I am currently studying statistics and 2D Ising models and noticed in my lecturer's notes the operators, acting in the spin space enter image description here

The text says that this is identity $2^N\times 2^N$ matrix. I don't understand how this operator acts on the vector of spins. Usually there is only tensor convolution by index, but here are fixed spins in a row, for example:


How should I change their value if they are fixed?

Someone, please explain the action, for example, for the simple operator $\delta_{\sigma_1'\sigma_1}$ and for spin row $\{\uparrow\}$. Should it work like this? $$\delta_{\sigma_1'\sigma_1}\{\sigma_1\}=\delta_{\sigma_1'\uparrow}\{\uparrow\}+\delta_{\sigma_1'\downarrow}\{\downarrow\}$$

What I've even got? And also how can this matrix be $2^N\times 2^N$ if the row is made of $N$ spins? I am probably too sleepy and just forgot how tensors work...


1 Answer 1


Ok, probably I guessed how it should work. At first we have only one vector in our space where operator act. This is not the vector of spins ${\sigma_1,\sigma_2,\ldots,\sigma_n}$. This is the vector of all possible states (every element of this vector is spin vector) with any order that we will fix initially (equivalence class), for example for 3 spins: $$\vec{s}=\left(\begin{array}{c} \{\uparrow,\uparrow,\uparrow\} \\ \{\uparrow,\uparrow,\downarrow\} \\ \{\uparrow,\downarrow,\uparrow\} \\ \{\uparrow,\downarrow,\downarrow\} \\ \{\downarrow,\uparrow,\uparrow\} \\ \{\downarrow,\uparrow,\downarrow\} \\ \{\downarrow,\downarrow,\uparrow\} \\ \{\downarrow,\downarrow,\downarrow\} \\ \end{array}\right)=\left(\begin{array}{c} \vec{\sigma}[1] \\ \vec{\sigma}[2] \\ \vec{\sigma}[3] \\ \vec{\sigma}[4] \\ \vec{\sigma}[5] \\ \vec{\sigma}[6] \\ \vec{\sigma}[7] \\ \vec{\sigma}[8] \\ \end{array}\right)$$ And in this case matrix $2^3\times 2^3$ will correctly work: $$\vec{s'}=E_{8\times 8}\left(\begin{array}{c} \{\uparrow,\uparrow,\uparrow\} \\ \{\uparrow,\uparrow,\downarrow\} \\ \{\uparrow,\downarrow,\uparrow\} \\ \{\uparrow,\downarrow,\downarrow\} \\ \{\downarrow,\uparrow,\uparrow\} \\ \{\downarrow,\uparrow,\downarrow\} \\ \{\downarrow,\downarrow,\uparrow\} \\ \{\downarrow,\downarrow,\downarrow\} \\ \end{array}\right)=\left(\begin{array}{c} \{\uparrow,\uparrow,\uparrow\} \\ \{\uparrow,\uparrow,\downarrow\} \\ \{\uparrow,\downarrow,\uparrow\} \\ \{\uparrow,\downarrow,\downarrow\} \\ \{\downarrow,\uparrow,\uparrow\} \\ \{\downarrow,\uparrow,\downarrow\} \\ \{\downarrow,\downarrow,\uparrow\} \\ \{\downarrow,\downarrow,\downarrow\} \\ \end{array}\right)=\vec{s}$$ So we numbered the sequence of states (spinvectors) and marked elements as $\vec{\sigma}[i]$. It's not the component of $\vec{\sigma}$, it's whole spinvector at the i-th place of sequence. We want to write: $$\vec{\sigma}'[i]=\delta_{ij}\vec{\sigma}[j]$$ And then to move from terms of $\vec{\sigma}[i]$ to normal spinvectors with their own indexes. So we have to number states not by artificial order but by their own spins. In terms of components, for example, for 2 spins (sigmavector has two compoments) we can write: $$\sigma'_{ik}=\delta_{ij}\delta_{kl}\sigma_{jl}$$ Let's check it: $$\sigma'_{\uparrow\uparrow}=\delta_{\uparrow j}\delta_{\uparrow l}\sigma_{jl}=\sigma_{\uparrow\uparrow}$$ And the same logic with another indexes, it really works like identity operator. It can look absurd for some operators, because we can get smth like $$\sigma'_{\uparrow\uparrow}=\uparrow\downarrow$$ but as we saw earlier the index $_{\uparrow\uparrow}$ is only necessary to mark the order of spinvectors in $\vec{s}$. It will only mean that the state where spins were $\uparrow\uparrow$ has now spins $\uparrow\downarrow$.

May be I couldn't understand the stupid stuff but teachers notes looked like they supposed acting on spinvectors $1\times N$ and it confused me.


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