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

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.

  • 2
    $\begingroup$ Are you sure the signs of the terms in the off diagonal elements are correct? $\endgroup$ May 22, 2018 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, 2018 at 19:20
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    $\begingroup$ With those signs, it is not hermitian. It should be. $\endgroup$
    – secavara
    May 22, 2018 at 19:20
  • $\begingroup$ and the eigenvalues are not real. $\endgroup$ May 22, 2018 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. $\endgroup$ May 23, 2018 at 1:02

1 Answer 1


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.

  • $\begingroup$ What is the state is not pure, i.e. Tr$(\rho)^2 \ne \hbox{Tr}\rho$? $\endgroup$ May 22, 2018 at 19:36
  • $\begingroup$ @ZeroTheHero I do not understand your comment? Which part does it refer to? $\endgroup$ May 22, 2018 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? $\endgroup$ May 22, 2018 at 20:06
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    $\begingroup$ 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). $\endgroup$ May 22, 2018 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. $\endgroup$ May 22, 2018 at 20:41

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