Given a 2-dimensional Hilbert space, quantum states can be expressed as $2\times 2$ density matrices. In terms of the Pauli matrices, or Bloch representation, they can be written as \begin{equation} \rho=\frac{1}{2}\left(\mathbb{I}+\vec{r}\cdot\vec{\sigma}\right) \end{equation} where $\vec{r}$ is the Bloch vector in the Bloch sphere and $|\vec{r}|\leq1$.

QUESTION 1: Are there $2\times 2$ matrices that are not valid quantum states?

QUESTION 2: Are there singular $2\times 2$ matrices that are valid quantum states?


1 Answer 1

  1. Yes. Not all $2 \times 2$ matrices have trace 1, but all density matrices do. There are also lots of other special properties of density matrices, like being Hermitian.
  2. Yes. For example, consider a pure state $\rho = | \psi \rangle \langle \psi |$. This is a rank 1 matrix, because it's just projection onto $|\psi \rangle$, so it is not invertible.
  • $\begingroup$ How about mixed states? Are there singular $2\times 2$ matrices that are valid mixed states? $\endgroup$
    – Janus
    Feb 14, 2016 at 5:57
  • $\begingroup$ @Janus. There are no 2 by 2 singular mixed states, but this is a special case for a two-dimensional system. (You can see this by diagonalizing the density matrix and noting that for a mixed state, there must be at least two non-zero eigenvalues.) $\endgroup$
    – march
    Feb 14, 2016 at 6:01
  • $\begingroup$ @march Thanks. So, say I have a random $2\times 2$ nonsingular positive definite matrix with trace 1, am I assured that it is a valid quantum state? Also, are you saying that for higher dimensional systems, there are $n\times n$ singular matrices with trace 1 that are valid quantum states? $\endgroup$
    – Janus
    Feb 14, 2016 at 6:12
  • $\begingroup$ @Janus. Yes to both of your questions. If you want details, you'll probably want to ask a new question. $\endgroup$
    – march
    Feb 14, 2016 at 6:13
  • $\begingroup$ Try the book "Geometry of quantum states" $\endgroup$
    – XXDD
    Feb 14, 2016 at 8:24

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