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I'm quite familiar with rotation in quantum/classical mechanics. I know rotation for an operator $O$ or state $|\psi \rangle$ acts like:

$$O \rightarrow R O R^{-1} \\ |\psi \rangle \rightarrow R |\psi \rangle $$

However, I don't understand how to apply this to the second quantized operator for example $c_{i \sigma} c_{j \sigma}^\dagger$ where $\sigma$ is spin index. I mean naively I can rotate this like an operator and so acts like:

$$c_{i \sigma} c_{j \sigma}^\dagger \rightarrow R c_{i \sigma} R^{-1} R c_{j \sigma}^\dagger R^{-1}$$ where $R \in SU(2)$, but then $c$ is in a different space from $R$. Maybe I'm supposed to do this? $$R c_{i \sigma} R^{-1} = c_{i R\sigma R^{-1}}$$

And if the above is true, then I have another confusion: in quantum mechanics, say two particle state, the only rotation invariant state is the singlet $\frac{1}{\sqrt{2}} |\uparrow \downarrow \rangle - | \downarrow \uparrow \rangle$ with $S(S+1) = 0$, and so analogously I think this operator is also rotational invariant

$$c_{i \uparrow} c_{j \downarrow} -c_{i \downarrow} c_{j \uparrow} $$

but this is not the only rotational invariant operator? Because it seems like $c_{i \sigma} c_{j \sigma}^\dagger$ is also invariant.. Is my thinking right?

In summary, I may be confused about how rotation acts in Fock space (for second quantized operator) versus Hilbert space (for spin state).

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2 Answers 2

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Let's focus, like your question, only on rotations in spin-space (the generalization to real-space is straight-forward). Then we first choose a basis of operators in 2nd quantization, let's say $c_{i, \uparrow}, c_{i, \downarrow}$ and their conjugates, which maintain the canonical anti-commutation relations $$ \{ c_{i,\sigma}, c^{\dagger}_{j,\sigma '} \} = \delta_{i,j} \delta_{\sigma, \sigma '}$$ A rotation in spin-space will be a linear transformation within this basis $$ R_{S_i} c_{i,\sigma} R^{-1}_{S_i} = a_r c_{i,\sigma} + b_r c_{i, \bar{\sigma}} $$ where $\bar{\sigma}$ is the opposite spin to $\sigma$. How to carry out this rotation is straight-forward from the construction of the spin operators in second quantization $$ S_{i}^{\alpha} = \frac{\hbar}{2} \sum_{\lambda, \lambda '} c^{\dagger}_{i,\lambda} \sigma^{\alpha}_{\lambda, \lambda '}c_{i,\lambda '}$$ with $\alpha = x,y,z$ for different spin operations at the point $i$, and $\sigma^{\alpha}$ is the respective Pauli matrix. From this, you can explicitly calculate how does each operator transform, as $R_{S_i} = e^{-i \sum_\alpha \theta_\alpha S^{\alpha}/\hbar}$.

The singlet you wrote is the only rotationaly invariant state for two spins. The point is that in second quantization we don't restrict ourselves from the beginning to a fixed number of particles, and allow them to change. We can construct, however, operators that are invariant under local spin rotations. For example $c^{\dagger}_{i,\uparrow}c_{i, \uparrow} + c^{\dagger}_{i,\downarrow}c_{i, \downarrow}$ is such an operator, as it just counts the number of particles at site $i$, and their rotation is immaterial to it.

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  • $\begingroup$ Thanks very much! To make sure I understand this, let's consider a simple rotation around z-axis by $\pi/2$ and so let $R_s = e^{-i \pi/2 S_z}$. From quantum mechanics, we expect for example $S_x \rightarrow S_y$. And you are saying the rotation acts in $c_{\sigma}$ space and leaves the pauli matrix $\sigma$ in the spin operator alone (this was my confusion, I thought we are transforming the matrix rather than c). However I tried to carry this out explicitly but I dont see how to simply things properly (I will write out the steps below), could you point out what Im missing here? Thanks again! $\endgroup$
    – MoreConfi
    Commented Jun 4, 2020 at 19:58
  • $\begingroup$ And also I'm not sure why you said the state is only rotation invariant for two spins. You are referring to there is no $l = 0$ state for multiple spin additions? But I guess for multiple spins we can have spin singlet pairs? Sorry I realize this is a lot of more work, I'm def also happy with a textbook reference if you have one in mind! $\endgroup$
    – MoreConfi
    Commented Jun 4, 2020 at 20:15
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    $\begingroup$ my comment about the singlet was the following: if you add together two spins, then the state you wrote is the $l=0$ one. If you add three spins there is no $l=0$ state. If you add four spins, you get several different $l=0$ states (2 such states, to be precise), which none of which is what you wrote, simply because what you wrote portrays two spins. etc. $\endgroup$
    – user245141
    Commented Jun 5, 2020 at 4:05
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Based on yu-v's answer, consider a simple rotation around z-axis by 𝜋/2 and so let

$$R_S = e^{-i \pi/4 c^\dagger_\alpha \sigma_{\alpha \beta} c_\beta}$$

The answer we expect is $S_x\rightarrow S_y$ or $1/2 (c^\dagger_\uparrow c_\downarrow + c^\dagger_\downarrow c_\uparrow ) \rightarrow 1/2 (-i c^\dagger_\uparrow c_\downarrow + i c^\dagger_\downarrow c_\uparrow )$

To carry out this calculation, we need to expand the exponential as

$$R_S = 1 -i \pi/4 c^\dagger_\alpha \sigma_{\alpha \beta} c_\beta + (-i \pi/4 c^\dagger_\alpha \sigma_{\alpha \beta} c_\beta)^2 +...\\ = 1 - i \pi/4 (c^\dagger_\uparrow c_\uparrow - c^\dagger_\downarrow c_\downarrow ) - \pi^2/32 (c^\dagger_\uparrow c_\uparrow + c^\dagger_\downarrow c_\downarrow+ 2 c^\dagger_\downarrow c^\dagger_\uparrow c_\downarrow c_\uparrow )+...$$

I see to be a faithful representation of the spin operator, we need the single occupancy constraint so the quarctic term should be zero, but there are extra quadratic terms that are higher order with coefficient of $\pi^n$ in n-th order expansion and I'm not sure how to simplify it.

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    $\begingroup$ To do this calcaultion you don't need to expand the exponential in such a way, just use Baker-Hausdorf formula and see that $e^{i \theta S^z}c_{\uparrow} e^{-i\theta S^z} = e^{-i\theta /2}c_\uparrow$ so you get that for $\pi/2$ rotation $c^{\dagger}_{\uparrow}c_{\downarrow} \to i c^{\dagger}_\uparrow c_\downarrow$. But more generally - convince yourself that the spin representation via the second-quantized operators gives the correct $[S_i, S_j] = i\epsilon_{ijk}S_k$ and that guarantees that all rotations will work as they should. $\endgroup$
    – user245141
    Commented Jun 5, 2020 at 4:01

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