# Off-diagonal elements of density matrix, measurement of coherence?

I have a ensemble of systems and each system is made of a single one-dimensional quantum harmonic oscillator. Suppose all systems in the ensemble are in the following quantum state

$$|\Psi\rangle = \dfrac{1}{\sqrt{2}}\left(|\psi_0\rangle e^{-iE_0 t/\hbar} + |\psi_1\rangle e^{-iE_1t/\hbar}\right)$$

where $|\psi_0\rangle$ and $|\psi_1\rangle$ are eigenstates of the Hamiltonian, which correspond to the ground state and the first excited state respectively. $E_0$ and $E_1$ are the eigenvalues associated with $|\psi_0\rangle$ and $|\psi_1\rangle$, repectively. Now if we use $|\psi_0\rangle$ and $|\psi_1\rangle$ as a set of orthonormal basis functions, the density matrix of this ensemble can be calculated to be

$$\hat{\rho} = |\Psi\rangle \langle\Psi|=\dfrac{1}{2} \left[ \begin{array}{cc} 1&e^{i(E_1-E_0)t/\hbar}\\ e^{-i(E_1-E_0)t/\hbar}&1\\ \end{array} \right]$$

Now my question is: Does the off-diagonal density matrix means that there is coherence between $|\psi_0\rangle$ and $|\psi_1\rangle$? If there is, can we measure this coherence via some experimental approach?

When it comes to decoherency it is often said, that the off-diagonal elements of the density matrix are responsible for coherency. They vanish iff there is no coherency. I spent a lot of time trying to understand this, and I'm still not shure if my explanation is correct. Let's see:

Imagine we'd be able to clone an ensemble discribed by $\rho$ before making measurements on it. This magic technique would allow us to obtain probabilities for measurement results without destroying the original ensemble. In the first run, we would determine the probability for $|\psi_0\rangle$ and $|\psi_1\rangle$ which should be $\langle \psi_0 | \rho | \psi_0 \rangle$ and $\langle\psi_1|\rho|\psi_1\rangle$ respectively. For the next experiments we change the measuring device to measure another observable. Lets assume that the set of orthonormal eigenvectors of this second observable contains an element of the form $$|\sigma\rangle := \alpha |\psi_0\rangle + \beta|\psi_1\rangle.$$ If there is no coherency in the ensemble, we would expect the probability for $|\sigma\rangle$ beeing $$|\alpha|^2 \langle\psi_0|\rho|\psi_0\rangle + |\beta|^2 \langle \psi_1 | \rho| \psi_1 \rangle.$$ But if our experiments show a different probability, one would likely say: "bazinga, there is is some quantum mystery (aka. coherency) going on". Quantum theory says, that the probability for $|\sigma\rangle$ is $$\langle \sigma |\rho |\sigma \rangle = |\alpha|^2 \langle\psi_0|\rho|\psi_0\rangle + |\beta|^2 \langle \psi_1 | \rho| \psi_1 \rangle + 2\Re\big( \alpha^*\beta \langle \psi_0|\rho|\psi_1\rangle\big),$$ which differs from the above iff the off-diagonal element $\langle\psi_0|\rho|\psi_1\rangle$ does not vanish. Summing up we see decoherency if the density matrix has non vanishing off-diagonal elements.

There is another point, I'd like to mention in this context. Lets say our density matrix has all vanishing off-diagonal elements. Does this mean, that there is no coherency in the ensemble? I think the right answer to this question is "no, but...". Lets take a closer look. For a density matrix $$\rho = \sum_k p_k |\varphi_k\rangle\langle\varphi_k|,$$ where $p_k$ is the probability for a system beeing in the state $|\varphi_k\rangle$, vanishing off-diagonal elements only mean that $$0 = \sum_k p_k \langle \psi_0 | \varphi_k \rangle \langle \varphi_k | \psi_1 \rangle,$$ i.e. the elements $p_k \langle \psi_0 | \varphi_k \rangle \langle \varphi_k | \psi_1 \rangle$ sum up to zero. To surely say that there is no coherency, we need all elements $p_k \langle \psi_0 | \varphi_k \rangle \langle \varphi_k | \psi_1 \rangle$ being zero, but this is not what follows from vanishing off-diagonal elements. The key point is, that we have no chance to distinguish between different ensembles which are described by the same density matrix. So for our experiments, an ensemble where all $p_k \langle \psi_0 | \varphi_k \rangle \langle \varphi_k | \psi_1 \rangle$ sum up to zero behaves exactly like an ensembly where all $p_k \langle \psi_0 | \varphi_k \rangle \langle \varphi_k | \psi_1 \rangle$ vanish, i.e. an example without coherent states. Summarizing we can say, that an ensemble described by a density matrix with vanishing off-diagonal elements behaves like if there would be no coherent states. There may still be coherent states in our ensemble, but we have no chance to detect them.

You are correct. The off-diagonals represent coherence between the two possibilities.

It can indeed be measured. Consider measuring in the +/- bais. IE acting on your density matrix with the operator $$|+\rangle \langle+|$$ where:

$$|+\rangle = (|0\rangle + |1\rangle) / \sqrt2$$.

The expectation value of this operator is given by Tr($$\rho |+\rangle \langle+|$$).

Putting this into your example you notice that the result of the measurement oscillates over time, with a frequency equal to the energy difference. If you instead had the density matrix with no off-diagonals this quantity would be "flat" (unchaning in time, value 1/2).

Your Qubit could be in the state $$|0\rangle$$, in which case the operator above has an expectation value of 1/2. If it was in the state $$|1\rangle$$ again the value is 1/2. If we had a machine that tossed a coin and made either $$|0\rangle$$ or $$|1\rangle$$ then the density matrix spat out by the machine is yours but with no off-diagonals, and in this case the expected outcome of the measurement is again 1/2.

But your state is not a random choice of the two options in that way, its a quantum superposition so the probabilities of the two possible cases don't just add in the normal way, which is what the off-diagonals keep track of.