0
$\begingroup$

$$\rho = \begin{bmatrix} 1/2 & (1+i)/{\sqrt{2}} \\ (1+i)/{\sqrt{2}} & 1/2 \end{bmatrix} $$

Can this matrix represent a a density operator of some single qubit state? I'm a little confused on the terminology, does a superposition of single qubit pure states count as a single qubit state.

The trace of $\rho$ is 1 so this could be some state. ${\rho}^2 \neq \rho$ so it is not a pure state.

$\endgroup$
  • 2
    $\begingroup$ Are you sure the signs of the terms in the off diagonal elements are correct? $\endgroup$ – By Symmetry May 22 '18 at 19:09
  • $\begingroup$ The signs are correct. Are you saying that the signs are not correct if this is to be a density matrix of an operator? $\endgroup$ – samlanader May 22 '18 at 19:20
  • 5
    $\begingroup$ With those signs, it is not hermitian. It should be. $\endgroup$ – secavara May 22 '18 at 19:20
  • $\begingroup$ and the eigenvalues are not real. $\endgroup$ – ZeroTheHero May 22 '18 at 19:34
  • 1
    $\begingroup$ The signs CANNOT be correct. 1. Your $\rho$ as it is now is not hermitian; 2. The eigenvalues of your $\rho$ are not real; 3. Even assuming a change in sign to make the matrix hermitian: $\rho'=\left( \begin{array}{cc} \frac{1}{2} & \frac{1+i}{\sqrt{2}} \\ \frac{1-i}{\sqrt{2}} & \frac{1}{2} \\ \end{array} \right)$, this $\rho'$ does not have non-negative only eigenvalues. Neither $\rho$ nor $\rho'$ can be physical density matrices. $\endgroup$ – ZeroTheHero May 23 '18 at 1:02
2
$\begingroup$

There are three conditions to be met for some operator to be a valid density matrix:

  • $\mathop{\mathrm{Tr}} \rho = 1$,
  • $\rho^\dagger = \rho$ and
  • $\rho \ge 0$.

Any operator fulfilling these properties will describe a (not necessarily pure) generalized state of a quantum system. I use the term generalized here to avoid confusion with the common identification of (rays of) vectors from a Hilbert space with states of a quantum system.

A qubit is a quantum system with two basis states, so any two-by-two density matrix can be interpreted as the generalized state of a qubit in some basis. Exactly those density matrices that additionally fulfil $\rho^2 = \rho$ correspond to pure states (that is, states that can be represented by state vectors).

As you also ask about nomenclature: Be careful with the word superposition. Any normalized linear superposition $\alpha\left|A\right> + \beta \left|B\right>$ of state vectors $\left|A\right>$ and $\left|B\right>$ is again a pure state of the system, it will have the density matrix $\rho = \big( \alpha\left|A\right> + \beta \left|B\right>\big)\big( \alpha^*\left<A\right| + \beta^* \left<B\right| \big)$ with the property $\rho^2 = \rho$. With a density matrix there is another way a system can be "in both states", namely, they can have some classical probability $p_A$ to be in state $\left|A\right>$ resp. $p_B$ to be in state $\left|B\right>$, this is not called superposition and gets you a mixed generalized state: $\rho = p_A \left|A\right>\left<A\right| + p_B \left|B\right>\left<B\right|$. A linear combination of density matrices will only be a density matrix if the coefficients are real, positive and sum two one. The density matrix corresponding to a superposition of state vectors will not be a superposition of the density matrices corresponding to the state vectors.

$\endgroup$
  • $\begingroup$ What is the state is not pure, i.e. Tr$(\rho)^2 \ne \hbox{Tr}\rho$? $\endgroup$ – ZeroTheHero May 22 '18 at 19:36
  • $\begingroup$ @ZeroTheHero I do not understand your comment? Which part does it refer to? $\endgroup$ – Sebastian Riese May 22 '18 at 19:39
  • $\begingroup$ A qubit is a quantum system with two states, so any two-by-two density matrix can be interpreted as the state of a qubit in some basis.. What if $\rho$ does not describe a pure state? $\endgroup$ – ZeroTheHero May 22 '18 at 20:06
  • 1
    $\begingroup$ Then the density matrix describes a mixed state of the qubit. This is totally legitimate and physical (e.g. when coupling the qubit to a thermal bath or an environment which is then traced out). $\endgroup$ – Sebastian Riese May 22 '18 at 20:37
  • $\begingroup$ I could edit to say "generalized state" to be more explicit that I am using the more general concept of state here, not a state in the sense "state vector" but "density matrix". Although I say in the sentence before that the described state is not necessarily pure, so there should be no confusion. $\endgroup$ – Sebastian Riese May 22 '18 at 20:41

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.