I know that density matrices are one way to describe a quantum system. We can write the density matrix of a system as:

$$ \rho = \sum_{\alpha} p_\alpha |\psi_\alpha\rangle\langle \psi_\alpha |$$

I'm a little lost on how to interpret this. I'm consider the following example where I have $| \psi \rangle = c_0 |0 \rangle + c_1 |1 \rangle$.

In this case, my density matrix can be written as :

$$\rho = |\psi \rangle \langle \psi |$$ I interpret this as: this is a pure state (since I can write it as a single outer product) and, in this basis, the population in the state $|\psi\rangle$ is 1.

However, if I now expand this out, I get:

$$ \rho = |c_0|^2 |0 \rangle\langle 0 | + |c_1|^2 |1 \rangle\langle 1 | + c_0 c_1^* |0 \rangle\langle 1 | + c_1 c_0^* |1 \rangle\langle 0 |$$

In this case, I interpret the diagonal elements to be the populations of the $| 0 \rangle$ and $| 1 \rangle$ state, but the rest, I'm not entirely sure.

My main question: How do I interpret a density matrix? What do the "coherences" on the off diagonal tell me about my system? Can I interpret this as some sort of mixture that, in this basis, tells me I have some of my system in the $|0 \rangle$ state and some in the $| 1 \rangle$ state?

  • $\begingroup$ No. ___________ $\endgroup$ – Norbert Schuch Feb 3 at 11:13

You have off-diagonal components because you are writing your density matrix in the basis of $|0 \rangle$ and $|1 \rangle$. Note that there exists some change-of-basis matrix $U$ such that $U \left[ \begin{smallmatrix} |c_0|^2 & c_0c_1^{\ast} \\ c_0^{\ast} c_{1} & |c_{1}|^2 \end{smallmatrix} \right] U^{\dagger} = \left[ \begin{smallmatrix} 1 & 0 \\ 0 & 0 \end{smallmatrix} \right]$.

Purity of your density matrix is a basis-independent statement. The easiest way to check for purity is to calculate $\mathrm{Tr}[\rho^2]$, which in your case is equal to $1$ (and so means your state is pure).

The fact that you have off-diagonals in the $\{|0\rangle, | 1 \rangle \}$ isn't really that significant. It just means that $|\psi\rangle$ is a superposition of these basis elements.

  • 1
    $\begingroup$ So is there no physical meaning to what a coherence is? (the off diagonal terms in a certain basis). If I coupled my system to a bath and the coherences in the |0> |1> basis go to 0, does this mean my system will now be in a statistical mixture of |0><0| and |1><1|? $\endgroup$ – Jlee523 Feb 3 at 3:47
  • $\begingroup$ Coherences take on physical meaning in the same way that the basis takes on physical meaning. Your system-bath coupling singles out the $|0\rangle, |1\rangle$ basis as remaining stable against decoherence, so the coherences tell you what information from the density matrix will be lost. Indeed, the system would become a statistical mixture of $|0\rangle\langle 0|$ and $|1\rangle\langle 1|$. More physical meaning could come from an energy difference between $|0\rangle$ and $|1\rangle$; then the relative phase of $c_0$ and $c_1$ will evolve with time leading to measurable interference effects. $\endgroup$ – Quantum Mechanic May 19 at 16:43

Very briefly, representing the state as a matrix you are making a choice of basis. In that basis/representation, the diagonal elements tell you the probability of finding the state in each possible outcome, when measuring it in that basis.

The off-diagonal elements, generally referred to as coherences, tell you about the probabilities of finding different outcomes when measuring in different measurement bases.

For example, for qubit density matrix has the form $\rho=\begin{pmatrix}p_1&c\\\bar c&p_2\end{pmatrix}$, where $p_1,p_2$ are the probabilities of observing first and second outcome in the "computational basis", while the coherence term $c\in\mathbb C$ tells you how the state looks like in different measurement bases. More specifically, it tells you that $\langle \sigma_x\rangle = 2\Re(c)$ and $\langle \sigma_y\rangle=2\Im(c)$, where $\sigma_x,\sigma_y$ are the standard Pauli matrices. These numbers can be reduced to probabilities of the possible outcomes in the measurement bases defined by the eigenvectors of these two matrices.


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.