3
$\begingroup$

I was studying the Hubbard model, where we define the spin operator $\vec{S} = \frac{1}{2} c^\dagger \vec{\sigma} c$, where the creation and annihilation operators are both vectors of the form $c^\dagger = (c^\dagger_\uparrow, c^\dagger_\downarrow)$ (I've used $\hbar = 1$). This spin operator is of course the same as the spin operator that we use in other lattice systems, like the Heisenberg model or the Ising model. In fact, when we write the spin operators in the Heisenberg models in terms of fermions, we use precisely this formula. However in these models, we also denote $\vec{S} = \frac{\vec{\sigma}}{2}$.

Is there any deeper reason why both these forms are possible, besides the fact that the spin commutation relations must satisfy $[S_i, S_j] = \epsilon_{ijk} S_k$? Should the number operator $n = c^\dagger c$ necessarily be 1 on the Heisenberg/Ising models? Under what conditions is the formula $\vec{S} = \frac{\vec{\sigma}}{2}$ true?

$\endgroup$
1

1 Answer 1

1
$\begingroup$

I think the difference between the representation of the spin operator in the Hubbard vs. the Ising models is due to the fact that in the Hubbard model, one tries to describe the situation of spin-1/2 fermions sitting in a lattice, whereas, in the Ising model, the spins are distinguishable and hence have neither fermionic nor bosonic symmetry. In general, the second-quantized representation of one-body operators follows the same structure for either fermions or bosons, e.g.,

$O_1 = \sum_{pq} \langle p | o_1 | q \rangle a_p^+ a_q,$

where $|p\rangle$ and $|q\rangle$ represent the single-particle states with which we are describing the problem and $a_p^+ (a_p)$ is the creation (annihilation) operator that obeys either the fermionic or bosonic (anti)commutation relations.

For both the Hubbard and Ising models, the single-particle states are the spin-1/2 states, $|\alpha\rangle \equiv |1/2,1/2\rangle$ and $|\beta\rangle \equiv |1/2,-1/2\rangle$. If we follow the above recipe for calculating a one-body operator, we can evaluate the form of each component of $\vec{S}$ using the usual raising and lowering operators $S_+$ and $S_-$.

$S_{x} = \langle \alpha | s_x |\alpha\rangle a_\alpha^+ a_\alpha + \langle \alpha | s_x |\beta\rangle a_\alpha^+ a_\beta + \langle \beta | s_x |\alpha\rangle a_\beta^+ a_\alpha + \langle \beta | s_x |\beta\rangle a_\beta^+ a_\beta = \frac{\hbar}{2} a_\alpha^+ a_\beta + \frac{\hbar}{2} a_\beta^+ a_\alpha$

$S_{y} = \langle \alpha | s_y |\alpha\rangle a_\alpha^+ a_\alpha + \langle \alpha | s_y |\beta\rangle a_\alpha^+ a_\beta + \langle \beta | s_y |\alpha\rangle a_\beta^+ a_\alpha + \langle \beta | s_y |\beta\rangle a_\beta^+ a_\beta = -i\frac{\hbar}{2} a_\alpha^+ a_\beta + i\frac{\hbar}{2} a_\beta^+ a_\alpha$

$S_{z} = \langle \alpha | s_z |\alpha\rangle a_\alpha^+ a_\alpha + \langle \alpha | s_z |\beta\rangle a_\alpha^+ a_\beta + \langle \beta | s_z |\alpha\rangle a_\beta^+ a_\alpha + \langle \beta | s_z |\beta\rangle a_\beta^+ a_\beta = \frac{\hbar}{2} a_\alpha^+ a_\alpha - \frac{\hbar}{2} a_\beta^+ a_\beta$

The above expressions coincide with how it is usually written in terms of the Pauli spin matrices, namely, $\vec{S} = \frac{\hbar}{2} a^+ \vec{\sigma} a$. In fact, in the Hubbard model (and other fermionic/bosonic models), we likely would have other quantum numbers related to the particle's spatial wave function (such as momentum $\vec{k}$ of a plane wave or an orbital label $p$), thus the more complete form of the spin operator for these situations would be $\vec{S} = \sum_{pq} \delta_{pq} \frac{\hbar}{2} a_p^+ \vec{\sigma} a_q$, where the $\delta_{pq}$ is explicitly included to show that the spin is diagonal in the spatial degrees of freedom. Note how the form of this operator is exactly that of the generic one-body operator $O_1$ and the action of the operator on a single-particle Hilbert space is subsumed in the matrix element of the second-quantized operator. So in fact, this form is not necessarily a result of the algebraic properties of spin, rather, it is simply the form the spin operator adopts in second-quantization for indistinguishable particles.

On the other hand, for the classical/distinguishable spins used in the Ising or Heisenberg models, there is no concept of particle creation or destruction (although spins can be interconverted between up and down), and hence no need to introduce Fock space operators, and there is no particle permutation symmetry that necessitates the construction of Fock space operators with definite (anti)commutation relations. So we can just proceed with usual one-body quantum mechanics and express our operators in a matrix representation in the relevant basis of $|\alpha\rangle$ and $|\beta\rangle$ as we did above. But here, there is no need to tack on the $a_p^+a_q$ term as that arises purely from considering the permutation symmetry of the particles. So, I'm not sure there is any need to reason that the Hubbard formula reduces to the Ising formula when $a^+a = 1$.

On the other hand, there are ways of converting the distinguishable spin problems into the elegant language of second-quantization. The ingredient that's missing is the concept of Fock space; you need an operator that can act on vacuum to generate a spin state. These type of operators can be introduced by resorting to a bosonic algebra (Schwinger representation) or a fermionic algebra (Jordan-Wigner representation), and hence, the construction of $n$-body operators like $S$ (1-body) or $S^2$ (2-body) proceeds according to the rules of many-body theory. I'm not aware of any kind of second-quantized algebra used in distinguishable spin problems that does not invoke an indistinguishable particle description.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.