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

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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?










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  • 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
$$H_rsdot=langle r |H|srangle$$
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?










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
$$H_rsdot=langle r |H|srangle$$
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?










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
$$H_rsdot=langle r |H|srangle$$
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






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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

















  • 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



$$ mathbf A otimes mathbf B =beginbmatrixa_11mathbf B &cdots &a_1nmathbf B \vdots &ddots &vdots \a_m1mathbf B &cdots &a_mnmathbf B endbmatrix. $$



We apply this to the $i^th$ term and then loop over all $i$ to achieve the desired matrix.






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














$textbfQuick hint :$



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.






share|cite|improve this answer




















  • 2





    That was very helpful. Thank you.

    – Luqman Saleem
    Dec 29 '18 at 22:48










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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



$$ mathbf A otimes mathbf B =beginbmatrixa_11mathbf B &cdots &a_1nmathbf B \vdots &ddots &vdots \a_m1mathbf B &cdots &a_mnmathbf B endbmatrix. $$



We apply this to the $i^th$ term and then loop over all $i$ to achieve the desired matrix.






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



$$ mathbf A otimes mathbf B =beginbmatrixa_11mathbf B &cdots &a_1nmathbf B \vdots &ddots &vdots \a_m1mathbf B &cdots &a_mnmathbf B endbmatrix. $$



We apply this to the $i^th$ term and then loop over all $i$ to achieve the desired matrix.






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



$$ mathbf A otimes mathbf B =beginbmatrixa_11mathbf B &cdots &a_1nmathbf B \vdots &ddots &vdots \a_m1mathbf B &cdots &a_mnmathbf B endbmatrix. $$



We apply this to the $i^th$ term and then loop over all $i$ to achieve the desired matrix.






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



$$ mathbf A otimes mathbf B =beginbmatrixa_11mathbf B &cdots &a_1nmathbf B \vdots &ddots &vdots \a_m1mathbf B &cdots &a_mnmathbf B endbmatrix. $$



We apply this to the $i^th$ term and then loop over all $i$ to achieve the desired matrix.







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









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

















  • 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














$textbfQuick hint :$



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.






share|cite|improve this answer




















  • 2





    That was very helpful. Thank you.

    – Luqman Saleem
    Dec 29 '18 at 22:48















2














$textbfQuick hint :$



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.






share|cite|improve this answer




















  • 2





    That was very helpful. Thank you.

    – Luqman Saleem
    Dec 29 '18 at 22:48













2












2








2







$textbfQuick hint :$



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.






share|cite|improve this answer















$textbfQuick hint :$



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.







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









SunyamSunyam

6141311




6141311







  • 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

















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