Retrospect

3.3Observables

Follows §3-6, §3-7 of Peres. About 17 minutes. You need: Lesson 3.2. Matrices are introduced here from scratch.

So far the outcomes of a test have had names: plus and minus, up and down, horizontal and vertical. Physicists like numbers. This lesson writes a number on each outcome, and finds that the test and its numbers together are captured by a single mathematical object, a matrix, whose properties answer the questions an experimenter would ask.

Numbers on outcomes

Take a complete test and attach a real number to each of its NN outcomes: a1,a2,…,aNa_1, a_2, \dots, a_N. Sometimes the numbers suggest themselves. A Stern–Gerlach magnet deflects atoms by amounts that correspond to +μ+\mu, 00 and −μ-\mu of magnetic moment, so those are the natural labels. Sometimes they are arbitrary: for the two exits of a polarising crystal we might as well write +1+1 and −1-1.

Once that is done, each run of the test yields a number. We then say, loosely, that we have measured something, and call the something an observable, AA. Remember from lesson 1.5 what this does and does not mean. There was no value of AA sitting in the system beforehand. A quantum measurement is nothing more than a test whose outcomes carry numerical labels, and the only values AA can ever show are the labels a1,…,aNa_1, \dots, a_N.

Do the measurement many times on systems all prepared in the state v\mathsf v. If er\mathsf e_r is the vector for the rrth outcome, rule H says that outcome occurs with probability ∣⟨er, v⟩∣2|\ip{\mathsf e_r}{\mathsf v}|^2. The average of the numbers obtained, called the mean or expectation value and written ⟨A⟩\langle A\rangle, is each label weighted by how often it turns up:

⟨A⟩  =  ∑rar ∣⟨er, v⟩∣2.\langle A\rangle \;=\; \sum_r a_r\,|\ip{\mathsf e_r}{\mathsf v}|^2 .
a₂a₁1½−2−1012
A in the x, y basis
xy
x1.000.00
y0.00−1.00
probabilities
p₁75%
p₂25%
mean ⟨A⟩0.50a₁p₁ + a₂p₂
spread ΔA0.87|a₁ − a₂| √(p₁p₂)
sample mean–no photons yet
test axis β0°
1.0
−1.0
state θ30°
Figure 1. A polarisation test with a number written on each exit. Turn the test’s axes, change the numbers, turn the state, and measure. Each photon gives one of the two numbers and never anything between; the mean is where the average of many settles. Find a state for which the spread is zero.

The same thing as a matrix

That formula for the mean mixes together two quite separate things: the state v\mathsf v, which belongs to the preparation, and the axes er\mathsf e_r with their labels ara_r, which belong to the test. A little algebra pulls them apart. Write out the inner product in components, ⟨er, v⟩=∑n(er)n‾ vn\ip{\mathsf e_r}{\mathsf v} = \sum_n \overline{(e_r)_n}\,v_n, substitute, and collect everything that does not involve v\mathsf v:

⟨A⟩  =  ∑m,nvˉm Amn vn,whereAmn  =  ∑rar (er)m (er)n‾.\langle A\rangle \;=\; \sum_{m,n} \bar v_m\,A_{mn}\,v_n, \qquad\text{where}\qquad A_{mn} \;=\; \sum_r a_r\,(e_r)_m\,\overline{(e_r)_n} .

The numbers AmnA_{mn}, arranged in a square with mm labelling rows and nn columns, form a matrix. It contains everything about the test and its labels, and nothing about the state. The mean is then written compactly as

⟨A⟩  =  ⟨v, Av⟩.\langle A\rangle \;=\; \ip{\mathsf v}{A\mathsf v} .
Refresher: what a matrix does to a vector

A matrix is a square table of numbers. It acts on a vector to give another vector: the mmth component of the result is row mm of the table multiplied, entry by entry, into the vector and summed. For two components,

(abcd)(xy)=(ax+bycx+dy).\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix}.

Two matrices are multiplied by letting them act one after the other: ABAB means “apply BB, then AA”. The order matters. In general ABAB and BABA are different matrices, and this fact, which looks like a nuisance, turns out to be the mathematical heart of the subject.

Because the labels ara_r are real, the matrix has a special symmetry: flipping it across its diagonal and conjugating every entry gives it back, Amn=Anm‾A_{mn} = \overline{A_{nm}}. Such a matrix is called Hermitian. Every observable is represented by a Hermitian matrix. The converse is true as well, as we shall see: every Hermitian matrix represents an observable.

Three examples worth knowing by heart

Label the exits of the three standard polarisation tests +1+1 and −1-1. Working out AmnA_{mn} from the formula, with components taken along the horizontal and vertical axes, gives

σ1=(100−1)horizontal / vertical,σ2=(0110)diagonal / anti-diagonal,σ3=(0−ii0)right / left circular.\underset{\text{horizontal / vertical}}{\sigma_1 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}}, \qquad \underset{\text{diagonal / anti-diagonal}}{\sigma_2 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}}, \qquad \underset{\text{right / left circular}}{\sigma_3 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}} .

These three are known as the Pauli matrices. Their means in a state are the three numbers s1s_1, s2s_2, s3s_3 of the last lesson: the coordinates of the state on the sphere. (They are numbered here in the order of the tests. Most books call the same three σz\sigma_z, σx\sigma_x and σy\sigma_y, after the directions of a magnet rather than the settings of a crystal.)

Yes-or-no questions

The simplest observables are those whose labels are all 0 or 1: yes or no. For a single state v\mathsf v, the question “does the system pass the test for v\mathsf v?” is represented by the matrix

Pv=vv†,which acts asPv u=v ⟨v, u⟩.P_{\mathsf v} = \mathsf v\mathsf v^\dagger, \qquad\text{which acts as}\qquad P_{\mathsf v}\,\mathsf u = \mathsf v\,\ip{\mathsf v}{\mathsf u} .

It takes any vector u\mathsf u and returns its shadow along v\mathsf v, which is why it is called a projector. Casting a shadow twice is the same as casting it once, so P2=PP^2 = P. And any observable is a sum of projectors on its outcomes, each weighted by its label:

A=∑rar Pr.A = \sum_r a_r\,P_r .

Eigenvectors: the states with a sure answer

Apply the matrix AA to one of the vectors that define it. Using the formula above, PrP_r leaves er\mathsf e_r alone and every other projector kills it, so

A er=ar er.A\,\mathsf e_r = a_r\,\mathsf e_r .

The matrix does not turn this vector at all; it only stretches it, by the factor ara_r. A vector that a matrix merely stretches is called an eigenvector of the matrix, and the stretch factor is the corresponding eigenvalue. (Eigen is German for “own”: these are the matrix’s own special directions.)

So the dictionary between the laboratory and the algebra reads:

  • the possible results of a measurement are the eigenvalues of its matrix;
  • the states for which the result is certain are its eigenvectors. Prepare the system in er\mathsf e_r and the measurement gives ara_r every time;
  • conversely, if a state gives the same result every time, it is an eigenvector. The spread of the results about their mean, ΔA\Delta A, defined by (ΔA)2=⟨A2⟩−⟨A⟩2(\Delta A)^2 = \langle A^2\rangle - \langle A\rangle^2, is zero exactly for eigenvectors.

Now run the dictionary backwards. Suppose someone hands you a Hermitian matrix, with no test attached. It is a theorem of algebra that its eigenvalues are all real, and that its eigenvectors can be chosen to form an orthonormal basis. By rule G* that basis is a test that can be carried out, and the eigenvalues are real numbers to write on its outcomes. So the matrix defines an observable after all. To find out what a measurement of it can give, you find its eigenvalues.

For a two-by-two matrix this is a short calculation. The eigenvalues aa are the numbers for which A−a⋅1A - a\cdot\mathbb 1 squashes some vector to zero, and that happens when its determinant vanishes. For σ2\sigma_2:

det⁡(−a11−a)=a2−1=0⟹a=±1,with eigenvectors 12(11) and 12(1−1).\det\begin{pmatrix} -a & 1 \\ 1 & -a \end{pmatrix} = a^2 - 1 = 0 \quad\Longrightarrow\quad a = \pm1, \qquad\text{with eigenvectors } \tfrac{1}{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix} \text{ and } \tfrac{1}{\sqrt2}\begin{pmatrix}1\\-1\end{pmatrix}.

The results are ±1\pm1, and the sure states are diagonal and anti-diagonal, as they should be.

Tests that do not share axes

Two observables defined on the same test, with different labels on its outcomes, are both diagonal matrices when written along that test’s axes, and diagonal matrices can be multiplied in either order: AB=BAAB = BA. They are said to commute. The converse holds too. If two Hermitian matrices commute, there is a single set of axes that are eigenvectors of both, and so a single test that measures both at once.

If they do not commute, there is no such test. Try σ1\sigma_1 and σ2\sigma_2:

σ1σ2=(01−10),σ2σ1=(0−110)=− σ1σ2.\sigma_1\sigma_2 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \qquad \sigma_2\sigma_1 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = -\,\sigma_1\sigma_2 .

The difference AB−BAAB - BA is called the commutator and written [A,B][A, B]. It is not zero here, and that is the algebraic face of the trade-off in the last lesson: the horizontal-against-vertical test and the diagonal test have no eigenvectors in common, so no state is sure for both.

A sum is a new test

Add two Hermitian matrices and you get another, so A+BA + B is an observable. It is natural to assume you would measure it by measuring AA, measuring BB, and adding the results. If AA and BB do not commute, that is wrong. A+BA + B has its own eigenvectors, different from those of either, and so its own test; and its possible results are its eigenvalues, which need not be sums of theirs. For instance σ1+σ2\sigma_1 + \sigma_2 has eigenvalues ±2\pm\sqrt2, not the −2,0,+2-2, 0, +2 you would get by adding ±1\pm1 to ±1\pm1. Its test is a linear-polarisation test with axes at 22.5°.

This finally disposes of the puzzle of lesson 1.5. For three magnet directions 120° apart, the three matrices that represent the three measurements really do add up to the zero matrix: μ1+μ2+μ3=0\mu_1 + \mu_2 + \mu_3 = 0 is true as a statement about observables, and so about their means in any state. It is false as a statement about individual results, each of which is ±μ\pm\mu, because the three do not commute and their results were never meant to be added.

This lesson follows §3-6 and §3-7. The book goes on to functions of an observable (same test, new labels), to how the matrix changes when the axes are changed, and to several results about eigenvalues that are set as exercises.

What to keep

  • An observable is a complete test with a real number written on each outcome. A measurement of it yields one of those numbers.
  • It is represented by a Hermitian matrix A=∑rarPrA = \sum_r a_r P_r, and its mean in the state v\mathsf v is ⟨v, Av⟩\ip{\mathsf v}{A\mathsf v}.
  • The eigenvalues of AA are the possible results. Its eigenvectors are the states that give a sure result.
  • Every Hermitian matrix is an observable: its eigenvectors supply the test.
  • Observables that commute can be measured by one test. Those that do not have no common sure states, and their sum is measured by a different test from either.

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.

1The two outcomes of a test are labelled +1+1 and −1-1. A certain state gives the first with probability 0.75. What is the mean ⟨A⟩\langle A\rangle?

2Same test and same state, but the outcomes are relabelled 33 and 11. What is the mean now?

3An observable is represented by the matrix (2005)\begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix}. What can a measurement of it give?

4A system is prepared in an eigenvector of A, and A is measured many times. What is the spread of the results?

5What does (0110)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} do to the vector (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix}?

6AA and BB each have the eigenvalues +1+1 and −1-1, and they do not commute. What can a measurement of A+BA + B give?

7Why must the matrix of an observable be Hermitian?