Why Correlation Function of Order Parameter Density Affects Magnetization

[RedX]

Does anyone know why:

$$<m(k)m(p)>=(2 \pi)^3 \delta(k+p)|m(k)|^2$$

where m(k) and m(p) are the Fourier transform of the order parameter density and the angled brackets <> stand for an ensemble average?

For example, the magnetization M is given by:

$$M=<\int d^3r m(r)>$$

 

[stephenhky]

This is not necessarily true, because it depends on what kind of models you are talking about. The model describes the probability functional, and it is this probability that gives correlation function.

But what is "usually" true is that the delta(k+p) is often there because of the Fourier transform.

Where did you see this from?

 

[RedX]

This is not necessarily true, because it depends on what kind of models you are talking about. The model describes the probability functional, and it is this probability that gives correlation function.

But what is "usually" true is that the delta(k+p) is often there because of the Fourier transform.

Where did you see this from?

This comes from a textbook, chapter 16 of a book by Huang.

Huang doesn't mention the model other than that it's of a ferromagnet. He uses the relation to derive an "Ornstein-Zernike form".

Anyways, so you're saying that in general, to calculate a correlation, you have to take the trace with the exponential of the Hamiltonian to see what you get?

It seems a bit of mystery, that the correlation of product of order parameters <m(r)m(0)> in position space is complicated: $$\frac{e^{-r/\xi}}{r}$$, but in momentum space:

<m(k)m(p)>=delta (k+p) is so easy.

 

[stephenhky]

I am not at home right now and hence I cannot check Huang's book at the moment. Do you mean Kerson Huang's "Statistical Mechanics"?

To your question, my answer is yes.

And it is true that for paramagnet or the longitudinal component of a ferromagnet (true for the transverse component but the correlation goes to infinity),

$$\frac{e^{-r/\xi}}{r}$$ .

If you do the Fourier transform, you will get

$$\langle m(k) m(p) \rangle = \frac{\delta(p+k)}{k^2 + \frac{1}{\xi^2}}$$ ,

which is the well-known Ornstein-Zernike form.

 

[RedX]

I am not at home right now and hence I cannot check Huang's book at the moment. Do you mean Kerson Huang's "Statistical Mechanics"?

To your question, my answer is yes.

And it is true that for paramagnet or the longitudinal component of a ferromagnet (true for the transverse component but the correlation goes to infinity),

$$\frac{e^{-r/\xi}}{r}$$ .

If you do the Fourier transform, you will get

$$\langle m(k) m(p) \rangle = \frac{\delta(p+k)}{k^2 + \frac{1}{\xi^2}}$$ ,

which is the well-known Ornstein-Zernike form.

Yeah it came from Kerson Huang's Statistical Mechanics.

I can actually derive this:

$$\langle m(k) m(p) \rangle = \frac{\delta(p+k)}{k^2 + \frac{1}{\xi^2}}$$

from quantum field theory, as this just leads to the Yukawa potential for exchange of spin-0 mesons.

However, the course I'm taking is statistical mechanics, so I'm not exactly sure how Huang is deriving it in his book, mainly how

$$<m(k)m(p)> =(2 \pi)^3 \delta(k+p)|m(k)|^2$$

comes about.

 

[stephenhky]

Actually what I said is also true for statistical mechanics and it is the magnetic susceptibility of the ferromagnet. IN QFT, $$\xi$$ is the screening length of the Yukawa potential, but in statistical mechanics, it is the correlation length of the fluctuations.

I have the book at hand now. I don't think his expression is correct. The left hand side is an ensemble AVERAGE of function of m(k), but the right hand side is the value of a function of m(k). This does not make sense to me at all.

And (16.11) is only correct if $$\tilde{m}(k)$$ means the Fourier transform of the FLUCTUATION $$m(x)-m(0)$$, where m(0) is the average of the magnetization as stated in the text.

I would suggest reading Ma's "Modern Theory of Critical Phenomena" on this topic.

 

[RedX]

I would suggest reading Ma's "Modern Theory of Critical Phenomena" on this topic.

Thanks. I think this critical phenomena stuff is interesting. I did the standard calculation of calculating the critical exponents of the van der waals model, and of the ising model in the mean-field approximation (calculating them for the van der waals was slightly hard, but the ising model was easy to calculate), and they are the same values. seems strange that so many disjoint phenomena have the same parameters. i'll have a look at the book.

 

[stephenhky]

Ya! it is the most interesting thing in critical phenomena - universality. I did it with the Heisenberg model with LGW functional at that time.

 

[RedX]

Now that I think about it,

$$<m(k)m(p)>=(2\pi)^3\delta(k+p)|m(k)|^2$$

does make sense in the context of 2nd quantization and quantum field theory.

Here you interpret m(k), the Fourier coefficient of the magnetization density field, as the creation and destruction operators of the magnon. That is, m(k) is a "ladder operator" that you see in the harmonic oscillator in QM, and m(k) creates a quanta of magnons.

So the physical interpretation of:

<m(k)m(p)>

is to create a magnon of momentum p from the lattice, and then annihilate a magnon of momentum k from the lattice. Only if p is equal to k will this not be equal to zero, and hence the delta function. This is a "propagator".

I think the statistical mechanics book wants to use a result from quantum field theory, but understandably doesn't want to explain it. However, it should at least say that this result comes from quantum field theory. Or at least the instructor should say that, which she didn't for my class.

 

[saaskis]

I think the statistical mechanics book wants to use a result from quantum field theory, but understandably doesn't want to explain it. However, it should at least say that this result comes from quantum field theory. Or at least the instructor should say that, which she didn't for my class.

Classical statistical field theory is already good enough. The field Hamiltonian is something like
$$\beta H = \int dx \left[(\nabla m)^2+M^2 m^2\right] \sim \sum_p m(-p) (p^2+M^2)m(p),$$
which is the simplest rotationally invariant Hamiltonian available. The partition function is
$$Z = \int Dm(p) \exp\left(-\sum_p m(-p) (p^2+M^2)m(p)\right),$$
giving the propagator
$$G(p,k) = \int dx dx' e^{ipx+ikx'} \langle m(x) m(x') \rangle \sim \delta(p+k) \frac{1}{p^2+M^2}.$$
Here the average is given by
$$\langle A(x') \rangle = \int Dm(x) \exp\left(-\beta H[\nabla m(x),m(x)]\right) A(x').$$
The derivation of the functional integral does not require quantization: it can be derived from the discrete lattice model e.g. by a Hubbard-Stratonovich transformation or by simply defining the field m(x) as a spatial average of values in the lattice, the averaging done such that the field varies slowly over the scale of the lattice constant a. And most importantly, statistical field theories come with a natural UV cutoff $$\Lambda \sim 1/a$$.

 

[stephenhky]

I agree with saaskis that there should be no second quantization here, at least for the theory of ferromagnets here. And M(x), as saaskis said, is the spatial average of magnetization at a point, which can be derived from a more microscopic model with [tex]\psi[\tex] and [tex]\psi^{\dag}[\tex], the second-quantized wavefunction of the particle, with Hubbard-Stratonovich transformation. However, M(x) itself is not second quantized. And the [tex]\langle m(k) m(p) \rangle[\tex] should be the [tex]G(p,k)[\tex] mentioned here.