# Relationships between elements in the density matrix

Say we have a density matrix for a spin-1 system given by:

$$\hat \rho = \left[\begin{array} A \hat \rho_{00}& \hat \rho_{01} & \hat \rho_{02} \\ \hat \rho_{10} & \hat \rho_{11} & \hat \rho_{12} \\ \hat \rho_{20} & \hat \rho_{21} & \hat \rho_{22} \end{array}\right]$$

The diagonal entries $$\rho_{00}$$, $$\rho_{11}$$ and $$\rho_{22}$$ describe the probability of the system being in the $$|0 \rangle$$, $$|1 \rangle$$ and $$|2 \rangle$$ state.

The off-diagonal elements of a density matrix are referred to as coherences and they describe phase relationships between states that can develop.

Is there some relationship between the coherences of $$\hat \rho$$ and the diagonal entries of $$\hat \rho$$?

My intuition leads me to believe there is a relationship. For example, if a system has a non-zero coherence say $$\rho_{01} = \langle 0 | \hat \rho | 1 \rangle$$ (and its conjugate). Say the system underwent a process that reduced the probability of being in the $$|0 \rangle$$ and $$|1 \rangle$$ state, then $$\rho_{00}$$ and $$\rho_{11}$$ would reduce, would we not then also expect $$\rho_{01}$$ (and its conjugate) to reduce as well?

• The relationships between the coefficients of $\rho$ are the ones which comes from the definition of a density matrix : $\rho^\dagger = \rho$, $\operatorname{Tr}(\rho) =\rho$ and $\rho \geq 0$. May 17, 2021 at 20:29
• @SolubleFish Hmm, so the trace of the density matrix should be equal to 1 (probability conservation). And we know that off-diagonal elements are equal to their conjugates (i.e. $\hat \rho^{\dagger} = \hat \rho$) as you said. However, I don't see how either of these statements give you a relationship between the coherences of $\hat \rho$ and the diagonal elements of $\hat \rho$. Could you elaborate? Thanks May 17, 2021 at 20:36

Consider a finite dimensional complex Hilbert space $$H$$ of dimension $$d$$ equipped with an inner product denoted by $$(\cdot,\cdot)$$ and let $$\rho$$ be a generic density operator, i.e. a positive semi-definite operator with unit trace. This enables us to write

$$\rho = (\rho^{1/2})^\dagger \,\rho^{1/2} \quad .$$

Define $$\rho_{nm} \equiv (v_n,\rho\, v_m)$$, where $$V=\{v_n\}$$ is a complete orthonormal basis in $$H$$. Using the Cauchy-Schwarz inequality, we can show that the off-diagonal elements of the density matrix are bounded by the diagonal elements: \begin{align} |\rho_{nm}| &= |(v_n,\,(\rho^{1/2})^\dagger \,\rho^{1/2}\, v_m) | =|(\rho^{1/2}\,v_n,\,\rho^{1/2}\, v_m) |\\ &\leq \sqrt{(\rho^{1/2}\, v_n,\,\rho^{1/2}\, v_n)\, (\rho^{1/2}\, v_m,\,\rho^{1/2}\, v_m)} \\ &= \sqrt{(v_n,\,\rho \,v_n) \, (v_m,\,\rho \,v_m)} \quad , \end{align}

which eventually yields

$$|\rho_{nm}|^2 \leq \rho_{nn}\, \rho_{mm} \quad. \tag{*}$$

The above inequality holds for all elements $$v_n,v_m \in V$$. Note that in particular the off-diagonal elements vanish if one of the corresponding diagonal elements is zero.

Regarding the purity, which we will define as $$\pi(\rho) \equiv \mathrm{Tr}\rho^2 \quad .$$ The purity shows how 'mixed' / 'pure' a density operator is and it is bounded by $$d^{-1}\leq \pi(\rho) \leq 1$$ for all density operators $$\rho$$. The lower bound is reached for maximally mixed states and the upper bound for pure states. Most importantly, the value of $$\pi(\rho)$$ is independent of the basis in which you express the density operator as a matrix.

But since every density operator is diagonalizable, there always exists a basis in which there are no off-diagonal elements. For example, a pure state as well as a maximally mixed state each have a basis in which all off-diagonal elements are zero, while their corresponding purities differ. So just from the fact that the off-diagonal elements (in a given basis) are zero or non-zero you don't know, in general, anything about the purity of a density operator.

In other words, the purity of a density operator is basis-independent, in contrast to the appearance of off-diagonal elements.

That being said, we still can conclude two things from equation $$(*)$$, namely:

$$\mathrm{Tr} \rho^2 \leq \left(\mathrm{Tr} \rho\right)^2 = 1$$ and further that the equality in $$(*)$$ holds for all $$v_n, v_m \in V$$ if only if $$\rho$$ is a pure state.

To show the latter statement, we first suppose that $$|\rho_{nm}|^2 = \rho_{nn}\, \rho_{mm}$$ holds for all $$v_n,v_m \in V$$. Then

$$\mathrm{Tr}\rho^2 = \sum\limits_{nm} |\rho_{nm}|^2 = \sum\limits_{nm} \rho_{nn} \, \rho_{mm} = \left(\mathrm{Tr}\rho\right)^2 = 1 \quad ,$$

which shows that $$\rho$$ is pure. For the reverse direction, we recall that for a pure density operator $$\rho^\psi$$ there exists by definition a unit vector $$\psi \in H$$ s.t. $$\rho^\psi = \psi \psi^\dagger$$. Expanding $$\psi$$ in the orthonormal basis $$V$$:

$$\psi = \sum\limits_n c_n \, v_n$$

with $$c_n \equiv (v_n,\psi)$$ and $$\sum\limits_n |c_n|^2 = 1$$ shows that $$\rho^\psi_{nm} = c_n\, \bar{c}_m$$ and thus $$|\rho^\psi_{nm}|^2 = |c_n|^2\, |c_m|^2 = \rho^\psi_{nn}\, \rho^\psi_{mm} \quad ,$$

which completes the proof.

Consequently, for a mixed density operator $$\sigma$$ we find that there exists at least one pair $$v_i,v_j \in V$$ such that:

$$|\sigma_{ij}|^2 < \sigma_{ii}\, \sigma_{jj} \quad ,$$

while for a pure state $$\rho$$

$$|\rho_{nm}|^2 = \rho_{nn}\, \rho_{mm} \quad \forall\, v_n,v_m \in V$$ holds.

It is stressed that we chose an arbitrary orthonormal basis and hence the above (in)equalities hold for any such basis (specifically the eigenbasis). All in all, we see that equation $$(*)$$ also serves as a criterion to decide whether a given density operator expressed in an arbitrary complete orthonormal basis is pure or mixed.

• Thank you, that was insightful. Just to clarify, is there some relationship between the inequality $|\rho_{nm}|^2 \leq \rho_{nn} \rho_{mm}$ and the pureness of the system? For example, a mixed state (or impure state) has the relation $\mathrm{Tr}[\rho^{2}] < 1$, where the value of $\mathrm{Tr}[\rho^{2}]$ tends closer towards 0 the more mixed the state is. Is there some relation between the mixedness of a state and the inequality that you have derived (i.e. does a more mixed state lead to smaller value of $|\rho_{nm}|^{2}$)? May 17, 2021 at 22:20
• @NahPlsMan I've edited the answer. Please double check everything and let me know if everything is fine and whether you have further questions. May 18, 2021 at 11:51
• Is there a reference that I can refer to for the inequality you derived? May 21, 2021 at 13:15
• @NahPlsMan Sorry, I don't know any particular book... But this is very well-known, see for example this homework sheet. Exercise $6$ this answer is basically exercise 6. May 21, 2021 at 13:21
• @NahPlsMan Another homework sheet. Both courses are on quantum optics... It seems that this is a fairly common exercise there. But as I said, I don't know any reference, sorry. Hope this helps, tho. May 21, 2021 at 13:32