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?

  • $\begingroup$ Usnig instead of Using. Sorry ... $\endgroup$
    – user46925
    Jun 4, 2015 at 21:32
  • $\begingroup$ @igael you know you can edit my question right? $\endgroup$
    – Lagrangian
    Jun 5, 2015 at 9:45
  • $\begingroup$ embarrassed by the 2 points retribution :) $\endgroup$
    – user46925
    Jun 5, 2015 at 9:47

2 Answers 2


The average entropy of a part of a state is computed e.g. the following papers:





(See also http://arxiv.org/abs/quant-ph/0407049, where these references are taken from.)

  • $\begingroup$ I encoutered these articles, but I don't see how they obtain $P(p_1, ..., p_{N_A})$ and how they evaluate the integral to obatin $\mathbb{E}(S_A)$. $\endgroup$
    – Lagrangian
    Jun 11, 2015 at 16:32

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?

  • $\begingroup$ I really mean random states chosen uniformly at random on the unit sphere in $\mathbb{C}^{N}$. I will correct this in my question, thanks for pointing that out. $\endgroup$
    – Lagrangian
    Jun 5, 2015 at 9:52
  • $\begingroup$ Why would the reduced density matrix be upper triangular? A density matrix as always Hermitian. $\endgroup$
    – Lagrangian
    Jun 5, 2015 at 16:07
  • $\begingroup$ Oh yes you are right- I'm sorry I wasn't careful enough there. I'll change it above. I hope now it's correct. $\endgroup$
    – Soliton
    Jun 5, 2015 at 17:30
  • $\begingroup$ Thanks a lot for the explanation, but I think that your answer is more or less a reformulation of my question. Maybe I wasn't clear enough about what I already got. My problem is actually finding the relation $\gamma_i(a)$ and computing the said integral. $\endgroup$
    – Lagrangian
    Jun 5, 2015 at 17:42
  • $\begingroup$ Ok sorry about that. I was hoping a concrete approach might help. $\endgroup$
    – Soliton
    Jun 5, 2015 at 19:17

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