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Physics LibreTexts

4.1: Mixed States

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First, we recall some properties of the trace:

  • Tr(aA)=aTr(A),
  • Tr(A+B)=Tr(A)+Tr(B).

Also remember that we can write the expectation value of A as

A=Tr(|ψψ|A),

where |ψ is the state of the system. It tells us everything there is to know about the system. But what if we don’t know everything?

As an example, consider that Alice prepares a qubit in the state |0 or in the state |+=(|0+|1)/2 depending on the outcome of a balanced (50:50) coin toss. How does Bob describe the state before any measurement? First, we cannot say that the state is 12|0+12|+, because this is not normalized!

The key to the solution is to observe that the expectation values must behave correctly. The expectation value A is the average of the eigenvalues of A for a given state. If the state is itself a statistical mixture (as in the example above), then the expectation values must also be averaged. So for the example, we require that for any A

A=12A0+12A+=12Tr(|00|A)+12Tr(|++|A)=Tr[(12|00|+12|++|)A]Tr(ρA),

where we defined

ρ=12|00|+12|++|.

The statistical mixture is therefore properly described by an operator, rather than a simple vector. We can generalize this as

ρ=kpk|ψkψk|,

where the pk are probabilities that sum up to one (kpk=1) and the |ψk are normalized states (not necessarily complete or orthogonal). Since ρ acts as a weight, or a density, in the expectation value, we call it the density operator. We can diagonalize ρ to find the spectral decomposition

ρ=jλj|λjλj|,

where {|λj}j forms a complete orthonormal basis, 0λj1, and jλj=1. We can also show that ρ is a positive operator:

ψ|ρ|ψ=jkcjckλj|ρ|λk=jkcjckλj|lλl|λlλlλk=jklcjckλlλjλlλlλk=jklcjckλlδjlδlk=lλl|cl|20

In general, an operator ρ is a valid density operator if and only if it has the following three properties:

  1. ρ=ρ,
  2. Tr(ρ)=1,
  3. ρ0.

The density operator is a generalization of the state of a quantum system when we have incomplete information. In the special case where one of the pj=1 and the others are zero, the density operator becomes the projector |ψjψj|. In other words, it is completely determined by the state vector |ψj. We call these pure states. The statistical mixture of pure states giving rise to the density operator is called a mixed state.

The unitary evolution of the density operator can be derived directly from the Schrödinger equation it|ψ=H|ψ:

idρdt=iddtjpj|ψjψj|=ij{dpjdt|ψjψj|+pj[(ddt|ψj)ψj|+|ψj(ddtψj|)]}=iρt+HρρH=[H,ρ]+iρt

This agrees with the Heisenberg equation for operators, and it is sometimes known as the Von Neumann equation. In most problems the probabilities pj have no explicit time-dependence, and tρ=0.


This page titled 4.1: Mixed States is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pieter Kok via source content that was edited to the style and standards of the LibreTexts platform.

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