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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?

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    $\begingroup$ 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 $\rho \geq 0$. $\endgroup$ May 17, 2021 at 20:29
  • $\begingroup$ @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 $\endgroup$
    – NahPlsMan
    May 17, 2021 at 20:36

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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.

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    $\begingroup$ 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 $|\rho_{nm}|^{2}$)? $\endgroup$
    – NahPlsMan
    May 17, 2021 at 22:20
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    $\begingroup$ @NahPlsMan I've edited the answer. Please double check everything and let me know if everything is fine and whether you have further questions. $\endgroup$ May 18, 2021 at 11:51
  • $\begingroup$ Is there a reference that I can refer to for the inequality you derived? $\endgroup$
    – NahPlsMan
    May 21, 2021 at 13:15
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    $\begingroup$ @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. $\endgroup$ May 21, 2021 at 13:21
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    $\begingroup$ @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. $\endgroup$ May 21, 2021 at 13:32

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