# Expectation for many-body system

1. Mar 20, 2015

### JorisL

Hi,

I'll start by sketching the specific model I'm looking at, a quantum spin chain.
This is defined as N spins (2 basis states) on a 1D lattice i.e. the sites are a subset of $\mathbb{Z}$.
Then we defined the state-space as
$$\{ \psi\in\mathcal{H}_N \text{ with } ||\psi|| = 1\}\cong S^d \subset \mathbb{R}^{d+1}$$
Some clarifications are needed here first $\mathcal{H}_N = \otimes^N \mathcal{H}_s$ where $\mathcal{H}_s\cong \mathcal{C}^2$ is the single-spin state space.
The isomorphism above with the d-sphere is clear (to me) with $d\equiv 2\cdot 2^N$.

Now I can get to the crucial part, we put a measure on the d-sphere naturally we pick the surface measure $d\sigma(\psi)\text{ on }S^d$. The defining property given too us is
$$\mathbb{E}(f) = \int_{S^d} d\sigma(\psi) f(\psi) = \int_{\mathbb{R}^{d+1}} d\sigma(\psi)\, \delta( ||\psi||-1 ) \, f\left(\frac{\psi}{||\psi||}\right)$$

Where f is a function on the d-sphere or the (d+1) dimensional real space.
This all seems okay so far.

As an example the following function was defined $f(\psi) = \langle\psi|O|\psi\rangle = \langle O \rangle_\psi = \text{Tr}\left(O|\psi\rangle\langle\psi|\right)$.
Then we can apply the definition above
$$\mathbb{E}(\langle O\rangle_\psi) = \text{Tr}\left(O|\psi\rangle\langle\psi|\right) = \text{Tr}\left(O\int d\sigma(\psi)|\psi\rangle\langle\psi|\right)$$

Now the claim is made that this is equal to $\text{tr}(O)$ which uses the normalized trace $\text{tr}(\cdot) \equiv d^{-1}\text{Tr}(\cdot)$.

I don't get how they arrive at this claim. It is mentioned that we need to use that the measure $d\sigma$ is invariant under all unitary $U$.
The lecturer made the remark that $\int d\sigma(\psi)|\psi\rangle\langle\psi| ={1\!\!1}_d$.

This seems logical but where did the normalization of the trace come from if that's true?

I've been thinking about this stuff for hours now but I can't resolve my difficulties with the idea.

Thanks,

Joris

P.S.: I'm not very familiar with measure theory

P.P.S.:
The nice thing is that if the above holds, we can interpret the expectation as that of a thermal ensemble in the limit $\beta\downarrow 0$.
Which in turn will be used when talking about the "Eigenstate Thermalization Hypothesis".

2. Mar 24, 2015

### wabbit

I do not get the same result so the following may be incorrect. Perhaps it might be helpful neverthelesss... or perhaps not:)

Taking coordinates, $\int d\sigma(\psi)|\psi\rangle\langle\psi| =\mathbb{1}_d$ amounts to $\int x_i x_j d\sigma =\delta_{ij}$, which is true by symmetry for $i\neq j$. For $i=j$, symmetry says it does not depend on i, and computing $d\cdot\int x_i^2 d\sigma =\int \sum x_i^2d\sigma =1$ yields 1/d times the identity matrix.

There is a correction to make though, $d=2\cdot 2^N$ here is the total dimension, the dimension of the sphere is $d-1$.

Last edited: Mar 24, 2015
3. Mar 24, 2015

### JorisL

Logically that makes sense, however I'm wondering where we should use the invariance of the measure under unitary transformations.

About the sphere, I had 2 contradicting sources. The info given during the lecture and slides from an older talk he had given.
I assumed the slides were correct since he taught without any notes so an error can get in rather quickly. (focus was on the identity first since that has thrown me off, hard)

I'll try to look at it later today and get back to you if it worked out.

Joris

4. Mar 24, 2015

### wabbit

The symmetry argument I use is in essence invariance of the measure under rotations of the sphere, so this might be related.
Other than that the physics in your post is way over my head, so I may have misinterpreted something in the translation.