How can I explicitly express the Ising Hamiltonian in matrix form?
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
$$H_{rs}dot{=}langle r |H|srangle$$
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
$$H_{rs}dot{=}langle r |H|srangle$$
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
$$H_{rs}dot{=}langle r |H|srangle$$
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
$$H_{rs}dot{=}langle r |H|srangle$$
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
8,42122438
8,42122438
asked Dec 29 '18 at 21:41
Luqman 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
$$ mathbf {A} otimes mathbf {B} ={begin{bmatrix}a_{11}mathbf {B} &cdots &a_{1n}mathbf {B} \vdots &ddots &vdots \a_{m1}mathbf {B} &cdots &a_{mn}mathbf {B} end{bmatrix}}. $$
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
$textbf{Quick hint :}$
Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_{i}^{alpha} rightarrowunderset{N_{}^{text{th}}text{order direct/kronecker product of identity and pauli matrices}}{mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesmathbb{I}_{}^{}}$$
which makes (for $i < j$):
$$sigma_{i}^{alpha}sigma_{j}^{beta} stackrel{i < j}{rightarrow} mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesunderset{j_{}^{text{th}} text{position}}{sigma_{}^{beta}}otimes cdotsotimesmathbb{I}_{}^{}$$
with standard definition of direct/Kronecker product of matrices
Note also that Kronecker product is associative.
$mathbb{I}$ 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
2
active
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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
$$ mathbf {A} otimes mathbf {B} ={begin{bmatrix}a_{11}mathbf {B} &cdots &a_{1n}mathbf {B} \vdots &ddots &vdots \a_{m1}mathbf {B} &cdots &a_{mn}mathbf {B} end{bmatrix}}. $$
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
$$ mathbf {A} otimes mathbf {B} ={begin{bmatrix}a_{11}mathbf {B} &cdots &a_{1n}mathbf {B} \vdots &ddots &vdots \a_{m1}mathbf {B} &cdots &a_{mn}mathbf {B} end{bmatrix}}. $$
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
$$ mathbf {A} otimes mathbf {B} ={begin{bmatrix}a_{11}mathbf {B} &cdots &a_{1n}mathbf {B} \vdots &ddots &vdots \a_{m1}mathbf {B} &cdots &a_{mn}mathbf {B} end{bmatrix}}. $$
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
$$ mathbf {A} otimes mathbf {B} ={begin{bmatrix}a_{11}mathbf {B} &cdots &a_{1n}mathbf {B} \vdots &ddots &vdots \a_{m1}mathbf {B} &cdots &a_{mn}mathbf {B} end{bmatrix}}. $$
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
InertialObserver
1,632517
1,632517
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
$textbf{Quick hint :}$
Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_{i}^{alpha} rightarrowunderset{N_{}^{text{th}}text{order direct/kronecker product of identity and pauli matrices}}{mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesmathbb{I}_{}^{}}$$
which makes (for $i < j$):
$$sigma_{i}^{alpha}sigma_{j}^{beta} stackrel{i < j}{rightarrow} mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesunderset{j_{}^{text{th}} text{position}}{sigma_{}^{beta}}otimes cdotsotimesmathbb{I}_{}^{}$$
with standard definition of direct/Kronecker product of matrices
Note also that Kronecker product is associative.
$mathbb{I}$ 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 |
$textbf{Quick hint :}$
Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_{i}^{alpha} rightarrowunderset{N_{}^{text{th}}text{order direct/kronecker product of identity and pauli matrices}}{mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesmathbb{I}_{}^{}}$$
which makes (for $i < j$):
$$sigma_{i}^{alpha}sigma_{j}^{beta} stackrel{i < j}{rightarrow} mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesunderset{j_{}^{text{th}} text{position}}{sigma_{}^{beta}}otimes cdotsotimesmathbb{I}_{}^{}$$
with standard definition of direct/Kronecker product of matrices
Note also that Kronecker product is associative.
$mathbb{I}$ 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 |
$textbf{Quick hint :}$
Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_{i}^{alpha} rightarrowunderset{N_{}^{text{th}}text{order direct/kronecker product of identity and pauli matrices}}{mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesmathbb{I}_{}^{}}$$
which makes (for $i < j$):
$$sigma_{i}^{alpha}sigma_{j}^{beta} stackrel{i < j}{rightarrow} mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesunderset{j_{}^{text{th}} text{position}}{sigma_{}^{beta}}otimes cdotsotimesmathbb{I}_{}^{}$$
with standard definition of direct/Kronecker product of matrices
Note also that Kronecker product is associative.
$mathbb{I}$ is $2 times 2$ identity matrix and $sigma_{}^{}$'s are standard Pauli matrices.
$textbf{Quick hint :}$
Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_{i}^{alpha} rightarrowunderset{N_{}^{text{th}}text{order direct/kronecker product of identity and pauli matrices}}{mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesmathbb{I}_{}^{}}$$
which makes (for $i < j$):
$$sigma_{i}^{alpha}sigma_{j}^{beta} stackrel{i < j}{rightarrow} mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesunderset{j_{}^{text{th}} text{position}}{sigma_{}^{beta}}otimes cdotsotimesmathbb{I}_{}^{}$$
with standard definition of direct/Kronecker product of matrices
Note also that Kronecker product is associative.
$mathbb{I}$ 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
Sunyam
6041311
6041311
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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