Do "cross terms" in a state operator necessarily correspond to a coherent superposition?

Question

In his discussion (Chapter 9.2-9.3) of the measurement problem, Ballentine says "any terms [in the state operator $\rho$] that [are] nondiagonal [in terms of having "mixed projectors" $|\alpha_r\rangle\langle \alpha_s|$ where the $\alpha_i$ are macroscopic indicator variables, and so these vectors are orthogonal] would correspond to coherent superpositions of macroscopically distinct “indicator position” eigenvectors". This quote is from the bottom of page 237 in the 2nd edition.

My question is what is meant by "correspond to". Correspond how? Ballentine is uncharacteristically imprecise here. I think the crux of my issue is in wondering whether the following is true?

Theorem. Suppose that a composite system has a state operator $\rho$ which includes "cross term" projectors from an orthonormal basis (but not necessarily the entire basis). The composite system is then in a state involving a coherent superposition of these orthonormal elements. That is, some element of the state operator will arise from a $|\psi \rangle$ which is a coherent superposition of orthonormal basis states (i.e., $\rho = c_{\psi}|\psi \rangle \langle \psi| + \sum_j c_{\chi_j}|\chi_j \rangle \langle \chi|$ where the $|\chi_j \rangle$ may or may not involve coherent superpositions.

Answer 1

Consider an orthonormal basis $|n\rangle$ (leaving aside macroscopicity of the basis states as it is not essential to this answer). Any state can be expanded as $$\rho=\sum_{mn}\rho_{mn}|m\rangle\langle n|.$$ The presence of a nonzero term $\rho_{mn}$ for $m\neq m$ implies that the state has a "coherence" in this basis. This can be inspected by computing the overlap of the state with the pure superposition state $$|m,n;\phi\rangle\equiv \frac{|m\rangle+e^{i\phi}|n\rangle}{\sqrt{2}}.$$ We look for the maximum overlap over all $\phi$, in case $\rho_{mn}=\rho_{nm}^*$ is complex: $$O(m,n;\rho)\equiv\max_\phi \langle m,n;\phi|\rho|m,n,\phi\rangle=\max_\phi \frac{\rho_{mn}e^{i\phi}+\rho_{nm}e^{-i\phi}+\rho_{mm}+\rho_{nn}}{2}=|\rho_{mn}|+\frac{\rho_{mm}+\rho_{nn}}{2}.$$ Positivity of the density matrix ($|\rho_{mn}|^2\leq \rho_{mm}\rho_{nn}$) implies that maximum overlap $O(m,n;\rho)$ is itself maximized for a state $\rho$ that is exactly equal to $|m,n;\phi\rangle$. Rewriting our expression as $|\rho_{mn}|=O(m,n;\rho)-\frac{\rho_{mm}+\rho_{nn}}{2}$, this means that [the state's overlap with a pure, equal-magnitude superposition of the two basis states in question] minus [the average probability of finding the state in either of the basis states] tells us [the magnitude of the coefficient of the off-diagonal element], and vice versa. If $|\rho_{mn}|=1/2$, the state must actually be a pure state of the form $|m,n;\phi\rangle$ (because $|\rho_{mn}|^2\leq \rho_{mm}\rho_{nn}$ by positivity and $\rho_{mm} \rho_{nn}\leq 1$ by normalization mean that $|\rho_{mn}|=1/2$ implies $\rho_{mm} =\rho_{nn}= 1/2$, which mean that $\rho_{ii}=0$ for $i\neq m,n$, which mean that $\rho_{ij}= 0$ for $i\neq m,n$, which mean that the density matrix only has support over the $2\times 2$ subspace, and you can check that it must be pure... you'll never see a density matrix with an off-diagonal elements greater than $1/2$, in any basis). If $0<|\rho_{mn}|\leq 1/2$, we know that there is some contribution to the state by a state of the form of $|m,n;\phi\rangle$ but that contribution may be small.

Did we prove that such a contribution must be there? Not quite, but remember we can always express $\rho$ in any basis we choose. Let's keep the original basis but change two of the basis states, $|m\rangle$ and $|n\rangle$, to $|m,n;\phi\rangle$ and $|m,n;\pi+\phi\rangle$, where the latter two are still orthonormal. Let's further choose $\phi$ to be the angle that achieves the maximum above. The probability of finding the state in either of the two new basis states is the same as the probability of finding it in either of the two old basis states, $\rho_{mm}+\rho_{nn}$. In this basis, $O(m,n;\rho)$ is exactly the expansion coefficient for the state $|m,n;\phi\rangle\langle m,n;\phi|$ and the contribution from $|m,n;\pi+\phi\rangle\langle m,n;\pi+\phi|$ is the same but with a change of sign: $-|\rho_{mn}|+\frac{\rho_{mm}+\rho_{nn}}{2}$. So we can always find a basis in which the pure superposition state features, given that $O(m,n;\rho)\neq 0$. But wait, that can be nonzero even if $|\rho_{mn}|=0$, so long as there is a nonzero probability of finding the state in either $|m\rangle$ or $|n\rangle$, because $\rho_{mm}+\rho_{nn}>0$ in that case! And we should have anticipated that, because the identity matrix can be written in any basis and have contributions from any projector we want. What's special in the $|\rho_{mn}|>0$ case is that the overlap of the state with $|m,n;\phi\rangle$ is greater than the overlap with $|m,n;\phi+\pi\rangle$ $\rho_{mn}\neq 0$. In our new basis, the bigger the difference between these [diagonal] components, the greater the value of $|\rho_{mn}|$ and thus the greater the contribution of the pure superposition state $|m,n;\phi\rangle$. What is really cool is that the maximum contribution can only come by another contribution being zero. For a state to be really pure and have a large contribution from $|m,n;\phi\rangle$, it must have a small contribution from $|m,n;\phi+\pi\rangle$ and vice versa.

---

Now let's briefly return to macroscopicity (macroscopicness?) and quantumness. If two states $|m\rangle$ and $|n\rangle$ are macroscopically distinguishable (i.e., if they correspond to two states that are very different like a light switch being on and off), then the size of the component $|\rho_{mn}|$ tells us something about how much of a macroscopic superposition is present in the state. There are many, many different ways in the literature of quantifying this coherence, specifically because it depends on the basis you choose, but everyone will agree that a large $|\rho_{mn}|$ means a large contribution to the state from a superposition of macroscopically distinguishable states.

---

Summary and addition: a nonzero term $\rho_{mn}$ for $m\neq n$ implies the state has a nonzero overlap with some state of the form $|m,n;\phi\rangle$ and thus can be expressed as $\rho=O |m,n;\phi\rangle \langle m,n;\phi|+(1-O)\rho^\prime$ for some $\rho^\prime$. But a state of the latter form does not necessarily have coherence, because it could be the maximally mixed state (for example). Instead of nebulously defining coherence in terms of the off-diagonal elements, we should look to the origins of the term. In interferometry, coherence of your signal means your signal oscillates in a stable manner with respect to some reference oscillator. Here, the interference comes about if we look at the overlap between our state and some state $|m,n;\phi\rangle$, with the result (as calculated before) $O(m,n;\phi;\rho)=\langle m,n;\phi|\rho|m,n;\phi\rangle=\Re(\rho_{mn}e^{i\phi})+\frac{\rho_{mm}+\rho_{nn}}{2},$ with $\Re$ signifying the real part. The quality of the interference signal is dictated by the amplitude of the oscillation relative to the average value of the signal, which is known as the interference visibility: $$V=\frac{\max_\phi O(m,n;\phi;\rho)-\min_\phi O(m,n;\phi;\rho)}{\max_\phi O(m,n;\phi;\rho)+\min_\phi O(m,n;\phi;\rho)}=\frac{|\rho_{mn}|}{\rho_{mm}+\rho_{nn}}.$$ Visibility greater than zero is what we refer to as coherence and that is why the off-diagonal element signifies coherence. The reason this is considered basis dependent is because we are using $|m,n;\phi\rangle$ as our reference. In a double-slit experiment, the two states in superposition are the paths that the photon takes to get to the final screen, and the oscillations are of the intensity detected with respect to spatial position or of the electric field with time. The only state that is coherent in no basis is the maximally mixed state, coherence is unambiguous in a $2\times 2$ system, and for anything else people can only agree that there exists nonzero coherence with respect to a particular basis.

Answer 2

After reading the comments, I want to make an answer based on what I think the question is asking. I hope this would be useful. Consider a set of states $\{ \psi_i \}$ and two density matrices:

$$ \rho_1 = \sum_i p_i \vert \psi_i \rangle \langle \psi_i \vert, \hspace{0.5 cm} \rho_2 = \vert \psi \rangle \langle \psi \vert$$

where $\psi$ is some linear combination of the basis states. Now even though $\vert \psi \rangle$ is some linear combination of states $\rho_2$ is a pure density state, whereas $\rho_1$ is not. The mathematical reason is that $\rho_2$ is a rank 1 operator which means the dimension of the image of the operator. From the definition, it is easy to deduce this fact by writing the matrices out in this basis:

$$ \rho_1 = \begin{bmatrix} p_1 & 0 & \dots & 0 \\ 0 & p_2 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & p_N \end{bmatrix}, \hspace{0.5 cm} \rho_2 = \begin{bmatrix} 0 & 0 & \dots & 0 \\ 0 & 0 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & 1 \end{bmatrix}$$

It is now clear $\rho_2^2 = \rho_2$. The upshot of this argument is that $\textbf{if}$ a pure density matrix is written in a basis where it's diagonal, there must be one entry in the diagonal entry. Transforming to this to another state will induce off-diagonal elements which causes the quantumness. However, if there are more than 1 non-zero entries in a diagonal basis, it must be a mixed density operator which encapsulates our classical uncertainty.

Answer 3

Let $|\psi \rangle$ = $\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle$, and choose the the orthonormal basis $|0\rangle, |1\rangle$. Then $|\psi \rangle $ is clearly not entangled but $| \psi \rangle \langle \psi |$ has off-diagonal terms. I think Ballentine is just making the simple observation that if we fix a basis and represent a density matrix in this basis, then the off-diagonal terms signify that there is some superposition in this basis.