How to perform a basis change of single-particle states in occupation number representationn?

Question

Let's say I have an electron with spin states $|\uparrow\rangle$ and $|\downarrow\rangle$. If I want to transform these states into the basis $| + \rangle = (|\uparrow\rangle + |\downarrow\rangle ) /\sqrt{2} $, $| - \rangle = (|\uparrow\rangle - |\downarrow\rangle ) /\sqrt{2}$, I can use the unitary matrix \begin{equation} U = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \newline 1 & -1 \end{pmatrix}. \end{equation} This is pretty straight forward for one electron but I don't know what I'm supposed to do in occupation number representation. In the one-electron Hilbert space I would just write $|10\rangle_\updownarrow = |\uparrow \rangle$, $|01\rangle_\updownarrow = |\downarrow \rangle$, $|10\rangle_\leftrightarrow = |+ \rangle$, $|01\rangle_\leftrightarrow = |- \rangle$ and use the exact same transformation matrix $U$. My problem arises when I have states like $|... 10\rangle$ and $|... 01\rangle$, where '$...$' stands for some other occupations in other states and I also want to transform these states, so I also need to transform all other states like \begin{equation} |\color{gray}{10101}10 \rangle \rightarrow \frac{1}{\sqrt 2} (|\color{gray}{10101}10 \rangle + |\color{gray}{10101}01 \rangle ) \end{equation} and so on. Is there a smart way to do this other than searching for all pairs of states where the first $N-2$ occupations are equal and applying the transformation to these pairs of states? Maybe I can construct a unitary transformation matrix by using the creation and annihilation operators $c_\uparrow$, $c_\downarrow$, $c_\uparrow^\dagger$ and $c_\downarrow^\dagger$?

Answer 1

What might be more convenient is defining how the creation and annihilation operators transform under the unitary transformation $U$. In your example, let $\hat U$ be the many-body unitary operator that takes you from the basis where $S_z$ is diagonal (spin in the $z$ direction) to the basis where $S_x$ is diagonal. To obtain the correct single-particle transformation we must have $$\hat U\ \hat c^\dagger_{i, \sigma} \ \hat U^{-1} = \sum_{\sigma'} U_{\sigma, \sigma'} \ \hat c^\dagger_{i, \sigma'}, $$ where $\sigma \in \{\uparrow, \downarrow\}$ is the spin (in the $z$-basis) and $i$ is any other index you have in the system. The above expression can be verified by seeing that it gives the right transformation for the states $c^\dagger_{i, \sigma} |0\rangle$ (you'll need to use that the vacuum is, of course, invariant $U|0\rangle = |0\rangle$).

Using the above expression you can figure out how any many-body state transforms: $$ \hat U c^\dagger_{i_1, \sigma_1} c^\dagger_{i_2, \sigma_2} \dots c^\dagger_{i_N, \sigma_N}|0\rangle = \hat U c^\dagger_{i_1, \sigma_1}\hat U^{-1} \hat Uc^\dagger_{i_2, \sigma_2}\hat U^{-1} \hat U \dots \hat U^{-1} \hat U c^\dagger_{i_N, \sigma_N} \hat U^{-1} |0\rangle \\ = \sum_{\sigma'_1, \sigma'_2, \dots, \sigma'_N} U_{\sigma_1, \sigma'_1} U_{\sigma_2, \sigma'_2} \dots U_{\sigma_N, \sigma'_N} c^\dagger_{i_1, \sigma'_1} c^\dagger_{i_2, \sigma'_2} \dots c^\dagger_{i_N, \sigma'_N}|0\rangle. $$