# Reduced density matrix

1. May 23, 2015

### JorisL

Good afternoon all,

I'm investigating typical values of entropy for a subsystem of a 1D (non-interacting) spin chain.
Most of the problem is essentially solved
I've shown that a typical pure state of the entire chain is close (trace norm) to the state $\Omega_S$ when reduced.
$$\Omega_S = \text{Tr}_E \frac{1\!\!1}{d_U}$$
Here $d_U=\text{dim}\mathcal{H}_U$ is the Hilbert space of the entire chain which we further on split as $\mathcal{H}_U = \mathcal{H}_S\otimes \mathcal{H}_E$.
I've essentially followed Section V in the paper by S. Popescu et al. with some sidesteps to understand every single step.

By the Fannes-Audenaert inequality we have that the von Neumann entropy of $\text{Tr}_E |\phi\rangle\langle\phi|$ is close to $S[ \Omega_S]$.

Now I'm not entirely sure how to find $S[ \Omega_S]$. I think I've found 3 approaches to this.
Which would be the nicest to use? (it's from the perspective of mathematical physics)
1. Direct calculation of the partial trace $\Omega_S$ and diagonalize
2. Use the thermal canonical principle from the same paper
$$H_S = \sum\limits_{i=1}^N \sigma_i^{(z)}$$
Now if $S\ll U$ I figured $H_S\ll k_BT$ and thus $\Omega_S \approx \frac{1\!\!1_S}{d_S}$ which leads to $S[\Omega_S] = log{N}$ with N the number of spins in the subsystem S.
I'm not entirely certain I can do this, but haven't found any immediate problems here
3. Describe the spins as fermions with $|\downarrow\rangle$ the absence of a particle while $|\uparrow\rangle$ the presence of a particle. Then we can use creation and annihilation operators and the machinery of the CAR-algebra description. I haven't entertained this method a lot but I'm wondering if I can use this?

So the main question is the following, is approach 2 valid? If it is I'm done.
If it is not, what can I best do. Direct calculation or mapping the system to a fermion chain?

Thanks,

Joris

2. May 28, 2015

### Staff: Admin

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. May 29, 2015

### suma

first, i am not sure but i have some question on this.

why is that the trace is being taken on the dimension of hilbert space to calculate Ωs, isn't hilbert space a number rather than a matrix. trace acts on matrix.

i think the best to check all of 3 options and see whether they agree, then you would know whether 2nd theory is valid.

let us know what you got

4. May 29, 2015

### JorisL

One usually is introduced to state vectors first. These are the simplest states you can have.
However in an experiment you sometimes don't know what your initial state is with certainty.
If you have for example an experiment where you prepare the system in state $|\psi_1\rangle$ or $|\psi_2\rangle$ with probabilities $p_1$ and $p_2$ respectively, you turn toward the density matrix formalism.
You introduce a density matrix $\rho = p_1|\psi_1\rangle\langle\psi_1| + p_2|\psi_2\rangle\langle\psi_2|$.
Your expectation values are now $\text{Tr} \rho A$ for observables A. (You can check that for a so-called pure state $\rho = |\psi\rangle\langle\psi |$ you get back to $\langle\psi |A|\psi\rangle$ using the cyclicity of the trace)
A little bit extra information on wikipedia

And in a mathematical approach we often call $\langle\cdot\rangle$ a state, now it's possible to show a one-to-one correspondence between these matrices and such states.

This approach is used in for example when you look at thermal states (as also mentioned above :-) ).
I know that density matrices are introduced in the first chapters of Ballentine if you like more information.
Weinberg also introduces them but his treatment is short at best, understandable because he's more focused on deriving things from symmetries.

Last edited: May 29, 2015
5. Jun 6, 2015

### JorisL

Ok I've been able to solve this, if I'm not making a grave mistake.
So for a spin-chain of length N we have $d_U = 2^N = d_S\cdot d_E$
Now the partial trace $\Omega_S = \text{Tr}_E \frac{1\!\!1}{d_U} = \frac{1\!\!1}{d_S}$

So we can easily calculate $S[\Omega_S] = \log d_S$.
For a spin-chain $d_S = 2^L$ if we look at the first L spins (of a total chain of length N). So $S[\Omega_S] = L\cdot \log 2$

Now we first use our result $\langle ||\Omega_S - \rho_S||_1 \rangle \leq \sqrt{\frac{d_S^2}{d_U}}$ which gives us that a typical state is close to the state we calculated.
We use this in the Fannes-Audenaert inequality
$\langle |S(\Omega_S)-S(\rho_S)|\rangle \leq \langle ||\Omega_S - \rho_S||_1\rangle - \langle ||\Omega_S - \rho_S||_1 \log ||\Omega_S - \rho_S||_1 \rangle$
$\langle |S(\Omega_S)-S(\rho_S)|\rangle \leq \langle \sqrt{\frac{d_S^2}{d_U}} \rangle - \langle \sqrt{\frac{d_S^2}{d_U}} \log \sqrt{\frac{d_S^2}{d_U}}\rangle$

We know that whenever $d_S^2 = 2^{2L} \ll d_U = 2^N$ the a typical (reduced) state $\rho_S$ is close to the simple state $\Omega_S$.
This condition also makes sure that the entropies of these states are close to each other.
The condition reduces to $L\ll N/2$ which coincides with the assertion $S\ll U$ and idea 2. The result is the same (I made an error in the original post $\log N$ should be $\log d_S = L\cdot \log 2$) but I'm not entirely convinced about the exponential reducing to the unit matrix due to signs etc.

The third method I haven't tried. Maybe I will look into this in a month or two when my finals are done.

Joris

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