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By definition of the density matrix for example the density matrix of $|0\rangle$ state (pure state) is:

$$\rho=|0\rangle \langle 0| = \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix}$$

But there is a "rule of thumb" (which I found here) for distinguishing pure state and mixed state: the bigger the off-diagonal elements the purer the state. It looks confusing for me, because for pure states ($|0\rangle$ and $|1\rangle$) the off-diagonal elements always zero.

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2 Answers 2

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You can determine whether a state is pure or mixed by considering the purity $\gamma$ which is defined as the trace (i.e. the sum of diagonal entries) of the density matrix squared.

\begin{equation} \gamma = Tr[\rho^2] \end{equation}

If $\gamma =1$ then the state described by $\rho$ is pure. If $\gamma$ is less than 1 then the state is mixed. Density matrices are in some sense more general than statevectors as they allow you to describe statistical mixtures of quantum states in addition to pure states.

For a discussion of meaning of the off-diagonal elements see this previous post. I won't reproduce the discussion here but basically the off diagonal elements tell you whether your state is a quantum superposition of basis states or only a statistical mixture of basis states.

In the case of $|0\rangle$ the state is pure but is not a superposition of the $\{|0\rangle, |1\rangle\}$ basis vectors. The rule of thumb is roughly testing whether we have a quantum superposition vs a classical mixture. Cases like $|0\rangle$ where the state is pure but not a superposition are edge cases where the rule doesn't work. However by checking the purity $\gamma$ we can always know for sure.

The rule works for examples like the maximally mixed state $\rho_{max} = \frac{1}{2} I$ where the off diagonal elements are 0. This indicates there is merely classical uncertainty about the state and no quantum superposition going on.

Using the purity test above we can easily see that $\rho=|0\rangle\langle0|$ is clearly a pure state.

\begin{equation} \gamma = Tr[|0\rangle\langle0|^2]= Tr[|0\rangle\langle0|] = 1 \end{equation}

For the case of $|0\rangle\langle0|$ this test is a bit unecessary as every density matrix constructed from taking the outer product of a statevector $|\psi\rangle$ with itself $(|\psi\rangle\langle \psi|)$ is a pure state.

In general a density matrix can expressed as a sum of outer products (pure states) weighted by classical probabilities $p_i$. Here these probabilities $p_i$ must obviously sum to 1.

\begin{equation} \rho = \sum_i p_i |\psi_i\rangle\langle\psi_i| \end{equation}

This form is what allows density matrices to describe mixed states. Note that here one or more of the $|\psi_i\rangle$ states could be a quantum superposition.

For pure states including $|0\rangle\langle0|$ there is no classical uncertainty about the state of the system. In other words the state is not mixed. For pure states we don't have to bother with the $p_i$ terms and the density matrix is just the single outer product $|\psi\rangle\langle \psi|$.

Hope this helps.

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  • $\begingroup$ thank you!) could you please tell me, how can I distinguish between superposition and statistical mixture by off-diagonal elements values? $\endgroup$
    – Curious
    Jun 17, 2023 at 16:44
  • $\begingroup$ I think in general your rule of thumb is useful for making such a distinction. There are just some cases like $|0\rangle\langle0|$ which seem to me to be exceptions. If the off diagonal entries are non-zero then the state must be a superposition of some kind. If the off diagonal elements are 0 then the state is likely to be mixed. I'm not an expert on this sort of thing so maybe someone else can give you a more definitive answer. $\endgroup$
    – Callum
    Jun 17, 2023 at 20:34
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The "rule of thumb" that you are referring to is correct, but must be quantified. If we label elements of the density matrix by $i$ for the row and $j$ for the column, the condition for being a pure state is that $$|\rho_{ij}|^2=\rho_{ii}\rho_{jj}.$$ This will always be true for $i=j$, while any mixed state will have $|\rho_{ij}|^2<\rho_{ii}\rho_{jj}$ for some $i\neq j$, so we can quantify purity by "how big" the off-diagonal elements $\rho_{ij}$ are (off-diagonal means $i\neq j$).

Notice that the right-hand side is bounded by the product of the diagonal elements corresponding to row $i$ and column $j$. In the example of $|0\rangle\langle 0|$ given above, $\rho_{11}=0$; this means that we can only ever have $|\rho_{01}|^2\leq 1\cdot 0=0$, so our pure state condition can be satisfied even with $\rho_{01}=0$.

The point is that we are interested in "how big" the off-diagonal elements are relative to the diagonal elements. When the diagonal elements are both $1/2$, the off-diagonal elements must have magnitude $1/2$, while other diagonal elements sustain pure states with smaller off-diagonal elements. This observation holds for states in any dimension.

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