2
$\begingroup$

Consider a Haar random quantum state $|\psi\rangle$. Note that

$$\rho =\mathbb{E}[|\psi\rangle\langle \psi|] = \frac{\mathbb{I}_{n}}{2^{n}}.$$

$\mathbb{I}_n$ is the identity operator on $n$ qubits. Now consider what happens when we trace out the last (or the $n^{\text{th}}$) qubit of $|\psi\rangle$. Denote that density matrix by $\rho^{[n-1]}$. Note that the partial trace is a linear operator. Hence,

$$ \rho^{[n-1]} = \mathbb{E}\left[\text{Tr}_n( |\psi\rangle\langle \psi|)\right] = \text{Tr}_n\left(\mathbb{E}\left[ |\psi\rangle\langle \psi|\right]\right) = \frac{\mathbb{I}_{n-1}}{2^{n-1}}$$

Clearly, the rank of this matrix is $2^{n-1}$ and the eigenvalues are each equal to $\frac{1}{2^{n-1}}$.


However, we could also write $|\psi\rangle$ in terms of its Schmidt decomposition:

$$|\psi\rangle = \alpha_1 |\phi_1 \rangle |\chi_1\rangle + \alpha_2 |\phi_2 \rangle |\chi_2\rangle,$$

where $|\phi_1 \rangle$ and $|\phi_2 \rangle$ are orthonormal vectors over $n-1$ qubits and $|\chi_1\rangle$ and $|\chi_2\rangle$ are orthonormal vectors over $1$ qubit. Clearly,

$$\text{Tr}_n( |\psi\rangle\langle \psi|) = |\alpha_1|^{2} |\phi_1 \rangle \langle \phi_1 | + |\alpha_2|^{2} |\phi_2 \rangle \langle \phi_2 |. $$

Clearly, the rank of this density matrix is $2$ and the eigenvalues are exactly $|\alpha_1|^{2} $ and $|\alpha_2|^{2}$. Since they have to sum to $1$, the eigenvalues are certainly large, and at least one of them is much larger than $\frac{1}{2^{n-1}}$.


How can both of these statements be simultaneously true? That is, how can the probabilistic average of a bunch of rank $2$ matrices suddenly give rise to a matrix of rank $2^{n-1}$? Rank of a sum of matrices is sub-additive, so it is certainly mathematically possible, but I am missing the qualitative intuition.

Furthermore, for a typical Haar random quantum state $|\psi\rangle$, is there anything we can say about how large or small $|\alpha_1|^{2} $ and $|\alpha_2|^{2}$ are and what the eigenvectors look like?

$\endgroup$
5
  • $\begingroup$ I don't quite see the problem. That summing matrices you can increase the rank isn't surprising: consider e.g. a sum of orthogonal rank-1 projectors. You are summing rank-2 operators which act in a higher-dimensional space, so the result will have a much higher rank. Geometrically, each operator only acts nontrivially on a 2dim plane, but the sum of the operators will act nontrivially on the full space (where "trivially" here means as the identity, without scaling) $\endgroup$
    – glS
    Nov 12 at 14:08
  • $\begingroup$ Makes sense! Is there a way to compute the eigenvalues and eigenvectors for a typical reduced density matrix? $\endgroup$
    – BlackHat18
    Nov 12 at 17:07
  • $\begingroup$ what do you mean with "typical" here? If you mean a general way to express the eigenvalues after partial trace in terms of eigenvalues before it, I'm not aware of any such relation. I'm not sure any simple such relation could exist: for example, partial tracing a rank-1 projection could result in all sorts of eigenvalues for the reduced state, depending on the degree of entanglement of the initial state, which isn't reflected in the eigenvalues of the operator, but rather in the Schmidt coefficients (i.e. the singular values of the "unvectorised" ket state) $\endgroup$
    – glS
    Nov 12 at 17:25
  • $\begingroup$ Sorry I mistyped. Is there a way to compute the eigenvalues and eigenvectors for a reduced density matrix obtained by tracing out one qubit (or some qubits) of a Haar random state, on an average? For example, equation 26 here (arxiv.org/pdf/1208.0692.pdf) indicates that if we only look at $n/4$ qubits, at most, and trace out $3n/4$ qubits, at least, then the Schmidt coefficients are very close to being uniformly distributed. $\endgroup$
    – BlackHat18
    Nov 12 at 18:01
  • $\begingroup$ Any state is a probabilistic average of rank 1 matrices – pure states. $\endgroup$
    – Danylo Y
    Nov 13 at 12:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.