I'm confused between two representations of the density operator in quantum statistical mechanics.

In the first case, we have :

$$\hat{\rho}=\sum_i P_i|\psi_i\rangle\langle \psi_i|$$

In the second case, we have :

$$\hat{\rho}=\sum_i \sum_kP_{ik}|\phi_i\rangle\langle \phi_k|$$

Are these two cases equivalent ?

For example, in case of a mixed state, I know the density operator would not contain any superposition terms. For example, in case of a pure state $2$ level system, $\hat{\rho}=|\psi\rangle\langle\psi|=a|0\rangle\langle 0|+b|1\rangle\langle 0|+c|0\rangle\langle 1|+d|1\rangle\langle 1|$

In the mixed state however, $\hat{\rho}=a|0\rangle\langle 0|+b|1\rangle\langle 1|$

According to this, the first representation seems to be the correct one. However, we can always expand the kets in some basis set, and doing that, we would obtain cross terms, as we did for the pure state. Perhaps if we decompose the states $|0\rangle$ and $|1\rangle$ into some other basis, maybe we would obtain cross terms, even in case of the mixed state

So, does it mean that both the representations are equivalent, and you can derive the second one, by just plugging $|\psi_i\rangle=\sum_k a_k |\phi_k\rangle$ in the first method ? Moreover, I think we can combine $a_k$ terms with the $P_i$ terms to get $P_{ik}$ terms.

So we can get the second representation from the first.

However, this creates a small problem. For example, I know that a pure state has only a single term in the density matrix. However, that would be true only in case of the first representation.

In this representation, the pure state would have a single term, while mixed states would have multiple terms. In case of the mixed ensemble, there would be no superposition terms $|a\rangle\langle b|$ since the index goes over only $i$.

Hence we can easily distinguish between pure and mixed states.

In the second representation, we have expanded our $|\psi\rangle$ in terms of $|\phi_k\rangle$'s. Hence in both the pure state and the mixed state expressions, we would have multiple terms. Moreover, the mixed state would also have superposition terms since the index goes over both $i$ and $k$.

How then, can we differentiate between pure and mixed states ? How can we then say, that the density matrix of the pure state, has only one single terms in it, and that $P_i=1$.

So,the only way we can differentiate between the density operators for pure and mixed states is by converting into the first expression ? Or can we differentiate even if both are represented through the second representation ?


1 Answer 1


The coefficients $P_{ik}$ can be zero in the second expression. They don't have to be nonzero just because the sum runs over them. The actual value depends on your density operator and the basis that you chose. The representations are equivalent and it is as you stated yourself, you obtain the second one by insertion of the linear expansion with respect to the other basis.

You can also switch to the the first representation based on the second by diagonalization of the nondiagonal density matrix. This is nothing but a change of basis.

You can differentiate pure and mixed states by calculation of the trace of powers of the density operator or the density matrix. The density operator of a pure state is idempotent i.e. $$ \hat \rho \hat \rho = \hat \rho \\ tr(\hat \rho \hat \rho) = tr(\hat \rho )=1 $$ For mixed states this no longer holds and the trace of powers of rho are less than one and $$\hat \rho_{mixed} \hat \rho_{mixed} \neq \hat \rho_{mixed} $$ These properties do not depend on the representation and you can use any density operator representation to check them.

For a quick check you can also look at the matrix elements. If your density operator is based on a pure state then the following holds for the matrix elements, $$\begin{aligned} |c_{ij}| &= \sqrt{|c_{ii}|*|c_{jj}|}\\ c_{ij} &= c_{ji}^* \end{aligned}$$ this means that off-diagonal elements cannot be zero if the diagonal elements aren't zero themselves for a pure state.

  • $\begingroup$ Ah so basically statements like the pure state must have a single non-zero term in the density matrix are not rigorous and subject to the way we have represented the density matrix. A more rigorous way of defining pure and mixed states would be by checking the trace as you have mentioned. Is that correct ? $\endgroup$
    – RayPalmer
    Commented Jun 12, 2022 at 15:27
  • $\begingroup$ @RayPalmer Yes, that is correct. $\endgroup$
    – Hans Wurst
    Commented Jun 12, 2022 at 17:07

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