State vector vs density operator

Question

We formulate quantum mechanics using language of state vectors. One alternative formulation is possible using density operator or density matrix. Why we are doing this alternative approach? Is the approach of state vector is not sufficient? Is the density matrix approach works more efficiently?

Answer 1

State vectors define pure quantum states of a system, and, for an isolated system, evolve with time following the Schrödinger equation (in the Schrödinger picture; in the Heisenberg picture they are constant.

Density matrices define classical statistical mixtures of pure quantum states. These can arise in two ways:

1. When we have incomplete information about a pure quantum state and only know that it is in pure state $A$ with probability $p_A$, pure state B with probability $p_B$ and so on. This is the classic situation described by the Wigner's Friend thought experiment, which I describe more fully in my answer here;

2. You have a big population (ensemble) of identical systems in different, random, pure quantum states such that, if you choose a system at random, it has probability $p_A$ to be in state $A$, $p_B$ to be in pure state $B$ and so on.

A density matrix $\rho$ evolves following the Liouville-von Neumann equation:

$$i \hbar \frac{\partial \rho}{\partial t} = [\hat{H},\rho]\tag{1}$$

where $\hat{H}$ is the quantum system's Hamiltonian. When we make a measurement with an observable $\hat{\mathscr{O}}$ we find all the moments of the statistical distribution of the measurement (and, therefore, the probability density function for the measurement outcome, since this is define by the set of its moments) with the formula:

$$M_n={\rm tr}\left(\rho\,\hat{\mathscr{O}}^n\right)\tag{2}$$

Although it looks a bit obscure, it can be shown that the above formulas are equivalent to the following intuitively obvious procedure:

1. For each random state $\psi_j(0)$ in the population at time $t=0$, work out its evolution with time by the Schrödinger equation;

2. When we get to time $t$, we work out the moments $M_j(n)=\left<\psi_j(t)|\hat{\mathscr{O}}^n|\psi_j(t)\right>$ for each of the pure states $\psi_j$ present in the population;

3. We now combine the moments according to the laws of classical probability to find the overall moments of the measurement outcome statistics when we sample a system and measure it with $\hat{\mathscr{O}}$ at random. Our overall moments are $M_n=\sum\limits_jp_j\,M_j(n)$, where $p_j$ are the relative frequencies (probabilities) of the pure states in the mixture.

Instead of doing this messy procedure, we simply form the density matrix

$$\rho = \sum_j p_j \,\left|\left.\psi_j\right>\right. \left<\left.\psi_j\right|\right.\tag{3}$$

and use (1) and (2) to work out what our measurement statistics are.

So the density matrix is really simply our wonted quantum mechanics applied in a particular way. In particular, I do not like the name "matrix" - the density "matrix" is still state information and does not represent a tensor on, a linear functional of or linear transformation of the quantum state space or any of the other things that the word "matrix" more wontedly implies. It's written as a matrix just because, by (1) and (2), this is the form that we can get statistics out of most simply.

Lastly, note that given a set of quantum states $\psi_j$, the density matrix in (3) is in general highly non unique. That is, there are many different probability sets $p_j$ (co-efficients) in (3) that will lead to the same density matrix. So, in forming the density matrix, and thus "losing track of" which pure state $\psi_j$ applies to which particle in the mixture, we have lost information. But, since the density matrix procedures defined by (1), (2) and (3) are equivalent to calculating the full probability distributions of measurement outcomes, then the density matrix defines all that can be found out about the mixture by measurement. It defines all that we can ever know by experiment. The loss of pure state information thus results in an, in general, non zero von Neumann entropy:

$$S=-{\rm tr}\left(\rho\,\log\rho\right)$$

for our system, which is the broadening-to-quantum-mechanical-systems of the classical Gibbs entropy. Indeed if we choose $\psi_j$ that are orthonormal in the mixture (equivalently, if we diagonalise the density matrix, which is Hermitian positive definite and thus always diagonalisable), then the von Neumann entropy reduces to -$\sum\limits_j\,p_j\log p_j$, which should look very familiar to you.

Answer 2

Density matrix is NECESSARY when the quantum system is not in a PURE state, i.e. it cannot be described by a wave-function. Such a case is when a beam of freely moving electrons is not spin-polarized, i.e. there is no direction in space along which all the electrons have the same spin-projection.

(For finding whether the electrons are polarized, and on which axis, we have to divide the set of electrons in subsets, and for each subset we have to measure the spin projection on some direction in space, e.g. for one subset the x-projection, for another subset the y-projection, and for a third subset the z-projection). The analysis of the distribution of the results can show if the beam is polarized along some direction in space, or not. If the beam is non-polarized, that means that we have a mixture. For instance, part of the electrons are polarized along the direction Z with spin-up, and part with spin-down. In this case we cannot write a wave-function as

$$\lvert\Psi \rangle = C_1 \lvert\text{spin up}\rangle + C_2 \lvert\text{spin down}\rangle$$

We can write a density matrix with the diagonal elements equal to the probability of spin up, respectively of spin down, and the off-diagonal elements equal to zero.

Good luck !

Answer 3

The density matrix formalism follows naturally from a state vector formalism when we average over a part of variables. It is not a "more general", but a more common approach in practical applications. Sometimes it is the only possible formulation due to experimental restrictions.

Answer 4

As others have noted state vectors only describe pure states, whereas density matrices can describe mixed states as well. There are however two different density matrices and literature usually confounds both (it is also happening in this page).

At the one hand we have the statistical mechanics density matrix introduced by von Neumann via

$$ \rho = \sum_i P_i \,\left| \Psi_i \rangle \langle \Psi_i \right|$$

This is a quantum analog of the distribution function used in classical statistical mechanics. The system is in some pure state $\left| \Psi_i \rangle \right.$ with $P_i$ giving the probability. In this approach the probabilities are interpreted as measuring our ignorance over the real quantum state.

At the other hand we have the quantum mechanical density matrix introduced by Landau and Dirac via

$$ \rho = \sum_{\alpha\beta} w_{\alpha\beta} \,\left| \alpha \rangle \langle \beta\right|$$

This doesn't have counterpart on classical mechanics and is interpreted as the "more general description of the state of a quantum system". This is the density matrix we use to describe open quantum systems as for instance cats :-) The diagonal components $w_{\alpha\alpha}$ measure the weight of the base $\left| \alpha \rangle \right.$ in the mixed state, whereas the off-diagonal components $w_{\alpha\beta}$ with $\beta\neq\alpha$ measure the quantum correlations between states. In this approach the existence of correlations is the reason why the quantum state is not described by a state vector.