How can I explicitly express the Ising Hamiltonian in matrix form?
Clash Royale CLAN TAG#URR8PPP
I am reading this book about numerical methods in physics. It has the following question:
Consider the Ising Hamiltonian defined as following $$H=-sum_ i=1^N-1
sigma_i^x sigma_ i+1 ^x + h sum_ i=1 ^N sigma_i^z$$
Write a program that computes the $2^N times 2^N$ matrix for
different $N$
From quantum mechanics, I know that any operator can be expressed in matrix form as follows
$$H_rsdot=langle r |H|srangle$$
where $|irangle$ are any basis.
My question is what basis can I take for this Hamiltonian? Is there any other way to write this Hamiltonian in matrix form?
condensed-matter quantum-spin computational-physics ising-model spin-chains
add a comment |
I am reading this book about numerical methods in physics. It has the following question:
Consider the Ising Hamiltonian defined as following $$H=-sum_ i=1^N-1
sigma_i^x sigma_ i+1 ^x + h sum_ i=1 ^N sigma_i^z$$
Write a program that computes the $2^N times 2^N$ matrix for
different $N$
From quantum mechanics, I know that any operator can be expressed in matrix form as follows
$$H_rsdot=langle r |H|srangle$$
where $|irangle$ are any basis.
My question is what basis can I take for this Hamiltonian? Is there any other way to write this Hamiltonian in matrix form?
condensed-matter quantum-spin computational-physics ising-model spin-chains
Comments are not for extended discussion; this conversation has been moved to chat.
– rob♦
Dec 30 '18 at 14:15
add a comment |
I am reading this book about numerical methods in physics. It has the following question:
Consider the Ising Hamiltonian defined as following $$H=-sum_ i=1^N-1
sigma_i^x sigma_ i+1 ^x + h sum_ i=1 ^N sigma_i^z$$
Write a program that computes the $2^N times 2^N$ matrix for
different $N$
From quantum mechanics, I know that any operator can be expressed in matrix form as follows
$$H_rsdot=langle r |H|srangle$$
where $|irangle$ are any basis.
My question is what basis can I take for this Hamiltonian? Is there any other way to write this Hamiltonian in matrix form?
condensed-matter quantum-spin computational-physics ising-model spin-chains
I am reading this book about numerical methods in physics. It has the following question:
Consider the Ising Hamiltonian defined as following $$H=-sum_ i=1^N-1
sigma_i^x sigma_ i+1 ^x + h sum_ i=1 ^N sigma_i^z$$
Write a program that computes the $2^N times 2^N$ matrix for
different $N$
From quantum mechanics, I know that any operator can be expressed in matrix form as follows
$$H_rsdot=langle r |H|srangle$$
where $|irangle$ are any basis.
My question is what basis can I take for this Hamiltonian? Is there any other way to write this Hamiltonian in matrix form?
condensed-matter quantum-spin computational-physics ising-model spin-chains
condensed-matter quantum-spin computational-physics ising-model spin-chains
edited Dec 29 '18 at 23:02
Norbert Schuch
8,62322438
8,62322438
asked Dec 29 '18 at 21:41
Luqman SaleemLuqman Saleem
249111
249111
Comments are not for extended discussion; this conversation has been moved to chat.
– rob♦
Dec 30 '18 at 14:15
add a comment |
Comments are not for extended discussion; this conversation has been moved to chat.
– rob♦
Dec 30 '18 at 14:15
Comments are not for extended discussion; this conversation has been moved to chat.
– rob♦
Dec 30 '18 at 14:15
Comments are not for extended discussion; this conversation has been moved to chat.
– rob♦
Dec 30 '18 at 14:15
add a comment |
2 Answers
2
active
oldest
votes
For completeness I'll summarize the answer here. After a fun conversation in the comments, we saw that it will be more illuminating to write
$$H=-sum_ i=1^N-1
sigma_i^x sigma_ i+1 ^x + h sum_ i=1 ^N sigma_i^z $$
as
$$ H=-sum_ i=1^N-1
sigma_i^x sigma_ i+1 ^x + hleft( sum_ i=1 ^N-1 1_1 otimes cdots 1_i-1 otimes sigma_i^z otimes 1_i+1 otimes cdots otimes 1_N-1 right)
+ h ( 1_1 otimes cdots otimes 1_N-1 otimes sigma^z_N) $$
where it is understood that (to prevent clutter)
$$ sigma_i^x sigma_i+1^x = 1_1 otimes cdotsotimes 1_i-1 otimes sigma^x_i otimes sigma^x_i+1 otimes 1_i+2 otimes cdots otimes 1_N $$
and the subscripts are there to denote nothing more than the position of insertion (i.e. they are all $2times 2$ identity)
We then create a matrix from the direct product of matrices $A,B$ where $[A] = m times n$ and $[B] = p times q$ matrices
$$ mathbf A otimes mathbf B =beginbmatrixa_11mathbf B &cdots &a_1nmathbf B \vdots &ddots &vdots \a_m1mathbf B &cdots &a_mnmathbf B endbmatrix. $$
We apply this to the $i^th$ term and then loop over all $i$ to achieve the desired matrix.
Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
– Norbert Schuch
Dec 29 '18 at 22:59
You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
– InertialObserver
Dec 29 '18 at 23:13
Much better, but for consistency you should also add the identities in the first term!
– Norbert Schuch
Dec 29 '18 at 23:13
Will do, gonna do that now
– InertialObserver
Dec 29 '18 at 23:14
1
I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
– InertialObserver
Dec 29 '18 at 23:31
|
show 3 more comments
$textbfQuick hint :$
Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_i^alpha rightarrowundersetN_^textthtextorder direct/kronecker product of identity and pauli matricesmathbbI_^otimescdotsotimesunderseti_^textth textpositionsigma_^alphaotimescdotsotimesmathbbI_^$$
which makes (for $i < j$):
$$sigma_i^alphasigma_j^beta stackreli < jrightarrow mathbbI_^otimescdotsotimesunderseti_^textth textpositionsigma_^alphaotimescdotsotimesundersetj_^textth textpositionsigma_^betaotimes cdotsotimesmathbbI_^$$
with standard definition of direct/Kronecker product of matrices
Note also that Kronecker product is associative.
$mathbbI$ is $2 times 2$ identity matrix and $sigma_^$'s are standard Pauli matrices.
2
That was very helpful. Thank you.
– Luqman Saleem
Dec 29 '18 at 22:48
add a comment |
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2 Answers
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2 Answers
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For completeness I'll summarize the answer here. After a fun conversation in the comments, we saw that it will be more illuminating to write
$$H=-sum_ i=1^N-1
sigma_i^x sigma_ i+1 ^x + h sum_ i=1 ^N sigma_i^z $$
as
$$ H=-sum_ i=1^N-1
sigma_i^x sigma_ i+1 ^x + hleft( sum_ i=1 ^N-1 1_1 otimes cdots 1_i-1 otimes sigma_i^z otimes 1_i+1 otimes cdots otimes 1_N-1 right)
+ h ( 1_1 otimes cdots otimes 1_N-1 otimes sigma^z_N) $$
where it is understood that (to prevent clutter)
$$ sigma_i^x sigma_i+1^x = 1_1 otimes cdotsotimes 1_i-1 otimes sigma^x_i otimes sigma^x_i+1 otimes 1_i+2 otimes cdots otimes 1_N $$
and the subscripts are there to denote nothing more than the position of insertion (i.e. they are all $2times 2$ identity)
We then create a matrix from the direct product of matrices $A,B$ where $[A] = m times n$ and $[B] = p times q$ matrices
$$ mathbf A otimes mathbf B =beginbmatrixa_11mathbf B &cdots &a_1nmathbf B \vdots &ddots &vdots \a_m1mathbf B &cdots &a_mnmathbf B endbmatrix. $$
We apply this to the $i^th$ term and then loop over all $i$ to achieve the desired matrix.
Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
– Norbert Schuch
Dec 29 '18 at 22:59
You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
– InertialObserver
Dec 29 '18 at 23:13
Much better, but for consistency you should also add the identities in the first term!
– Norbert Schuch
Dec 29 '18 at 23:13
Will do, gonna do that now
– InertialObserver
Dec 29 '18 at 23:14
1
I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
– InertialObserver
Dec 29 '18 at 23:31
|
show 3 more comments
For completeness I'll summarize the answer here. After a fun conversation in the comments, we saw that it will be more illuminating to write
$$H=-sum_ i=1^N-1
sigma_i^x sigma_ i+1 ^x + h sum_ i=1 ^N sigma_i^z $$
as
$$ H=-sum_ i=1^N-1
sigma_i^x sigma_ i+1 ^x + hleft( sum_ i=1 ^N-1 1_1 otimes cdots 1_i-1 otimes sigma_i^z otimes 1_i+1 otimes cdots otimes 1_N-1 right)
+ h ( 1_1 otimes cdots otimes 1_N-1 otimes sigma^z_N) $$
where it is understood that (to prevent clutter)
$$ sigma_i^x sigma_i+1^x = 1_1 otimes cdotsotimes 1_i-1 otimes sigma^x_i otimes sigma^x_i+1 otimes 1_i+2 otimes cdots otimes 1_N $$
and the subscripts are there to denote nothing more than the position of insertion (i.e. they are all $2times 2$ identity)
We then create a matrix from the direct product of matrices $A,B$ where $[A] = m times n$ and $[B] = p times q$ matrices
$$ mathbf A otimes mathbf B =beginbmatrixa_11mathbf B &cdots &a_1nmathbf B \vdots &ddots &vdots \a_m1mathbf B &cdots &a_mnmathbf B endbmatrix. $$
We apply this to the $i^th$ term and then loop over all $i$ to achieve the desired matrix.
Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
– Norbert Schuch
Dec 29 '18 at 22:59
You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
– InertialObserver
Dec 29 '18 at 23:13
Much better, but for consistency you should also add the identities in the first term!
– Norbert Schuch
Dec 29 '18 at 23:13
Will do, gonna do that now
– InertialObserver
Dec 29 '18 at 23:14
1
I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
– InertialObserver
Dec 29 '18 at 23:31
|
show 3 more comments
For completeness I'll summarize the answer here. After a fun conversation in the comments, we saw that it will be more illuminating to write
$$H=-sum_ i=1^N-1
sigma_i^x sigma_ i+1 ^x + h sum_ i=1 ^N sigma_i^z $$
as
$$ H=-sum_ i=1^N-1
sigma_i^x sigma_ i+1 ^x + hleft( sum_ i=1 ^N-1 1_1 otimes cdots 1_i-1 otimes sigma_i^z otimes 1_i+1 otimes cdots otimes 1_N-1 right)
+ h ( 1_1 otimes cdots otimes 1_N-1 otimes sigma^z_N) $$
where it is understood that (to prevent clutter)
$$ sigma_i^x sigma_i+1^x = 1_1 otimes cdotsotimes 1_i-1 otimes sigma^x_i otimes sigma^x_i+1 otimes 1_i+2 otimes cdots otimes 1_N $$
and the subscripts are there to denote nothing more than the position of insertion (i.e. they are all $2times 2$ identity)
We then create a matrix from the direct product of matrices $A,B$ where $[A] = m times n$ and $[B] = p times q$ matrices
$$ mathbf A otimes mathbf B =beginbmatrixa_11mathbf B &cdots &a_1nmathbf B \vdots &ddots &vdots \a_m1mathbf B &cdots &a_mnmathbf B endbmatrix. $$
We apply this to the $i^th$ term and then loop over all $i$ to achieve the desired matrix.
For completeness I'll summarize the answer here. After a fun conversation in the comments, we saw that it will be more illuminating to write
$$H=-sum_ i=1^N-1
sigma_i^x sigma_ i+1 ^x + h sum_ i=1 ^N sigma_i^z $$
as
$$ H=-sum_ i=1^N-1
sigma_i^x sigma_ i+1 ^x + hleft( sum_ i=1 ^N-1 1_1 otimes cdots 1_i-1 otimes sigma_i^z otimes 1_i+1 otimes cdots otimes 1_N-1 right)
+ h ( 1_1 otimes cdots otimes 1_N-1 otimes sigma^z_N) $$
where it is understood that (to prevent clutter)
$$ sigma_i^x sigma_i+1^x = 1_1 otimes cdotsotimes 1_i-1 otimes sigma^x_i otimes sigma^x_i+1 otimes 1_i+2 otimes cdots otimes 1_N $$
and the subscripts are there to denote nothing more than the position of insertion (i.e. they are all $2times 2$ identity)
We then create a matrix from the direct product of matrices $A,B$ where $[A] = m times n$ and $[B] = p times q$ matrices
$$ mathbf A otimes mathbf B =beginbmatrixa_11mathbf B &cdots &a_1nmathbf B \vdots &ddots &vdots \a_m1mathbf B &cdots &a_mnmathbf B endbmatrix. $$
We apply this to the $i^th$ term and then loop over all $i$ to achieve the desired matrix.
edited Dec 29 '18 at 23:30
answered Dec 29 '18 at 22:43
InertialObserverInertialObserver
2,278623
2,278623
Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
– Norbert Schuch
Dec 29 '18 at 22:59
You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
– InertialObserver
Dec 29 '18 at 23:13
Much better, but for consistency you should also add the identities in the first term!
– Norbert Schuch
Dec 29 '18 at 23:13
Will do, gonna do that now
– InertialObserver
Dec 29 '18 at 23:14
1
I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
– InertialObserver
Dec 29 '18 at 23:31
|
show 3 more comments
Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
– Norbert Schuch
Dec 29 '18 at 22:59
You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
– InertialObserver
Dec 29 '18 at 23:13
Much better, but for consistency you should also add the identities in the first term!
– Norbert Schuch
Dec 29 '18 at 23:13
Will do, gonna do that now
– InertialObserver
Dec 29 '18 at 23:14
1
I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
– InertialObserver
Dec 29 '18 at 23:31
Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
– Norbert Schuch
Dec 29 '18 at 22:59
Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
– Norbert Schuch
Dec 29 '18 at 22:59
You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
– InertialObserver
Dec 29 '18 at 23:13
You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
– InertialObserver
Dec 29 '18 at 23:13
Much better, but for consistency you should also add the identities in the first term!
– Norbert Schuch
Dec 29 '18 at 23:13
Much better, but for consistency you should also add the identities in the first term!
– Norbert Schuch
Dec 29 '18 at 23:13
Will do, gonna do that now
– InertialObserver
Dec 29 '18 at 23:14
Will do, gonna do that now
– InertialObserver
Dec 29 '18 at 23:14
1
1
I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
– InertialObserver
Dec 29 '18 at 23:31
I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
– InertialObserver
Dec 29 '18 at 23:31
|
show 3 more comments
$textbfQuick hint :$
Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_i^alpha rightarrowundersetN_^textthtextorder direct/kronecker product of identity and pauli matricesmathbbI_^otimescdotsotimesunderseti_^textth textpositionsigma_^alphaotimescdotsotimesmathbbI_^$$
which makes (for $i < j$):
$$sigma_i^alphasigma_j^beta stackreli < jrightarrow mathbbI_^otimescdotsotimesunderseti_^textth textpositionsigma_^alphaotimescdotsotimesundersetj_^textth textpositionsigma_^betaotimes cdotsotimesmathbbI_^$$
with standard definition of direct/Kronecker product of matrices
Note also that Kronecker product is associative.
$mathbbI$ is $2 times 2$ identity matrix and $sigma_^$'s are standard Pauli matrices.
2
That was very helpful. Thank you.
– Luqman Saleem
Dec 29 '18 at 22:48
add a comment |
$textbfQuick hint :$
Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_i^alpha rightarrowundersetN_^textthtextorder direct/kronecker product of identity and pauli matricesmathbbI_^otimescdotsotimesunderseti_^textth textpositionsigma_^alphaotimescdotsotimesmathbbI_^$$
which makes (for $i < j$):
$$sigma_i^alphasigma_j^beta stackreli < jrightarrow mathbbI_^otimescdotsotimesunderseti_^textth textpositionsigma_^alphaotimescdotsotimesundersetj_^textth textpositionsigma_^betaotimes cdotsotimesmathbbI_^$$
with standard definition of direct/Kronecker product of matrices
Note also that Kronecker product is associative.
$mathbbI$ is $2 times 2$ identity matrix and $sigma_^$'s are standard Pauli matrices.
2
That was very helpful. Thank you.
– Luqman Saleem
Dec 29 '18 at 22:48
add a comment |
$textbfQuick hint :$
Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_i^alpha rightarrowundersetN_^textthtextorder direct/kronecker product of identity and pauli matricesmathbbI_^otimescdotsotimesunderseti_^textth textpositionsigma_^alphaotimescdotsotimesmathbbI_^$$
which makes (for $i < j$):
$$sigma_i^alphasigma_j^beta stackreli < jrightarrow mathbbI_^otimescdotsotimesunderseti_^textth textpositionsigma_^alphaotimescdotsotimesundersetj_^textth textpositionsigma_^betaotimes cdotsotimesmathbbI_^$$
with standard definition of direct/Kronecker product of matrices
Note also that Kronecker product is associative.
$mathbbI$ is $2 times 2$ identity matrix and $sigma_^$'s are standard Pauli matrices.
$textbfQuick hint :$
Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_i^alpha rightarrowundersetN_^textthtextorder direct/kronecker product of identity and pauli matricesmathbbI_^otimescdotsotimesunderseti_^textth textpositionsigma_^alphaotimescdotsotimesmathbbI_^$$
which makes (for $i < j$):
$$sigma_i^alphasigma_j^beta stackreli < jrightarrow mathbbI_^otimescdotsotimesunderseti_^textth textpositionsigma_^alphaotimescdotsotimesundersetj_^textth textpositionsigma_^betaotimes cdotsotimesmathbbI_^$$
with standard definition of direct/Kronecker product of matrices
Note also that Kronecker product is associative.
$mathbbI$ is $2 times 2$ identity matrix and $sigma_^$'s are standard Pauli matrices.
edited Dec 29 '18 at 23:37
answered Dec 29 '18 at 22:46
SunyamSunyam
6141311
6141311
2
That was very helpful. Thank you.
– Luqman Saleem
Dec 29 '18 at 22:48
add a comment |
2
That was very helpful. Thank you.
– Luqman Saleem
Dec 29 '18 at 22:48
2
2
That was very helpful. Thank you.
– Luqman Saleem
Dec 29 '18 at 22:48
That was very helpful. Thank you.
– Luqman Saleem
Dec 29 '18 at 22:48
add a comment |
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– rob♦
Dec 30 '18 at 14:15