental chiral representations of SU(2)Two-Component Spinor Index PlacementSpinor bundle of a spin Manifold is a Clifford bundleUnderstanding of Spin(n) and SO(n)Spinor chiral transformation by $psi to gamma^5 psi$Why do the $Gamma$ matrices behave like vectors and tensors in the spinor representation of SO groups?Can we 'build' spinor structure not only from a Riemann Manifold but 'extract it' also from another algebraic structures?Spinors - Groups and Double Cover of Lorentz Group
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spinor representation
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)What are spinors mathematically?Fundamental chiral representations of SU(2)Two-Component Spinor Index PlacementSpinor bundle of a spin Manifold is a Clifford bundleUnderstanding of Spin(n) and SO(n)Spinor chiral transformation by $psi to gamma^5 psi$Why do the $Gamma$ matrices behave like vectors and tensors in the spinor representation of SO groups?Can we 'build' spinor structure not only from a Riemann Manifold but 'extract it' also from another algebraic structures?Spinors - Groups and Double Cover of Lorentz Group
$begingroup$
I am trying to study Special unitary group of order 2 and some textbooks mention objects transform under special unitary group are called Spinors then How can we represent a spinor using matrix ?
group-theory lie-groups spin-geometry
asked Mar 20 at 17:24
robinrobin
234
$endgroup$
|
show 1 more comment
$begingroup$
I am trying to study Special unitary group of order 2 and some textbooks mention objects transform under special unitary group are called Spinors then How can we represent a spinor using matrix ?
group-theory lie-groups spin-geometry
asked Mar 20 at 17:24
robinrobin
234
$endgroup$
$begingroup$
In quantum mechanics spinors are represented as column vectors. The thing that makes them spinors is the way the components of the column vector mix when a coordinate rotation occurs.
$endgroup$
– Spencer
Mar 20 at 17:54
$begingroup$
So according to you for SU (2) Spinor is a 2 by 1 matrix , isn't it ? @ Spencer
$endgroup$
– robin
Mar 20 at 18:21
$begingroup$
Yes a spinor can be represented that way.
$endgroup$
– Spencer
Mar 20 at 18:34
$begingroup$
Can spinors be 2 by 2 under SU (2) @ Spencer
$endgroup$
– robin
Mar 20 at 18:44
$begingroup$
Did I manage to answer your question?
$endgroup$
– Spencer
Mar 20 at 20:09
|
show 1 more comment
$begingroup$
I am trying to study Special unitary group of order 2 and some textbooks mention objects transform under special unitary group are called Spinors then How can we represent a spinor using matrix ?
group-theory lie-groups spin-geometry
asked Mar 20 at 17:24
robinrobin
234
$endgroup$
I am trying to study Special unitary group of order 2 and some textbooks mention objects transform under special unitary group are called Spinors then How can we represent a spinor using matrix ?
group-theory lie-groups spin-geometry
group-theory lie-groups spin-geometry
asked Mar 20 at 17:24
robinrobin
234
asked Mar 20 at 17:24
robinrobin
234
asked Mar 20 at 17:24
robinrobin
234
asked Mar 20 at 17:24
robinrobin
234
asked Mar 20 at 17:24
robinrobin
234
234
$begingroup$
In quantum mechanics spinors are represented as column vectors. The thing that makes them spinors is the way the components of the column vector mix when a coordinate rotation occurs.
$endgroup$
– Spencer
Mar 20 at 17:54
$begingroup$
So according to you for SU (2) Spinor is a 2 by 1 matrix , isn't it ? @ Spencer
$endgroup$
– robin
Mar 20 at 18:21
$begingroup$
Yes a spinor can be represented that way.
$endgroup$
– Spencer
Mar 20 at 18:34
$begingroup$
Can spinors be 2 by 2 under SU (2) @ Spencer
$endgroup$
– robin
Mar 20 at 18:44
$begingroup$
Did I manage to answer your question?
$endgroup$
– Spencer
Mar 20 at 20:09
|
show 1 more comment
$begingroup$
In quantum mechanics spinors are represented as column vectors. The thing that makes them spinors is the way the components of the column vector mix when a coordinate rotation occurs.
$endgroup$
– Spencer
Mar 20 at 17:54
$begingroup$
So according to you for SU (2) Spinor is a 2 by 1 matrix , isn't it ? @ Spencer
$endgroup$
– robin
Mar 20 at 18:21
$begingroup$
Yes a spinor can be represented that way.
$endgroup$
– Spencer
Mar 20 at 18:34
$begingroup$
Can spinors be 2 by 2 under SU (2) @ Spencer
$endgroup$
– robin
Mar 20 at 18:44
$begingroup$
Did I manage to answer your question?
$endgroup$
– Spencer
Mar 20 at 20:09
$begingroup$
In quantum mechanics spinors are represented as column vectors. The thing that makes them spinors is the way the components of the column vector mix when a coordinate rotation occurs.
$endgroup$
– Spencer
Mar 20 at 17:54
$begingroup$
In quantum mechanics spinors are represented as column vectors. The thing that makes them spinors is the way the components of the column vector mix when a coordinate rotation occurs.
$endgroup$
– Spencer
Mar 20 at 17:54
$begingroup$
So according to you for SU (2) Spinor is a 2 by 1 matrix , isn't it ? @ Spencer
$endgroup$
– robin
Mar 20 at 18:21
$begingroup$
So according to you for SU (2) Spinor is a 2 by 1 matrix , isn't it ? @ Spencer
$endgroup$
– robin
Mar 20 at 18:21
$begingroup$
Yes a spinor can be represented that way.
$endgroup$
– Spencer
Mar 20 at 18:34
$begingroup$
Yes a spinor can be represented that way.
$endgroup$
– Spencer
Mar 20 at 18:34
$begingroup$
Can spinors be 2 by 2 under SU (2) @ Spencer
$endgroup$
– robin
Mar 20 at 18:44
$begingroup$
Can spinors be 2 by 2 under SU (2) @ Spencer
$endgroup$
– robin
Mar 20 at 18:44
$begingroup$
Did I manage to answer your question?
$endgroup$
– Spencer
Mar 20 at 20:09
$begingroup$
Did I manage to answer your question?
$endgroup$
– Spencer
Mar 20 at 20:09
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
$SU(2)$ can be defined as the group of determinant-$1$ transformations which leaves,
$$ | z_1 |^2 + |z_2|^2 = 1,$$
invariant.
The transformations themselves act on the two complex numbers $z_1$ and $z_2$. It is common to represent $z_1$ and $z_2$ as the two components of a complex column vector,
$$ left(beginarray z_1 \ z_2 endarrayright),$$
but there is nothing to prevent us from representing these complex numbers in a $2times 2$ matrix.
In particular the actual elements of $SU(2)$ can be thought of as spinors. From wikipedia we have the group elements of $SU(2)$ can be represented as
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) qquad |alpha|^2 + |beta|^2 = 1,$$
notice that this matrix is determined by only two complex numbers which satisfy the same normalization as the $z_1$ and $z_2$ indicated previously. In fact this matrix is completely determined by the spinor in its first column.
To see that the representation is really faithful let us compare the action of group elements when they act on our spinors.
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) left( beginarray z_1 \ z_2endarray right) = left( beginarray alpha z_1 - barbeta z_1 \ beta z_1 + baralpha z_2endarrayright)$$
Now consider
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) left( beginarray z_1 & - barz_2 \ z_2 & barz_1 endarrayright)$$
$$ = left( beginarray alpha z_1 - barbeta z_2 & -alpha barz_2 - barbeta barz_1 \ beta z_1 + baralpha z_2 & -beta barz_2 + baralpha barz_1 endarrayright) $$
$$ = left( beginarray alpha z_1 - barbeta z_2 & overline -baralpha z_2 - beta z_1 \ beta z_1 + baralpha z_2 & overline-barbeta z_2 + alpha z_1 endarrayright) $$
$$ = left( beginarray alpha z_1 - barbeta z_2 & -overlineleft( baralpha z_2 + beta z_1right) \ beta z_1 + baralpha z_2 & overline alpha z_1 -barbeta z_2 endarrayright) $$
$$ = left( beginarray alpha z_1 - barbeta z_2 & -overlineleft( beta z_1 + baralpha z_2 right) \ beta z_1 + baralpha z_2 & overline alpha z_1 -barbeta z_2 endarrayright) $$
You can see that the $z_1$ and $z_2$ are mixed in the same way for both transformations.
It doesn't cause any difficulty if we allow the quadratic form $|z_1|^2+|z_2|^2$ to equal any real number.
The transformation matrices would remain the same.
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) qquad |alpha|^2 + |beta|^2 = 1,$$
The spinors would still be expressible as $2times 2$ matrices.
$$ left( beginarray z_1 & - barz_2 \ z_2 & barz_1 endarrayright)$$
answered Mar 20 at 19:32
SpencerSpencer
8,74012156
$endgroup$
$begingroup$
"SU(2) can be defined as the group of determinant-1 transformations which leaves, |z1|2+|z2|2=1, invariant.", But i read in a book that matrix which preserve quadratic form ( |z1|2+|z2|2 ) invarient is called SU (2) Can you please clarify me @ Spencer
$endgroup$
– robin
Mar 21 at 13:28
$begingroup$
Can we represent a vector defined over the field of real numbers in spinor notation
$endgroup$
– robin
Mar 21 at 14:01
$begingroup$
With regards to the definition of SU(2). It doesn't change the matrices, but it does change your space of spinors slightly. I will amend my answer to accommodate this.
$endgroup$
– Spencer
Mar 21 at 15:21
$begingroup$
With regards to real numbers, you certainly can define spinors that way. If a spinors is described by a pair of complex numbers, then it can be described by 4 real numbers. Your SU(2) matrices would have to be 4x4 in that case.
$endgroup$
– Spencer
Mar 21 at 15:22
1
$begingroup$
@Robin You can use the Pauli matrices to represent a vector in 2x2 form - see the section "Pauli vector" in en.wikipedia.org/wiki/Pauli_matrices
$endgroup$
– iSeeker
Mar 25 at 15:10
|
show 3 more comments
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$SU(2)$ can be defined as the group of determinant-$1$ transformations which leaves,
$$ | z_1 |^2 + |z_2|^2 = 1,$$
invariant.
The transformations themselves act on the two complex numbers $z_1$ and $z_2$. It is common to represent $z_1$ and $z_2$ as the two components of a complex column vector,
$$ left(beginarray z_1 \ z_2 endarrayright),$$
but there is nothing to prevent us from representing these complex numbers in a $2times 2$ matrix.
In particular the actual elements of $SU(2)$ can be thought of as spinors. From wikipedia we have the group elements of $SU(2)$ can be represented as
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) qquad |alpha|^2 + |beta|^2 = 1,$$
notice that this matrix is determined by only two complex numbers which satisfy the same normalization as the $z_1$ and $z_2$ indicated previously. In fact this matrix is completely determined by the spinor in its first column.
To see that the representation is really faithful let us compare the action of group elements when they act on our spinors.
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) left( beginarray z_1 \ z_2endarray right) = left( beginarray alpha z_1 - barbeta z_1 \ beta z_1 + baralpha z_2endarrayright)$$
Now consider
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) left( beginarray z_1 & - barz_2 \ z_2 & barz_1 endarrayright)$$
$$ = left( beginarray alpha z_1 - barbeta z_2 & -alpha barz_2 - barbeta barz_1 \ beta z_1 + baralpha z_2 & -beta barz_2 + baralpha barz_1 endarrayright) $$
$$ = left( beginarray alpha z_1 - barbeta z_2 & overline -baralpha z_2 - beta z_1 \ beta z_1 + baralpha z_2 & overline-barbeta z_2 + alpha z_1 endarrayright) $$
$$ = left( beginarray alpha z_1 - barbeta z_2 & -overlineleft( baralpha z_2 + beta z_1right) \ beta z_1 + baralpha z_2 & overline alpha z_1 -barbeta z_2 endarrayright) $$
$$ = left( beginarray alpha z_1 - barbeta z_2 & -overlineleft( beta z_1 + baralpha z_2 right) \ beta z_1 + baralpha z_2 & overline alpha z_1 -barbeta z_2 endarrayright) $$
You can see that the $z_1$ and $z_2$ are mixed in the same way for both transformations.
It doesn't cause any difficulty if we allow the quadratic form $|z_1|^2+|z_2|^2$ to equal any real number.
The transformation matrices would remain the same.
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) qquad |alpha|^2 + |beta|^2 = 1,$$
The spinors would still be expressible as $2times 2$ matrices.
$$ left( beginarray z_1 & - barz_2 \ z_2 & barz_1 endarrayright)$$
answered Mar 20 at 19:32
SpencerSpencer
8,74012156
$endgroup$
$begingroup$
"SU(2) can be defined as the group of determinant-1 transformations which leaves, |z1|2+|z2|2=1, invariant.", But i read in a book that matrix which preserve quadratic form ( |z1|2+|z2|2 ) invarient is called SU (2) Can you please clarify me @ Spencer
$endgroup$
– robin
Mar 21 at 13:28
$begingroup$
Can we represent a vector defined over the field of real numbers in spinor notation
$endgroup$
– robin
Mar 21 at 14:01
$begingroup$
With regards to the definition of SU(2). It doesn't change the matrices, but it does change your space of spinors slightly. I will amend my answer to accommodate this.
$endgroup$
– Spencer
Mar 21 at 15:21
$begingroup$
With regards to real numbers, you certainly can define spinors that way. If a spinors is described by a pair of complex numbers, then it can be described by 4 real numbers. Your SU(2) matrices would have to be 4x4 in that case.
$endgroup$
– Spencer
Mar 21 at 15:22
1
$begingroup$
@Robin You can use the Pauli matrices to represent a vector in 2x2 form - see the section "Pauli vector" in en.wikipedia.org/wiki/Pauli_matrices
$endgroup$
– iSeeker
Mar 25 at 15:10
|
show 3 more comments
$begingroup$
$SU(2)$ can be defined as the group of determinant-$1$ transformations which leaves,
$$ | z_1 |^2 + |z_2|^2 = 1,$$
invariant.
The transformations themselves act on the two complex numbers $z_1$ and $z_2$. It is common to represent $z_1$ and $z_2$ as the two components of a complex column vector,
$$ left(beginarray z_1 \ z_2 endarrayright),$$
but there is nothing to prevent us from representing these complex numbers in a $2times 2$ matrix.
In particular the actual elements of $SU(2)$ can be thought of as spinors. From wikipedia we have the group elements of $SU(2)$ can be represented as
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) qquad |alpha|^2 + |beta|^2 = 1,$$
notice that this matrix is determined by only two complex numbers which satisfy the same normalization as the $z_1$ and $z_2$ indicated previously. In fact this matrix is completely determined by the spinor in its first column.
To see that the representation is really faithful let us compare the action of group elements when they act on our spinors.
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) left( beginarray z_1 \ z_2endarray right) = left( beginarray alpha z_1 - barbeta z_1 \ beta z_1 + baralpha z_2endarrayright)$$
Now consider
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) left( beginarray z_1 & - barz_2 \ z_2 & barz_1 endarrayright)$$
$$ = left( beginarray alpha z_1 - barbeta z_2 & -alpha barz_2 - barbeta barz_1 \ beta z_1 + baralpha z_2 & -beta barz_2 + baralpha barz_1 endarrayright) $$
$$ = left( beginarray alpha z_1 - barbeta z_2 & overline -baralpha z_2 - beta z_1 \ beta z_1 + baralpha z_2 & overline-barbeta z_2 + alpha z_1 endarrayright) $$
$$ = left( beginarray alpha z_1 - barbeta z_2 & -overlineleft( baralpha z_2 + beta z_1right) \ beta z_1 + baralpha z_2 & overline alpha z_1 -barbeta z_2 endarrayright) $$
$$ = left( beginarray alpha z_1 - barbeta z_2 & -overlineleft( beta z_1 + baralpha z_2 right) \ beta z_1 + baralpha z_2 & overline alpha z_1 -barbeta z_2 endarrayright) $$
You can see that the $z_1$ and $z_2$ are mixed in the same way for both transformations.
It doesn't cause any difficulty if we allow the quadratic form $|z_1|^2+|z_2|^2$ to equal any real number.
The transformation matrices would remain the same.
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) qquad |alpha|^2 + |beta|^2 = 1,$$
The spinors would still be expressible as $2times 2$ matrices.
$$ left( beginarray z_1 & - barz_2 \ z_2 & barz_1 endarrayright)$$
answered Mar 20 at 19:32
SpencerSpencer
8,74012156
$endgroup$
$begingroup$
"SU(2) can be defined as the group of determinant-1 transformations which leaves, |z1|2+|z2|2=1, invariant.", But i read in a book that matrix which preserve quadratic form ( |z1|2+|z2|2 ) invarient is called SU (2) Can you please clarify me @ Spencer
$endgroup$
– robin
Mar 21 at 13:28
$begingroup$
Can we represent a vector defined over the field of real numbers in spinor notation
$endgroup$
– robin
Mar 21 at 14:01
$begingroup$
With regards to the definition of SU(2). It doesn't change the matrices, but it does change your space of spinors slightly. I will amend my answer to accommodate this.
$endgroup$
– Spencer
Mar 21 at 15:21
$begingroup$
With regards to real numbers, you certainly can define spinors that way. If a spinors is described by a pair of complex numbers, then it can be described by 4 real numbers. Your SU(2) matrices would have to be 4x4 in that case.
$endgroup$
– Spencer
Mar 21 at 15:22
1
$begingroup$
@Robin You can use the Pauli matrices to represent a vector in 2x2 form - see the section "Pauli vector" in en.wikipedia.org/wiki/Pauli_matrices
$endgroup$
– iSeeker
Mar 25 at 15:10
|
show 3 more comments
$begingroup$
$SU(2)$ can be defined as the group of determinant-$1$ transformations which leaves,
$$ | z_1 |^2 + |z_2|^2 = 1,$$
invariant.
The transformations themselves act on the two complex numbers $z_1$ and $z_2$. It is common to represent $z_1$ and $z_2$ as the two components of a complex column vector,
$$ left(beginarray z_1 \ z_2 endarrayright),$$
but there is nothing to prevent us from representing these complex numbers in a $2times 2$ matrix.
In particular the actual elements of $SU(2)$ can be thought of as spinors. From wikipedia we have the group elements of $SU(2)$ can be represented as
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) qquad |alpha|^2 + |beta|^2 = 1,$$
notice that this matrix is determined by only two complex numbers which satisfy the same normalization as the $z_1$ and $z_2$ indicated previously. In fact this matrix is completely determined by the spinor in its first column.
To see that the representation is really faithful let us compare the action of group elements when they act on our spinors.
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) left( beginarray z_1 \ z_2endarray right) = left( beginarray alpha z_1 - barbeta z_1 \ beta z_1 + baralpha z_2endarrayright)$$
Now consider
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) left( beginarray z_1 & - barz_2 \ z_2 & barz_1 endarrayright)$$
$$ = left( beginarray alpha z_1 - barbeta z_2 & -alpha barz_2 - barbeta barz_1 \ beta z_1 + baralpha z_2 & -beta barz_2 + baralpha barz_1 endarrayright) $$
$$ = left( beginarray alpha z_1 - barbeta z_2 & overline -baralpha z_2 - beta z_1 \ beta z_1 + baralpha z_2 & overline-barbeta z_2 + alpha z_1 endarrayright) $$
$$ = left( beginarray alpha z_1 - barbeta z_2 & -overlineleft( baralpha z_2 + beta z_1right) \ beta z_1 + baralpha z_2 & overline alpha z_1 -barbeta z_2 endarrayright) $$
$$ = left( beginarray alpha z_1 - barbeta z_2 & -overlineleft( beta z_1 + baralpha z_2 right) \ beta z_1 + baralpha z_2 & overline alpha z_1 -barbeta z_2 endarrayright) $$
You can see that the $z_1$ and $z_2$ are mixed in the same way for both transformations.
It doesn't cause any difficulty if we allow the quadratic form $|z_1|^2+|z_2|^2$ to equal any real number.
The transformation matrices would remain the same.
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) qquad |alpha|^2 + |beta|^2 = 1,$$
The spinors would still be expressible as $2times 2$ matrices.
$$ left( beginarray z_1 & - barz_2 \ z_2 & barz_1 endarrayright)$$
answered Mar 20 at 19:32
SpencerSpencer
8,74012156
$endgroup$
$SU(2)$ can be defined as the group of determinant-$1$ transformations which leaves,
$$ | z_1 |^2 + |z_2|^2 = 1,$$
invariant.
The transformations themselves act on the two complex numbers $z_1$ and $z_2$. It is common to represent $z_1$ and $z_2$ as the two components of a complex column vector,
$$ left(beginarray z_1 \ z_2 endarrayright),$$
but there is nothing to prevent us from representing these complex numbers in a $2times 2$ matrix.
In particular the actual elements of $SU(2)$ can be thought of as spinors. From wikipedia we have the group elements of $SU(2)$ can be represented as
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) qquad |alpha|^2 + |beta|^2 = 1,$$
notice that this matrix is determined by only two complex numbers which satisfy the same normalization as the $z_1$ and $z_2$ indicated previously. In fact this matrix is completely determined by the spinor in its first column.
To see that the representation is really faithful let us compare the action of group elements when they act on our spinors.
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) left( beginarray z_1 \ z_2endarray right) = left( beginarray alpha z_1 - barbeta z_1 \ beta z_1 + baralpha z_2endarrayright)$$
Now consider
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) left( beginarray z_1 & - barz_2 \ z_2 & barz_1 endarrayright)$$
$$ = left( beginarray alpha z_1 - barbeta z_2 & -alpha barz_2 - barbeta barz_1 \ beta z_1 + baralpha z_2 & -beta barz_2 + baralpha barz_1 endarrayright) $$
$$ = left( beginarray alpha z_1 - barbeta z_2 & overline -baralpha z_2 - beta z_1 \ beta z_1 + baralpha z_2 & overline-barbeta z_2 + alpha z_1 endarrayright) $$
$$ = left( beginarray alpha z_1 - barbeta z_2 & -overlineleft( baralpha z_2 + beta z_1right) \ beta z_1 + baralpha z_2 & overline alpha z_1 -barbeta z_2 endarrayright) $$
$$ = left( beginarray alpha z_1 - barbeta z_2 & -overlineleft( beta z_1 + baralpha z_2 right) \ beta z_1 + baralpha z_2 & overline alpha z_1 -barbeta z_2 endarrayright) $$
You can see that the $z_1$ and $z_2$ are mixed in the same way for both transformations.
It doesn't cause any difficulty if we allow the quadratic form $|z_1|^2+|z_2|^2$ to equal any real number.
The transformation matrices would remain the same.
$$ left( beginarray alpha & - barbeta \ beta & baralpha endarrayright) qquad |alpha|^2 + |beta|^2 = 1,$$
The spinors would still be expressible as $2times 2$ matrices.
$$ left( beginarray z_1 & - barz_2 \ z_2 & barz_1 endarrayright)$$
answered Mar 20 at 19:32
SpencerSpencer
8,74012156
edited Apr 1 at 23:48
answered Mar 20 at 19:32
SpencerSpencer
8,74012156
answered Mar 20 at 19:32
SpencerSpencer
8,74012156
answered Mar 20 at 19:32
SpencerSpencer
8,74012156
8,74012156
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"SU(2) can be defined as the group of determinant-1 transformations which leaves, |z1|2+|z2|2=1, invariant.", But i read in a book that matrix which preserve quadratic form ( |z1|2+|z2|2 ) invarient is called SU (2) Can you please clarify me @ Spencer
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– robin
Mar 21 at 13:28
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Can we represent a vector defined over the field of real numbers in spinor notation
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– robin
Mar 21 at 14:01
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With regards to the definition of SU(2). It doesn't change the matrices, but it does change your space of spinors slightly. I will amend my answer to accommodate this.
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– Spencer
Mar 21 at 15:21
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With regards to real numbers, you certainly can define spinors that way. If a spinors is described by a pair of complex numbers, then it can be described by 4 real numbers. Your SU(2) matrices would have to be 4x4 in that case.
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– Spencer
Mar 21 at 15:22
1
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@Robin You can use the Pauli matrices to represent a vector in 2x2 form - see the section "Pauli vector" in en.wikipedia.org/wiki/Pauli_matrices
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– iSeeker
Mar 25 at 15:10
|
show 3 more comments
$begingroup$
"SU(2) can be defined as the group of determinant-1 transformations which leaves, |z1|2+|z2|2=1, invariant.", But i read in a book that matrix which preserve quadratic form ( |z1|2+|z2|2 ) invarient is called SU (2) Can you please clarify me @ Spencer
$endgroup$
– robin
Mar 21 at 13:28
$begingroup$
Can we represent a vector defined over the field of real numbers in spinor notation
$endgroup$
– robin
Mar 21 at 14:01
$begingroup$
With regards to the definition of SU(2). It doesn't change the matrices, but it does change your space of spinors slightly. I will amend my answer to accommodate this.
$endgroup$
– Spencer
Mar 21 at 15:21
$begingroup$
With regards to real numbers, you certainly can define spinors that way. If a spinors is described by a pair of complex numbers, then it can be described by 4 real numbers. Your SU(2) matrices would have to be 4x4 in that case.
$endgroup$
– Spencer
Mar 21 at 15:22
1
$begingroup$
@Robin You can use the Pauli matrices to represent a vector in 2x2 form - see the section "Pauli vector" in en.wikipedia.org/wiki/Pauli_matrices
$endgroup$
– iSeeker
Mar 25 at 15:10
$begingroup$
"SU(2) can be defined as the group of determinant-1 transformations which leaves, |z1|2+|z2|2=1, invariant.", But i read in a book that matrix which preserve quadratic form ( |z1|2+|z2|2 ) invarient is called SU (2) Can you please clarify me @ Spencer
$endgroup$
– robin
Mar 21 at 13:28
$begingroup$
"SU(2) can be defined as the group of determinant-1 transformations which leaves, |z1|2+|z2|2=1, invariant.", But i read in a book that matrix which preserve quadratic form ( |z1|2+|z2|2 ) invarient is called SU (2) Can you please clarify me @ Spencer
$endgroup$
– robin
Mar 21 at 13:28
$begingroup$
Can we represent a vector defined over the field of real numbers in spinor notation
$endgroup$
– robin
Mar 21 at 14:01
$begingroup$
Can we represent a vector defined over the field of real numbers in spinor notation
$endgroup$
– robin
Mar 21 at 14:01
$begingroup$
With regards to the definition of SU(2). It doesn't change the matrices, but it does change your space of spinors slightly. I will amend my answer to accommodate this.
$endgroup$
– Spencer
Mar 21 at 15:21
$begingroup$
With regards to the definition of SU(2). It doesn't change the matrices, but it does change your space of spinors slightly. I will amend my answer to accommodate this.
$endgroup$
– Spencer
Mar 21 at 15:21
$begingroup$
With regards to real numbers, you certainly can define spinors that way. If a spinors is described by a pair of complex numbers, then it can be described by 4 real numbers. Your SU(2) matrices would have to be 4x4 in that case.
$endgroup$
– Spencer
Mar 21 at 15:22
$begingroup$
With regards to real numbers, you certainly can define spinors that way. If a spinors is described by a pair of complex numbers, then it can be described by 4 real numbers. Your SU(2) matrices would have to be 4x4 in that case.
$endgroup$
– Spencer
Mar 21 at 15:22
1
1
$begingroup$
@Robin You can use the Pauli matrices to represent a vector in 2x2 form - see the section "Pauli vector" in en.wikipedia.org/wiki/Pauli_matrices
$endgroup$
– iSeeker
Mar 25 at 15:10
$begingroup$
@Robin You can use the Pauli matrices to represent a vector in 2x2 form - see the section "Pauli vector" in en.wikipedia.org/wiki/Pauli_matrices
$endgroup$
– iSeeker
Mar 25 at 15:10
|
show 3 more comments
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$begingroup$
In quantum mechanics spinors are represented as column vectors. The thing that makes them spinors is the way the components of the column vector mix when a coordinate rotation occurs.
$endgroup$
– Spencer
Mar 20 at 17:54
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So according to you for SU (2) Spinor is a 2 by 1 matrix , isn't it ? @ Spencer
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– robin
Mar 20 at 18:21
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Yes a spinor can be represented that way.
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– Spencer
Mar 20 at 18:34
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Can spinors be 2 by 2 under SU (2) @ Spencer
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– robin
Mar 20 at 18:44
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Did I manage to answer your question?
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– Spencer
Mar 20 at 20:09