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What is the difference between vector space and dual space?
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I read that in Dirac notation, kets are elements of a vector space and bras are elements of the dual space. My question is, what is the difference between vector space and dual space, and why are bras elements of the dual space?
I hope to answer me quickly. Thank you.
linear-algebra definition mathematical-physics quantum-mechanics dual-spaces
edited 3 hours ago
Bernard
110k635102
asked 4 hours ago
yana ghanim
332
add a comment |Â
up vote
6
down vote
favorite
I read that in Dirac notation, kets are elements of a vector space and bras are elements of the dual space. My question is, what is the difference between vector space and dual space, and why are bras elements of the dual space?
I hope to answer me quickly. Thank you.
linear-algebra definition mathematical-physics quantum-mechanics dual-spaces
edited 3 hours ago
Bernard
110k635102
asked 4 hours ago
yana ghanim
332
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I read that in Dirac notation, kets are elements of a vector space and bras are elements of the dual space. My question is, what is the difference between vector space and dual space, and why are bras elements of the dual space?
I hope to answer me quickly. Thank you.
linear-algebra definition mathematical-physics quantum-mechanics dual-spaces
edited 3 hours ago
Bernard
110k635102
asked 4 hours ago
yana ghanim
332
I read that in Dirac notation, kets are elements of a vector space and bras are elements of the dual space. My question is, what is the difference between vector space and dual space, and why are bras elements of the dual space?
I hope to answer me quickly. Thank you.
linear-algebra definition mathematical-physics quantum-mechanics dual-spaces
edited 3 hours ago
Bernard
110k635102
asked 4 hours ago
yana ghanim
332
edited 3 hours ago
Bernard
110k635102
edited 3 hours ago
Bernard
110k635102
edited 3 hours ago
Bernard
110k635102
110k635102
asked 4 hours ago
yana ghanim
332
asked 4 hours ago
yana ghanim
332
asked 4 hours ago
yana ghanim
332
332
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
A vector space over a field $mathbbF$ is a set $V$ with operations $+$ and $cdot$ satisfying the vector space axioms. Given a vector space $V$, it's dual space $V^star$ is defined as $mathrmHom(V,mathbbF)$, i.e. the set of all linear maps(functionals) between the vector space and its underlying field(considered as an own vector space in this case).
Normally, the Dirac notation $langle v|wrangle$ is a representation of a scalar product and the Bra and Ket correspond on a low level to vectors input in this scalar product. Note, that for the complex scalar product, order of the arguments is important due to its hermitian nature and its semi-bilinearity, i.e. that it is conjugate linear w.r.t. to one argument. A classical way is to define the standard complex scalar product to be conjugate linear in the second argument. In Bra-Ket-notation, you usually reverse the order of the arguments of the complex scalar product, i.e. $langle x,yrangle=langle y|xrangle$ resulting in conjugate linearity in the first argument.
The key thing is that there is a strong correspondence between scalar products and members of the dual space, i.e. linear functionals.
There is a way that any functional corresponds in a one-to-one fashion to a representation using the scalar product(of the associated vector space). This is known as Riesz representation theorem.
More precisely, you may look at a vector $vin V$ and suppose that $langlecdot,cdotrangle$ is an associated scalar product(turning $V$ into a euclidean/unitary space in the real/complex case). Then the map $varphi_v:wmapstolangle w,vrangle$ is a member of the dual space $V^star$ and the theorem says that any linear functional $psiin V^star$ can be written as such a $varphi_v$ uniquely.
Thus, you may convert a scalar product between two vectors into an application of a linear functional to another vector, pulling the problem statement into the realm of dual spaces, where you have other mathematical possibilities to tackle various questions.
EDIT: Note, that the definition of $varphi_v$ of course depends also on the argument which is assumed to be conjugate linear, in this case the second, as linearity in the first is needed to make $varphi_v$ a linear map(check this).
answered 4 hours ago
zzuussee
1,057319
Thanks for the edit, how silly of me to write a name in lower case.
– zzuussee
3 hours ago
thank you for your answer.... ...I understand that that with ket we work in linear space...but when we built a relation as you write"Æv:w↦⟨w,v⟩ is a member of the dual space V⋆" between ket and its conjugate bra ,we will have dual space
– yana ghanim
2 hours ago
@yanaghanim I don't really know what you're asking or why you re-decided in accepting the answer but I would like to resolve any issues you might have with the concept.
– zzuussee
2 hours ago
thank you ...I want to now what the meaning of dual space...why we use it with bra not with ket???
– yana ghanim
2 hours ago
1
@yanaghanim I've edited my answer to add some more information. For the complex scalar product, order of the arguments is important due to its hermitian nature and its semi-bilinearity, i.e. that it is conjugate linear w.r.t. to one argument. Note that the Bra-Ket-notation usually reverses the order of the arguments of the complex scalar product, i.e. $langle x,yrangle=langle y|xrangle$. This is more a notational detail. It is more fruitful, I think, to understand the nature of the correspondence between the dual space and scalar product.
– zzuussee
2 hours ago
add a comment |Â
up vote
1
down vote
For the finite dimensional case we have for a column vector $u$:
$$
u = vert u rangle\
u^+= langle u vert
$$
where $u^+$ is the transposed, complex conjugated, thus adjugated, vector.
It is a linear form from $V$ to the scalar field.
If $V$ is a vector space then $V^*$, consisting of all linear forms of $V$, is a linear vector space as well.
answered 3 hours ago
mvw
30.1k22150
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
A vector space over a field $mathbbF$ is a set $V$ with operations $+$ and $cdot$ satisfying the vector space axioms. Given a vector space $V$, it's dual space $V^star$ is defined as $mathrmHom(V,mathbbF)$, i.e. the set of all linear maps(functionals) between the vector space and its underlying field(considered as an own vector space in this case).
Normally, the Dirac notation $langle v|wrangle$ is a representation of a scalar product and the Bra and Ket correspond on a low level to vectors input in this scalar product. Note, that for the complex scalar product, order of the arguments is important due to its hermitian nature and its semi-bilinearity, i.e. that it is conjugate linear w.r.t. to one argument. A classical way is to define the standard complex scalar product to be conjugate linear in the second argument. In Bra-Ket-notation, you usually reverse the order of the arguments of the complex scalar product, i.e. $langle x,yrangle=langle y|xrangle$ resulting in conjugate linearity in the first argument.
The key thing is that there is a strong correspondence between scalar products and members of the dual space, i.e. linear functionals.
There is a way that any functional corresponds in a one-to-one fashion to a representation using the scalar product(of the associated vector space). This is known as Riesz representation theorem.
More precisely, you may look at a vector $vin V$ and suppose that $langlecdot,cdotrangle$ is an associated scalar product(turning $V$ into a euclidean/unitary space in the real/complex case). Then the map $varphi_v:wmapstolangle w,vrangle$ is a member of the dual space $V^star$ and the theorem says that any linear functional $psiin V^star$ can be written as such a $varphi_v$ uniquely.
Thus, you may convert a scalar product between two vectors into an application of a linear functional to another vector, pulling the problem statement into the realm of dual spaces, where you have other mathematical possibilities to tackle various questions.
EDIT: Note, that the definition of $varphi_v$ of course depends also on the argument which is assumed to be conjugate linear, in this case the second, as linearity in the first is needed to make $varphi_v$ a linear map(check this).
answered 4 hours ago
zzuussee
1,057319
Thanks for the edit, how silly of me to write a name in lower case.
– zzuussee
3 hours ago
thank you for your answer.... ...I understand that that with ket we work in linear space...but when we built a relation as you write"Æv:w↦⟨w,v⟩ is a member of the dual space V⋆" between ket and its conjugate bra ,we will have dual space
– yana ghanim
2 hours ago
@yanaghanim I don't really know what you're asking or why you re-decided in accepting the answer but I would like to resolve any issues you might have with the concept.
– zzuussee
2 hours ago
thank you ...I want to now what the meaning of dual space...why we use it with bra not with ket???
– yana ghanim
2 hours ago
1
@yanaghanim I've edited my answer to add some more information. For the complex scalar product, order of the arguments is important due to its hermitian nature and its semi-bilinearity, i.e. that it is conjugate linear w.r.t. to one argument. Note that the Bra-Ket-notation usually reverses the order of the arguments of the complex scalar product, i.e. $langle x,yrangle=langle y|xrangle$. This is more a notational detail. It is more fruitful, I think, to understand the nature of the correspondence between the dual space and scalar product.
– zzuussee
2 hours ago
add a comment |Â
up vote
5
down vote
accepted
A vector space over a field $mathbbF$ is a set $V$ with operations $+$ and $cdot$ satisfying the vector space axioms. Given a vector space $V$, it's dual space $V^star$ is defined as $mathrmHom(V,mathbbF)$, i.e. the set of all linear maps(functionals) between the vector space and its underlying field(considered as an own vector space in this case).
Normally, the Dirac notation $langle v|wrangle$ is a representation of a scalar product and the Bra and Ket correspond on a low level to vectors input in this scalar product. Note, that for the complex scalar product, order of the arguments is important due to its hermitian nature and its semi-bilinearity, i.e. that it is conjugate linear w.r.t. to one argument. A classical way is to define the standard complex scalar product to be conjugate linear in the second argument. In Bra-Ket-notation, you usually reverse the order of the arguments of the complex scalar product, i.e. $langle x,yrangle=langle y|xrangle$ resulting in conjugate linearity in the first argument.
The key thing is that there is a strong correspondence between scalar products and members of the dual space, i.e. linear functionals.
There is a way that any functional corresponds in a one-to-one fashion to a representation using the scalar product(of the associated vector space). This is known as Riesz representation theorem.
More precisely, you may look at a vector $vin V$ and suppose that $langlecdot,cdotrangle$ is an associated scalar product(turning $V$ into a euclidean/unitary space in the real/complex case). Then the map $varphi_v:wmapstolangle w,vrangle$ is a member of the dual space $V^star$ and the theorem says that any linear functional $psiin V^star$ can be written as such a $varphi_v$ uniquely.
Thus, you may convert a scalar product between two vectors into an application of a linear functional to another vector, pulling the problem statement into the realm of dual spaces, where you have other mathematical possibilities to tackle various questions.
EDIT: Note, that the definition of $varphi_v$ of course depends also on the argument which is assumed to be conjugate linear, in this case the second, as linearity in the first is needed to make $varphi_v$ a linear map(check this).
answered 4 hours ago
zzuussee
1,057319
Thanks for the edit, how silly of me to write a name in lower case.
– zzuussee
3 hours ago
thank you for your answer.... ...I understand that that with ket we work in linear space...but when we built a relation as you write"Æv:w↦⟨w,v⟩ is a member of the dual space V⋆" between ket and its conjugate bra ,we will have dual space
– yana ghanim
2 hours ago
@yanaghanim I don't really know what you're asking or why you re-decided in accepting the answer but I would like to resolve any issues you might have with the concept.
– zzuussee
2 hours ago
thank you ...I want to now what the meaning of dual space...why we use it with bra not with ket???
– yana ghanim
2 hours ago
1
@yanaghanim I've edited my answer to add some more information. For the complex scalar product, order of the arguments is important due to its hermitian nature and its semi-bilinearity, i.e. that it is conjugate linear w.r.t. to one argument. Note that the Bra-Ket-notation usually reverses the order of the arguments of the complex scalar product, i.e. $langle x,yrangle=langle y|xrangle$. This is more a notational detail. It is more fruitful, I think, to understand the nature of the correspondence between the dual space and scalar product.
– zzuussee
2 hours ago
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
A vector space over a field $mathbbF$ is a set $V$ with operations $+$ and $cdot$ satisfying the vector space axioms. Given a vector space $V$, it's dual space $V^star$ is defined as $mathrmHom(V,mathbbF)$, i.e. the set of all linear maps(functionals) between the vector space and its underlying field(considered as an own vector space in this case).
Normally, the Dirac notation $langle v|wrangle$ is a representation of a scalar product and the Bra and Ket correspond on a low level to vectors input in this scalar product. Note, that for the complex scalar product, order of the arguments is important due to its hermitian nature and its semi-bilinearity, i.e. that it is conjugate linear w.r.t. to one argument. A classical way is to define the standard complex scalar product to be conjugate linear in the second argument. In Bra-Ket-notation, you usually reverse the order of the arguments of the complex scalar product, i.e. $langle x,yrangle=langle y|xrangle$ resulting in conjugate linearity in the first argument.
The key thing is that there is a strong correspondence between scalar products and members of the dual space, i.e. linear functionals.
There is a way that any functional corresponds in a one-to-one fashion to a representation using the scalar product(of the associated vector space). This is known as Riesz representation theorem.
More precisely, you may look at a vector $vin V$ and suppose that $langlecdot,cdotrangle$ is an associated scalar product(turning $V$ into a euclidean/unitary space in the real/complex case). Then the map $varphi_v:wmapstolangle w,vrangle$ is a member of the dual space $V^star$ and the theorem says that any linear functional $psiin V^star$ can be written as such a $varphi_v$ uniquely.
Thus, you may convert a scalar product between two vectors into an application of a linear functional to another vector, pulling the problem statement into the realm of dual spaces, where you have other mathematical possibilities to tackle various questions.
EDIT: Note, that the definition of $varphi_v$ of course depends also on the argument which is assumed to be conjugate linear, in this case the second, as linearity in the first is needed to make $varphi_v$ a linear map(check this).
answered 4 hours ago
zzuussee
1,057319
A vector space over a field $mathbbF$ is a set $V$ with operations $+$ and $cdot$ satisfying the vector space axioms. Given a vector space $V$, it's dual space $V^star$ is defined as $mathrmHom(V,mathbbF)$, i.e. the set of all linear maps(functionals) between the vector space and its underlying field(considered as an own vector space in this case).
Normally, the Dirac notation $langle v|wrangle$ is a representation of a scalar product and the Bra and Ket correspond on a low level to vectors input in this scalar product. Note, that for the complex scalar product, order of the arguments is important due to its hermitian nature and its semi-bilinearity, i.e. that it is conjugate linear w.r.t. to one argument. A classical way is to define the standard complex scalar product to be conjugate linear in the second argument. In Bra-Ket-notation, you usually reverse the order of the arguments of the complex scalar product, i.e. $langle x,yrangle=langle y|xrangle$ resulting in conjugate linearity in the first argument.
The key thing is that there is a strong correspondence between scalar products and members of the dual space, i.e. linear functionals.
There is a way that any functional corresponds in a one-to-one fashion to a representation using the scalar product(of the associated vector space). This is known as Riesz representation theorem.
More precisely, you may look at a vector $vin V$ and suppose that $langlecdot,cdotrangle$ is an associated scalar product(turning $V$ into a euclidean/unitary space in the real/complex case). Then the map $varphi_v:wmapstolangle w,vrangle$ is a member of the dual space $V^star$ and the theorem says that any linear functional $psiin V^star$ can be written as such a $varphi_v$ uniquely.
Thus, you may convert a scalar product between two vectors into an application of a linear functional to another vector, pulling the problem statement into the realm of dual spaces, where you have other mathematical possibilities to tackle various questions.
EDIT: Note, that the definition of $varphi_v$ of course depends also on the argument which is assumed to be conjugate linear, in this case the second, as linearity in the first is needed to make $varphi_v$ a linear map(check this).
answered 4 hours ago
zzuussee
1,057319
edited 1 hour ago
answered 4 hours ago
zzuussee
1,057319
answered 4 hours ago
zzuussee
1,057319
answered 4 hours ago
zzuussee
1,057319
1,057319
Thanks for the edit, how silly of me to write a name in lower case.
– zzuussee
3 hours ago
thank you for your answer.... ...I understand that that with ket we work in linear space...but when we built a relation as you write"Æv:w↦⟨w,v⟩ is a member of the dual space V⋆" between ket and its conjugate bra ,we will have dual space
– yana ghanim
2 hours ago
@yanaghanim I don't really know what you're asking or why you re-decided in accepting the answer but I would like to resolve any issues you might have with the concept.
– zzuussee
2 hours ago
thank you ...I want to now what the meaning of dual space...why we use it with bra not with ket???
– yana ghanim
2 hours ago
1
@yanaghanim I've edited my answer to add some more information. For the complex scalar product, order of the arguments is important due to its hermitian nature and its semi-bilinearity, i.e. that it is conjugate linear w.r.t. to one argument. Note that the Bra-Ket-notation usually reverses the order of the arguments of the complex scalar product, i.e. $langle x,yrangle=langle y|xrangle$. This is more a notational detail. It is more fruitful, I think, to understand the nature of the correspondence between the dual space and scalar product.
– zzuussee
2 hours ago
add a comment |Â
Thanks for the edit, how silly of me to write a name in lower case.
– zzuussee
3 hours ago
thank you for your answer.... ...I understand that that with ket we work in linear space...but when we built a relation as you write"Æv:w↦⟨w,v⟩ is a member of the dual space V⋆" between ket and its conjugate bra ,we will have dual space
– yana ghanim
2 hours ago
@yanaghanim I don't really know what you're asking or why you re-decided in accepting the answer but I would like to resolve any issues you might have with the concept.
– zzuussee
2 hours ago
thank you ...I want to now what the meaning of dual space...why we use it with bra not with ket???
– yana ghanim
2 hours ago
1
@yanaghanim I've edited my answer to add some more information. For the complex scalar product, order of the arguments is important due to its hermitian nature and its semi-bilinearity, i.e. that it is conjugate linear w.r.t. to one argument. Note that the Bra-Ket-notation usually reverses the order of the arguments of the complex scalar product, i.e. $langle x,yrangle=langle y|xrangle$. This is more a notational detail. It is more fruitful, I think, to understand the nature of the correspondence between the dual space and scalar product.
– zzuussee
2 hours ago
Thanks for the edit, how silly of me to write a name in lower case.
– zzuussee
3 hours ago
Thanks for the edit, how silly of me to write a name in lower case.
– zzuussee
3 hours ago
thank you for your answer.... ...I understand that that with ket we work in linear space...but when we built a relation as you write"Æv:w↦⟨w,v⟩ is a member of the dual space V⋆" between ket and its conjugate bra ,we will have dual space
– yana ghanim
2 hours ago
thank you for your answer.... ...I understand that that with ket we work in linear space...but when we built a relation as you write"Æv:w↦⟨w,v⟩ is a member of the dual space V⋆" between ket and its conjugate bra ,we will have dual space
– yana ghanim
2 hours ago
@yanaghanim I don't really know what you're asking or why you re-decided in accepting the answer but I would like to resolve any issues you might have with the concept.
– zzuussee
2 hours ago
@yanaghanim I don't really know what you're asking or why you re-decided in accepting the answer but I would like to resolve any issues you might have with the concept.
– zzuussee
2 hours ago
thank you ...I want to now what the meaning of dual space...why we use it with bra not with ket???
– yana ghanim
2 hours ago
thank you ...I want to now what the meaning of dual space...why we use it with bra not with ket???
– yana ghanim
2 hours ago
1
1
@yanaghanim I've edited my answer to add some more information. For the complex scalar product, order of the arguments is important due to its hermitian nature and its semi-bilinearity, i.e. that it is conjugate linear w.r.t. to one argument. Note that the Bra-Ket-notation usually reverses the order of the arguments of the complex scalar product, i.e. $langle x,yrangle=langle y|xrangle$. This is more a notational detail. It is more fruitful, I think, to understand the nature of the correspondence between the dual space and scalar product.
– zzuussee
2 hours ago
@yanaghanim I've edited my answer to add some more information. For the complex scalar product, order of the arguments is important due to its hermitian nature and its semi-bilinearity, i.e. that it is conjugate linear w.r.t. to one argument. Note that the Bra-Ket-notation usually reverses the order of the arguments of the complex scalar product, i.e. $langle x,yrangle=langle y|xrangle$. This is more a notational detail. It is more fruitful, I think, to understand the nature of the correspondence between the dual space and scalar product.
– zzuussee
2 hours ago
add a comment |Â
up vote
1
down vote
For the finite dimensional case we have for a column vector $u$:
$$
u = vert u rangle\
u^+= langle u vert
$$
where $u^+$ is the transposed, complex conjugated, thus adjugated, vector.
It is a linear form from $V$ to the scalar field.
If $V$ is a vector space then $V^*$, consisting of all linear forms of $V$, is a linear vector space as well.
answered 3 hours ago
mvw
30.1k22150
add a comment |Â
up vote
1
down vote
For the finite dimensional case we have for a column vector $u$:
$$
u = vert u rangle\
u^+= langle u vert
$$
where $u^+$ is the transposed, complex conjugated, thus adjugated, vector.
It is a linear form from $V$ to the scalar field.
If $V$ is a vector space then $V^*$, consisting of all linear forms of $V$, is a linear vector space as well.
answered 3 hours ago
mvw
30.1k22150
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For the finite dimensional case we have for a column vector $u$:
$$
u = vert u rangle\
u^+= langle u vert
$$
where $u^+$ is the transposed, complex conjugated, thus adjugated, vector.
It is a linear form from $V$ to the scalar field.
If $V$ is a vector space then $V^*$, consisting of all linear forms of $V$, is a linear vector space as well.
answered 3 hours ago
mvw
30.1k22150
For the finite dimensional case we have for a column vector $u$:
$$
u = vert u rangle\
u^+= langle u vert
$$
where $u^+$ is the transposed, complex conjugated, thus adjugated, vector.
It is a linear form from $V$ to the scalar field.
If $V$ is a vector space then $V^*$, consisting of all linear forms of $V$, is a linear vector space as well.
answered 3 hours ago
mvw
30.1k22150
edited 3 hours ago
answered 3 hours ago
mvw
30.1k22150
answered 3 hours ago
mvw
30.1k22150
answered 3 hours ago
mvw
30.1k22150
30.1k22150
add a comment |Â
add a comment |Â
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