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This problem is related to the absence of symmetry breaking in the finite-size quantum Ising model. In an infinite system, the ground state is twice degenerate if $\lambda > 1$. Therefore, states with broken symmetry $\hat{\sigma}_i^x\rightarrow -\hat{\sigma}_i^x$ symmetry are possible for which $\langle\hat{\sigma}_i^x\rangle\neq 0$. For any system of finite size, the ground state is non-degenerate, so it must be symmetric with respect to the transformation $\hat{\sigma}_i^x\rightarrow -\hat{\sigma} _i^x$, and it is impossible to have $\langle\hat{\sigma}_i^x\rangle\neq 0$. An indication of symmetry breaking at $N =\infty$ in finite-size systems is the exponential smallness of the energy gap, $E_1 - E_0\sim O(e^{-Nc(\lambda)})$ if $\lambda > 1$.

So, if $\lambda > 1$, I would advise you to perform calculations for the modified Hamiltonian $$ \hat{H}_1 = \hat{H} - h_1 \sum_{i = 1}^N\hat{\sigma}^x_i, $$ with the parameter $h_1$ satisfying the conditions $E_1 - E_0 \ll h_1 \ll E_2 - E_0$.

A short but fairly complete discussion of symmetry breaking in the quantum Ising model can be found in he first chapter of the Franchini's book

This problem is related to the absence of symmetry breaking in the finite-size quantum Ising model. In an infinite system, the ground state is twice degenerate if $\lambda > 1$. Therefore, states with broken symmetry $\hat{\sigma}_i^x\rightarrow -\hat{\sigma}_i^x$ are possible for which $\langle\hat{\sigma}_i^x\rangle\neq 0$. For any system of finite size, the ground state is non-degenerate, so it must be symmetric with respect to the transformation $\hat{\sigma}_i^x\rightarrow -\hat{\sigma} _i^x$, and it is impossible to have $\langle\hat{\sigma}_i^x\rangle\neq 0$. An indication of symmetry breaking at $N =\infty$ in finite-size systems is the exponential smallness of the energy gap, $E_1 - E_0\sim O(e^{-Nc(\lambda)})$ if $\lambda > 1$.

So, if $\lambda > 1$, I would advise you to perform calculations for the modified Hamiltonian $$ \hat{H}_1 = \hat{H} - h_1 \sum_{i = 1}^N\hat{\sigma}^x_i, $$ with the parameter $h_1$ satisfying the conditions $E_1 - E_0 \ll h_1 \ll E_2 - E_0$.

A short but fairly complete discussion of symmetry breaking in the quantum Ising model can be found in he first chapter of the Franchini's book

This problem is related to the absence of symmetry breaking in the finite-size quantum Ising model. In an infinite system, the ground state is twice degenerate if $\lambda > 1$. Therefore, states with broken $\hat{\sigma}_i^x\rightarrow -\hat{\sigma}_i^x$ symmetry are possible for which $\langle\hat{\sigma}_i^x\rangle\neq 0$. For any system of finite size, the ground state is non-degenerate, so it must be symmetric with respect to the transformation $\hat{\sigma}_i^x\rightarrow -\hat{\sigma} _i^x$, and it is impossible to have $\langle\hat{\sigma}_i^x\rangle\neq 0$. An indication of symmetry breaking at $N =\infty$ in finite-size systems is the exponential smallness of the energy gap, $E_1 - E_0\sim O(e^{-Nc(\lambda)})$ if $\lambda > 1$.

So, if $\lambda > 1$, I would advise you to perform calculations for the modified Hamiltonian $$ \hat{H}_1 = \hat{H} - h_1 \sum_{i = 1}^N\hat{\sigma}^x_i, $$ with the parameter $h_1$ satisfying the conditions $E_1 - E_0 \ll h_1 \ll E_2 - E_0$.

A short but fairly complete discussion of symmetry breaking in the quantum Ising model can be found in he first chapter of the Franchini's book

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Gec
  • 6.2k
  • 2
  • 13
  • 31

This problem is related to the absence of symmetry breaking in the finite-size quantum Ising model. In an infinite system, the ground state is twice degenerate if $\lambda > 1$. Therefore, states with broken symmetry $\hat{\sigma}_i^x\rightarrow -\hat{\sigma}_i^x$ are possible for which $\langle\hat{\sigma}_i^x\rangle\neq 0$. For any system of finite size, the ground state is non-degenerate, so it must be symmetric with respect to the transformation $\hat{\sigma}_i^x\rightarrow -\hat{\sigma} _i^x$, and it is impossible to have $\langle\hat{\sigma}_i^x\rangle\neq 0$. An indication of symmetry breaking at $N =\infty$ in finite-size systems is the exponential smallness of the energy gap, $E_1 - E_0\sim O(e^{-Nc(\lambda)})$ if $\lambda > 1$.

So, if $\lambda > 1$, I would advise you to perform calculations for the modified Hamiltonian $$ \hat{H}_1 = \hat{H} - h_1 \sum_{i = 1}^N\hat{\sigma}^x_i, $$ with the parameter $h_1$ satisfying the conditions $E_1 - E_0 \ll h_1 \ll E_2 - E_0$.

A short but fairly complete discussion of symmetry breaking in the quantum Ising model can be found in he first chapter of the Franchini's book