@@ -422,3 +422,206 @@ states, again as expected due to the ratio of the interaction matrix
422422elements and the single-particle energies.
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425+ *** error: more than one subsection in a slide (insert missing !split):
426+ % !split
427+ \subsection{Quantum computing and solving the eigenvalue problem for the Lipkin model}
428+
429+ We will study a simpler variant of the Lipkin model without the $W$-term and a total spin of $J=1$ only as maximum value of the spin.
430+ This corresponds to a system with $N=2$ particles (fermions in our case).
431+ Our Hamiltonian is given by the quasispin operators (see below)
432+ \[
433+ \hat{H} = \epsilon\hat{J}_z -\frac{1}{2}V(\hat{J}_+\hat{J}_++\hat{J}_-\hat{J}_-).
434+ \]
435+
436+ As discussed earlier
437+ the quasispin operators act like lowering and raising angular momentum
438+ operators.
439+
440+ With these properties we can calculate the Hamiltonian
441+ matrix for the Lipkin model by computing the various matrix elements
442+ \begin{equation}
443+ \langle JJ_z|H|JJ_z'\rangle,
444+ \end{equation}
445+ where the non-zero elements are given by
446+ \[
447+ \begin{split}
448+ \langle JJ_z|H|JJ_z'\rangle & = \epsilon J_z\\
449+ \langle JJ_z|H|JJ_z'\pm 2\rangle & = \langle JJ_z\pm 2|H|JJ_z'\rangle \\ &= -\frac{1}{2}VC,
450+ \end{split}
451+ \]
452+ where $C$ is the Clebsch-Gordan coefficients (from the raising and lowering operators) one gets when
453+ $J_{\pm}^2$ operates on the state $|JJ_z\rangle$. Using the above
454+ definitions we can calculate the exact solution to the Lipkin model.
455+
456+ With the $V$-interaction terms, we obtain the following Hamiltonian matrix
457+ \begin{equation}
458+ \begin{pmatrix}-\epsilon & 0 & -V\\
459+ 0&0&0\\
460+ -V&0&\epsilon
461+ \end{pmatrix}
462+ \end{equation}
463+
464+ \subsection{Quantum Circuit, rewriting the Lipkin model in terms of Pauli matrices}
465+
466+ Before we proceed however, we need to rewrite the quasispin operators in terms of Pauli spin matrices/operators.
467+
468+ We take the liberty here of reminding you of some of the derivations done previously.
469+
470+ We defined the number operator as
471+ \[
472+ N=\sum_{n\sigma}a^\dagger_{n\sigma}a_{n\sigma},
473+ \]
474+
475+ which commutes with the Lipkin Hamiltonian. This can be seen by
476+ examining the Lipkin model Hamiltonian and noticing that the one-body
477+ part simply counts particles while the two-body term moves particles
478+ in pairs. Thus, the Hamiltonian conserves particle number. To find
479+ more symmetries we rewrote the Lipkin Hamiltonian in terms of $SU(2)$ quasispin
480+ operators
481+ \begin{align}
482+ H = \epsilon J_z + \frac{1}{2}V(J^2_++J^2_-),
483+ \end{align}
484+ via the mappings
485+ \[
486+ J_z=\sum_{n}j_z^{(n)},
487+ \]
488+ and
489+ \[
490+ J_\pm=\sum_nj^{(n)}_{\pm},
491+ \]
492+ where we have the onebody operators
493+ \[
494+ j_z^{(n)}=\frac{1}{2}\sum_{\sigma}\sigma a^\dagger_{n\sigma}a_{n\sigma},
495+ \]
496+ and
497+ \[
498+ j^{(n)}_{\pm}=a^\dagger_{n\pm}a_{n\mp}.
499+ \]
500+
501+ These operators obey the $SU(2)$ commutation relations
502+
503+ \[
504+ [J_+,J_-]=2J_z,
505+ \]
506+ and
507+ \[
508+ [J_z,J_\pm]=\pm J_\pm.
509+ \]
510+ Here the ladder operators are defined as $J_{\pm}= J_x\pm iJ_y$. With this rewriting, we can see that the total spin operator $J^2$, which is defined as
511+ \[
512+ J^2= J^2_x+J^2_y+J^2_z =
513+ \frac{1}{2}\{J_+,J_-\}+J_z^2,
514+ \]
515+ commutes with the Hamiltonian since the Hamiltonian.
516+ We note also that the rotation operator
517+ \[
518+ R=e^{i\phi J_z},
519+ \]
520+ commutes with the Hamiltonian, which can be explained as follows. Writing $J_z$ as
521+ \[
522+ J_z=\frac{1}{2}(N_+-N_-),
523+ \]
524+ where $N_\pm=\sum_{n\pm}a^\dagger_{n\pm}a_{n\pm}$, allows us to see that it measures half the difference between the number of particles in the upper and lower levels. Thus, the possible eigenvalues $r$ of the signature operator are
525+ \begin{align}
526+ r=+1, & j_z=2n \\
527+ r=+i, & j_z=2n+\frac{1}{2} \\
528+ r=-1, & j_z=2n+1 \\
529+ r=-i, & j_z=2n+\frac{3}{2} \\
530+ \end{align}
531+
532+ for $n\in\mathbb{Z}$. Note that $r$ is real or imaginary if the number
533+ of particles $N$ is even or odd, respectively. Since, as discussed
534+ above, the Lipkin Hamiltonian conserves $N$, $r$ cannot jump between
535+ being real and imaginary. Additionally, because particles must be
536+ moved in pairs, and $J_z$ measures half the difference between
537+ particles in the upper and lower levels, $j_z$ can only change by as
538+
539+ \[
540+ j_z\rightarrow \frac{1}{2}[(N_+\pm 2n)-(N_-\mp 2n)]
541+ \]
542+ or $j_z\rightarrow J_z\pm2n$.
543+
544+ To solve the Lipkin model with a quantum computer, the first step is
545+ to map the system to a set of qubits. We will restrict ourselves here
546+ to the half-filled case where the number of particles $N$ equals the
547+ degeneracy of the states $\Omega$. One could assign each possible
548+ state $(n,\sigma)$ a qubit such that the qubit being in the state
549+ $\vert 1\rangle$ or $\vert 0\rangle$ would imply that the state
550+ $(n,\sigma)$ is occupied or unoccupied, respectively. This mapping
551+ scheme (which we will call occupation mapping) requires 2$\Omega$
552+ qubits.
553+
554+ However, because there are only two energy levels in the Lipkin model,
555+ any other natural mapping is possible. In this mapping scheme (which
556+ we will call level mapping) each doublet ($(n,+1)$, $(n,-1)$) would be
557+ assigned a qubit such that the qubit being in the state $\vert 0\rangle$ or
558+ $\vert 1\rangle$ would imply that the particle is in the $(n,+1)$ or $(n,-1)$
559+ state, respectively. Note that these are the only two possible
560+ configurations of the doublet as we are restricting ourselves to the
561+ half-filled case and the Lipkin Hamiltonian only moves particles
562+ between energy levels, not degenerate states. Thus the level mapping
563+ only requires $\Omega$ qubits which is half that of the occupation
564+ mapping.
565+
566+ The Hamiltonian takes the form
567+
568+ \begin{align}
569+ H=\epsilon J_z + \frac{1}{2}V(J^2_++J^2_-).
570+ \end{align}
571+
572+ Plugging the mapping from the total $J$ operators to the individual one-body $j$ operators yields
573+
574+ \begin{align}
575+ H &= \epsilon\sum_{n}j_z^{(n)} + \frac{1}{2}V\left[\left(\sum_nj^{(n)}_{+}\right)^2+\left(\sum_nj^{(n)}_{-}\right)^2\right]
576+ \\
577+ &= \epsilon\sum_{n}j_z^{(n)} + \frac{1}{2}V\sum_{n,m}\left(j^{(n)}_+j^{(m)}_++j^{(n)}_-j^{(m)}_-\right)
578+ \\
579+ &= \epsilon\sum_{n}j_z^{(n)} + 2V\sum_{n<m}\left(j^{(n)}_xj^{(m)}_x-j^{(n)}_yj^{(m)}_y\right),
580+ \end{align}
581+ where we have used the definitions
582+ \[
583+ j_{\pm}^{(n)}=j_x^{(n)}\pm ij_y^{(n)}.
584+ \]
585+ To convert to Pauli matrices, we make the transformations
586+ \[
587+ j_x^{(n)} \rightarrow X_n/2,
588+ \]
589+ and
590+ \[
591+ j_y^{(n)} \rightarrow Y_n/2,
592+ \]
593+ and finally
594+ \[
595+ j_z^{(n)} \rightarrow Z_n/2,
596+ \]
597+ which preserve the above $SU(2)$ commutation relations.
598+ The factor of $1/2$
599+ is due to the eigenvalues of the Pauli matrices being $\pm 1$
600+ while we are dealing with spin $1/2$ particles.
601+
602+ This transforms our Hamiltonian into
603+ \[
604+ H=\frac{1}{2}\epsilon\sum_{k=1}^nZ_k+\frac{1}{2}V\sum_{n\neq j=1}^N(X_kX_j-Y_kY_j).
605+ \]
606+
607+ With this form, we can clearly see that the first (one-body) term in
608+ the Hamiltonian returns the energy $-\epsilon/2$ or $+\epsilon/2$ if
609+ the qubit representing the particle of a doublet is in the ground
610+ ($\vert 1\rangle$) or excited ($\vert 0\rangle$) state,
611+ respectively. The action of the second (two-body) term in the
612+ Hamiltonian can be determined by noting that
613+ \begin{align}
614+ \frac{1}{2}(XX-YY)\vert 00\rangle &= \vert 11\rangle,
615+ \\
616+ \frac{1}{2}(XX-YY)\vert 01\rangle &= 0,
617+ \\
618+ \frac{1}{2}(XX-YY)\vert 10\rangle &= 0,
619+ \\
620+ \frac{1}{2}(XX-YY)\vert 11\rangle &= \vert 00\rangle.
621+ \end{align}
622+
623+ That is, the two-body term moves a pair of particles between the
624+ ground states $\vert 00\rangle$ and the excited states $\vert
625+ 11\rangle$ of their respective doublets.
626+
627+
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