How can I explicitly express the Ising Hamiltonian in matrix form?












2














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?










share|cite|improve this question
























  • Comments are not for extended discussion; this conversation has been moved to chat.
    – rob
    Dec 30 '18 at 14:15
















2














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?










share|cite|improve this question
























  • Comments are not for extended discussion; this conversation has been moved to chat.
    – rob
    Dec 30 '18 at 14:15














2












2








2


1





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?










share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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


















  • 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










2 Answers
2






active

oldest

votes


















3














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.






share|cite|improve this answer























  • 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



















2














$textbf{Quick hint :}$



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.






share|cite|improve this answer



















  • 2




    That was very helpful. Thank you.
    – Luqman Saleem
    Dec 29 '18 at 22:48











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f451075%2fhow-can-i-explicitly-express-the-ising-hamiltonian-in-matrix-form%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














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.






share|cite|improve this answer























  • 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
















3














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.






share|cite|improve this answer























  • 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














3












3








3






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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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


















  • 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











2














$textbf{Quick hint :}$



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.






share|cite|improve this answer



















  • 2




    That was very helpful. Thank you.
    – Luqman Saleem
    Dec 29 '18 at 22:48
















2














$textbf{Quick hint :}$



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.






share|cite|improve this answer



















  • 2




    That was very helpful. Thank you.
    – Luqman Saleem
    Dec 29 '18 at 22:48














2












2








2






$textbf{Quick hint :}$



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.






share|cite|improve this answer














$textbf{Quick hint :}$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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














  • 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


















draft saved

draft discarded




















































Thanks for contributing an answer to Physics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f451075%2fhow-can-i-explicitly-express-the-ising-hamiltonian-in-matrix-form%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Сан-Квентин

Алькесар

Josef Freinademetz