The expectation value of entanglement entropy of composite system in a random pure state I'm trying to compute the expectation value of entanglement entropy of composite system in a random pure state, but I'm running into some problems.
The system we are considering is composed of two subsystems $\mathcal{H} = \mathcal{H}_A \otimes \mathcal{H}_B$ with dimensions $N_A$ and $_B$. 
Let's say that system A is the smaller of the two: $N_A \leq N_B$. 
We are considering random pure states $|\psi\rangle \in \mathcal{H}$ these are generated as follows:

For basis $\{e_i\}$ of $\mathcal{H}_A$ and basis $\{f_j\}$ of
  $\mathcal{H}_B$ we van write $$ |\psi\rangle =
> \sum_{i=1}^{N_A}\sum_{j=1}^{N_a} \Psi_{ij} |e_i\rangle \otimes
> |f_j\rangle. $$ $\Psi_{ij}$ can be seen as the coordinates of a point
  on the unit sphere $S^{N_AN_B-1}$ in $\mathbb{C}^{N_ANB}$.  So for
  each $\psi$ there is a corresponding point on the unit sphere. It is
  this point that is chosen uniformly at random.

Equivalently we can construct the random states as $U|\psi_\rangle$ where U is a random unitary matrix chosen with the Haar measure.
The reduced density matrix of A is $\rho_A = \text{Tr}_B |\psi\rangle\langle\psi|$ with corresponding entanglement entropy $S_A (\psi) = -\text{Tr} \rho_A \log \rho_A$.
I want to compute the expectation value of $S_A$, given by 
$$
\mathbb{E}(S_A) = \int_{S^{(N_AN_B-1)}} d\sigma(\psi) S_A(\psi),
$$
where $d\sigma(\psi)$ is the uniform measure on the unit sphere $S^{(N_AN_B-1)}$.
I tried two different things: 
Using the Schmidt decomposition
Every state $\psi$ can be Schmidt decomposed: there exist orthonormal families $\{e_1, e_2, ..., e_{N_A}\}\in \mathcal{H}_A$ and $\{f_1, f_2, ..., f_{N_A}\} \in \mathcal{H}_B$ and real numbers $c_1, c_2, ..., c_{N_A} \geq 0$ with $\sum_i c_i^2 = 1$ such that
$$
|\psi \rangle = \sum_{i=1}^{N_A} c_i |e_i\rangle \otimes |f_i\rangle.
$$
The entanglement entropy in this case is given by $ S_A (\psi) = \sum_i c_i ^2 \log c_i ^2 $.
I thought I could generate a random state by taking a random Schmidt decomposition, by which I mean, take all $c_i$ uniformly with $\sum_i c_i^2 = 1$, take a random orthonormal basis of $\mathcal{H}_A$ (using a random unitary matrix with the Haar measure to generate one from some fixed basis) and a random orthogonal family in $\mathcal{H}_B$ (again using a random unitary matrix U with the Haar measure to generate one, but since we would only care about the $N_A$ first collums I guess I should adapt the measure in some way to compensate for this).
I fear however that this is not correct: I have no dependence on the choice of orthonormal families so when computing the expectation value the integrals over the unitary matrices would just be trivial.
So my first question is: Do my "random Schmidt decomposed states" coincide with (normal) random states? And if not, why?
Usnig a uniform measure on the unit sphere
My second try (which I didn't complete yet) was just to use the uniform measure on the unit sphere as described above.
Using this I could give a probability density of $\rho_A = \Psi\Psi^\dagger$ and then I could write $\rho_A = U\Lambda U^\dagger$ with U some unitary matrix and $\Lambda = \text{diag}(p_1, p_2, ..., p_{N_A})$. 
I could then give a probability density for $ \Lambda$ as 
$$P(p_1, p_2, ..., p_{N_A}) = \int d\sigma (U) P(U\Lambda U^\dagger) $$
where $d\sigma (U)$ is the Haar measure.
But I'm stuck a bit with this. 
Once I find this I could conmute the expectation value as
$$
\mathbb{E}(S_A) = -\int dp_1dp_2, ... dp_{N_A} P(p_1, p_2, ..., p_{N_A}) \sum_i p_i \log p_i
$$
My second question is
Is this a correct way to do it? Can anyone help me with the parametrisation of $\Psi$ in terms of angles on the unit sphere, or with another method to obtain $P(p_1,...,p_{N_A})$ and maybe some of the subsequent integrals?
I found something in this article, but most of the steps are a bit vague to me.
Should this kind of question rather be posted in the math stack exchange? I reposted it over there since its actually a technical question on the math and there isn't so much physics involved. Should I remove it here?
 A: The average entropy of a part of a state is computed e.g. the following papers:
http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.71.1291
http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.72.1148
http://journals.aps.org/pre/abstract/10.1103/PhysRevE.52.5653
http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.77.1
(See also http://arxiv.org/abs/quant-ph/0407049, where these references are taken from.)
A: I'm not sure what you mean by choosing a random pure state uniformly. 
Do you want:
a) Each $|\Psi\rangle$ is chosen uniformly in $\mathcal{H}$, so $|\Psi\rangle = \frac{1}{\sqrt{N_A N_B}}\sum_{i=1}^{N_A} \sum_{j=1}^{N_B} |e_i\rangle \otimes |f_j\rangle$ and $\rho = |\Psi\rangle\langle\Psi|$ 
b) The density matrix is chosen uniformly in each pure state, so $\rho = \frac{1}{N_A N_B} \sum_{i=1}^{N_A} \sum_{j=1}^{N_B} |e_i\rangle \otimes |f_j\rangle \langle e_i| \otimes \langle f_j|$ 
c) As I understood from your two approaches you actually want $|\Psi\rangle$ to be chosen uniformly but random out of $\mathcal{H}$, right? (I guess I just wanted to make sure that there is a distinction between the uniform choice and uniform random choice...) If so, then see my answers below.
To your questions: 
Schmidt decomposition: Here is it not clear to me how to define the probability measure of your state in order to get a uniform distribution, since the choice of basis vectors for the decomposition is not so transparent. 
Bloch sphere: I didn't quite understand who you would pick this unitary matrix U to construct $\rho_A$ and on which you probability P depends on? 
I would do the following: 
Use the Kronecker-basis 
$$\mathcal{H} = \langle\{ e_1\otimes f_1, e_1\otimes f_2, ..., e_1 \otimes f_{N_B}, e_2 \otimes f_1, ..., e_{N_A} \otimes f_{N_B} \}\rangle =: \langle \{ \Psi_1, ..., \Psi_{N_A N_B} \} \rangle $$
$$= \{ \sum_{i=1}^{N_A N_B} a_i \Psi_i\ | \sum_{i=1}^{N_A N_B} |a_i|^2 = 1\} = \{ \sum_{i=1}^{N_A N_B} a_i \Psi_i\ | a= (a_1, ..., a_{N_A N_B}) \in S^{N_A N_B -1} \}.$$
So for a random uniform choice of an element out of $\mathcal{H}$ I would understand a uniform choice of $a= (a_1, ..., a_{N_A N_B}) \in S^{N_A N_B -1}$, hence use a as a random variable with values in $S^n$ ($n = N_A N_B -1$) and measure $d\mu(a) = \frac{1}{|S^n|} d\Omega_n$ where $|S^n|$ is the area of the n-dimensional sphere and $d\Omega_n$ the usual volume element of it. The reduced density matrix is then some combination of $a_k$'s and you'd have to look for his eigenvalues $\{\lambda_i(a)\}_{i=1}^{N_A}$ in order to calculate $S_A(a) = - \sum_{i=1}^{N_A} \lambda_i(a) log \lambda_i(a)$.
The formula for the expectation value of the entropy would then be
$$E[S_A(a)] = \int_{S^n} S_A(a) d\mu(a)$$
where the integration is now quite a mess, depending on what you get for the eigenvalues.
Does that make sense?
