# Z=Tr(Exp(-bH)) notation

1. Dec 25, 2007

### 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?

2. Dec 26, 2007

### mjsd

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.

3. Dec 26, 2007

### 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 Boltzman 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

4. Dec 26, 2007

### 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)

5. Dec 27, 2007

### 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

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

Last edited: Dec 27, 2007
6. Dec 27, 2007

### 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?

Last edited: Dec 27, 2007
7. Dec 27, 2007

### mjsd

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.

8. Dec 27, 2007

### christianjb

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

9. Dec 27, 2007

### 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.

10. Dec 27, 2007

### 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

Last edited: Dec 27, 2007
11. Dec 28, 2007

### 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?

12. Dec 28, 2007

### 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) here).
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 $$e^{-\beta \hat{H}}$$.

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

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

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

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

Thus,

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

Last edited: Dec 28, 2007
13. Dec 31, 2007

### 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.