Say we have some state $$|\psi\rangle=\frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)$$ it is in a quantum superposition of $|0\rangle$ and $|1\rangle$. Its density matrix is $$\rho=\begin{pmatrix}\frac 1 2 & \frac i 2\\ -\frac i 2 & \frac 1 2\end{pmatrix}$$ If we measure $\rho$ we get some classical density matrix of the form $$\rho=\begin{pmatrix}\frac 1 2 & 0\\ 0 & \frac 1 2\end{pmatrix}$$

Is there some nice intepretation of what these cross terms mean, other than just arrising from the phase of the state?

  • $\begingroup$ Remember that you can always find a basis in which the density matrix is diagonal. Hence, any interpretation you give it must be basis dependent, not physical. $\endgroup$ – user1936752 Dec 26 '18 at 19:57
  • $\begingroup$ @user1936752 while this is true, this is also true of individual diagonal entries, which have the physical interpretation of populations. $\endgroup$ – ZeroTheHero Dec 26 '18 at 20:11
  • $\begingroup$ @ZeroTheHero, I see. I thought there was no physical interpretation of the diagonal entries either (since they change with change of basis). But perhaps I am wrong here $\endgroup$ – user1936752 Dec 26 '18 at 20:22
  • $\begingroup$ @user1936752 you are not wrong: the off-diagonal matrix elements (the so-called coherences) do depend on the basis but so do the diagonal entries, which are the populations in a given basis. Whereas the coherences can be complex, the populations are of course real. The off-diagonals are historically less clear. $\endgroup$ – ZeroTheHero Dec 26 '18 at 21:54
  • 1
    $\begingroup$ Compute the expectation value of a 2x2 operator A with no diagonal terms, by tracing it with ρ. $\endgroup$ – Cosmas Zachos Dec 26 '18 at 23:21

I’m assuming that by cross-terms you mean off diagonal terms.

First, when talking about off-diagonal terms we clearly mean in a fixed basis. For example, in an eigenbasis of $\rho$ your state has no off-diagonal elements whatsoever.

Second, what you say about measuring clearly holds when performing measurements in that basis ($|0\rangle, |1\rangle$ in your basis). Or more precisely when we measure an observable that has the same eigenbasis.

Coming to the physical meaning of these off-diagonal terms, they are also called ‘coherences’. They are responsible for the interference effects of a quantum particle. In particular those effects that make them sometime look like waves. The prototypical experiment that reveal quantum coherence is the double slit experiment.

The following thought experiment has been put forward by Ahronov et al. to describe the essence of the quantum coherence and the double slit experiment.

Imagine to prepare the following state

$$ |\Psi\rangle = ( |\phi\rangle_L + e^{i \theta} |\phi\rangle_R)/\sqrt{2} $$

Where $|\phi\rangle_{L,R}$ are Gaussian wavepackets centered around $L, R$ (left, right) with spread much smaller than their separation. This requirement assures that the two states are essentially orthogonal. The left (right) wave packet is prepared with momentum $p$ ($-p$) such that it travels to the right (left). If you measure position at $t=0$, you will find two Gaussian blobs around $L$ and $R$ and the phase $\theta$ is not observable. From this experiment alone the state cannot be distinguished from a classical state.

Now let evolve the state until the time where the wavepackets collide. At this point it turns out that if you measure position the phase $\theta$ becomes observable! It’s like the particle interfered with itsealf, pretty much like a wave.

If you repeat the experiment many times with different phases $\theta$, and every time measure the position of the particle (for example the particle hits a screen), you will indeed see a figure of interference forming on the screen.

  • $\begingroup$ Nice answer: your simplified experiment is basically describing an interferometer with two arms. No? $\endgroup$ – Bruce Greetham Dec 27 '18 at 10:44
  • $\begingroup$ @BruceGreetham Thank you, it’s actually a thought experiment made up by Ahronov. I will post a reference when I find it. $\endgroup$ – lcv Dec 27 '18 at 11:10

Actually there is no clear physical interpretation of the density matrix itself if viewed from all angles. To see this you need to look at pure states and mixed states separately.

(The following first two paragraphs are for the sake of completeness and anybody familiar with the matter can basically skip them.)

About pure states

In the case of pure states the interpretation translates from their representation as Hilbert space vectors or $|\text{ket} \rangle$ states. Since the density matrix $\rho$ is constructed as $$ \rho = |\psi\rangle\langle\psi| $$ from a pure state $|\psi\rangle$ the interpretation of the matrix representation becomes apparent immediatly if look at a superposition like $$ |\psi\rangle = \frac{1}{\sqrt{2}}\left(|0\rangle + e^{i\theta} |1\rangle\right) \quad \Rightarrow \quad \rho_\psi = \frac12 \left(|0\rangle\langle 0| + e^{-i\theta}|0\rangle\langle 1| + e^{i\theta}|1\rangle\langle 0| + |1\rangle\langle 1|\right) $$ using quantum information notation for a qubit state. A measurement corresponds to a projection to some eigenstate and taking the trace from the resulting state afterwards. This basically yields the probability to measure the eigenstate given the initial state. If you for example use the eigenstate $|0\rangle$ of this basis you get $$ p_0 = \text{trace}\{\rho_\psi |0\rangle\langle 0|\} = \langle 0|\rho_\psi |0\rangle. $$ Here we see that the diagonal terms $|0\rangle\langle 0|$ and $|1\rangle\langle 1|$ give the probability of being in the eigenstates of that basis. A complete measurement of all possible eigenstates of the system yields the result that was posted in your answer. If you measure some other state it will give you a combination of the diagonal elements, where the trace yields the actual probability to measure that state. So the measurement with $|\psi\rangle$ itself yields the probability $p_\psi = 1$ as should be expected.

About mixed states

The information about whether the state was coherent is lost in the measurement of a single probability, hence the measurement of a completely mixed state $$ \rho_\text{mixed} = \frac12 \left(|0\rangle\langle 0| + |1\rangle\langle 1|\right) $$ yields the same result as above when measuring with the basis states. From this it seems like the off diagonal terms give a complete indication of whether the state is a coherent superposition or not. But the picture gets murkier if you consider partially mixed states.

Mixed states are usually introduced as some classical statistical mixture of the states before the measurement. In the qubit example consider the 'classical' probability $p$ to measure the state $|0\rangle\langle 0|$ within the mixture. Then the full mixture can be represented as $$ \rho = p |0\rangle\langle 0| + (1-p) |1\rangle\langle 1|, $$ where $p=\frac12$ gives you the completely mixed state. Any other $p \in [0,1]$ makes it seem like there is some partial pure state within the mixed state. This interpretation, however, is problematic given the mathematical structure of mixed states.

Mixed states can be expressed by any convex combination of pure states

The set of quantum states for a given Hilbert space constitutes a convex set with the pure states on the boundary of that set (for a qubit this is somewhat captured by the Bloch sphere, where the only the states with radius 1 are pure states). That means only the pure states have a unique representation, as any inner state of the set can be expressed by a convex combination of any subset of pure states. Even if we say that there is some clear physical interpretation to the pure states, it is not clear at all what it means that any mixed state can be expressed by an arbitrary collection of pure states. Here is where the simple picture of off-diagonal elements as an indication for coherence breaks down. Those elements indeed represent some kind of coherence if you're working with a specific basis of states, but the physical interpretation as simply a measure of coherence isn't entirely justified.

About entangled states

The picture becomes even murkier if we look at the one particle density matrix of an entangled state. Consider a maximally entangled two particle state $$ \rho_{ab} = \frac{1}{\sqrt{2}} \left(|00\rangle + |11\rangle \right). $$ This state is obviously coherent and pure, which is also seen in the density matrix of the complete system. You can look at the states of the individual subsystems by calculating the partial trace (here shown for subsystem $a$): $$ \rho_{a} = \text{trace}_b\{\rho_{ab}\} = \langle 0_b | \rho_{ab}| 0_b \rangle + \langle 1_b | \rho_{ab}| 1_b \rangle = \frac12 \left( |0_a\rangle\langle 0_a| + |1_a\rangle\langle 1_a| \right). $$ The outcome is a completely mixed state (same for subsystem $b$), which is somewhat surprising given the above assertion that pure states are located on the boundary of the set of states and the completely mixed state is at its center. While this might come down to the mathematical operation of reducing the dimension of the Hilbert space and thus neglecting information about the whole system, it adds even more problems as to how exactly one could interpret off-diagonal elements, especially if you consider partially entangled states.


I haven't given too much thought to these quantities but...

The vanishing of the off diagonal terms is associated with quantum decoherence so the presence of nonzero off diagonal terms is an indication that the system still is coherent to some degree.

The presence of off diagonal terms in the energy basis implies (suggests?) that the system is not stationary since $\rho$ (likely?) will not commute with $H$ in that case.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.