In statistical physics (mechanics), the transition from Maxwell-Boltzmann statistics to Bose-Einstein and Fermi-Dirac statistics was motivated by classically inexplicable phenomena such as Bose-Einstein condensation and stability of atoms with even numbers of electrons. In the new statistics, we have a different counting of microstates based on the assumption that the involved particles are indistinguishable (that is how the relevant partition functions are calculated). I want to know why the set of Quantum states (density operators) accommodates indistinguishability better than the set of classical states (probability distributions). For purposes of this question please feel free to assume finite dimensionality and discreteness of spaces. I'm looking for an answer with the following flavor:

Classically the convex set of states is a simplex. Any point in the interior of the simplex has a unique decomposition in terms of the pure states (that lie on the corners of the simplex). Why is this uniqueness attributed to distinguishability of classical states and possibility of joint measurements (commutativity). In contrast in Quantum Statistics, the convex set of states (density operators) is not a simplex. It is known that any mixed state has infinitely many decompositions in terms of pure states. How is this related to indistinguishability of states? How are states in Quantum Information Theory related to micro and macro states in Bose-Einstein and Fermi-Dirac statistics?


1 Answer 1


I think one thing that might help to understand this better is the characterisation of the set of possible realisations of a given state $\rho$ in terms of convex decompositions of pure states.

Let $\rho\in\mathrm{Lin}(\mathcal H)$ be some quantum state, represented as a density matrix acting in some finite-dimensional Hilbert space $\mathcal H$, and suppose we can write $$\rho = \sum_i p_i \mathbb P(u_i) = \sum_j q_j\mathbb P(v_j),$$ for some probabilities vectors $(p_i)_i,(q_j)_j$, thus such that $\sum_i p_i=\sum_i q_i=0$ and $p_i,q_i\in[0,1]$, with $u_i,v_i\in\mathcal H$ unit vectors, and where I used the shorthand notation $\mathbb P(u)\equiv|u\rangle\!\langle u|$. Then there is some unitary $V$ such that $$\sqrt{p_i} u_i = \sum_j V_{ij} \sqrt{q_j} v_j.$$ Furthermore, this is an iff condition, which thus characterises all possible ways to "interpret" a state $\rho$ as a mixture of pure states.

Proof of the characterisation

Proving the above characterisation is relatively straightforward. Start by noticing that any sum of the form $\sum_i p_i \mathbb P(u_i)$ can be "purified" by writing it as $$\sum_i p_i \mathbb P(u_i) = \operatorname{Tr}_2\left(\mathbb P\left[\sum_i \sqrt{p_i} (u_i\otimes \tilde u_i)\right]\right)$$ for some (any) set of orthonormal vectors $\tilde u_i$ living in some purifying space. Therefore correspondingly to the two different convex decompositions of $\rho$ we get the equality of the corresponding purifications: $$ \operatorname{Tr}_2\left(\mathbb P\left[\sum_i \sqrt{p_i} (u_i\otimes \tilde u_i)\right]\right) = \operatorname{Tr}_2\left(\mathbb P\left[\sum_i \sqrt{q_i} (v_i\otimes \tilde v_i)\right]\right).$$ with $\{\tilde u_i\}_i,\{\tilde v_i\}_i$ orthonormal vectors. Because purifications of the same state always differ by a local unitary operation, there must be some unitary $V$ acting on the purifying space such that $$\sum_i\sqrt{p_i} (u_i\otimes\tilde u_i) = \sum_i \sqrt{q_i} (v_i\otimes V\tilde v_i).$$ Taking inner product on second space wrt $\tilde u_i$, we then get $$\sqrt{p_i} u_i = \sum_j \sqrt{q_j} V_{ij} v_j,$$ defining $V_{ij}\equiv \langle \tilde u_i,V\tilde v_j\rangle$.

Physical interpretation of different decompositions

Note that the above also gives an operation interpretation of the different decompositions: these describe the different pure states that can be obtained with some probabilities, depending on what (projective) measurement is performed on the purifying space.

So, to summarise, the reason a (mixed) quantum state can be realised as different convex combinations of pure states is that each such decomposition corresponds to a different operation performed on the purifying space. One needs to observe that a mixed state can always be thought of as a piece of a larger pure state. Depending on what operation/measurement is performed on the part of the state that is not observed, one can get different decompositions.

As an example, consider the state $$\rho = \frac13\begin{pmatrix}1&0\\0&2\end{pmatrix}.$$ Its straightforward purification is $\frac{1}{\sqrt3}(|0,0\rangle+\sqrt2|1,1\rangle)$. However, using the above results with $V=H$ (the Hadamard unitary matrix), one also finds the decomposition $$\rho = \frac{1}{2} \mathbb P\left[\left(\frac{|0\rangle+\sqrt2|1\rangle}{\sqrt3}\right)\right] + \frac{1}{2}\mathbb P\left[\left(\frac{|0\rangle-\sqrt2|1\rangle}{\sqrt3}\right)\right]. $$ Note how the weights are more mixed than the other one. The physical interpretation of this is that this decomposition describes the information gained about the system when measuring the ancilla in the $|\pm\rangle$ basis, which is not the optimal one to read the information about the correlations between system and ancilla.


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