One state, different ways to describe it
Send single photons, all prepared the same way, at a polarising filter set horizontal: three out of four get through. Turn the filter to 45° and send more photons prepared exactly as before: now more than nine out of ten get through. Is the photon "mostly horizontal" or "mostly diagonal"? And which set of numbers is the real description of it?
The previous essay attached one complex number, an amplitude, to each way an event could happen. This essay collects the amplitudes that belong to a system into a single object, its state, and asks what changes when we describe that state using a different set of alternatives. The answer turns out to be the same as for an arrow drawn on a page and described along two different pairs of axes: the arrow does not change, its coordinates do, and each set of coordinates answers a different question.
Polarisers, one photon at a time
Light is a wave of electric field, and in a polarised beam the field oscillates along one fixed direction across the beam, its polarisation. A polarising filter has a direction of its own. Classical optics says what fraction of a beam gets through: if the light is polarised at angle $\theta$ and the filter is set at $\varphi$, the transmitted intensity is a fraction $\cos^2(\theta - \varphi)$ of the original. This is Malus's law, known since the early nineteenth century.
Now turn the light down until photons arrive one at a time, and put a photon counter behind the filter. A photon cannot be partly transmitted; each one either gets through whole or is absorbed. What Malus's law becomes is a probability: each photon gets through with probability $\cos^2(\theta - \varphi)$, and over many photons that reproduces the classical fraction. Experiments add one more fact. A photon that does get through leaves polarised along the filter's direction, whatever its polarisation was before.
A state is an arrow
Everything above fits one simple description. Represent the photon's polarisation by an arrow of length 1 pointing at angle $\theta$. The amplitude for it to pass a filter at angle $\varphi$ is the arrow's component along the filter's direction, $\cos(\theta - \varphi)$, and the probability is its square, as in the double-slit essay. The arrow is the photon's state: once we know it, we can compute the probabilities for any filter we might choose.
A more useful device than a filter is a polarising beam splitter, a crystal that sends light polarised one way straight on and light polarised at right angles off to the side. With one set horizontal (H) and vertical (V), every photon comes out one way or the other, and never both. H and V form a complete list of alternatives for this question. For a photon polarised at $\theta$, the state can then be written as two numbers, its amplitudes for the two exits:
$$(a_H,\; a_V) = (\cos\theta,\; \sin\theta).$$The squares add to 1, since $\cos^2\theta + \sin^2\theta = 1$: the photon certainly leaves by one exit or the other. Such a complete set of alternatives is called a basis, and the amplitudes are the state's components in that basis. Writing a state in a particular basis is also called choosing a representation, which is why later essays refer back to this one as the essay on representations.
The same arrow, different axes
Nothing forces us to set the beam splitter horizontal. Turned by 45°, it has exits for diagonal (D, at 45°) and antidiagonal (A, at 135°) polarisation, and those are another complete pair of alternatives. The same photon described in this basis has components equal to the arrow's shadows on the new axes:
$$\begin{aligned}a_D &= \cos(\theta - 45°),\\[4pt] a_A &= \cos(\theta - 135°).\end{aligned}$$The two pairs of numbers look nothing alike, yet they describe exactly the same photon. Neither is the state. The state is the arrow, and each basis gives a set of coordinates for it. What a basis does is fit a question. The H/V components answer "which exit of a horizontal splitter?" directly; the D/A components answer the diagonal question directly. Either list can answer the other's question too, with a little arithmetic, as the worked example shows.
Worked example: a photon at 30°
In the H/V basis. The components are $\cos 30° = 0.866$ and $\sin 30° = 0.500$. Squaring: $P(H) = 0.750$ and $P(V) = 0.250$. That is the three-in-four of the opening.
In the D/A basis. The components are $\cos(-15°) = 0.966$ and $\cos(-105°) = -0.259$. Squaring: $P(D) = 0.933$ and $P(A) = 0.067$. That is the more-than-nine-in-ten. Both pairs of probabilities add to one, as they must.
Asking the H question of the D/A list. Suppose all we had written down was the D/A description, $(0.966, -0.259)$, and someone asked for $P(H)$. The arrow for "horizontal" is itself a state, and in the D/A basis it has components $(\cos(-45°), \cos(-135°)) = (0.707, -0.707)$. The amplitude for our photon to be found horizontal is the overlap of the two arrows, computed from either list by multiplying matching components and adding:
$$\begin{aligned} a_H &= (0.966)(0.707)\\[4pt] &\quad + (-0.259)(-0.707)\\[4pt] &= 0.683 + 0.183 = 0.866. \end{aligned}$$That is $\cos 30°$ again, and $P(H) = 0.750$. The answer does not depend on which basis we happened to write the state in. That is the whole content of saying the two lists describe one state.
Check: nothing depends on the axes. The overlap of two arrows is their dot product, which is the cosine of the angle between them. Rotating the axes changes every component but not that angle, so every probability survives any change of basis.
Optional: why the entries must be allowed to be complex
So far every component has been a real number. Light can also be circularly polarised: the field rotates as the wave travels. Such a photon passes a filter at any angle with probability ½, so no real arrow in the plane describes it. Its H/V components are $(1/\sqrt2,\; i/\sqrt2)$: equal sizes, but a quarter-turn of phase between them.
Check it against the D exit: the overlap with the diagonal arrow, which has H/V components $(1/\sqrt2, 1/\sqrt2)$, is $\tfrac12 + \tfrac{i}{2}$, whose squared size is ½. So $P(H) = P(V) = P(D) = P(A) = \tfrac12$. By these tests the circular photon looks like an even mixture of everything. It is not: a circular analyser, which has exits for right- and left-handed circular light, sends every one of these photons out of the same exit. The relative phase between components is physical information.
With complex components the overlap recipe needs one adjustment: take the complex conjugate of one list's entries before multiplying. The formalism essay makes this precise.
When the list gets longer
Polarisation needs only two alternatives. Most systems need more. Think back to the electron arriving at the screen in the double-slit experiment. The question "where does it land?" has a separate alternative for every detector window across the screen, and the state, described in that basis, is a long list of amplitudes, one per window. Make the windows narrower and the list becomes a continuous function of position, one complex number for every $x$. That function is the wavefunction, and the next essay is about it.
The same electron could be described in other bases. Listing its amplitudes for each possible momentum gives a different function, carrying the same state. For an electron bound in an atom, the natural list is one amplitude for each energy level, Bohr's levels among them. As with the photon, none of these descriptions is the electron; each is the state written in the alternatives of one particular question.
A note on history: two mechanics that were one
This is why quantum mechanics was discovered twice. In 1925 Werner Heisenberg, soon joined by Max Born and Pascual Jordan, built a mechanics out of square arrays of numbers, each entry labelled by a pair of energy levels of the atom: matrix mechanics. In 1926 Erwin Schrödinger produced a wave equation for a function spread through space: wave mechanics. They looked like rival theories with nothing in common. Within months Schrödinger and others showed they give identical predictions. In the language of this essay, one theory wrote everything in the basis of energy levels and the other in the basis of positions. Paul Dirac and Jordan then developed the general rules for moving between any two descriptions, and the two mechanics became two coordinate systems for a single theory.
Two questions
1. Adding a filter lets more through. Photons polarised horizontally meet a vertical filter. What fraction gets through? Now insert a filter at 45° between the source and the vertical filter. What fraction gets through both?
Worked answer
Without the middle filter: $\cos^2 90° = 0$. None.
With it: a horizontal photon passes the 45° filter with probability $\cos^2 45° = \tfrac12$, and those that pass leave polarised at 45°. Each of those then meets the vertical filter at 45° to its polarisation and passes with probability $\tfrac12$. Overall, $\tfrac12\times\tfrac12 = \tfrac14$.
The tempting wrong turn is to reason that a filter can only remove photons, so adding one cannot increase what gets through. That would be right if the filter only sorted photons that already had definite H or V labels. It does not. A photon that passes the middle filter is in a new state, the 45° arrow, and described in the H/V basis that state has a vertical component of $0.707$.
2. Partly horizontal, or sometimes horizontal? A friend reads the worked example and says: "So a photon at 30° is really horizontal three-quarters of the time and vertical one-quarter of the time; we just don't know which." Test this against the diagonal splitter. If it were true, what fraction would leave by the D exit? What fraction actually does?
Worked answer
If each photon were definitely H (probability ¾) or definitely V (¼), each would reach the diagonal splitter as an H or V arrow. Both H and V are at 45° to the D axis, so either kind leaves by D with probability $\cos^2 45° = \tfrac12$. The total would be $\tfrac34\cdot\tfrac12 + \tfrac14\cdot\tfrac12 = \tfrac12$. The actual photons at 30° leave by D with probability $0.933$.
So the friend's picture makes a wrong prediction. A photon at 30° is not a hidden choice between H and V; it is a single definite state that is neither of the two H/V alternatives. The H/V components are amplitudes, and amplitudes combine before squaring, exactly as the two slits did in the previous essay. Treating them as probabilities of a hidden label throws away the cross term, and with it the 0.933.
Further reading
- Dirac, Principles of Quantum Mechanics (4th ed.), Ch. I §2 "The Polarization of Photons" and §4 "Superposition and Indeterminacy": the photon-and-filter argument this essay follows, stated with great economy. Ch. III "Representations" develops the idea of describing one state in many bases in full.
- Feynman, Lectures on Physics, Vol. III, Ch. 8 "The Hamiltonian Matrix", §8-1 "Amplitudes and vectors", §8-2 "Resolving state vectors" and §8-3 "What are the base states of the world?": the analogy between a state and an arrow, taken further.
- Born, Atomic Physics (Dover, 8th ed.), Ch. V §3 "Matrix Mechanics" and §4 "Wave Mechanics"; in §8 he describes the two as "two forms of the same content".
Where this leaves us
The photon at 30° is neither "mostly horizontal" nor "mostly diagonal". It is one state, an arrow, and the two sets of numbers are its coordinates in two bases. Each set answers its own question directly, and either one can answer any question, with the same result. A state is not the same thing as the numbers used to describe it.
For a photon's polarisation the list is two numbers long. For an electron's position it becomes a function, a complex number at every point of space, and that function has to change in time as the electron moves. What does such a function tell us, what does its size at a point mean, and what equation governs how it evolves? That is the subject of the next essay.