Density matrix

The formalism of density operators and matrices was introduced by John von Neumann[6] in 1927 and independently, but less systematically by Lev Landau[7][8] and Felix Bloch[9] in 1927 and 1946 respectively.

In quantum mechanics, the state of a quantum system is represented by a state vector, denoted ${\displaystyle |\psi \rangle }$ (and pronounced ket). A quantum system with a state vector ${\displaystyle |\psi \rangle }$ is called a pure state. However, it is also possible for a system to be in a statistical ensemble of different state vectors: For example, there may be a 50% probability that the state vector is ${\displaystyle |\psi _{1}\rangle }$ and a 50% chance that the state vector is ${\displaystyle |\psi _{2}\rangle }$. This system would be in a mixed state. The density matrix is especially useful for mixed states, because any state, pure or mixed, can be characterized by a single density matrix.[10]^:102

A mixed state is different from a quantum superposition. The probabilities in a mixed state are classical probabilities (as in the probabilities one learns in classic probability theory / statistics), unlike the quantum probabilities in a quantum superposition. In fact, a quantum superposition of pure states is another pure state, for example ${\displaystyle |\psi \rangle =(|\psi _{1}\rangle +|\psi _{2}\rangle )/{\sqrt {2}}}$. In this case, the coefficients ${\displaystyle 1/{\sqrt {2}}}$ are not probabilities, but rather probability amplitudes.[10]^:81

Example: light polarization

The incandescent light bulb (1) emits completely random polarized photons (2) with mixed state density matrix
${\displaystyle {\begin{bmatrix}0.5&0\\0&0.5\\\end{bmatrix}}}$ .
After passing through vertical plane polarizer (3), the remaining photons are all vertically polarized (4) and have pure state density matrix
${\displaystyle {\begin{bmatrix}1&0\\0&0\\\end{bmatrix}}}$ .

An example of pure and mixed states is light polarization. Photons can have two helicities, corresponding to two orthogonal quantum states, ${\displaystyle |R\rangle }$ (right circular polarization) and ${\displaystyle |L\rangle }$ (left circular polarization). A photon can also be in a superposition state, such as ${\displaystyle (|R\rangle +|L\rangle )/{\sqrt {2}}}$ (vertical polarization) or ${\displaystyle (|R\rangle -|L\rangle )/{\sqrt {2}}}$ (horizontal polarization). More generally, it can be in any state ${\displaystyle \alpha |R\rangle +\beta |L\rangle }$ (with ${\displaystyle |\alpha |^{2}+|\beta |^{2}=1}$), corresponding to linear, circular, or elliptical polarization. If we pass ${\displaystyle (|R\rangle +|L\rangle )/{\sqrt {2}}}$ polarized light through a circular polarizer which allows either only ${\displaystyle |R\rangle }$ polarized light, or only ${\displaystyle |L\rangle }$ polarized light, intensity would be reduced by half in both cases. This may make it seem like half of the photons are in state ${\displaystyle |R\rangle }$ and the other half in state ${\displaystyle |L\rangle }$. But this is not correct: Both ${\displaystyle |R\rangle }$ and ${\displaystyle |L\rangle }$ photons are partly absorbed by a vertical linear polarizer, but the ${\displaystyle (|R\rangle +|L\rangle )/{\sqrt {2}}}$ light will pass through that polarizer with no absorption whatsoever.

However, unpolarized light (such as the light from an incandescent light bulb) is different from any state like ${\displaystyle \alpha |R\rangle +\beta |L\rangle }$ (linear, circular, or elliptical polarization). Unlike linearly or elliptically polarized light, it passes through a polarizer with 50% intensity loss whatever the orientation of the polarizer; and unlike circularly polarized light, it cannot be made linearly polarized with any wave plate because randomly oriented polarization will emerge from a wave plate with random orientation. Indeed, unpolarized light cannot be described as any state of the form ${\displaystyle \alpha |R\rangle +\beta |L\rangle }$ in a definite sense. However, unpolarized light can be described with ensemble averages, e.g. that each photon is either ${\displaystyle |R\rangle }$ with 50% probability or ${\displaystyle |L\rangle }$ with 50% probability. The same behavior would occur if each photon was either vertically polarized with 50% probability or horizontally polarized with 50% probability (since a vertically or horizontally polarized state can be expressed as a linear combination of ${\displaystyle |R\rangle }$ and ${\displaystyle |L\rangle }$ states).

Therefore, unpolarized light cannot be described by any pure state, but can be described as a statistical ensemble of pure states in at least two ways (the ensemble of half left and half right circularly polarized, or the ensemble of half vertically and half horizontally linearly polarized). These two ensembles are completely indistinguishable experimentally, and therefore they are considered the same mixed state. One of the advantages of the density matrix is that there is just one density matrix for each mixed state, whereas there are many statistical ensembles of pure states for each mixed state. Nevertheless, the density matrix contains all the information necessary to calculate any measurable property of the mixed state.

Where do mixed states come from? To answer that, consider how to generate unpolarized light. One way is to use a system in thermal equilibrium, a statistical mixture of enormous numbers of microstates, each with a certain probability (the Boltzmann factor), switching rapidly from one to the next due to thermal fluctuations. Thermal randomness explains why an incandescent light bulb, for example, emits unpolarized light. A second way to generate unpolarized light is to introduce uncertainty in the preparation of the system, for example, passing it through a birefringent crystal with a rough surface, so that slightly different parts of the beam acquire different polarizations. A third way to generate unpolarized light uses an EPR setup: A radioactive decay can emit two photons traveling in opposite directions, in the quantum state ${\displaystyle (|R,L\rangle +|L,R\rangle )/{\sqrt {2}}}$. The two photons together are in a pure state, but if you only look at one of the photons and ignore the other, the photon behaves just like unpolarized light.

More generally, mixed states commonly arise from a statistical mixture of the starting state (such as in thermal equilibrium), from uncertainty in the preparation procedure (such as slightly different paths that a photon can travel), or from looking at a subsystem entangled with something else.[10]^:101–106

For a finite-dimensional function space, the most general density operator is of the form

${\displaystyle \rho =\sum _{j}p_{j}|\psi _{j}\rangle \langle \psi _{j}|,}$

where the coefficients ${\displaystyle p_{j}}$ are non-negative and add up to one, and ${\displaystyle \textstyle |\psi _{j}\rangle \langle \psi _{j}|}$ is an outer product written in bra-ket notation. This represents a mixed state, with probability ${\displaystyle p_{j}}$ that the system is in the pure state ${\displaystyle \textstyle |\psi _{j}\rangle }$.

For the above example of unpolarized light, the density operator equals

${\displaystyle \rho ={\tfrac {1}{2}}|R\rangle \langle R|+{\tfrac {1}{2}}|L\rangle \langle L|,}$

where ${\displaystyle \textstyle |L\rangle }$ is the left-circularly-polarized photon state and ${\displaystyle \textstyle |R\rangle }$ is the right-circularly-polarized photon state.

Let ${\displaystyle A}$ be an observable of the system, and suppose the ensemble is in a mixed state such that each of the pure states ${\displaystyle \textstyle |\psi _{j}\rangle }$ occurs with probability ${\displaystyle p_{j}}$. Then the corresponding density operator equals

${\displaystyle \rho =\sum _{j}p_{j}|\psi _{j}\rangle \langle \psi _{j}|.}$

The expectation value of the measurement can be calculated by extending from the case of pure states (see Measurement in quantum mechanics):

${\displaystyle \langle A\rangle =\sum _{j}p_{j}\langle \psi _{j}|A|\psi _{j}\rangle =\sum _{j}p_{j}\operatorname {tr} \left(|\psi _{j}\rangle \langle \psi _{j}|A\right)=\sum _{j}\operatorname {tr} \left(p_{j}|\psi _{j}\rangle \langle \psi _{j}|A\right)=\operatorname {tr} \left(\sum _{j}p_{j}|\psi _{j}\rangle \langle \psi _{j}|A\right)=\operatorname {tr} (\rho A),}$

where ${\displaystyle \operatorname {tr} }$ denotes trace. Thus, the familiar expression ${\displaystyle \langle A\rangle =\langle \psi |A|\psi \rangle }$ for pure states is replaced by

${\displaystyle \langle A\rangle =\operatorname {tr} (\rho A)}$

for mixed states.

Moreover, if ${\displaystyle A}$ has spectral resolution

${\displaystyle A=\sum _{i}a_{i}|a_{i}\rangle \langle a_{i}|=\sum _{i}a_{i}P_{i},}$

where ${\displaystyle P_{i}=|a_{i}\rangle \langle a_{i}|}$, the corresponding density operator after the measurement is given by

${\displaystyle \;\rho '=\sum _{i}P_{i}\rho P_{i}.}$

Note that the above density operator describes the full ensemble after measurement. The sub-ensemble for which the measurement result was the particular value ${\displaystyle a_{i}}$ is described by the different density operator

${\displaystyle \rho _{i}'={\frac {P_{i}\rho P_{i}}{\operatorname {tr} [\rho P_{i}]}}.}$

This is true assuming that ${\displaystyle \textstyle |a_{i}\rangle }$ is the only eigenket (up to phase) with eigenvalue ${\displaystyle a_{i}}$; more generally, ${\displaystyle P_{i}}$ in this expression would be replaced by the projection operator into the eigenspace corresponding to eigenvalue ${\displaystyle a_{i}}$.

More generally, suppose ${\displaystyle \Phi }$ is a function that associates to each observable ${\displaystyle A}$ a number ${\displaystyle \Phi (A)}$, which we may think of as the "expectation value" of ${\displaystyle A}$. If ${\displaystyle \Phi }$ satisfies some natural properties (such as giving positive values on positive operators), then there is a unique density matrix ${\displaystyle \rho }$ such that

${\displaystyle \Phi (A)=\operatorname {tr} (\rho A)}$

for all ${\displaystyle A}$.[1] That is to say, any reasonable "family of expectation values" can be represented by a density matrix. This observation suggests that density matrices are the most general notion of a quantum state.

The von Neumann entropy ${\displaystyle S}$ of a mixture can be expressed in terms of the eigenvalues of ${\displaystyle \rho }$ or in terms of the trace and logarithm of the density operator ${\displaystyle \rho }$. Since ${\displaystyle \rho }$ is a positive semi-definite operator, it has a spectral decomposition such that ${\displaystyle \rho =\textstyle \sum _{i}\lambda _{i}|\varphi _{i}\rangle \langle \varphi _{i}|}$, where ${\displaystyle |\varphi _{i}\rangle }$ are orthonormal vectors, ${\displaystyle \lambda _{i}>0}$, and ${\displaystyle \textstyle \sum \lambda _{i}=1}$. Then the entropy of a quantum system with density matrix ${\displaystyle \rho }$ is

${\displaystyle S=-\sum _{i}\lambda _{i}\ln \lambda _{i}=-\operatorname {tr} (\rho \ln \rho ).}$

This definition implies that the von Neumann entropy of any pure state is zero.[15]^:217 If ${\displaystyle \rho _{i}}$ are states that have support on orthogonal subspaces, then the von Neumann entropy of a convex combination of these states,

${\displaystyle \rho =\sum _{i}p_{i}\rho _{i},}$

is given by the von Neumann entropies of the states ${\displaystyle \rho _{i}}$ and the Shannon entropy of the probability distribution ${\displaystyle p_{i}}$:

${\displaystyle S(\rho )=H(p_{i})+\sum _{i}p_{i}S(\rho _{i}).}$

When the states ${\displaystyle \rho _{i}}$ do not have orthogonal supports, the sum on the right-hand side is strictly greater than the von Neumann entropy of the convex combination ${\displaystyle \rho }$.[10]^:518

Given a density operator ${\displaystyle \rho }$ and a projective measurement as in the previous section, the state ${\displaystyle \rho '}$ defined by the convex combination

${\displaystyle \rho '=\sum _{i}P_{i}\rho P_{i},}$

which can be interpreted as the state produced by performing the measurement but not recording which outcome occurred,[16]^:159 has a von Neumann entropy larger than that of ${\displaystyle \rho }$, except if ${\displaystyle \rho =\rho '}$. It is however possible for the ${\displaystyle \rho '}$ produced by a generalized measurement, or POVM, to have a lower von Neumann entropy than ${\displaystyle \rho }$.[10]^:514

Another motivation for considering density matrices comes from consideration of systems and their subsystems. Suppose we have two quantum systems, described by Hilbert spaces ${\displaystyle {\mathcal {H}}_{1}}$ and ${\displaystyle {\mathcal {H}}_{2}}$. The composite system is then the tensor product ${\displaystyle {\mathcal {H}}_{1}\otimes {\mathcal {H}}_{2}}$ of the two Hilbert spaces. Suppose now that the composite system is in a pure state ${\displaystyle \psi \in {\mathcal {H}}_{1}\otimes {\mathcal {H}}_{2}}$. If ${\displaystyle \psi }$ happens to have the special form ${\displaystyle \psi =\psi _{1}\otimes \psi _{2}}$, then we may reasonably say that the state of the first subsystem is ${\displaystyle \psi _{1}}$. In this case, we say that the two systems are not entangled. In general, however, ${\displaystyle \psi }$ will not decompose as a single tensor product of vectors in ${\displaystyle {\mathcal {H}}_{1}}$ and ${\displaystyle {\mathcal {H}}_{2}}$. If ${\displaystyle \psi }$ cannot be decomposed as a single tensor product of states in the component systems, we say that the two systems are entangled. In that case, there is no reasonable way to associate a pure state ${\displaystyle \psi _{1}\in {\mathcal {H}}_{1}}$ to the state ${\displaystyle \psi \in {\mathcal {H}}_{1}\otimes {\mathcal {H}}_{2}}$.[1]

If, for example, we have a wave function ${\displaystyle \psi (x_{1},x_{2})}$ describing the state of two particles, there is no natural way to construct a wave function (i.e., pure state) ${\displaystyle \psi _{1}(x_{1})}$ that describes the states of the first particle—unless ${\displaystyle \psi (x_{1},x_{2})}$ happens to be a product of a function ${\displaystyle \psi _{1}(x_{1})}$ and a function ${\displaystyle \psi _{2}(x_{2})}$.

The upshot of the preceding discussion is that even if the total system is in a pure state, the various subsystems that make it up will typically be in mixed states. Thus, the use of density matrices is unavoidable.

On the other hand, whether the composite system is in a pure state or a mixed state, we can perfectly well construct a density matrix that describes the state of ${\displaystyle {\mathcal {H}}_{1}}$. Denote the density matrix of the composite system of two systems by ${\displaystyle \rho }$. Then the state of, say, ${\displaystyle {\mathcal {H}}_{1}}$, is described by a reduced density operator, given by taking the "partial trace" of ${\displaystyle \rho }$ over ${\displaystyle {\mathcal {H}}_{2}}$.[1]

If the state of ${\displaystyle {\mathcal {H}}_{1}\otimes {\mathcal {H}}_{2}}$ happens to be a density matrix of the special form ${\displaystyle \rho =\rho _{1}\otimes \rho _{2}}$ where ${\displaystyle \rho _{1}}$ and ${\displaystyle \rho _{2}}$ are density matrices on ${\displaystyle {\mathcal {H}}_{1}}$ and ${\displaystyle {\mathcal {H}}_{2}}$, then the partial trace of ${\displaystyle \rho }$ with respect to ${\displaystyle {\mathcal {H}}_{2}}$ is just ${\displaystyle \rho _{1}}$. A typical ${\displaystyle \rho }$ will not be of this form, however.

Just as the Schrödinger equation describes how pure states evolve in time, the von Neumann equation (also known as the Liouville–von Neumann equation) describes how a density operator evolves in time (in fact, the two equations are equivalent, in the sense that either can be derived from the other.) The von Neumann equation dictates that[17][18]

${\displaystyle i\hbar {\frac {\partial \rho }{\partial t}}=[H,\rho ]~,}$

where the brackets denote a commutator.

Note that this equation only holds when the density operator is taken to be in the Schrödinger picture, even though this equation seems at first look to emulate the Heisenberg equation of motion in the Heisenberg picture, with a crucial sign difference:

${\displaystyle i\hbar {\frac {dA^{(H)}}{dt}}=-[H,A^{(H)}]~,}$

where ${\displaystyle A^{(H)}(t)}$ is some Heisenberg picture operator; but in this picture the density matrix is not time-dependent, and the relative sign ensures that the time derivative of the expected value ${\displaystyle \langle A\rangle }$ comes out the same as in the Schrödinger picture.[1]

If the Hamiltonian is time-independent, the von Neumann equation can be easily solved to yield

${\displaystyle \rho (t)=e^{-iHt/\hbar }\rho (0)e^{iHt/\hbar }.}$

For a more general Hamiltonian, if ${\displaystyle G(t)}$ is the wavefunction propagator over some interval, then the time evolution of the density matrix over that same interval is given by

${\displaystyle \rho (t)=G(t)\rho (0)G(t)^{\dagger }.}$

The density matrix operator may also be realized in phase space. Under the Wigner map, the density matrix transforms into the equivalent Wigner function,

${\displaystyle W(x,p){\stackrel {\mathrm {def} }{=}}{\frac {1}{\pi \hbar }}\int _{-\infty }^{\infty }\psi ^{*}(x+y)\psi (x-y)e^{2ipy/\hbar }\,dy.}$

The equation for the time evolution of the Wigner function is then the Wigner-transform of the above von Neumann equation,

${\displaystyle {\frac {\partial W(q,p,t)}{\partial t}}=-\{\{W(q,p,t),H(q,p)\}\},}$

where ${\displaystyle H(q,p)}$ is the Hamiltonian, and ${\displaystyle \{\{\cdot ,\cdot \}\}}$ is the Moyal bracket, the transform of the quantum commutator.

The evolution equation for the Wigner function is then analogous to that of its classical limit, the Liouville equation of classical physics. In the limit of vanishing Planck's constant ${\displaystyle \hbar }$ , ${\displaystyle W(q,p,t)}$ reduces to the classical Liouville probability density function in phase space.

The classical Liouville equation can be solved using the method of characteristics for partial differential equations, the characteristic equations being Hamilton's equations. The Moyal equation in quantum mechanics similarly admits formal solutions in terms of quantum characteristics, predicated on the ∗−product of phase space, although, in actual practice, solution-seeking follows different methods.

Density matrices are a basic tool of quantum mechanics, and appear at least occasionally in almost any type of quantum-mechanical calculation. Some specific examples where density matrices are especially helpful and common are as follows:

• Quantum decoherence theory typically involves non-isolated quantum systems developing entanglement with other systems, including measurement apparatuses. Density matrices make it much easier to describe the process and calculate its consequences. Quantum decoherence explains why a system interacting with an environment transitions from being a pure state, exhibiting superpositions, to a mixed state, an incoherent combination of classical alternatives. This transition is fundamentally reversible, as the combined state of system and environment is still pure, but for all practical purposes irreversible, as the environment is a very large and complex quantum system, and it is not feasible to reverse their interaction. Decoherence is thus very important for explaining the classical limit of quantum mechanics, but cannot explain wave function collapse, as all classical alternatives are still present in the mixed state, and wave function collapse selects only one of them.[19]
• Similarly, in quantum computation, quantum information theory, and other fields where state preparation is noisy and decoherence can occur, density matrices are frequently used. Quantum tomography is a process by which, given a set of data representing the results of quantum measurements, a density matrix consistent with those measurement results is computed.[20][21]
• When analyzing a system with many electrons, such as an atom or molecule, an imperfect but useful first approximation is to treat the electrons as uncorrelated or each having an independent single-particle wavefunction. This is the usual starting point when building the Slater determinant in the Hartree–Fock method. If there are ${\displaystyle N}$ electrons filling the ${\displaystyle N}$ single-particle wavefunctions ${\displaystyle |\psi _{i}\rangle }$, then the collection of ${\displaystyle N}$ electrons together can be characterized by a density matrix ${\displaystyle \sum _{i=1}^{N}|\psi _{i}\rangle \langle \psi _{i}|}$.

It is now generally accepted that the description of quantum mechanics in which all self-adjoint operators represent observables is untenable.[22][23] For this reason, observables are identified with elements of an abstract C*-algebra A (that is one without a distinguished representation as an algebra of operators) and states are positive linear functionals on A. However, by using the GNS construction, we can recover Hilbert spaces which realize A as a subalgebra of operators.

Geometrically, a pure state on a C*-algebra A is a state which is an extreme point of the set of all states on A. By properties of the GNS construction these states correspond to irreducible representations of A.

The states of the C*-algebra of compact operators K(H) correspond exactly to the density operators, and therefore the pure states of K(H) are exactly the pure states in the sense of quantum mechanics.

The C*-algebraic formulation can be seen to include both classical and quantum systems. When the system is classical, the algebra of observables become an abelian C*-algebra. In that case the states become probability measures, as noted in the introduction.