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I have a question regarding the Holstein-Primakoff representation.

In the HP-representation we define the spin operators in terms of bosonic creation and annihilation operators.

$$ S_j^+ = \sqrt{2S - n_j} a_j \\ S_j^{^-} = a^\dagger_j\sqrt{2S - n_j} \\ S^z_j = S - n_j $$

Where $a_j$ and $n_j$ are operators. When we derive the magnon dispersion relation, we make the assumption that

$$ \frac{\langle n_j \rangle}{\langle S^z \rangle}\ll 1 $$

which is fine. As far as I understand this only means that we assume that most of the spins are pointing along the z-direction. However, when we go further in the derivation, we do a series expansion in $n_j/S$, so that i.e.

$$ S_j^+ = \sqrt{2S}\sqrt{1- \frac{n_j}{2S}+...} \approx \sqrt{2S}\sqrt{1- \frac{n_j}{2S}} $$

This is where I don't understand. Say that we are in a spin $1/2$-system. In that case we can for one site have at most 1 magnon excitation and $S=\pm 1/2$, so for each individual site $n_j/2S$ is not much smaller than one.

My question is then, how can we justify that the series expansion makes sense in the operators at each individual site? Will the contributions from each site when summed up not contribute because

$$ \frac{\langle n_j \rangle}{\langle S^z \rangle}\ll 1 $$

or have I misunderstood something?

Any thoughts would be much appreciated.

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2 Answers 2

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The assumption is, that the spin $S$ is a large parameter. A conjecture that is apparently not valid for $S=1/2$. The expansion is in $1/S$, which is assumend to be close to zero.

$$ S^+_j = \sqrt{2S-n_j}a^\dagger_j = \sqrt{2S}\sqrt{1-\frac{n_j}{2S}}a^\dagger_j\approx\sqrt{2S}\cdot\left(1-\frac{n_j}{4S}\right)a^\dagger_j$$ The second term, being of order $S^{-1/2}$, is neglected.

Assuming $S$ to be large, amounts to a semiclassical approximation. In this limit the relative uncertainty of the spin-operators becomes vanishingly small. (Use the spin-algebra to see this.) $$ \frac{\Delta S^i\Delta S^j}{S^2} \longrightarrow 0$$ The working hypothesis is, that low-energy excitations are realized as small deviations around the fully aligned groundstate. For small spin, e.g $S=1/2$ there is no way to only slightly deviate from let's say $S^z_i=+1/2$ Which brings us back to your objection/observation, that the expansion is not valid in the 'small-spin' limit.

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  • $\begingroup$ Thank you very much for the answer. It seems then that I have not really understood what S is. If we have one electron at each site in a lattice, is S not 1/2. Or is S the sum of all the spins so that if we have N filled lattice points S = N*(1/2)? $\endgroup$
    – camzor00
    Mar 1, 2014 at 11:59
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    $\begingroup$ @camzor00 $S$ is the spin on each site. For example if you consider a chain of spin 1/2 particles (electrons), then $S=1/2$ $\endgroup$
    – Nephente
    Mar 1, 2014 at 15:19
  • $\begingroup$ Okay, so this model is not applicable to a lattice (or string) of electrons? Isn't this what we usually consider in a condensed matter context? I apologize for being slow, but I guess I am still confused as to which physical systems this approximation is valid for. $\endgroup$
    – camzor00
    Mar 1, 2014 at 16:45
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    $\begingroup$ @camzor00 Yes. It is not clear why this expansion should hold for spin 1/2. Apparently, it does surprisingly well in 2 and 3 dimensions. There may be techniques better suited for 1d low spin systems, but i have to refer you to the literature for that, since I'm far from an expert. $\endgroup$
    – Nephente
    Mar 2, 2014 at 14:12
  • $\begingroup$ @camzor00 If my answer was helpful to you, please consider giving it an upvote. Thank you. $\endgroup$
    – Nephente
    Mar 5, 2014 at 7:40
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It is possible and consistent if the states are a superposition


Holstein Pirmakoff Transformation:

At a particular site

$S_{+}=(\hbar \sqrt{2s-a^+a}) a$

here $S_+$ increases the spin of the Spin Eigenstate.

On the R.H.S., we have $a$ as a bosonic annihilation operator.

  1. Its action on a definite number state is to reduce the number by 1
  2. $a^{+}a < 2s$
  3. Bosonic Quasiparticles being like Simple harmonic Oscillator have oscillation frequencies
  4. In Heisenberg Picture, Oscillations of $a$ appear as Oscillations of $S_{+}$.
  5. Oscillations of bosonic operators correspond to Spin Oscillations.

Assertion 1:

The Spin 1 state is the ground state of bosonic Operator $a$

Proof:

$S_{+}|1\rangle = 0$

i.e. $a|1\rangle = 0$

Assertion 2:

The Spin -1 state has occupation number 1 in bosonic space

$S_z |-1\rangle =-\frac{\hbar}{2}$

$S_z=\hbar(\frac{1}{2}-a^+a)$

doing the algebra, $a^+a=1$

Assertion 3

for $| \psi \rangle=|1> + \epsilon|-1\rangle, \epsilon \approx 0 \implies \langle a^+a \rangle \approx 0$

$\langle \psi | a^+a| \psi\rangle = \langle \psi |a^+a|1\rangle + \epsilon \langle \psi |a^+a|-1\rangle $

$\langle \psi |a^+a|1\rangle =0$ Since Spin 1 state is the bosonic groudstate.

So,

$\langle \psi | a^+a| \psi\rangle =\epsilon \langle \psi |a^+a|-1\rangle \approx \epsilon C \approx 0 $

Assertion 3 shows that there are spin states which have occupation number really close to 0, as close to 0 as any real number can get.

Going further, let's look at what an oscillation at single site would look like:

$|\psi(t)\rangle = |1\rangle + \alpha e^{i\omega t}|-1\rangle, \alpha \approx 0$

Now if we have a lattice, the angular phase change rate $\omega$ ( angular velocity) could get coupled to specific spatial phase change rate k ( angular wavenumber).

let's construct a specific example:

$|\psi(x,t)\rangle = |1\rangle + \alpha e^{i(\omega t-kx)}|-1\rangle, \alpha \approx 0$

All of these states exist without violating the low occupation number approximation. Hence there is no contradiction.

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  • $\begingroup$ This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review $\endgroup$
    – Miyase
    Jan 8, 2023 at 15:20
  • $\begingroup$ Doesn't it answer the question? The small mean number operator to S ratio means "small" enough deviations in local wavefunction from the ground state . $\endgroup$ Jan 8, 2023 at 15:35
  • $\begingroup$ Read the comment to the end: "that don't require clarification from the asker". Perhaps your answer contains relevant information, but it could use a bit more proof, or at least pointers to useful documentation to support what you claim. $\endgroup$
    – Miyase
    Jan 8, 2023 at 17:49

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