Retrospect

3.4Mixtures and the density matrix

Follows §3-8 of Peres. About 16 minutes. You need: Lesson 3.3.

Pure states are the best case: the most that can be known. Real sources are sloppier. A lamp, an oven, a polariser that is slightly off: none of them prepares a pure state. This last lesson builds the tool for describing any preparation at all, and finds that it forgets a surprising amount about where the light came from.

A recipe with a coin in it

Suppose a preparation is not fully pinned down. Each time, with probability p1p_1 it produces the pure state u1\mathsf u_1, with probability p2p_2 the state u2\mathsf u_2, and so on. The uα\mathsf u_\alpha can be any normalised vectors; they need not be orthogonal. Lesson 1.3 called this a mixture, and the coin-tossing source there was one.

What will a measurement of an observable AA give on average? In the pure state uα\mathsf u_\alpha the mean is ⟨uα, A uα⟩\ip{\mathsf u_\alpha}{A\,\mathsf u_\alpha}. Averaging again over the coin,

⟨A⟩  =  ∑αpα ⟨uα, A uα⟩.\langle A\rangle \;=\; \sum_\alpha p_\alpha\,\ip{\mathsf u_\alpha}{A\,\mathsf u_\alpha} .

As in the last lesson, the trick is to separate what belongs to the preparation from what belongs to the test. Everything about the preparation can be gathered into one matrix,

ρ  =  ∑αpα uαuα†,\rho \;=\; \sum_\alpha p_\alpha\,\mathsf u_\alpha\mathsf u_\alpha^\dagger ,

a weighted sum of the projectors on the ingredients. It is called the density matrix. In terms of it the mean takes a form that is the same for every observable:

⟨A⟩  =  Tr⁡(ρA).\langle A\rangle \;=\; \Tr(\rho A) .
Refresher: the trace of a matrix

The trace of a square matrix, written Tr⁡\Tr, is the sum of the entries on its diagonal. Two properties are used constantly. The trace of a sum is the sum of the traces. And inside a trace the order of a product can be switched: Tr⁡(AB)=Tr⁡(BA)\Tr(AB) = \Tr(BA), even when ABAB and BABA are different matrices.

A consequence: the trace does not depend on which axes the matrix is written along. It is a property of the matrix itself, and it equals the sum of the eigenvalues.

Three facts about ρ\rho follow from how it was built.

  • It is Hermitian, and Tr⁡ρ=1\Tr\rho = 1, because the probabilities pαp_\alpha add up to 1.
  • The probability of outcome μ\mu of any test is Tr⁡(ρPμ)\Tr(\rho P_\mu), where PμP_\mu is the projector on that outcome. So every probability that can be measured is computed from ρ\rho.
  • A pure state is the special case of a single ingredient, ρ=uu†\rho = \mathsf u\mathsf u^\dagger, a projector, for which ρ2=ρ\rho^2 = \rho. In general Tr⁡ρ2\Tr\rho^2 lies between 1/N1/N and 1 and equals 1 only for pure states. It is a convenient measure of how pure a preparation is.

Two examples, with components along the horizontal and vertical axes:

ρ=(120012)coin toss between H and V,ρ=(12121212)every photon diagonal.\underset{\text{coin toss between H and V}}{\rho = \begin{pmatrix} \tfrac12 & 0 \\ 0 & \tfrac12 \end{pmatrix}}, \qquad\qquad \underset{\text{every photon diagonal}}{\rho = \begin{pmatrix} \tfrac12 & \tfrac12 \\ \tfrac12 & \tfrac12 \end{pmatrix}} .

The diagonal entries are the same: both give half and half in the horizontal-against-vertical test. The difference is in the off-diagonal corners, which hold the relative phase between the two components. The superposition has it; the mixture does not. That is lesson 1.3 written as two matrices.

Different recipes, the same light

Now work out ρ\rho for a different recipe: toss a coin and make the photon right circular or left circular. The two projectors are

12(1−ii1)and12(1i−i1),and half of each isρ=(120012).\tfrac12\begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix} \quad\text{and}\quad \tfrac12\begin{pmatrix} 1 & i \\ -i & 1 \end{pmatrix}, \qquad\text{and half of each is}\qquad \rho = \begin{pmatrix} \tfrac12 & 0 \\ 0 & \tfrac12 \end{pmatrix}.

The same matrix as the coin toss between horizontal and vertical. Every probability of every test is Tr⁡(ρP)\Tr(\rho P), and the two recipes have the same ρ\rho. So there is no test, of any kind, that distinguishes them. Not a difficult test that nobody has thought of: none.

HVDARL12
recipe 1
ρHV
H0.500.00
V0.000.50
Tr ρ²0.500completely random
recipe 2
ρHV
H0.500.00
V0.000.50
Tr ρ²0.500completely random
chance of the + outcome (H, D or R) in each test
testrecipe 1recipe 2
H/V50.0%50.0%
±45°50.0%50.0%
circular50.0%50.0%

The same density matrix. No experiment on these photons can tell the two recipes apart.

start from
Recipe 150% H + 50% V
50%
50%
off
Recipe 250% R + 50% L
50%
50%
off
Figure 1. Two recipes, each a weighted blend of up to three pure states. Each has its density matrix and its point inside the sphere of lesson 3.2. Try the presets, then build two different recipes of your own that land on the same point.

This is quite unlike mixing in the ordinary sense. A mixture of chemicals has one composition, and analysis will find it. A quantum mixture that is not pure can be made up in infinitely many ways, and once made, the way is unrecoverable. Someone handed megajoules of the light could never discover whether the coin had chosen between horizontal and vertical or between right and left, even though the two machines that made it look nothing like each other.

A pure state is the exception. It can be “mixed” in only one way, from itself.

K. Completeness of the quantum description. The density matrix ρ\rho completely specifies all the properties of an ensemble of quantum systems.

Rule K is the last of the eleven. It says that ρ\rho is not a summary that leaves something out: there is nothing else to know. The book shows later (chapter 6) that if recipes with the same ρ\rho could be told apart, the ability could be used to send signals faster than light.

One loose end from lesson 2.1. The two boxes there, one with exactly equal numbers of horizontal and vertical photons and one with exactly equal numbers of right and left, could be told apart. There is no conflict. Counting photons into a box is not the same recipe as tossing a coin for each: the photons in a counted box are not independent of one another. Any one photon drawn from either box has ρ=121\rho = \tfrac12\mathbb 1, but the box as a whole is a larger system with a density matrix of its own, and those differ.

The ball of all states

For polarisation, any density matrix can be written with the help of the three Pauli matrices of the last lesson:

ρ  =  12(1+s1σ1+s2σ2+s3σ3),sj=Tr⁡(ρ σj).\rho \;=\; \tfrac12\left(\mathbb 1 + s_1\sigma_1 + s_2\sigma_2 + s_3\sigma_3\right), \qquad s_j = \Tr(\rho\,\sigma_j) .

The three numbers sjs_j are the means of the three standard tests with their outcomes labelled ±1\pm1: the same s1,s2,s3s_1, s_2, s_3 as in lesson 3.2. For a pure state they locate a point on the surface of the sphere. For a mixture they locate a point inside it. The nearer the centre, the less is known, and the centre itself, s=0s = 0, is the random mixture of rule C: ordinary unpolarised light.

Mixing two preparations with weights λ\lambda and 1−λ1 - \lambda puts the point on the straight line between theirs, at the matching distance along. That is all the figure above does. And it makes the loss of information visible: a point inside the ball lies on infinitely many chords, and any of them could have been the recipe.

It also tells you how to find an unknown ρ\rho: measure the three means, on three batches of photons. Three numbers and you are done. In general a system with NN outcomes needs N+1N + 1 different tests. And there is a built-in check on the experiment: the eigenvalues of ρ\rho are 12(1±∣s∣)\tfrac12(1 \pm |s|), and probabilities cannot be negative. If the three measured numbers give a point outside the sphere, the fault is in the apparatus.

12?test

Eight rounds. Each gives two recipes for making photons: can any experiment tell their light apart?

Figure 2. Two recipes, described in words. Can any experiment tell the light of one from the light of the other? Work out each density matrix, or each point in the ball, before you answer.

Where you have got to

That completes the first three chapters of the book, which Peres calls “gathering the tools”. Look back over what the tools are.

Part I found that outcomes of experiments on single photons and atoms are random, that a superposition is a different thing from a mixture, that a single system cannot be made to reveal its state, and that a measurement does not read off a value that was already there. Part II turned those findings into rules about tests, and found that when paths cannot be told apart it is amplitudes, not probabilities, that add. Part III showed that amplitudes are the components of a vector, that every probability is the squared size of an inner product, that a measurement is a Hermitian matrix, and that any preparation whatever is a density matrix.

The contents page lists the eleven rules, A to K, in one place. You should now be able to say what each one means and which experiment it came from.

What comes next in the book: systems whose tests have infinitely many outcomes, such as the position of a particle, where the uncertainty relation between position and momentum is finally derived properly (chapter 4); systems made of two or more parts, where the really strange consequences live (chapters 5 to 7, including Bell’s theorem); how states change in time (chapter 8); and at the end, a hard look at the measuring apparatus itself (chapter 12).

This lesson follows §3-8. The appendix that closes the chapter, §3-9, is on Dirac’s notation, which the refresher in lesson 3.1 introduced.

What to keep

  • A preparation that yields the pure states uα\mathsf u_\alpha with probabilities pαp_\alpha is described by the density matrix ρ=∑αpα uαuα†\rho = \sum_\alpha p_\alpha\,\mathsf u_\alpha\mathsf u_\alpha^\dagger.
  • Every mean is Tr⁡(ρA)\Tr(\rho A) and every probability is Tr⁡(ρP)\Tr(\rho P). Tr⁡ρ=1\Tr\rho = 1, and Tr⁡ρ2=1\Tr\rho^2 = 1 only for a pure state.
  • Different recipes can give the same ρ\rho, and then no test can tell them apart. Rule K: ρ\rho is the complete description.
  • For polarisation ρ=12(1+s1σ1+s2σ2+s3σ3)\rho = \tfrac12(\mathbb 1 + s_1\sigma_1 + s_2\sigma_2 + s_3\sigma_3): pure states on the sphere, mixtures inside, unpolarised light at the centre.
  • Three measured means determine ρ\rho for polarisation; a point outside the sphere means an error.

Check yourself

7 questions. Each can be answered once, and the reasoning is shown either way. Work them out before you click: guessing teaches nothing.

1A source tosses a coin for each photon and makes it horizontal or vertical. Using ρ=121\rho = \tfrac12\mathbb 1, what is the probability that a photon from it passes a test for diagonal polarisation?

2What is Tr⁡ρ2\Tr\rho^2 for that source?

3A preparation has ρ=(0.75000.25)\rho = \begin{pmatrix} 0.75 & 0 \\ 0 & 0.25 \end{pmatrix} along the horizontal and vertical axes. Is it a pure state?

4One machine tosses a coin between horizontal and vertical. Another tosses a coin between right and left circular. A physicist is given a very large number of photons from one of them. How can she find out which?

5The three means for a beam are measured as s=(0.6, 0.6, 0.6)s = (0.6,\ 0.6,\ 0.6). What should the experimenter conclude?

6How many different tests are needed to determine the density matrix of a system with three outcomes, given unlimited copies?

7Where in the ball does a preparation sit if nothing at all is known about its polarisation?