Why is the Tr notation used in the Z=Tr(Exp(-bH)) equation for the Ising model?

[YaroslavVB]

Why is the above notation used instead of Sum_x Exp(-b H(x))? Does the "Tr" notation above have anything in common with the standard matrix trace? Are there any introductory textbooks which explain the reason "Trace" is used to replace the sum?

 

[mjsd]

Why is the above notation used instead of Sum_x Exp(-b H(x))? Does the "Tr" notation above have anything in common with the standard matrix trace? Are there any introductory textbooks which explain the reason "Trace" is used to replace the sum?

I believe your H(x) means the eigenvalues of H. and yes, the Tr means Trace: "the sum of the diagonal entries" of a matrix, in this case H. You can replace the sum of all eigenvalues with the Tr because
$$Tr(A)= \sum_i A_i$$
where $$A_i$$'s are the eigenvalues of a diagonalisable matrix A.
You can see this by changing the basis of A into its diagonal basis:
$$D= P^{-1} A P$$
where P is the change of basis matrix, D is the diagonal matrix containing all eigenvalues down its diagonal. so
$$\sum_i A_i = Tr(D) = Tr(P^{-1}AP) = Tr(PP^{-1}A) = Tr(A)$$

this explains why u can write Z as a Tr.

 

[YaroslavVB]

Sorry, I didn't specify the context -- I see this notation come when talking about Ising models. So Z is the partition functions, H(x) is the Hamiltonian (for it could be the number of adjacent pairs with aligned spins in a configuration x). Exp[-b H(x)] is the Boltzmann potential for a particular configuration x, so Sum_x Exp[-b H(x)] is the sum of Boltzmann potentials over all configurations. However, I sometimes see Sum_x replaced with Tr with no explanation, for instance, eq.1.4 in Cardy's "Scaling and Renormalization in Statistical Physics"

For a particular Hamiltonian the partition function could look like below (H now means magnetic field)
http://yaroslavvb.com/share/partition.PNG
(from Cardy again)

So the confusing part for me is why "Sum_x B(x)" is replaced by "Tr B" where B is the Boltzmann potential

 

[mjsd]

Yes, I knew your Z means partition function. the Sum_x B(x) replaced by Tr B is of the same reason as to why eq. 1.4 in Cardy can write Z = Tr Exp(-bH).
your quantum mechanical operator H, whatever it is, will contain all the definite states (say E_i). And you know that
$$Z= \sum_i e^{-\beta E_i}$$
so what I told you before still applies. The trace is taken over state space (ie. basis independent)

 

[christianjb]

Define
$$\hat{\rho}=e^{-\beta\hat{H}}$$
Then, because
$$\mathbf{I}=\sum_i |e_i><e_i|$$

$$\hat{\rho}=e^{-\beta\hat{H}}\sum_i |e_i><e_i|$$
and because if
$$\hat{A}|a_i>=a_i|a_i>$$
we have that
$$f(\hat{A})|a_i>=f(a_i)|a_i>$$
Thus,
$$\hat{\rho}=\sum_i e^{-\beta e_i}|e_i><e_i|$$

The trace is basis set independent and we may as well use the convenient basis set of the eigenvectors of the Hamiltonian

$$Tr[\hat{\rho}]=\sum_j<e_j|\rho|e_j>=\sum_j e^{-\beta e_j}=Z$$
Thus, we see that
$$Tr[\hat{\rho}]=Z$$

Also, it's easy to show that

$$<A>=Tr[\hat{\rho}\hat{A}]$$

 

[YaroslavVB]

Thanks for the hints. I'm still confused because I see places use "Ising model Hamiltonian" to denote a a function that gives the energy of a configuration, and what does it mean to find eigenstates of the energy function?

For instance first equation in page below seems to be a function that counts the number of aligned spins
http://www.nyu.edu/classes/tuckerman/stat.mech/lectures/lecture_26/node2.html

Also equation 31 in http://arxiv.org/abs/cond-mat/0010392v1 also seems to be a function that gives the energy of a configuration. Then the author talks about finding the spectrum of H, but again, how can you find a spectrum of a function that has different range and domain?

 

[mjsd]

Thanks for the hints. I'm still confused because I see places use "Ising model Hamiltonian" to denote a a function that gives the energy of a configuration, and what does it mean to find eigenstates of the energy function?

For instance first equation in page below seems to be a function that counts the number of aligned spins
http://www.nyu.edu/classes/tuckerman/stat.mech/lectures/lecture_26/node2.html

Also equation 31 in http://arxiv.org/abs/cond-mat/0010392v1 also seems to be a function that gives the energy of a configuration. Then the author talks about finding the spectrum of H, but again, how can you find a spectrum of a function that has different range and domain?

I could be wrong, but it sounds like you don't understand what is meant by finding eigenvalues and eigenstates of a Hamiltonian operator, H. Suppose you've started with some Hilbert Space in which H can act. H acting on a state

$$H |\psi(\vec a)\rangle = |\psi ' (\vec a)\rangle$$

it will take it to another state. $$\vec a$$ denotes a list of variables that $$\psi$$ depends on.
The eigenvalue problem is to find a set of all states (and their respective eigenvalues) that will satisfy

$$H |\Psi_i (\vec a)\rangle = E_i |\Psi_i (\vec a)\rangle$$

here $$E_i$$ is the eigenvalue corresponding to the eigenstate $$|\Psi_i\rangle$$. Note each $$|\Psi_i\rangle$$ can be very different with different range/domain. Finding the spectrum of H can mean finding $$E_i$$ or $$|\Psi_i\rangle$$ or both. Anyway, just in case you don't know, this is why they talk about "diagonalising" the matrix. hope I am not telling you something you already know.

 

[christianjb]

OK- I finally finished my post from last night (see above). (The server failed halfway through writing the post.)

 

[christianjb]

Oh- I guess I should add that $$\hat{\rho}$$ is called the 'density matrix'. (Sometimes its normalized such that its trace is unity.) If you do a search for 'density matrix' and 'partition function' you'll find a load of pages explaining their connection.

It's worth noting that the concise definition of the density matrix: $$\hat{\rho}=e^{-\beta\hat{H}}$$ is one of the most beautiful equations in statistical mechanics.

 

[YaroslavVB]

Thanks for the explanation, I understand what it means to diagonalize an operator, what was confusing is that when pointed to Hamiltonian of a 1d Ising model, I saw a quadratic form instead of an operator

I found a paper titled "Exact eigenvalues of the Ising Hamiltonian ... " (http://yaroslavvb.com/papers/dixon-exact.pdf)
There they write (eq.2) Hamiltonian of a 1d Ising as -J x'.A.x where x is the vector of spins, A is adjacency matrix of a path graph and J is coupling strength. Am I correct in understanding that spectrum of this Hamiltonian is the same as the spectrum of A?

I charted the eigenstates of a 5x1 Ising model based on this assumption here
http://www.yaroslavvb.com/research/qr/ising-eigenvectors/ising-eigenvectors.html

 

[YaroslavVB]

Suppose we have a two variable Ising system, x1,x2. The system has 4 states --
(-1,-1),(-1,1),(1,-1),(1,1). The states have the following density -- p(x1,x2)=Exp(-x1 x2)/Z

How would I find the Hamiltonian and it's spectrum for this system?

 

[Edgardo]

To show that $$\sum_{i=1}^{N} e^{-\beta E_{i}} = \mbox{Tr} \left(e^{-\beta \hat{H}} \right)$$ we consider the following points:

1) Functions of operators
Let $$\hat{A}$$ be some operator, $$|\Psi_{i} \rangle$$ its eigenstates and $$a_{i}$$ its eigenvalues, i.e. $$\hat{A} |\Psi_{i} \rangle = a_{i} |\Psi_{i} \rangle$$

Then functions of the operator $$\hat{A}$$ are calculated by

$$f(\hat{A}) = \sum_{i=1}^{N} f(a_i) |\Psi_{i} \rangle \langle \Psi_{i}|$$

(see Plenio, lecture notes on quantum mechanics 2002, page 51 eq. (1.86) http://www.lsr.ph.ic.ac.uk/~plenio/teaching.html ).
As an example consider the function $$f(x) = e^x$$. So we want to calculate
$$e^{\hat{A}}$$. But what is the meaning of such a thing as $$e^{\hat{A}}$$? (Note: That's an operator in the exponent.)

The answer is, take the formula

$$f(\hat{A}) = \sum_{i=1}^{N} f(a_i) |\Psi_{i} \rangle \langle \Psi_{i}|$$

Let's call this equation (1).

Since $$f(x) = e^x$$ it follows that $$f(\hat{A}) = e^{\hat{A}}$$ and $$f(a_{i})=e^{a_{i}}$$. Thus, we get from eq. (1)

$$e^{\hat{A}} = \sum_{i=1}^{N} e^{a_i} |\Psi_{i} \rangle \langle \Psi_{i}|$$ 2) Calculating $$e^{-\beta \hat{H}} }$$

Let's take the operator $$\hat{H}$$ with its eigenstates $$|\Psi_{i} \rangle$$ and its eigenvalues $$E_{i}$$, i.e.

$$\hat{H} |\Psi_{i} \rangle = E_{i} |\Psi_{i} \rangle$$

According to eq. (1) a function of the operator $$\hat{H}$$ is calculated by

$$f(\hat{H}) = \sum_{i=1}^{N} f(E_{i}) |\Psi_{i}\rangle \langle \Psi_{i}|$$

Again we take $$f(x)=e^x$$ from which it follows that $$f(\hat{H}) = e^{\hat{H}}$$ and $$f(E_{i}) = e^{E_i}$$.
Thus,

$$e^{\hat{H}} = \sum_{i=1}^{N} e^{E_i} |\Psi_i \rangle \langle \Psi_i|$$

or, slightly modified if we start from the eigenequation

$$-\beta \hat{H} |\Psi_{i} \rangle = -\beta E_{i} |\Psi_{i} \rangle$$

we get

$$e^{-\beta \hat{ H}} = \sum_{i=1}^{N} e^{-\beta E_i} |\Psi_i \rangle \langle \Psi_i|$$

Let us call this equation (2).3) Trace of the operator $$e^{-\beta \hat{H}}$$

The trace of an operator $$\hat{A}$$ is defined as

$$\mbox{Tr} ( \hat{A} ) = \sum_{i=1}^{N} \langle \Psi_i | \hat{A} | \Psi_i \rangle$$

where $$|\Psi_i \rangle$$ are the eigenstates of $$\hat{A}$$.

We are interested in the case where $$\hat{A}$$ equals $$e^{-\beta \hat{H}}$$.

$$\mbox{Tr} \left( e^{-\beta \hat{H}} \right) = \sum_{i=1}^{N} \langle \Psi_i | e^{-\beta \hat{H}} | \Psi_i \rangle$$

$$= \sum_{i=1}^{N} \langle \Psi_i | \underbrace{ \left( \sum_{k=1}^{N} e^{-\beta E_k} |\Psi_k \rangle \langle \Psi_k|\right) }_{\mbox{from <br /> <br /> eq. (2)}} | \Psi_i \rangle$$

$$= \sum_{i=1}^{N} \sum_{k=1}^{N} e^{-\beta E_k} \underbrace{\langle\Psi_i|\Psi_k \rangle}_{\delta_{ik}} \langle\Psi_i|\Psi_k \rangle = <br /> <br /> \sum_{i=1}^{N} e^{-\beta E_i}$$

(Note: $$\langle \Psi_i | \Psi_k \rangle = \delta_{ik}$$ since the eigenstates are orthonormal.)

Thus,

$$\sum_{i=1}^{N} e^{-\beta E_{i}} = \mbox{Tr} \left(e^{-\beta \hat{H}} \right)$$

 

[YaroslavVB]

Thanks for the reply, it now all makes sense. My original confusion was because half the places I was looking at talked about "quantum Ising model" and the other half "classical Ising model". In the latter case, "Hamiltonian" referred to the energy of a particular configuration, not a quantum operator.