BB #83: The Born Rule is Pythagorean

It’s actually obvious and might fall under the “Duh!” heading for some, but it only recently sunk in on me that the Born Rule is really just another case of the Pythagorean theorem. The connection is in the way the coefficients of a quantum superposition, when squared, must sum to unity (one).

For that matter, Special Relativity, which is entirely geometric, is yet another example of the Pythagorean theorem, but that’s another story. (One I’ve already told. See: SR #X6: Moving at Light Speed)

The obvious connection is the geometry behind how a quantum state projects onto the basis eigenvectors axes.

Let’s start by looking at the Pythagorean theorem. Essentially, it’s a measure of the distance between two points in a Euclidean space (a space we think of as “marked off” in a regular rectangular grid with a system of coordinates).

As we generally learn it in grade school, it’s the length of the hypotenuse (the long side) of a right triangle given the lengths of its sides. (They don’t tell you this in grade school but they’re sneaking you into elementary trigonometry. Pun accidental but not bad.) The formula we learn is:

\displaystyle\mathsf{hypotenuse}=\sqrt{(\mathsf{side}_{A})^2+(\mathsf{side}_{B})^2}

But, per the way Pythagoras (570-495 B.C.) demonstrated his theorem…

The Pythagorean theorem: The sum of the squares of the two short sides of a right triangle is equal to the square of the longest side (the hypotenuse).

…the actual formula is:

\displaystyle\mathsf{hypotenuse}^2=(\mathsf{side}_{A})^2+(\mathsf{side}_{B})^2

It’s just that the length-squared isn’t usually what we want — we want just the length. So, we take the square root of both sides to get the first formula.

§

In the diagram above, the Pythagorean formula is giving us the square of the distance from the lower-right point to the upper-left point. The two short sides of the right-triangle act as the X and Y axes and provide the coordinate system for the three points of the triangle.

[If the lower-left corner is (0,0), then the upper-left corner is (0,3) and the lower-right corner is (4,0).]

This notion of distance generalizes to any number of dimensions. For instance, in three-dimensional space, we have:

\displaystyle\mathsf{distance}=\sqrt{x^2+y^2+z^2}

The Pythagorean formula can be generalized to:

\displaystyle\mathsf{distance}=\sqrt{x_1^2+x_2^2+x_3^2+\cdots{x}_N^2}=\sqrt{\sum^{N}_{i=1}{x}_{i}^{2}}

The distance between two points in a Euclidean space with any number of dimensions is the square root of the sum of the squares of the “sides” involved. (The “sides” in this case being the difference between the projections of each point on the respective axes. That is, how far you travel on each axis going from point A to point B.)

§

The Pythagorean theorem underpins trigonometry in how it forces a ratio between the two sides. For a given length along the hypotenuse (for a given Euclidean distance), the sum of the squares of the sides has to equal the square of the hypotenuse (distance). So, if one side is longer, another side must be shorter. The squares must always add up to the given value.

We can simplify things by setting the distance to one. The square of one is one (and the square root of one is one), so now the squares of the sides only need to sum to one:

\displaystyle{x}^2+{y}^2={1.0}

In three dimensions:

\displaystyle{x}^2+{y}^2+{z}^2={1.0}

And so on for as many dimensions as required. The requirement is that the squares of the distances along each axis need to sum to one.

§

Now recall the canonical example of a quantum superposition between two states:

\displaystyle|\Psi\rangle={c}_{0}|0\rangle+{c}_{1}|1\rangle

The two complex coefficients (c) must have squares that sum to one because the square of the coefficient gives us the probability of getting that state in a measurement. If both states are equally probable, then the coefficient squares must each be 1/2. Which means:

\displaystyle|\Psi\rangle=\tfrac{1}{\sqrt{2}}|0\rangle+\tfrac{1}{\sqrt{2}}|1\rangle

Because 1/√2 squares to 1/2.

If the superposition had three components:

\displaystyle|\Psi\rangle={c}_{1}|A\rangle+{c}_{2}|B\rangle+{c}_{3}|C\rangle

Then we require:

\displaystyle(c_1)^2+(c_2)^2+(c_3)^2=1

Which is exactly what we had above for three-dimensional space (given the distance is set to one). It should be apparent that, if all three are equally probable, then they all have a coefficient of 1/√3.

As above, the idea expands to any number of superposition components:

\displaystyle\sum^{N}_{i=0}({c}_{i})^2=1

The probabilities of the states must sum to one because the probability of getting one of those states (in a measurement) is 100%. There is always a definite result to a measurement.

§

And that’s basically it. The lead image (repeated below) illustrates graphically the two-state case just mentioned:

\displaystyle|\Psi\rangle=\tfrac{1}{\sqrt{2}}|0\rangle+\tfrac{1}{\sqrt{2}}|1\rangle

It shows the equal superposition of the |0〉 and |1〉 states with the state vector (purple) at 45° angle between the horizontal and vertical axes (which represent the two states — they are the eigenvectors for those states).

Therefore, the projection (red and green lines) of the state vector (purple line) onto those axes gives us the coefficient values for the two states.

For the X-axis: cos(45°)=0.7071 (and change). For the Y-axis: sin(45°)=the same thing.

Because 45° is halfway between the axes, so naturally the projections are equal.

Sine and cosine are simply ratios of the hypotenuse to a side where (as we’ve been doing here) the hypotenuse is set to one. In a right triangle with one 45° angle, the other angle is necessarily also 45°, making it an isosceles triangle (two sides with the same length).

This gives us a special version of the Pythagorean theorem:

\displaystyle{a^2}+{a^2}={2a^2}=1

If we divide both sides by two:

\displaystyle{a^2}=\frac{1}{2}

And then take the square root of both sides:

\displaystyle{a}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}

Which should look very familiar.

The Born Rule tells us that these projections of the state vector, when squared, are the probability factors for the respective states. With two equally probable states, the projections give us the value 1/√2 because of the geometry shown here, and the squaring is an artifact of the squares in the Pythagorean theorem.

§

Bottom line, the Born Rule is geometric. Specifically, Pythagorean geometry. People sometimes do find it strange we must square the superposition coefficients to get the probability factors. Turns out it’s because basic geometry.

And it’s hardly the only place where we find the square root of a sum of squares meaningful. For example, it’s part of how to calculate the standard deviation of a population, which is statistics, not geometry.

The basic concept turns up all sorts of places!

Stay geometric, my friends! Go forth and spread beauty and light.


About Wyrd Smythe

The canonical fool on the hill watching the sunset and the rotation of the planet and thinking what he imagines are large thoughts. View all posts by Wyrd Smythe

9 responses to “BB #83: The Born Rule is Pythagorean

  • Wyrd Smythe

    The notion of an inner product is also related to the Pythagorean theorem. Usually, we take the inner product of a pair of vectors, get a scalar value, and treat that as a measure of the two vectors’ orthogonality (a value of zero indicating they are orthogonal):

    \displaystyle\textbf{inner product}\equiv\langle{B}|{A}\rangle=\textit{scalar value}

    But if we take the inner product of a vector with itself we always get a non-zero value (because a vector can’t be orthogonal to itself), and that value is the square of the length of the vector.

    \displaystyle\langle{A}|{A}\rangle=\mathit{length^2}

    A math equation that gives the square of the length should look a little familiar. We get this because a common way of calculating the inner product is:

    \displaystyle\langle{B}|{A}\rangle={A_x}{B_x}+{A_y}{B_y}

    Assuming two-dimensional vectors A and B, each with x and y components. When we take the inner product of a vector with itself, we get:

    \displaystyle\langle{A}|{A}\rangle={A_x}{A_x}+{A_y}{A_y}={A_x}^2+{A_y}^2

    Which, ta da, is the Pythagorean theorem.

  • Mark Edward Jabbour

    Hmmm. Seems like you’re moving toward the “Tom Brady effect”? I.e we (humans) practice math sans formal education? Some, of course, better than others. Carry on, my friend. ~
    cheers

    • Wyrd Smythe

      Huh? I can’t follow your meaning. Tom Brady… the football player? I don’t know much about football (American or otherwise). I agree humans do math, or try to, without formal education (but many pick it up in other ways).

      I’m lost here, dude!…

  • Mark Edward Jabbour

    Yeah. Exactly! Wasn’t the “square of the … ” jingle from “My Fair Lady”? ~
    In other words, Tom Brady (the greatest quarterback of all time, i.e. GOAT) was doing high functioning math without the knowing the math?

    • Wyrd Smythe

      I don’t know what you’re referring to in My Fair Lady, sorry. Do you remember which song it was in?

      If you’re referring to Tom Brady’s brain “doing the math” to calculate a pass, I’m not sure how it connects with the post, but all brains do a natural form of calculus when we throw things, catch things, or do all sorts of physical things. Athletes of all kinds train their brains to be especially good at this and many have a natural talent for it. (FWIW: It’s not by the Pythagorean theorem but by stereoscopic vision that we judge distance.) The “math” our brains learn to do when it comes to moving objects of all kinds is more along the lines of differential calculus — time and space judgements with multiple variables. Not surprising that the really good ones stand out.

      I’m surprised you didn’t use Aaron Judge! (Oh, wait, it’s football season, isn’t it. The only sport I follow is baseball, so I hardly notice the others.)

  • Mark Edward Jabbour

    Elite NFL quarterbacks are the rarest of humans. However, they represent the evolution (highest?) of human capabilities. Some individuals have the ability to do high level math, without even knowing anything about the numbers. And then, of course, there is regression to the mean. ~
    Also then, the lowering of the mean. And so … Yeah … Brady’s wife is suing for divorce. So there is that …
    cheers my friend.

And what do you think?

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: