3.1States as vectors

Part II ended with tables of amplitudes that behave like the tables which turn one set of axes into another. Axes of what? This lesson supplies the answer, and with it the picture of a quantum state that physicists carry in their heads: an arrow in a space where every complete test provides a set of axes.

A test is a set of axes

On a map, a point is the same point whether you give its position as so far east and so far north, or as so far along two other directions at right angles. The point does not change; its coordinates do, and a small table of numbers converts one set of coordinates into the other.

The proposal is to read quantum tests in exactly that way.

• Choosing a complete test is choosing a set of axes at right angles to one another, one axis for each of its $N$ outcomes.
• The pure state prepared by outcome $m$ is the arrow of length 1 along the $m$th axis. Write it $\mathsf e_m$.
• A different complete test is a different set of axes, $\mathsf e_\mu$, in the same space, and the table of amplitudes $C_{\mu m}$ is the table that relates the two sets:

Read that line as: the state “outcome $m$ of the first test”, looked at from the second test’s axes, has the coordinate $C_{\mu m}$ along axis $\mu$. Its coordinate along an axis is its amplitude to give that outcome.

This is not a free choice of notation; it makes a claim that could have been false. If there are three tests, going from the first set of axes to the third directly must agree with going by way of the second. Work it through and the condition is precisely $\mathcal C_{rm} = \sum_\mu \Gamma_{r\mu}C_{\mu m}$, the rule for combining amplitudes that the last lesson reached from experiments. The geometry and the experiments agree.

A state is an arrow

If the special states $\mathsf e_m$ are arrows along the axes, the obvious next step is to allow arrows that point anywhere:

The numbers $v_m$ are the components of the vector $\mathsf v$ along the axes of the chosen test. Each one is the amplitude for the corresponding outcome, so the probability of outcome $m$ is $|v_m|^2$. Since some outcome must occur,

For an ordinary arrow, the sum of the squares of its components is the square of its length: that is Pythagoras’s theorem. So the statement that probabilities add up to 1 is the statement that the arrow has length 1. And the length of an arrow does not depend on which axes you measure it against, which is why the probabilities add up to 1 for every test at once.

In the figure the solid axes are the outcomes “horizontal” and “vertical”, and an arrow at angle $\theta$ has components $\cos\theta$ and $\sin\theta$: the photon is linearly polarised at $\theta$, and its probabilities are the $\cos^2\theta$ and $\sin^2\theta$ of lesson 1.2. Everything in that lesson was this picture.

Do not read too much into the fact that these axes look like directions in the laboratory. For photons, by coincidence, the two outcomes of a linear-polarisation test are also two directions at right angles in space. For a silver atom the two outcomes are “up” and “down” along the magnet, which are opposite directions in space, but they are still two axes at right angles in the space of states. The space of states is not the room. It is an abstract space with one dimension for each outcome of a complete test.

G. The superposition principle. Every vector, other than the zero vector, represents a pure state that can be prepared.

Rule G is a sweeping claim, and it is silent on how. It says that for any list of components you care to write down, some arrangement of apparatus prepares that state. Finding the arrangement is the experimenter’s problem, in the same way that Euclid says a circle of any radius exists and leaves the making of compasses to others.

When the same vector is referred to a different test’s axes, its components are converted by the same table of amplitudes:

Two lists of components, one state. Which list you use is a matter of convenience, like choosing between map grids.

Why the components have to be complex

If components were ordinary real numbers, the picture in Figure 1 would be the whole story, and the only polarisations would be linear ones. But lesson 1.3 showed light in which the vertical swing lags the horizontal one by a quarter of a cycle: circular polarisation. Its amplitudes for horizontal and vertical both have length $1/\sqrt2$, the same as for diagonal light. What differs is their relative angle.

The phase difference $\delta$ of lesson 1.3 is the angle between the two components, each drawn as an arrow in its own right. So a state vector is a list of $N$ complex numbers. It cannot be drawn on paper, even for $N = 2$, but its algebra is the algebra of ordinary arrows with one adjustment, which comes next.

Lengths and angles: the inner product

For ordinary arrows there is a single operation that yields both lengths and angles: the dot product, $\mathbf u\cdot\mathbf v = u_1v_1 + u_2v_2 + \dots$. An arrow dotted with itself gives its length squared; two arrows at right angles give zero.

With complex components that formula fails at the first hurdle: $v_m^{\,2}$ need not be positive, or even real, so $\sum v_m^{\,2}$ is useless as a length. What we want for the length is $\sum |v_m|^2$, and $|v_m|^2 = \bar v_m v_m$. So conjugate the first vector:

This is the inner product (or scalar product) of $\mathsf u$ and $\mathsf v$. It is a complex number. Its main properties are quickly checked from the definition.

• $\ip{\mathsf v}{\mathsf v} = \sum_m |v_m|^2$ is real and positive. Its square root is the length, or norm, of $\mathsf v$, written $\|\mathsf v\|$. A vector of norm 1 is called normalised.
• Swapping the two vectors conjugates the result: $\ip{\mathsf v}{\mathsf u} = \overline{\ip{\mathsf u}{\mathsf v}}$. So the size of the inner product is the same either way round.
• Two vectors are called orthogonal when their inner product is zero. This is the right angle of the space of states.
• For any two vectors, $|\ip{\mathsf u}{\mathsf v}|^2 \le \ip{\mathsf u}{\mathsf u}\,\ip{\mathsf v}{\mathsf v}$. This is the Schwarz inequality. For normalised vectors it says that $|\ip{\mathsf u}{\mathsf v}|^2$ lies between 0 and 1, which is just what a probability needs. Hold that thought for the next lesson.

Try it on the polarisations you know.

Horizontal and vertical are orthogonal. So are right and left circular, though you would never guess it from their components without the conjugate. These are the “opposite” pairs of lesson 1.3, and the word orthogonal was borrowed from here.

A set of $N$ normalised, mutually orthogonal vectors is an orthonormal basis: a set of axes. With respect to such a basis the component of a vector along an axis is its inner product with that axis,

the complex version of “the component is the shadow”. And the unitary tables of lesson 2.4 are exactly the tables that convert components between two orthonormal bases, just as rotations convert coordinates between two sets of perpendicular axes.

This lesson follows §3-1 and §3-2. The book proves the Schwarz inequality and a few related identities in the text and exercises; they are worth doing once.