Does a physical interpretation of density matrix off-diagonal terms exist?

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?

• 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. Dec 26, 2018 at 19:57
• @user1936752 while this is true, this is also true of individual diagonal entries, which have the physical interpretation of populations. Dec 26, 2018 at 20:11
• @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 Dec 26, 2018 at 20:22
• Compute the expectation value of a 2x2 operator A with no diagonal terms, by tracing it with ρ. Dec 26, 2018 at 23:21
• – glS
Sep 19, 2022 at 15:17

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. (see [1]) 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 the 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 as if the particle interfered with itself, pretty much like a wave.

If you repeat the experiment many times with different phases $$\theta$$, and every time you 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.

 Added 19/9/22

Mathematical Details

We work in a setting where there is a free particle (not subject to any potential) moving on the real line. Hence the Hamiltonian is $$H=a p^2$$ ($$a=1/(2m)$$). If you want to be pedantic the Hilbert space is $$\mathcal{H} = L^2(\mathbb{R})$$.

First consider as initial state a Gaussian wavepacket centered around $$x_0$$ moving with momentum $$p_0$$:

$$\psi(x)=\frac{e^{-\frac{(x-x_{0})^{2}}{4\sigma^{2}}}}{(2\pi\sigma^{2})^{1/4}}e^{ip_{0}x}.$$

Correctly $$|\psi(x)|^2$$ is a Gaussian centered around $$x_0$$ with standard deviation $$\sigma$$.

The probability distribution of position at time $$t$$ is

\begin{align} \left|\psi(x,t)\right|^{2} &=\frac{e^{-\frac{(x-x_{t})^{2}}{2\sigma_{t}^{2}}}}{\sqrt{2\pi\sigma_{t}^{2}}} \\ x_{t} &=x_{0}+2p_{0} a t \\ \sigma_{t} &=\sqrt{\sigma^{2}+\frac{a^{2}t^{2}}{\sigma^{2}}}. \end{align}

This means that if $$p_{0}>0$$ ($$p_{0}<0$$) the wavepacket travels to the right (left). Moreover the wavepacket spread a bit in time according to $$\sigma_t$$.

Now initialize the system in the following superposition. A left Gaussian packet centered in $$x_{0}=-L/2$$ with $$p_{0}=k_{0}$$ and standard deviation $$\sigma$$ and a right wavepacket at $$x_{0}=+L/2$$ and $$p_{0}=-k_{0}$$ (and same spread) with some relative phase $$\vartheta$$: $$|\psi\rangle=\frac{1}{\sqrt{2}}\left(|\phi\rangle_{L}+e^{i\vartheta}|\phi\rangle_{R}\right).$$ We assume $$L\gg\sigma$$ so that the states $$|\phi\rangle_{L}$$ and $$|\phi\rangle_{R}$$ are essentially orthogonal.

The probability density of observing the particle at position $$x$$ is $$\left|\psi(x)\right|^{2}=\frac{1}{2}\left(\left|\phi_{L}(x)\right|^{2}+\left|\phi_{R}(x)\right|^{2}+2\mathrm{Re}\left(\overline{\phi_{L}(x)}\phi_{R}(x)e^{i\vartheta}\right)\right).$$

So the term containing $$\vartheta$$ is essentially zero if $$L\gg\sigma$$ which we have assumed. Hence the phase $$\vartheta$$ is un-observable if one measures the position of the particle.

As time goes by the wave-packets travel towards each other. After a time $$t=T=L/(4k_{0}a)$$ both wave-packets are centered in zero (and spread a bit). The probability density of position at this time is $$\left|\psi(x,T)\right|^{2}=2\frac{e^{-\frac{x^{2}}{2\sigma_{T}^{2}}}}{\sqrt{2\pi\sigma_{T}^{2}}}\cos^{2}\left(k_{0}x-\vartheta/2\right).$$

Correctly the integral of the above is 1 plus exponentially small corrections in $$L$$ (since we have ignored the small overlap of the two initial wavepackets). In Eq.~(9) of [1] there is incorrectly a 4 in the above formula.

The phase $$\vartheta$$ manifest itself by shifting the interference pattern by $$\delta$$ $$\delta=\hbar\frac{\vartheta}{2k_{0}},$$ where we inserted back proper factor of $$\hbar$$ (the paper does not have the factor of 2).

References

• Nice answer: your simplified experiment is basically describing an interferometer with two arms. No? Dec 27, 2018 at 10:44
• @BruceGreetham Thank you, it’s actually a thought experiment made up by Ahronov. I will post a reference when I find it.
– lcv
Dec 27, 2018 at 11:10
• any update on that reference of Aharonov? Aug 25, 2021 at 11:53
• @ZeroTheHero I finally added the reference (I also updated the answer quite a bit)
– lcv
Sep 19, 2022 at 16:21
• Thanks for that! Sep 19, 2022 at 18:06

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.)

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.

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.
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.