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In particular, why do we need both of these to find the volume? And should I be thinking of it as an actual volume or not?

This Hilbert space volume is talked about in this paper. It says

There exist the unique, uniform measure $\nu$ on $P$ induced by the Haar measure on the group of unitary matrices $U(N)$. Integration over the set $P$ thus amounts to an integration of the corresponding angles and phases in N-dimensional complex space that determine the families of orthonormal projectors.

And

Since the simplex $\Delta$ is a subset of the ($N-1$)-dimensional hyperplane, there exist a natural measure on $\Delta$ which is defined as a usual normalized Lebesgue measure $\mathcal{L}_{N-1}$ on $R^{N-1}$.

The volume of the Hilbert space is then defined to be $\mu=\nu \times \mathcal{L}_{N-1}$, but why?

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  • $\begingroup$ Could you give a little bit more detail. $\endgroup$
    – MBN
    Commented May 19, 2011 at 19:44
  • $\begingroup$ @MBN I've added some more info. $\endgroup$
    – Calvin
    Commented May 19, 2011 at 20:12
  • $\begingroup$ Here is an arXiv link to the paper mentioned in the question(v4): arxiv.org/abs/quant-ph/9804024 $\endgroup$
    – Qmechanic
    Commented Feb 8, 2012 at 15:41

1 Answer 1

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1) First of all, $\mu$ is not a measure on the $N$-dimensional Hilbert space $H\cong\mathbb{C}^N $, but on the space $C$ of density operators $\rho:H\to H$, i.e., positive operators $\rho\in B(H)\equiv {\cal L}(H,H)$ with trace $\mathrm{tr}(\rho)=1$. Let us below sketch an argument for the choice of the measure $\mu$.

2) A positive operator $\rho$ can always be diagonalized

$$U^{-1}\rho U= \mathrm{diag}(\lambda_1, \ldots, \lambda_N) ,$$

by an unitary operator $U\in U(H)\cong U(N)$.

3) The positivity yields

$$\lambda_1, \ldots, \lambda_N\geq 0,$$

and the trace condition yields

$$1=\mathrm{tr}(\rho)=\sum_{i=1}^N \lambda_i.$$

4) It is not hard to show that if the eigenvalues $\lambda_1, \ldots, \lambda_N$ are all different (which is the generic situation; more precisely, below we shall say that this is true almost everywhere wrt. the Lebesgue measure $d^N\lambda$), then any other unitary operator $V\in U(H)$, that diagonalizes $\rho$, must be related to $U$ as

$$ VU^{-1}= P_{\pi} \ \mathrm{diag}(e^{i\varphi_1}, \ldots, e^{i\varphi_N}),\qquad \pi\in S_{N},\qquad \varphi_1, \ldots, \varphi_N \in \mathbb{R}, $$

where $P_{\pi}:H\to H$ is a permutation operator corresponding to a permutation $ \pi\in S_{N}$. So the unitary operator $U\in U(H)$ is unique modulo phase factors and permutations of the eigenvalues

$$(\lambda_1, \ldots, \lambda_N)\qquad\rightarrow \qquad(\lambda_{\pi(1)}, \ldots, \lambda_{\pi(N)}),\qquad \pi\in S_{N}.$$

We can thus generically identify a density operator $\rho$ with a unitary operator $U$ and a set of eigenvalues $(\lambda_1, \ldots, \lambda_N)$ modulo the above mentioned relations.

5) Note that there is not a preferred orthonormal basis in $H\cong\mathbb{C}^N$. Therefore it is natural to use the Haar measure $\nu$ on $U(H)\cong U(N)$. Hence, a natural class of measures $\mu$ on the space $C$ of density operators $\rho$ is,

$$\mu=\nu \times d^N\lambda\ f(\lambda_1, \ldots, \lambda_N) \ \delta(1-\sum_{i=1}^N \lambda_i)\ \prod_{j=1}^n \theta(\lambda_j),$$

where $\delta$ is the Dirac delta distribution; $\theta$ is the Heaviside step function; $d^N\lambda$ is the $N$-dimensional Lebesgue measure on the space $\mathbb{R}^N$ of real eigenvalues, and $f: \mathrm{R}^n\to [0,\infty[ $ is a totally symmetric

$$f(\lambda_{\pi(1)}, \ldots, \lambda_{\pi(N)})=f(\lambda_1, \ldots, \lambda_N),\qquad \pi\in S_{N},$$

measurable function, that is integrable

$$\int d^N\lambda \ f(\lambda_1, \ldots, \lambda_N) \ \delta(1-\sum_{i=1}^N \lambda_i)\prod_{j=1}^N \theta(\lambda_j) ~<~ \infty$$

on the simplex

$$\Delta_{N-1}=\left\{\vec{\lambda}\in \mathbb{R}^N \mid \lambda_1\geq 0 , \ldots, \lambda_N\geq 0; \sum_{i=1}^N \lambda_i=1\right\},$$

which is a compact set in $\mathbb{R}^N$.

6) Note that the Haar measure $\nu$ on the compact group $U(H)$ is finite, $\nu(U(H))<\infty$, and the volume of the simplex

$$ \mathrm{Vol}(\Delta_{N-1})= \int d^N\lambda \ \delta(1-\sum_{i=1}^N \lambda_i)\prod_{j=1}^N \theta(\lambda_j)~<~ \infty,$$

and hence the total volume $\mu(C)<\infty$ is finite. Therefore, one may normalize the measure by dividing with $\mu(C)$.

7) A choice $f(\vec{\lambda})=1$ is just a convenient choice, but not necessary. Another natural choice, from the perspective of the Gaussian unitary ensemble (GUE), is the square of the Vandermonde determinant $f(\vec{\lambda})=\Delta^2(\vec{\lambda})$.

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  • $\begingroup$ Thanks @Qmechanic, so $\mu$ is the volume of the set of density matrices rather than of the Hilbert space? Also, what is $S_N$ defined as? $\endgroup$
    – Calvin
    Commented May 25, 2011 at 17:26
  • $\begingroup$ @Calvin. 1) Almost correct, however $\mu$ is strictly speaking a measure, not a volume. 2) $S_N$ is the symmetric group. See en.wikipedia.org/wiki/Measure_%28mathematics%29 and en.wikipedia.org/wiki/Symmetric_group $\endgroup$
    – Qmechanic
    Commented May 25, 2011 at 18:11
  • $\begingroup$ Do you know why the paper I refer to in my question defines $\mu$ as a volume? $\endgroup$
    – Calvin
    Commented May 25, 2011 at 19:41
  • $\begingroup$ Dear @Calvin. It doesn't. The paper's terminology seems pretty consistent. But don't get too caught up about it, as it is borderline semantics. On one hand, the measure $\mu$ is a function from all the measurable subsets to $[0,\infty]$. On the other hand, a volume typically denotes the measure $\mu(A)$ of some particular subset $A$, typically the entire set $C$. $\endgroup$
    – Qmechanic
    Commented May 26, 2011 at 13:08

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