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LagrangeEuler

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- Thread starter LagrangeEuler
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- #1

LagrangeEuler

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- #2

DrDu

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- #3

LagrangeEuler

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I can't get trace in case without circular boundary condition.

- #4

DrDu

Science Advisor

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So what? The term <S_1|T|S_N> is missing. Ok.I can't get trace in case without circular boundary condition.

You get [itex] \sum_{S_1} \sum_{S_N} \langle S_1| T^{N-1} |S_N\rangle [/itex].

What happens when you decompose |S_1/N> into eigenfunctions of T?

- #5

LagrangeEuler

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I'm not quite sure what you think? One eigenvector is just ##|+1\rangle##.

- #6

LagrangeEuler

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\langle +|\hat{T}|+ \rangle & \langle +|\hat{T}|- \rangle \\

\langle -|\hat{T}|+ \rangle & \langle -|\hat{T}|- \rangle \end{array} \right)[/tex]

For ##|S_1/_N\rangle## there are two possibilities ##|+\rangle## or ##|-\rangle##.

But I don't know how to write that in some normal way and calculate.

- #7

DrDu

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- #8

LagrangeEuler

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[tex] \left( \begin{array}{ccc}

e^{j+h} & e^{-j} \\

e^{-j} & e^{j-h}

\end{array} \right) [/tex]

eigenvectors are

[tex] \left( \begin{array}{ccc}

\frac{1}{2}e^{-h}(-e^{2j}+e^{2h+2j}\pm \sqrt{4e^{2h}+e^{4j}-2e^{2h+4j}+e^{4h+4j}}) \\

1

\end{array} \right) [/tex]

- #9

DrDu

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What I meant is that you can write

[itex] U^T T U=\Lambda[/itex] with [itex]\Lambda[/itex] being a diagonal matrix of the eigenvalues [itex]\lambda_1[/itex] and [itex] \lambda_2[/itex]. I assume [itex]\lambda_1>\lambda_2[/itex] and call V_0 the matrix V with j=0.

Hence [itex]Z=Tr(V^{N-1}V_0)=Tr(U^T\Lambda^{N-1}UV_0)=Tr(\Lambda^{N-1} UV_0U^T)\approx

\lambda_1^{N-1} (UV_0U^T)_{11}[/itex].

Hence Z differs from the Z for periodic boundary conditions only by the factor [itex] (UV_0U^T)_{11}/\lambda_1[/itex]. As we are only interested in Log Z, this makes a term which is 1/N times smaller than Z. This is of the same order as the error resulting from the replacement of [itex]\Lambda^{N-1}[/itex] by [itex]\lambda_1^{N-1}[/itex].

[itex] U^T T U=\Lambda[/itex] with [itex]\Lambda[/itex] being a diagonal matrix of the eigenvalues [itex]\lambda_1[/itex] and [itex] \lambda_2[/itex]. I assume [itex]\lambda_1>\lambda_2[/itex] and call V_0 the matrix V with j=0.

Hence [itex]Z=Tr(V^{N-1}V_0)=Tr(U^T\Lambda^{N-1}UV_0)=Tr(\Lambda^{N-1} UV_0U^T)\approx

\lambda_1^{N-1} (UV_0U^T)_{11}[/itex].

Hence Z differs from the Z for periodic boundary conditions only by the factor [itex] (UV_0U^T)_{11}/\lambda_1[/itex]. As we are only interested in Log Z, this makes a term which is 1/N times smaller than Z. This is of the same order as the error resulting from the replacement of [itex]\Lambda^{N-1}[/itex] by [itex]\lambda_1^{N-1}[/itex].

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